Page 1 :
B. Sc. DEGREE PROGRAMME, (2019 Admission onwards), , MATHEMATICS, SIXTH SEMESTER, , CORE : MTS6B12 CALCULUS OF, MULTI VARIABLE, 5 hours/week, 4 Credits, 100 Marks [Int:20+Ext:80]

Page 2 :
SIXTH SEMESTER, MTS6 B12 CALCULUS OF MULTI VARIABLE, 5 hours/week, , 4 Credits, , 100 Marks [Int:20+Ext:80], , Aims, Objectives and Outcomes, The intention of the course is to extend the immensely useful ideas and notions such as limit,, continuity, derivative and integral seen in the context of function of single variable to function of, several variables. The corresponding results will be the higher dimensional analogues of what we, learned in the case of single variable functions. The results we develop in the course of calculus of, multivariable is extremely useful in several areas of science and technology as many functions that, arise in real life situations are functions of multivariable., The successful completion of the course will enable the student to, , , , , , , , , , , , , , , , , , Understand several contexts of appearance of multivariable functions and their, representation using graph and contour diagrams., Formulate and work on the idea of limit and continuity for functions of several variables., Understand the notion of partial derivative, their computation and interpretation., Understand chain rule for calculating partial derivatives., Get the idea of directional derivative, its evaluation, interpretation, and relationship with, partial derivatives., Understand the concept of gradient, a few of its properties, application and interpretation., Understand the use of partial derivatives in getting information of tangent plane and normal, line., Calculate the maximum and minimum values of a multivariable function using second, derivative test and Lagrange multiplier method., Find a few real life applications of Lagrange multiplier method in optimization problems., Extend the notion of integral of a function of single variable to integral of functions of two, and three variables., Address the practical problem of evaluation of double and triple integral using Fubini’s, theorem and change of variable formula., Realise the advantage of choosing other coordinate systems such as polar, spherical,, cylindrical etc. in the evaluation of double and triple integrals ., See a few applications of double and triple integral in the problem of finding out surface, area ,mass of lamina, volume, centre of mass and so on., Understand the notion of a vector field, the idea of curl and divergence of a vector field,, their evaluation and interpretation., Understand the idea of line integral and surface integral and their evaluations., Learn three major results viz. Green’s theorem, Gauss’s theorem and Stokes’ theorem of, multivariable calculus and their use in several areas and directions., , 55

Page 3 :
Syllabus, , Text, , Calculus: Soo T Tan Brooks/Cole, Cengage Learning ( 2010 ) ISBN 0­, 534­46579­X), , Module­I, (18 hrs), 13.1: Functions of two or more variables­ Functions of Two Variables,, Graphs of Functions of Two Variables, Level Curves, Functions of Three, Variables and Level Surfaces, 13.2: Limits and continuity­An Intuitive Definition of a Limit, existence and, non existence of limit, Continuity of a Function of Two Variables, Continuity, on a Set, continuity of polynomial and rational functions, continuity of, composite functions, Functions of Three or More Variables, The 𝜀 −, 𝛿 Definition of a Limit, 13.3: Partial Derivatives­ Partial Derivatives of Functions of Two Variables,, geometric interpretation, Computing Partial Derivatives, Implicit, Differentiation, Partial Derivatives of Functions of More Than Two, Variables, Higher­Order Derivatives, clairaut theorem, harmonic functions, 13.4: Differentials­ Increments, The Total Differential, interpretation, Error, in Approximating ∆z by 𝑑𝑧 [only statement of theorem1 required ; proof, omitted ] Differentiability of a Function of Two Variables, criteria,, Differentiability and Continuity, Functions of Three or More Variables, 13.5: The Chain rule­ The Chain Rule for Functions Involving One, Independent Variable, The Chain Rule for Functions Involving Two, Independent Variables, The General Chain Rule, Implicit Differentiation, Module­II, (16 hrs), 13.6: Directional Derivatives and Gradient vectors ­ The Directional, Derivative, The Gradient of a Function of Two Variables, Properties of the, Gradient, Functions of Three Variables, 13.7: Tangent Planes and Normal Lines­ Geometric Interpretation of the, Gradient, Tangent Planes and Normal Lines, Using the Tangent Plane of 𝑓 to, approximate the Surface 𝑧 = 𝑓(𝑥, 𝑦), 13.8: Extrema of Functions of two variables ­ Relative and Absolute, Extrema, Critical Points—Candidates for Relative Extrema, The Second, , 56

Page 4 :
Derivative Test for Relative Extrema, Finding the Absolute Extremum, Values of a Continuous Function on a Closed Set, 13.9: Lagrange Multipliers­ Constrained Maxima and Minima, The Method, of Lagrange Multipliers, Lagrange theorem, Optimizing a Function Subject to, Two Constraints, Module­III, (21 hrs), 14.1: Double integrals­ An Introductory Example, Volume of a Solid, Between a Surface and a Rectangle, The Double Integral Over a Rectangular, Region, Double Integrals Over General Regions, Properties of Double, Integrals, 14.2: Iterated Integrals­Iterated Integrals Over Rectangular Regions,, Fubini’s Theorem for Rectangular Regions, Iterated Integrals Over, Nonrectangular Regions, 𝑦­ simple and 𝑥­ simple regions, advantage of changing, the order of integration, , 14.3:Double integrals in polar coordinates­ Polar Rectangles, Double, Integrals Over Polar Rectangles, Double Integrals Over General Regions, r­, , simple region, method of evaluation, , 14.4: Applications of Double integral­ Mass of a Lamina, Moments and, Center of Mass of a Lamina, Moments of Inertia, Radius of Gyration of a, Lamina, 14.5: Surface Area­ Area of a Surface 𝑧 = 𝑓(𝑥, 𝑦) , Area of Surfaces with, Equations 𝑦 = 𝑔(𝑥, 𝑧) and 𝑥 = ℎ(𝑦, 𝑧), 14.6: Triple integrals­ Triple Integrals Over a Rectangular Box, definition,, method of evaluation as iterated integrals, Triple Integrals Over General, Bounded Regions in Space, Evaluating Triple Integrals Over General, Regions, evaluation technique, Volume, Mass, Center of Mass, and Moments of, Inertia, 14.7: Triple Integrals in cylindrical and spherical coordinates­ evaluation of, integrals in Cylindrical Coordinates, Spherical Coordinates, 14.8: Change of variables in multiple integrals­ Transformations, Change of, Variables in Double Integrals [only the method is required; derivation, omitted], illustrations, Change of Variables in Triple Integrals, , 57

Page 5 :
Module­IV, (25 hrs), 15.1: Vector Fields­ V.F. in two and three dimensional space, Conservative, Vector Fields, 15.2: Divergence and Curl­ Divergence­ idea and definition, Curl­ idea and, definition, 15.3: Line Integrals­ Line integral w.r.t. arc length ­motivation, basic idea and, definition, Line Integrals with Respect to Coordinate Variables, orientation of, curve Line Integrals in Space, Line Integrals of Vector Fields, 15.4: Independence of Path and Conservative Vector Fields­ path, independence through example, definition, fundamental theorem for line integral,, Line Integrals Along Closed Paths, work done by conservative vector field,, , Independence of Path and Conservative Vector Fields, Determining, Whether a Vector Field Is Conservative, test for conservative vector field, Finding a Potential Function, Conservation of Energy, 15.5: Green’s Theorem­ Green’s Theorem for Simple Regions, proof of, theorem for simple regions, finding area using line integral, Green’s Theorem for, More General Regions, Vector Form of Green’s Theorem, 15.6: Parametric Surfaces­Why We Use Parametric Surfaces, Finding, Parametric Representations of Surfaces, Tangent Planes to Parametric, Surfaces, Area of a Parametric Surface [derivation of formula omitted], 15.7: Surface Integrals­Surface Integrals of Scalar Fields, evaluation of, surface integral for surfaces that are graphs , [derivation of formula omitted;, only method required] Parametric Surfaces, evaluation of surface integral for, parametric surface, Oriented Surfaces, Surface Integrals of Vector Fields­, definition, flux integral, evaluation of surface integral for graph[method only],, Parametric Surfaces, evaluation of surface integral of a vector field for, parametric surface [method only], 15.8: The Divergence Theorem­divergence theorem for simple solid regions, (statement only), illustrations, Interpretation of Divergence, 15.9: Stokes Theorem­generalization of Green’s theorem –Stokes Theorem,, illustrations, Interpretation of Curl, , 58

Page 6 :
References:, 1 JoelHass, Christopher Heil & Maurice D. Weir : Thomas’ Calculus(14/e), Pearson(2018) ISBN 0134438981, , 2, 3, , Robert A Adams & Christopher Essex : Calculus: A complete Course, (8/e) Pearson Education Canada (2013) ISBN: 032187742X, Jon Rogawski: Multivariable Calculus Early Transcendentals (2/e) W. H., Freeman and Company(2012) ISBN: 1­4292­3187­4, , 4, 5, 6, 7, , Anton, Bivens & Davis : Calculus Early Transcendentals (10/e) John, Wiley & Sons, Inc.(2012) ISBN: 978­0­470­64769­1, James Stewart : Calculus (8/e) Brooks/Cole Cengage Learning(2016) ISBN:, 978­1­285­74062­1, Jerrold E. Marsden & Anthony Tromba :Vector Calculus (6/e) W. H., Freeman and Company ,New York(2012) ISBN: 978­1­4292­1508­4, Arnold Ostebee & Paul Zorn: Multivariable Calculus (2/e) W. H. Freeman, Custom Publishing, N.Y.(2008)ISBN: 978­1­4292­3033­9, , 59

Page 7 :
MODULE 1, Textbook : Calculus: Soo T Tan Brooks/Cole, Cengage Learning (2010) ISBN 0-534-46579-X), Sections 13.1-13.5

Page 8 :
1034, , Chapter 13 Functions of Several Variables, , 13.1, , Functions of Two or More Variables, Functions of Two Variables, Up to now, we have dealt only with functions of one variable. In practice, however, we, often encounter situations in which one quantity depends on two or more quantities., For example, consider the following:, The volume V of a right circular cylinder depends on its radius r and its height, h (V  pr 2h) ., The volume V of a rectangular box depends on its length l, width w, and height, h (V  lwh)., The revenue R from the sale of commodities A, B, C, and D at the unit prices of, 10, 14, 20, and 30 dollars, respectively, depends on the number of units x, y, z,, and w of commodities A, B, C, and D sold (R  10x  14y  20z  30w)., Just as we used a function of one variable to describe the dependency of one variable on another, we can use the notion of a function of several variables to describe, the dependency of one variable on several variables. We begin with the definition of a, function of two variables., , DEFINITION Function of Two Variables, Let D  {(x, y) 冟 x, y 僆 R} be a subset of the xy-plane. A function f of two variables is a rule that assigns to each ordered pair of real numbers (x, y) in D a, unique real number z. The set D is called the domain of f, and the set of corresponding values of z is called the range of f., , The number z is usually written z  f(x, y) . The variables x and y are independent, variables, and z is the dependent variable., As in the case of a function of a single variable, a function of two or more variables can be described verbally, numerically, graphically, or algebraically., , EXAMPLE 1 Home Mortgage Payments In a typical housing loan, the borrower, makes periodic payments toward reducing indebtedness to the lender, who charges, interest at a fixed rate on the unpaid portion of the debt. In practice, the borrower is, required to repay the lender in periodic installments, usually of the same size over a, fixed term, so that the loan (principal plus interest charges) is amortized at the end, of the term. Table 1 gives the monthly loan repayment on a loan of $1000, f(t, r),, where t is the term of the loan in years and r is the interest rate per annum (%/year), compounded monthly. Referring to the table, we see that the monthly installment, for a 30-year loan of $1000 when the current interest rate is 7%/year is given by, f(30, 7)  6.6530 (dollars). Therefore, if the amount borrowed is $350,000, the monthly, repayment is 350(6.6530), or $2328.55.

Page 9 :
13.1, , 1035, , Functions of Two or More Variables, , TABLE 1, Interest rate %/year, , Term of the loan (years), , t, , r, , 6, , 6 14, , 6 12, , 6 34, , 7, , 7 14, , 7 12, , 7 34, , 8, , 5, , 19.3328, , 19.4493, , 19.5661, , 19.6835, , 19.8012, , 19.9194, , 20.0379, , 20.1570, , 20.2764, , 10, , 11.1021, , 11.2280, , 11.3548, , 11.4824, , 11.6108, , 11.7401, , 11.8702, , 12.0011, , 12.1328, , 15, , 8.4386, , 8.5742, , 8.7111, , 8.8491, , 8.9883, , 9.1286, , 9.2701, , 9.4128, , 9.5565, , 20, , 7.1643, , 7.3093, , 7.4557, , 7.6036, , 7.7530, , 7.9038, , 8.0559, , 8.2095, , 8.3644, , 25, , 6.4430, , 6.5967, , 6.7521, , 6.9091, , 7.0678, , 7.2281, , 7.3899, , 7.5533, , 7.7182, , 30, , 5.9955, , 6.1572, , 6.3207, , 6.4860, , 6.6530, , 6.8218, , 6.9921, , 7.1641, , 7.3376, , 35, , 5.7019, , 5.8708, , 6.0415, , 6.2142, , 6.3886, , 6.5647, , 6.7424, , 6.9218, , 7.1026, , 40, , 5.5021, , 5.6774, , 5.8546, , 6.0336, , 6.2143, , 6.3967, , 6.5807, , 6.7662, , 6.9531, , Although the monthly installments based on a $1000 loan are displayed in the form, of a table for selected values of t and r in Example 1, an algebraic expression for computing f(t, r) also exists:, f(t, r) , , 10r, 12c1  a1 , , 0.01r 12t, b, d, 12, , But as in the case of a single variable, we are primarily interested in functions that can, be described by an equation relating the dependent variable z to the independent, variables x and y. Also, as in the case of a single variable, unless otherwise specified,, the domain of a function of two variables is the set of all points (x, y) for which, z  f(x, y) is a real number., , EXAMPLE 2 Let f(x, y)  x 2  xy  2y. Find the domain of f, and evaluate f(1, 2),, , f(2, 1), f(t, 2t), f(x 2, y), and f(x  y, x  y)., , Solution Since x 2  xy  2y is a real number whenever (x, y) is an ordered pair of, real numbers, we see that the domain of f is the entire xy-plane. Next, we have, f(1, 2)  12  (1)(2)  2(2)  3, f(2, 1)  22  (2)(1)  2(1)  4, f(t, 2t)  t 2  (t)(2t)  2(2t)  t 2  4t, f(x 2, y)  (x 2) 2  (x 2)(y)  2y  x 4  x 2y  2y, and, f(x  y, x  y)  (x  y)2  (x  y)(x  y)  2(x  y),  x 2  2xy  y 2  x 2  y 2  2x  2y,  2(y 2  xy  x  y)

Page 10 :
1036, , Chapter 13 Functions of Several Variables, y, , EXAMPLE 3 Find and sketch the domain of the function:, Test point (1, 0), , 1, 0, 1, , x, , 1 2 3, , x  y2, , a. f(x, y)  2y 2  x, , b. t(x, y) , , ln(x  y  1), yx, , Solution, a. f(x, y) is a real number provided that y 2  x  0. Therefore, the domain of f is, D  {(x, y) 冟 y 2  x  0}, , (a) The domain of f(x, y)  √ y 2  x, y, yx, 1, , Test point (1, 0), 0, , 1, , x, , 1, , x  y  1  0 1, , (b) The domain of g(x, y) , , ln(x  y  1), yx, , To sketch the region D, we first draw the curve y 2  x  0, or y 2  x, which, is a parabola (Figure 1a). Observe that this curve divides the xy-plane into two, regions: the points satisfying y 2  x  0 and the points satisfying y 2  x  0., To determine the region of interest, we pick a point in one of the regions, say,, the point (1, 0). Substituting the coordinates x  1 and y  0 into the inequality, y 2  x  0, we obtain 0  1  0, which is false. This shows that the test point, is not contained in the required region. Therefore, the region that does not contain, the test point together with the curve x  y 2 is the required domain (Figure 1a)., b. Because the logarithmic function is defined only for positive numbers, we must, have x  y  1  0. Furthermore, the denominator of the expression cannot be, zero, so y  x  0, or y  x. Therefore, the domain of t is, D  {(x, y) 冟 x  y  1  0 and, , FIGURE 1, , y  x}, , To sketch the domain of D, we first draw the graph of the equation, xy10, which is a straight line. The dashed line is used to indicate that points on the line, are not included in D. This line divides the xy-plane into two half-planes. If we, pick the test point (1, 0) and substitute the coordinates x  1 and y  0 into, the inequality x  y  1  0, we obtain 2  0, which is true. This computation, tells us that the upper half-plane containing the test point satisfies the inequality, x  y  1  0. Next, because y  x, all the points lying on the line y  x in this, half-plane must be excluded from D. Again, we indicate this with a dashed line, (Figure 1b)., , z, , (x, y, z), , z  f (x, y), , Graphs of Functions of Two Variables, Just as the graph of a function of one variable enables us to visualize the function, so, too does the graph of a function of two variables., , S, , 0, , DEFINITION Graph of a Function of Two Variables, y, x, , (x, y), , D, , FIGURE 2, The graph of f is the surface, S consisting of all points (x, y, z),, where z  f(x, y) and (x, y) 僆 D., , Let f be a function of two variables with domain D. The graph of f is the set, S  {(x, y, z) 冟 z  f(x, y), (x, y) 僆 D}, , Since each ordered triple (x, y, z) may be represented as a point in threedimensional space, R3, the set S is a surface in space (see Figure 2).

Page 11 :
13.1, , Functions of Two or More Variables, , 1037, , EXAMPLE 4 Sketch the graph of f(x, y)  29  x 2  y 2. What is the range of f ?, Solution The domain of f is D  {(x, y) 冟 x 2  y 2, tered at the origin. Writing z  f(x, y), we have, , 9}, the disk with radius 3, cen-, , z  29  x 2  y 2, z2  9  x 2  y2, or, x 2  y2  z2  9, The last equation represents a sphere of radius 3 centered at the origin. Since z  0,, we see that the graph of f is just an upper hemisphere (Figure 3). Furthermore, z must, be less than or equal to 3, so the range of f is [0, 3]., z, (0, 0, 3), , 0, V, , FIGURE 3, The graph of f(x, y)  29  x 2  y 2, is the upper hemisphere of radius 3,, centered at the origin., , (0, 3, 0), , (3, 0, 0), , y, , x, , Computer Graphics, The graph of a function of two variables can be sketched with the aid of a graphing, utility. In most cases the techniques that are used involve plotting the traces of a surface in the vertical planes x  k and y  k for equally spaced values of k. The program uses a “hidden line” routine that determines what parts of certain traces should, be eliminated to give the illusion of the surface in three dimensions. In the next example we sketch the graph of a function of two variables and then show a computergenerated version of it., , EXAMPLE 5 Let f(x, y)  x 2  4y 2., a. Sketch the graph of f., , b. Use a CAS to plot the graph of f., , Solution, a. We recognize that the graph of the function is the surface z  x 2  4y 2, which is, the elliptic paraboloid, x2, , 1, , y2, 1 2, a b, 2, , z, , Using the drawing skills developed in Section 11.6, we obtain the sketch shown, in Figure 4a.

Page 12 :
1038, , Chapter 13 Functions of Several Variables, , b. The computer-generated graph of f is shown in Figure 4(b)., z, z, , 4, 1, , 1, , 2, , 2, , 1, , 1, , 2, V, FIGURE 4, The graph of f(x, y)  x 2  4y 2, , 2, , y, , y, , x, , x, , (a), , (b), , Figure 5 shows the computer-generated graphs of several functions., z, , z, , y, , x, , y, , x, , cos (x 2  2y 2), (b) f (x, y)  ___________, 1  x 2  2y 2, , (a) f (x, y)  x 3  3y 2x, z, , z, , y, , FIGURE 5, Some computer-generated, graphs of functions of, two variables, , x, , y, x, 2, , (c) f (x, y)  x 2y 2e x  y, , 2, , (d) f (x, y)  ln ( x 2  2y 2  1)

Page 13 :
13.1, , Functions of Two or More Variables, , 1039, , Level Curves, We can visualize the graph of a function of two variables by using level curves. To, define the level curve of a function f of two variables, let z  f(x, y) and consider the, trace of f in the plane z  k (k, a constant), as shown in Figure 6a. If we project this, trace onto the xy-plane, we obtain a curve C with equation f(x, y)  k, called a level, curve of f (Figure 6b)., y, , z, z  f (x, y), , f (x, y)  k, C, , zk, 0, f (x, y)  k, , x, , y, 0, , C, , FIGURE 6, , (a) The level curve C with equation f(x, y)  k, is the projection of the trace of f in the, plane z  k onto the xy-plane., , x, , (b) The level curve C, , DEFINITION Level Curves, The level curves of a function f of two variables are the curves in the xy-plane, with equations f(x, y)  k, where k is a constant in the range of f., Notice that the level curve with equation f(x, y)  k is the set of all points in the, domain of f corresponding to the points on the surface z  f(x, y) having the same, height or depth k. By drawing the level curves corresponding to several admissible values of k, we obtain a contour map. The map enables us to visualize the surface represented by the graph of z  f(x, y): We simply lift or depress the level curve to see the, “cross sections” of the surface. Figure 7a shows a hill, and Figure 7b shows a contour, map associated with that hill., y, , z, , 800, 600, 400, 200, 0, , 400, , 600, , 200, 0, , y, x, , FIGURE 7, , (a) A hill, , 0, (b) A contour map of the hill, , x

Page 14 :
1040, , Chapter 13 Functions of Several Variables, , EXAMPLE 6 Sketch a contour map for the surface described by f(x, y)  x 2  y 2,, , using the level curves corresponding to k  0, 1, 4, 9, and 16., , Solution The level curve of f corresponding to each value of k is a circle x 2  y 2  k, of radius 1k, centered at the origin. For example, if k  4, the level curve is the circle with equation x 2  y 2  4, centered at the origin and having radius 2. The required, contour map of f comprises the origin and the four concentric circles shown in Figure 8a., The graph of f is the paraboloid shown in Figure 8b., z, , y, , z  x 2  y2, , k0, k1, k4, k9, k  16, x, , 0, y, x, V, , (a) Contour map for f (x, y)  x 2  y2, , FIGURE 8, , (b) The graph of z  x 2  y2, , EXAMPLE 7 Sketch a contour map for the hyperbolic paraboloid defined by, f(x, y)  y 2  x 2., Solution The level curve corresponding to each value of k is the graph of the equation y 2  x 2  k. For k  0 the level curves have equations, y2, x2, , 1, k, k, or, y2, ( 1k)2, , , , x2, ( 1k)2, , 1, , These curves are a family of hyperbolas with asymptotes y  x and vertices, (0, 1k) . For example, if k  4, then the level curve is the hyperbola, y2, x2, , 1, 4, 4, with vertices (0, 2)., If k  0, the level curves have equations y 2  x 2  k or x 2  y 2  k, which can, be put in the standard form, x2, ( 1k)2, , , , y2, ( 1k) 2, , 1, , and represent a family of hyperbolas with asymptotes y  x. The contour map comprising the level curves corresponding to k  0, 2, 4, 6, and 8 is sketched in, Figure 9a. The graph of z  y 2  x 2 is shown in Figure 9b.

Page 15 :
13.1, , y, , Functions of Two or More Variables, , 1041, , z, , k0, , 0, x, y, x, , k0, V, , FIGURE 9, , (a) Contour map for f (x, y)  y2  x 2, , (b) The graph of z  y2  x 2, , Figure 10 shows some computer-generated graphs of functions of two variables and, their corresponding level curves., , z, , z, , z, , x, y, y, y, , x, x, x 2  2y 2, (a) Graph of f (x, y)  cos ________, 4, , (, , ), , 2, , (c) Graph of f (x, y)  xye x  y, , (b) Graph of f (x, y)  y 4  8y 2  4x 2, , y, , y, , y, , x, , x 2  2y 2, (d) Level curves of f (x, y)  cos ________, 4, , (, , ), , FIGURE 10, The graphs of some functions and their level curves, , 2, , x, , x, , (e) Level curves of f (x, y)  y 4  8y 2  4x 2, , 2, , (f) Level curves of f (x, y)  xye x  y, , 2

Page 16 :
1042, , Chapter 13 Functions of Several Variables, , Aside from their use in constructing topographic maps of mountain ranges, level, curves are found in many areas of practical interest. For example, if T(x, y) denotes the, temperature at a location within the continental United States with longitude x and latitude y at a certain time of day, then the temperature at the point (x, y) is the height, (or depth) of the surface with equation z  T(x, y). In this context the level curve, T(x, y)  k is a curve superimposed on the map of the United States connecting all, points that have the same temperature at a given time (Figure 11). These level curves, are called isotherms. Similarly, if P(x, y) measures the barometric pressure at the location (x, y), then the level curves of the function P are called isobars. All points on an, isobar P(x, y)  k have the same barometric pressure at a given time., 50, , 40, , 30, 30, , 60, , 40, 70, , 50, 60, , 80, , 70, 80, , FIGURE 11, Isotherms: level curves connecting, points that have the same temperature, , 70, , 80, , 80, , Functions of Three Variables and Level Surfaces, A function f of three variables is a rule that assigns to each ordered triple (x, y, z) in a, domain D  {(x, y, z) 冟 x, y, z 僆 R} a unique real number w denoted by f(x, y, z). For, example, the volume V of a rectangular box of length x, width y, and height z can be, described by the function f defined by f(x, y, z)  xyz., , EXAMPLE 8 Find the domain of the function f defined by, f(x, y, z)  1x  y  z  xeyz, Solution f(x, y, z) is a real number provided that x  y  z  0 or, equivalently,, z x  y. Therefore, the domain of f is, D  {(x, y, z) 冟 z, , x  y}, , This is the half-space consisting of all points lying on or below the plane z  x  y., , Since the graph of a function of three variables is composed of the points, (x, y, z, w), where w  f(x, y, z), lying in four-dimensional space, we cannot draw the, graphs of such functions. But by examining the level surfaces, which are the surfaces, with equations, f(x, y, z)  k, , k, a constant, , we are often able to gain some insight into the nature of f.

Page 17 :
13.1, , Functions of Two or More Variables, , 1043, , EXAMPLE 9 Find the level surfaces of the function f defined by, , z, , f(x, y, z)  x 2  y 2  z 2, , k1, k4, k9, , Solution The required level surfaces of f are the graphs of the equations, x 2  y 2  z 2  k, where k  0. These surfaces are concentric spheres of radius 1k, centered at the origin (see Figure 12). Observe that f has the same value for all points, (x, y, z) lying on any such sphere., y, , x, , FIGURE 12, The level surfaces of, f(x, y, z)  x 2  y 2  z 2, corresponding to k  1, 4, 9, , 13.1, , CONCEPT QUESTIONS, 4. What is a level surface of a function of three variables? If, w  T(x, y, z) gives the temperature of a point (x, y, z) in, three-dimensional space, what does the level surface w  k, describe?, , 1. What is a function of two variables? Give an example of, one by stating its rule, domain, and range., 2. What is the graph of a function of two variables? Illustrate, with a sketch., 3. What is a level curve of a function of two variables? Illustrate with a sketch., , 13.1, , EXERCISES, , 1. Let f(x, y)  x 2  3xy  2x  3. Find, a. f(1, 2), b. f(2, 1), c. f(2h, 3k), d. f(x  h, y), e. f(x, y  k), 2. Let t(x, y) , , 2xy, 2x  3y 2, 2, , 9. t(x, y)  24  x 2  y 2, 2, , x, yz, , 12. t(x, y, z) , , . Find, , a. t(1, 2), c. t(u, √), e. t(u  √, √), , 10. h(x, y)  ln(xy  1), , 11. f(x, y, z)  29  x  y  z 2, 2, , 13. h(u, √, w)  tan u  √ cos w, , b. t(2, 1), d. t(2, a), , 1, , 14. f(x, y, z) , , 24  x  y 2  z 2, 2, , 3. Let f(x, y, z)  2x 2  2y 2  3z 2. Find, a. f(1, 2, 3), b. f(0, 2, 1), c. f(t, t, t), d. f(u, u  1, u  1), e. f(x, x, 2x), , In Exercises 15–22, find and sketch the domain of the function., , 4. Let t(r, s, t)  res>t. Find, a. t(2, 0, 3), c. t(1, 1, 1), e. t(r  h, s  k, t  l), , 17. f(u, √) , b. t(1, ln 3, 1), d. t(t, t, t), , 15. f(x, y)  1y  1x, , 7. f(u, √) , , u√, u√, , u  √2, , 20. h(x, y) , , ln(y  x), 1x  y  1, , 6. t(x, y)  x  2y  3, , 21. f(x, y, z)  29  x 2  y 2  z 2, , 8. h(x, y)  1x  2y, , 22. t(x, y, z) , , 2, , 2, , V Videos for selected exercises are available online at www.academic.cengage.com/login., , 24  x 2  y 2, z3, , xy, 2x  y, , 18. h(x, y)  1xy  1, , 19. f(x, y)  x ln y  y ln x, , In Exercises 5–14, find the domain and the range of the function., 5. f(x, y)  x  3y  1, , u√, 2, , 16. t(x, y) 

Page 18 :
1044, , Chapter 13 Functions of Several Variables, , In Exercises 23–30, sketch the graph of the function., 23. f(x, y)  4, , In Exercises 33–38, match the function with one of the graphs, labeled a through f., (b), , (a), , 24. f(x, y)  6  2x  3y, , z, , 25. f(x, y)  x 2  y 2, , z, , 26. t(x, y)  y 2, 27. h(x, y)  9  x 2  y 2, 28. f(x, y)  2x 2  y 2, 29. f(x, y) , , 1, 236  9x 2  36y 2, 2, , y, , y, x, , 30. f(x, y)  cos x, , x, , 31. The figure shows the contour map of a hill. The numbers in, the figure are measured in feet. Use the figure to answer the, questions below., , (d), , (c), , z, , z, , N, W, , E, , C, , S, 500, , 400, , S, , y, , 300, 200, , B, , x, y, , 100, x, , A, (f), , (e), , z, , z, , a. What is the altitude of the point on the hill corresponding to the point A? The point B?, b. If you start out from the point on the hill corresponding, to point A and move north, will you be ascending or, descending? What if you move east from the point on the, hill corresponding to point B?, c. Is the hill steeper at the point corresponding to A or at, the point corresponding to C? Explain., 32. A contour map of a function f is shown in the figure. Use it, to estimate the value of f at P and Q., y, , 4, , Q, 3, , 23.5, 18.4, 12.5, 8.0, , 35. f(x, y)  cos, , x, cos y, 2, , 36. f(x, y)  (x 2  y 2)ex, , 0, , 38. f(x, y)  , , graph of the function., x, , 39. f(x, y)  3x 2  3y 2  2, 2y2, , 40. f(x, y)  (4x 2  9y 2)ex, , P, , 2, , 34. f(x, y)  cos(x 2  y 2), , 33. f(x, y)  2x 2  y 3, , 41. f(x, y)  cos x  cos y, , 19.9, , 42. f(x, y) , 2, , 2y2, , x, 2(x 2  y 2), , cas In Exercises 39–42, use a computer or calculator to plot the, , 3.8, 3.6, 9.3, 14.5, 26.7, , y, , x, , 2y2, , 0, , 4, 4, , x, , 37. f(x)  ex, , 2, , 2, , y, , 4, , 1  2 sin(x 2  y 2), x 2  y2

Page 19 :
13.1, In Exercises 43–52, sketch the level curves f(x, y)  k of the, function for the indicated values of k., , 1045, , Functions of Two or More Variables, , (e), , (f), y, , y, , 43. f(x, y)  2x  3y; k  2, 1, 0, 1, 2, 44. f(x, y)  x 2  4y 2; k  0, 1, 2, 3, 4, 45. f(x, y)  xy; k  2, 1, 0, 1, 2, 46. f(x, y)  216  x 2  y 2; k  0, 1, 2, 3, 4, , 48. f(x, y)  y 2  x 2;, , k  2, 1, 0, 1, 2, , 49. f(x, y)  ln(x  y); k  2, 1, 0, 1, 2, , 24y2, , 57. f(x, y)  e12x, , x, 50. f(x, y)  ; k  2, 1, 0, 1, 2, y, 51. f(x, y)  y  x 2;, 52. f(x, y)  x  sin y;, , x, , x, , xy, 47. f(x, y) , ; k  2, 0, 1, 2, xy, , z, , k  2, 1, 0, 1, 2, k  2, 1, 0, 1, 2, , In Exercises 53–56, describe the level surfaces of the function., 53. f(x, y, z)  2x  4y  3z  1, 54. f(x, y, z)  2x 2  3y 2  6z 2, 55. f(x, y, z)  x 2  y 2  z 2, 56. f(x, y, z)  x 2  y 2  z  2, y, , In Exercises 57–62, match the graph of the surface with one of, the contour maps labeled a through f., , x, , 58. f(x, y)  x  y, , (b), , (a), y, , 2, , z, , y, , x, , x, , (d), , (c), , y, y, , y, , x, , 59. f(x, y)  cos2x  y, 2, , x, , 2, , z, , x, y, , x

Page 20 :
1046, , Chapter 13 Functions of Several Variables, , 60. f(x, y)  sin x  sin y, , 68. Find an equation of the level surface of f(x, y, z)  2x 2 , 3y 2  z that contains the point (1, 2, 3)., , z, , 69. Can two level curves of a function f of two variables x and y, intersect? Explain., 70. A level set of f is the set S  {(x, y) 冟 f(x, y)  k, where k is, in the range of f}. Let, y, , f(x, y)  e, , 0, x 2  y2  1, , if x 2  y 2  1, if x 2  y 2  1, , Sketch the level set of f for k  0 and 3., 71. Refer to Exercise 70. Let, , x, , 61. f(x, y)  sin(x  y), , f(x, y)  e, , z, , 1  2x 2  y 2, x 2  y2  1, , if x 2  y 2  1, if x 2  y 2  1, , (a) Sketch the graph of f and (b) describe the level set of f, for k  0, 12, 1, and 3., 72. Body Mass The body mass index (BMI) is used to identify,, evaluate, and treat overweight and obese adults. The BMI, value for an adult of weight w (in kilograms) and height h, (in meters) is defined to be, M  f(w, h) , , y, x, , 62. f(x, y)  ln(2x 2  y 2), , z, , w, h2, , According to federal guidelines, an adult is overweight if he, or she has a BMI value between 25 and 29.9 and is “obese”, if the value is greater than or equal to 30., a. What is the BMI of an adult who weighs in at 80 kg and, stands 1.8 m tall?, b. What is the maximum weight of an adult of height 1.8 m, who is not classified as overweight or obese?, 73. Poiseuille’s Law Poiseuille’s Law states that the resistance R,, measured in dynes, of blood flowing in a blood vessel of, length l and radius r (both in centimeters) is given by, R  f(l, r) , , y, , x, cas In Exercises 63–66, (a) use a computer or calculator to plot the, , graph of the function f, and (b) plot some level curves of f and, compare them with the graph obtained in part (a)., 63. f(x, y)  冟 x 冟  冟 y 冟, 65. f(x, y) , , xy(x 2  y 2), , 64. f(x, y) , , xy, 2x  y, 2, , 2, , x 2  y2, 2y2, , 66. f(x, y)  ye1x, , 67. Find an equation of the level curve of f(x, y)  2x 2  y 2, that contains the point (3, 4)., , kl, r4, , where k is the viscosity of blood (in dyne-sec/cm2). What is, the resistance, in terms of k, of blood flowing through an, arteriole with radius 0.1 cm and length 4 cm?, 74. Surface Area of a Human Body An empirical formula by, E.F. Dubois relates the surface area S of a human body (in, square meters) to its weight W (in kilograms) and its height, h (in centimeters). The formula, given by, S  0.007184W 0.425H 0.725, is used by physiologists in metabolism studies., a. Find the domain of the function S., b. What is the surface area of a human body that weighs, 70 kg and has a height of 178 cm?, 75. Cobb-Douglas Production Function Economists have found that, the output of a finished product, f(x, y), is sometimes, described by the function, f(x, y)  ax by 1b

Page 21 :
13.1, where x stands for the amount of money expended for labor,, y stands for the amount expended on capital, and a and b, are positive constants with 0  b  1., a. If p is a positive number, show that f( px, py)  pf(x, y)., b. Use the result of part (a) to show that if the amount of, money expended for labor and capital are both increased, by r percent, then the output is also increased by r percent., 76. Continuous Compound Interest If a principal of P dollars is, deposited in an account earning interest at the rate of r/year, compounded continuously, then the accumulated amount at, the end of t years is given by, A  f(P, r, t)  Pert, dollars. Find the accumulated amount at the end of 3 years, if $10,000 is deposited in an account earning interest at the, rate of 10%/year., 77. Home Mortgages Suppose a home buyer secures a bank loan, of A dollars to purchase a house. If the interest rate charged, is r/year and the loan is to be amortized in t years, then the, principal repayment at the end of i months is given by, B  f(A, r, t, i)  A c, , 1 1  12r 2 i  1, d, 1 1  12r 2 12t  1, , 0, , i, , 12t, , Suppose the Blakelys borrow $280,000 from a bank to, help finance the purchase of a house and the bank charges, interest at a rate of 6%/year, compounded monthly. If the, Blakelys agree to repay the loan in equal installments over, 30 years, how much will they owe the bank after the sixtieth, payment (5 years)? The 240th payment (20 years)?, 78. Wilson Lot-Size Formula The Wilson lot-size formula in economics states that the optimal quantity Q of goods for a, store to order is given by, Q  f(C, N, h) , , 2CN, B h, , where C is the cost of placing an order, N is the number of, items the store sells per week, and h is the weekly holding, cost for each item. Find the most economical quantity of, ten-speed bicycles to order if it costs the store $20 to place, an order and $5 to hold a bicycle for a week and the store, expects to sell 40 bicycles a week., 79. Force Generated by a Centrifuge A centrifuge is a machine, designed for the specific purpose of subjecting materials to a, sustained centrifugal force. The magnitude of a centrifugal, force F in dynes is given by, F  f(M, S, R) , , p2S 2MR, 900, , where S is in revolutions per minute (rpm), M is the mass in, grams, and R is the radius in centimeters. Find the centrifugal force generated by an object revolving at the rate of, 600 rpm in a circle of radius 10 cm. Express your answer, as a multiple of the force of gravity. (Recall that 1 gram of, force is equal to 980 dynes.), , Functions of Two or More Variables, , 1047, , 80. Temperature of a Thin Metal Plate A thin metal plate located in, the xy-plane has a temperature of, T(x, y) , , 120, 1  2x 2  y 2, , degrees Celsius at the point (x, y). Describe the isotherms, of T, and sketch those corresponding to T  120, 60, 40,, and 20., 81. International America’s Cup Class Drafted by an international, committee in 1989, the rules for the new International, America’s cup class includes a formula that governs the, basic yacht dimensions. The formula f(L, S, D) 42,, where, f(L, S, D) , , L  1.25S 1>2  9.80D 1>3, 0.388, , balances the rated length L (in meters), the rated sail area S, (in square meters) and the displacement D (in cubic meters)., All changes in the basic dimensions are tradeoffs. For, example, if you want to pick up speed by increasing the, sail area, you must pay for it by decreasing the length or, increasing the displacement, both of which slow the boat, down. Show that yacht A of rated length 20.95 m, rated sail, area 277.3 m2, and displacement 17.56 m3, and the longer, and heavier yacht B with L  21.87, S  311.78, and, D  22.48 both satisfy the formula., 82. Ideal Gas Law According to the ideal gas law, the volume V, of an ideal gas is related to its pressure P and temperature T, by the formula, V, , kT, P, , where k is a positive constant. Describe the level curves of, V, and give a physical interpretation of your result., 83. Newton’s Law of Gravitation According to Newton’s Law of, Gravitation a body of mass m 1 located at the origin of an, xyz-coordinate system attracts another body of mass m 2, located at the point (x, y, z) with a force of magnitude, given by, F, , Gm 1m 2, x  y2  z2, 2, , where G is the universal constant of gravitation. Describe, the level surfaces of F, and give a physical interpretation of, your result., 84. Equipotential Curves Consider the crescent-shaped region R (in, the following figure) that lies inside the disk, D1  {(x, y) 冟 (x  2)2  y 2, , 4}, , and outside the disk, D2  {(x, y) 冟 (x  1)2  y 2, , 1}

Page 22 :
1048, , Chapter 13 Functions of Several Variables, y, , where c is the speed of sound in still air, about 1100 ft/sec., (This phenomenon is called the Doppler effect.) Suppose, a railroad train is traveling at 100 ft/sec (approximately, 68 mph) in still air and the frequency of a note emitted by, the locomotive whistle is 500 Hz. What is the frequency of, the note heard by a passenger in a train moving at 50 ft/sec, in the opposite direction to the first train?, , 2, D1, ƒ  50, , 1, , ƒ  100, , R, , D2, 0, , 1, , 2, , 3, , x, , 4, , 86. A function f(x, y) is homogeneous of degree n if it satisfies, the equation f(tx, ty)  t nf(x, y) for all t. Show that, , 1, 2, , xy  y 2, 2x  y, , f(x, y) , , If the electrostatic potential along the inner circle is kept at, 50 volts and the electrostatic potential along the outer circle, is kept at 100 volts, then the electrostatic potential at any, point (x, y) in the region R is given by, f(x, y)  150 , , 200x, x 2  y2, , Show that the equipotential curves of f are arcs of circles, that have their centers on the positive x-axis and pass, through the origin. Sketch the equipotential curve corresponding to a potential of 75 volts., 85. The Doppler Effect Suppose that a sound with frequency f is, emitted by an object moving along a straight line with speed, u and that a listener is traveling along the same line in the, opposite direction with speed √. Then the frequency F heard, by the listener is given by, Fa, , is homogeneous of degree 1., In Exercises 87–90, determine whether the statement is true or, false. If it is true, explain why it is true. If it is false, give an, example to show why it is false., 87. f is a function of x and y if and only if for any two, points P1 (x 1, y1) and P2 (x 2, y2) in the domain of f,, f(x 1, y1)  f(x 2, y2) implies that P1(x 1, y1)  P2(x 2, y2)., 88. The equation x 2  y 2  z 2  4 defines at least two functions of x and y., 89. The level curves of a function f of two variables, f(x, y)  k,, exist for all values of k., 90. The level surfaces of the function f(x, y, z)  ax  by , cz  d consist of a family of parallel planes that are orthogonal to the vector n  ai  bj  ck., , c√, bf, cu, , 13.2 Limits and Continuity, An Intuitive Definition of a Limit, Figure 1 shows the graph of a function f of two variables. This figure suggests that, f(x, y) is close to the number L if the point (x, y) is close to the point (a, b)., z, f (x, y), L, (x, y, f (x, y)), , z  f (x, y), , FIGURE 1, The functional value f(x, y) is close, to L if (x, y) is close to (a, b)., , 0, x, , (a, b), , (x, y), , y

Page 23 :
13.2, , Limits and Continuity, , 1049, , DEFINITION Limit of a Function of Two Variables at a Point, Let f be a function that is defined for all points (x, y) close to the point (a, b), with the possible exception of (a, b) itself. Then the limit of f(x, y) as (x, y), approaches (a, b) is L, written, lim, , (x, y)→(a, b), , f(x, y)  L, , if f(x, y) can be made as close to L as we please by restricting (x, y) to be sufficiently close to (a, b)., , y, , (a, b), , 0, , FIGURE 2, There are infinitely many paths the, point (x, y) could take in approaching, the point (a, b)., , x, , At first glance, there appears to be little difference between this definition and the, definition of the limit of a function of one variable, with the exception that the points, (x, y) and (a, b) lie in the plane. But there are subtle differences. In the case of a function of one variable, the point x can approach the point x  a from only two directions: from the left and from the right. As a consequence, the function f has a limit L, as x approaches a if and only if f(x) approaches L from the left (lim x→a f(x)  L) and, from the right (lim x→a f(x)  L), a fact that we observed in Section 1.1., The situation is a little more complicated in the case of a function of two variables, because there are infinitely many ways in which we can approach a point (a, b) in the, plane (Figure 2). Thus, if f has a limit L as (x, y) approaches (a, b) , then f(x, y) must, approach L along every possible path leading to (a, b)., To see why this is true, suppose that, f(x, y) → L 1, , as, , (x, y) → (a, b), , f(x, y) → L 2, , as, , (x, y) → (a, b), , along a path C1 and that, along another path C2, where L 1  L 2. Then no matter how close (x, y) is to (a, b) ,, f(x, y) will assume values that are close to L 1 and also values that are close to L 2 depending on whether (x, y) is on C1 or on C2. Therefore, f(x, y) cannot be made as close as, we please to a unique number L by restricting (x, y) to be sufficiently close to (a, b);, that is, lim (x, y)→(a, b) f(x, y) cannot exist., An immediate consequence of this observation is the following criterion for demonstrating that a limit does not exist., , Technique for Showing That lim (x, y)→(a, b) f(x, y) Does Not Exist, If f(x, y) approaches two different numbers as (x, y) approaches (a, b) along two, different paths, then lim (x, y)→(a, b) f(x, y)  L does not exist., , EXAMPLE 1 Show that, , lim, , (x, y)→(0, 0), , x 2  y2, x 2  y2, , does not exist., , Solution The function f(x, y)  (x 2  y 2)>(x 2  y 2) is defined everywhere except at, (0, 0). Let’s approach (0, 0) along the x-axis (see Figure 3). On the path C1, y  0, so, lim, (x, y)→(0, 0), along C1, , f(x, y)  lim f(x, 0)  lim, x→0, , x→0, , x2, x2, ,  lim 1  1, x→0

Page 24 :
1050, , Chapter 13 Functions of Several Variables, , Next, let’s approach (0, 0) along the y-axis. On the path C2, x  0 (Figure 3), so, , y, , lim, (0, y), , (x, y)→(0, 0), along C2, , C2, , y→0, , FIGURE 3, A point on C1 has the form (x, 0), and, a point on C2 has the form (0, y)., , EXAMPLE 2 Show that, ,  lim (1)  1, y→0, , xy, , lim, , (x, y)→(0, 0), , x  y2, 2, , does not exist., , Solution The function f(x, y)  xy>(x 2  y 2) is defined everywhere except at (0, 0)., Let’s approach (0, 0) along the x-axis (Figure 4). On the path C1, y  0, so, lim, (x, y)→(0, 0), along C1, , y, , f(x, y)  lim f(x, 0)  lim, x→0, , x→0, , (x, x), C2, C3, , lim, , C1, (x, 0), , 0, x2, ,  lim 0  0, x→0, , Similarly, you can show that f(x, y) also approaches 0 as (x, y) approaches (0, 0) along, the y-axis, path C2 (Figure 4)., Now consider yet another approach to (0, 0), this time along the line y  x (Figure 4). On the path C3, y = x, so, , yx, , 0, , y2, , x, , (x, 0), , (0, y), , y→0, , y 2, , Since f(x, y) approaches two different numbers as (x, y) approaches (0, 0) along two, different paths, we conclude that the given limit does not exist., , C1, 0, , f(x, y)  lim f(0, y)  lim, , x, , FIGURE 4, f(x, y) → 0 as (x, y) → (0, 0) along, C1 and C2, but f(x, y) → 12 as, (x, y) → (0, 0) along C3, so, lim (x, y)→(0, 0) f(x, y) does not exist., , (x, y)→(0, 0), along C3, , f(x, y)  lim f(x, x)  lim, x→0, , x→0, , x2, x2  x2, ,  lim, , x→0, , 1, 1, , 2, 2, , Since f(x, y) approaches two different numbers as (x, y) approaches (0, 0) along two, different paths, we conclude that the given limit does not exist., The graph of f shown in Figure 5 confirms this result visually. Notice the ridge that, occurs above the line y  x because f(x, y)  12 for all points (x, y) on that line except, at the origin., z, y, , x, , FIGURE 5, xy, The graph of f(x, y)  2, x  y2, , Although the method of Examples 1 and 2 is effective in demonstrating when a, limit does not exist, it cannot be used to prove the existence of the limit of a function, at a point. Using this method, we would have to show that f(x, y) approaches a unique, number L as (x, y) approaches the point along every path, which is clearly an impossible task. Fortunately, the Limit Laws for a function of a single variable can be extended, to functions of two or more variables. For example, the Sum Law, the Product Law,, the Quotient Law, and so forth, all hold. So does the Squeeze Theorem.

Page 25 :
13.2, , Limits and Continuity, , 1051, , EXAMPLE 3 Evaluate, a. lim (x, y)→(1, 2) (x 3y 2  x 2y  x 2  2x  3y), b. lim (x, y)→(2, 4), , 8xy, B 2x  y, 3, , Solution, a. We have, lim, , (x, y)→(1, 2), , (x 3y 2  x 2y  x 2  2x  3y)  (1)3(2)2  (1) 2(2)  (1)2  2(1)  3(2), 7, , b. We have, 8xy, 8xy,  3, lim, (x, y)→(2, 4) B 2x  y, B (x, y)→(2, 4) 2x  y, 3, , lim, , , , 8(2)(4), 3, 1, 82, B 2(2)  4, 3, , The next example utilizes the Squeeze Theorem to show the existence of a limit., , EXAMPLE 4 Find, , 2x 2y, , lim, , x 2  y2, , (x, y)→(0, 0), , if it exists., , Solution Observe that the numerator of the rational function has degree 3, whereas, the denominator has degree 2. This suggests that when x and y are both close to zero,, the numerator is much smaller than the denominator, and we suspect that the limit, might exist and that it is equal to zero., To prove our assertion, we observe that y 2  0, so x 2>(x 2  y 2) 1. Therefore,, `, , 0, Let f(x, y)  0, t(x, y)  `, lim, , (x, y)→(0, 0), , f(x, y) , , 2x 2y, x2  y, , 2x 2y, x 2  y2, , lim, , (x, y)→(0, 0), , `, 2, , 2x 2 冟 y 冟, , 2冟y冟, , x 2  y2, , ` , and h(x, y)  2 冟 y 冟. Then, , 00, , and, , lim, , (x, y)→(0, 0), , h(x, y) , , lim, , (x, y)→(0, 0), , 2冟y冟  0, , By the Squeeze Theorem,, lim, , (x, y)→(0, 0), , t(x, y) , , lim, , (x, y)→(0, 0), , `, , 2x 2y, x 2  y2, , `0, , and this, in turn, implies that, lim, , (x, y)→(0, 0), , 2x 2y, x 2  y2, , 0, , Continuity of a Function of Two Variables, The definition of continuity for a function of two variables is similar to that for a function of one variable.

Page 26 :
1052, , Chapter 13 Functions of Several Variables, , DEFINITION Continuity at a Point, Let f be a function that is defined for all points (x, y) close to the point (a, b)., Then f is continuous at the point (a, b) if, lim, , (x, y)→(a, b), , f(x, y)  f(a, b), , Thus, f is continuous at (a, b) if f(x, y) approaches f(a, b) as (x, y) approaches (a, b), along any path. Loosely speaking, a function f is continuous at a point (a, b) if the graph, of f does not have a hole, gap, or jump at (a, b). If f is not continuous at (a, b), then f, is said to be discontinuous there. For example, the functions f, t, and h whose graphs, are shown in Figure 6 are discontinuous at the indicated points., z, , z, , z, (c, d, g(c, d)), , 0, , FIGURE 6, , 0, , 0, y, , (a, b), , x, , z  h(x, y), , z  g(x, y), , z  f (x, y), , (c, d), , x, , (a) f is not defined at (a, b)., , (b), , lim, , (x, y) → (c, d), , g(x, y)  g(c, d), , y, , y, x, (c), , lim, , (x, y) → (0, 0), , h(x, y) does not exist., , y, , Continuity on a Set, , ∂, (a, b), , N∂, , Let’s digress a little to introduce some terminology. We define the D-neighborhood, about (a, b) to be the set, Nd  {(x, y) 冟 2(x  a)2  (y  b)2  d}, x, , 0, , FIGURE 7, The d-neighborhood about (a, b), y, Interior, point, Boundary, point, , Thus, Nd is just the set of all points lying inside the circle of radius d centered at (a, b), (see Figure 7)., Let R be a plane region. A point (a, b) is said to be an interior point of R if there, exists a d-neighborhood about (a, b) that lies entirely in R (Figure 8). A point (a, b) is, called a boundary point of R if every d-neighborhood of R contains points in R and, also points not in R., A region R is said to be an open region if every point of R is an interior point of, R. A region is closed if it contains all of its boundary points. Finally, a region that contains some but not all of its boundary points is neither open nor closed. For example,, the regions, , R, 0, , x, , FIGURE 8, An interior point and a boundary point of R, , A  e(x, y) `, , y2, x2, ,  1f ,, 9, 4, , B  e(x, y) `, , y2, x2, , 9, 4, , 1f, , and, C  e(x, y) `, , y2, x2, , 9, 4, , y2, x2, 1; y  0f d e(x, y) `, ,  1; y  0f, 9, 4

Page 27 :
13.2, y, , y, , B, x, , (a) A is open., , 1053, , y, , A, 0, , Limits and Continuity, , C, x, , 0, , (b) B is closed., , 0, , x, , (c) C is neither open nor closed., , FIGURE 9, Every point in A is an interior point; B contains all of its boundary points;, C contains some but not all of its boundary points., , shown in Figure 9a–c are open, closed, and neither open nor closed, respectively., As we mentioned in Section 1.3, continuity is a “local” concept. The following definition explains what we mean by continuity on a region., , DEFINITION Continuity on a Region, Let R be a region in the plane. Then f is continuous on R if f is continuous at, every point (x, y) in R. If (a, b) is a boundary point, the condition for continuity is modified to read, lim, , (x, y)→(a, b), , f(x, y)  f(a, b), , where (x, y) 僆 R, that is, (x, y) is restricted to approach (a, b) along paths lying, inside R., , EXAMPLE 5 Show that the function f defined by f(x, y)  29  x 2  y 2 is contin-, , uous on the closed region R  {(x, y) 冟 x 2  y 2 9}, which is the set of all points lying, on and inside the circle of radius 3 centered at (0, 0) in the xy-plane., Solution Observe that the set R is precisely the domain of f. Now, if (a, b) is any interior point of R, then, lim, , (x, y)→(a, b), , f(x, y) , , lim, , (x, y)→(a, b), , 2, , lim, , 29  x 2  y 2, , (x, y)→(a, b), , (9  x 2  y 2), ,  29  a 2  b 2,  f(a, b), This shows that f is continuous at (a, b)., Next, if (c, d) is a boundary point of R and (x, y) is restricted to lie inside R, we, obtain, lim, , (x, y)→(c, d), , f(x, y)  f(c, d), , as before, thus showing that f is continuous at (c, d) as well., The graph of f is the upper hemisphere of radius 3 centered at the origin together, with the circle in the xy-plane having equation x 2  y 2  9. (See Figure 10.)

Page 28 :
1054, , Chapter 13 Functions of Several Variables, , z, 3, , z  f (x, y), , 0, , 3, , 3, , FIGURE 10, The graph of f(x, y)  29  x 2  y 2, has no holes, gaps, or jumps., , y, , x, , The following theorem summarizes the properties of continuous functions of, two variables. The proofs of these properties follow from the limit laws and will be, omitted., , THEOREM 1 Properties of Continuous Functions of Two Variables, If f and t are continuous at (a, b) , then the following functions are also continuous at (a, b)., a. f, , b. ft, , t, , c. cf, , c, a constant, , d. f>t t(a, b)  0, , A consequence of Theorem 1 is that polynomial and rational functions are continuous., A polynomial function of two variables is a function whose rule can be expressed, as a finite sum of terms of the form cx my n, where c is a constant and m and n are nonnegative integers. For example, the function f defined by, f(x, y)  2x 2y 5  3xy 3  8xy 2  3y  4, is a polynomial function in the two variables x and y. A rational function is the quotient of two polynomial functions. For example, the function t defined by, t(x, y) , , x 3  xy  y 2, x 2  y2, , is a rational function., , THEOREM 2 Continuity of Polynomial and Rational Functions, A polynomial function is continuous everywhere (that is, in the whole plane)., A rational function is continuous at all points in its domain (that is, at all points, where its denominator is defined and not equal to zero)., , EXAMPLE 6 Determine where the function is continuous:, a. f(x, y) , , xy(x 2  y 2), x y, 2, , 2, , b. t(x, y) , , 1, y  x2

Page 29 :
13.2, , Limits and Continuity, , 1055, , Solution, a. The function f is a rational function and is therefore continuous everywhere, except at (0, 0) , where its denominator is equal to zero (Figure 11)., b. The function t is a rational function and is continuous everywhere except along, the curve y  x 2, where its denominator is equal to zero (Figure 12)., z, , y, , x, , 0, , y, , x, , FIGURE 11, The graph of f has a hole at the origin., , xy (x 2  y 2_), (b) The graph of z  _________, x2  y2, , (a) The domain of f, , z, y, , y  x2, y, , FIGURE 12, As (x, y) approaches the curve y  x 2, from the region y  x 2, z  f(x, y), approaches infinity; as (x, y), approaches the curve y  x 2 from, the region y  x 2, z approaches, minus infinity., , x, , 0, , x, 1, (b) The graph of z  _______2, yx, , (a) The domain of t, , The next theorem tells us that the composite function of two continuous functions, is also a continuous function., , THEOREM 3 Continuity of a Composite Function, If f is continuous at (a, b) and t is continuous at f(a, b), then the composite function h  t ⴰ f defined by h(x, y)  t( f(x, y)) is continuous at (a, b)., , EXAMPLE 7 Determine where the function is continuous:, a. F(x, y)  sin xy, , b. G(x, y) , , 1, 2, , cos(2x 2  y 2), 1  2x 2  y 2

Page 30 :
1056, , Chapter 13 Functions of Several Variables, , Solution, a. We can view the function F as the composition t ⴰ f of the functions f and t, defined by f(x, y)  xy and t(t)  sin t. Thus,, F(x, y)  t( f(x, y))  sin ( f(x, y))  sin xy, Since f is continuous on the whole plane and t is continuous on (⬁, ⬁), we conclude that F is continuous everywhere. The graph of F is shown in Figure 13a., b. The function G is the quotient of p(x, y)  12 cos(2x 2  y 2) and q(x, y) , 1  2x 2  y 2. The function p in turn involves the composition of t(t)  12 cos t, and f(x, y)  2x 2  y 2. Since both f and t are continuous everywhere, we, see that p is continuous everywhere. The function q is continuous everywhere, as well and is never zero. Therefore, by Theorem 3, G is continuous everywhere., The graph of G is shown in Figure 13b., z, , z, , y, y, x, , x, , FIGURE 13, , cos (2x 2  y 2), (b) G(x, y)  ____________, is continuous everywhere., 1  2x 2  y 2, 1_, 2, , (a) F(x, y)  sin xy is continuous everywhere., , Functions of Three or More Variables, The notions of the limit of a function of three or more variables and that of the continuity of a function of three or more variables parallel those of a function of two variables. For example, if f is a function of three variables, then we write, lim, , (x, y, z)→(a, b, c), , f(x, y, z)  L, , to mean that there exists a number L such that f(x, y, z) can be made as close to L as, we please by restricting (x, y, z) to be sufficiently close to (a, b, c)., , EXAMPLE 8 Evaluate, , limp, , e2y (sin x  cos y), , (x, y, z)→( 2 , 0, 1), , 1  y2  z2, , Solution, limp, , (x, y, z)→( 2 , 0, 1), , e2y (sin x  cos y), 1y z, 2, , 2, , , , ., , e0[sin (p>2)  cos 0], 101, , A function f of three variables is continuous at (a, b, c) if, lim, , (x, y, z)→(a, b, c), , f(x, y, z)  f(a, b, c), , , , 2, 1, 2

Page 31 :
13.2, , Limits and Continuity, , 1057, , ln z, , EXAMPLE 9 Determine where f(x, y, z) , , is continuous., 21  x 2  y 2  z 2, Solution We require that z  0 and 1  x 2  y 2  z 2  0; that is, z  0 and, x 2  y 2  z 2  1. So f is continuous on the set {(x, y, z) 冟 x 2  y 2  z 2  1 and, z  0}, which is the set of points above the xy-plane and inside the upper hemisphere, with center at the origin and radius 1., , The E-D Definition of a Limit (Optional), The notion of the limit of a function of two variables given earlier can be made more, precise as follows., , DEFINITION Limit of f(x, y), Let f be a function of two variables that is defined for all points (x, y) on a disk, with center at (a, b) with the possible exception of (a, b) itself. Then, lim, , (x, y)→(a, b), , f(x, y)  L, , if for every e  0, there exists a d  0 such that, 冟 f(x, y)  L 冟  e, , Geometrically speaking, f has the limit L at (a, b) if given any e  0, we can find, a circle of radius d centered at (a, b) such that L  e  f(x, y)  L  e for all interior points (x, y)  (a, b) of the circle (Figure 14)., , z, L+´, f(x, y), , L, (x, y, f(x, y)), , L−´, , EXAMPLE 10 Prove that lim (x, y)→(a, b) x  a., Solution, , Let e  0 be given. We need to show that there exists a d  0 such that, , z  f(x, y), 0, y, x, , 0  2(x  a)2  (y  b)2  d, , whenever, , ∂, (a, b), , FIGURE 14, f(x, y) lies in the interval, (L  e, L  e) whenever, (x, y)  (a, b) is in the, d-neighborhood of (a, b)., , 冟 f(x, y)  a 冟  e, whenever (x, y)  (a, b) is in the d-neighborhood about (a, b). To find such a d, consider, 冟 f(x, y)  a 冟  冟 x  a 冟  2(x  a)2, , (x, y), , 2(x  a)2  (y  b)2, , Thus, if we pick d  e, we see that d  0 and that 2(x  a)2  (y  b)2  d implies, that 冟 f(x, y)  a 冟  e as was to be shown. Since e is arbitrary, the proof is complete., , EXAMPLE 11 Prove that, Solution, , lim, , (x, y)→(0, 0), , 2x 2y, x 2  y2, ,  0. (See Example 4.), , Let e  0 be given. Consider, 冟 f(x, y)  0 冟  `, , 2x 2y, x2  y, , `  2 冟 y 冟a, 2, , 2 冟 y 冟  22y 2, , x2, x 2  y2, , b, , (x, y)  (0, 0), , 22x 2  y 2, , If we pick d  e>2, then d  0, and 2x 2  y 2  d implies that 冟 f(x, y)  0 冟  e., Since e is arbitrary, the proof is complete.

Page 32 :
1058, , Chapter 13 Functions of Several Variables, , 13.2, , CONCEPT QUESTIONS, , 1. a. Explain what it means for a function f of two variables to, have a limit at (a, b)., b. Describe a technique that you could use to show that the, limit of f(x, y) as (x, y) approaches (a, b) does not exist., 2. Explain what it means for a function of two variables to be, continuous (a) at a point (a, b) and (b) on a region in the, plane., 3. Determine whether each function f is continuous or discontinuous. Explain your answer., a. f(P, T ) measures the volume of a balloon ascending into, the sky as a function of the atmospheric pressure P and, the air temperature T., , 13.2, , EXERCISES, , In Exercises 1–12, show that the limit does not exist., x y, 2, , 1., , lim, , (x, y)→(0, 0), , 3., , lim, , (x, y)→(0, 0), , 5., , lim, , (x, y)→(0, 0), , 7., , lim, , (x, y)→(1, 0), , 8., , lim, , (x, y)→(0, 0), , 9., , 3x  y, 2, , 17., , 2x  3xy  4y, 2, , (x, y)→(0, 0), , 2xy, 2x  y, , lim, , 4., , lim, , 6., , 4, , (x, y)→(0, 0), , 2, , x y, 2, , 23., 25., , x y z, , 2, , 22., , lim, , lim, , ex sin1 (y  x), , lim, , e2x, , (x, y)→(0, 1), , ln(x 2  3y), , lim, , 24., , x 2 sin p(2x  y), x, sin1 a b, y, x, 1, y, , (x, y)→(0, 1), , cos1 (x  2y), , lim, , x  2y 2  z 4, 2, , x  t 2, y  t 2, z  t., , 27., , lim, , (x, y)→(0, 0), , 29., , lim, , (x, y)→(0, 0), , xy, , lim, , lim, , (x, y)→11, 12 2, , 2, , y2, , (x, y)→(3, 4), , xy  yz  xz, xyz  3, [esin px  ln(cos p(y  z))], , In Exercises 27–30, use polar coordinates to find the limit. Hint:, If x  r cos u and y  r sin u, then (x, y) → (0, 0) if and only if, r → 0., , x  y3  z3, 3, , Hint: Approach (0, 0, 0) along the curve with parametric equations, , (x, y, z)→(0, 0, 0), , lim, , 20., , x, tan1 a b, y, , (x, y, z)→(0, 3, 1), , xz 2  2y 2, , lim, , e, xy1, , (x, y, z)→(1, 2, 3), , 26., , 2xyz, , lim, , lim, , (x, y)→(2, 1), , xy  yz  xz, , (x, y, z)→(0, 0, 0), , 12., , 21., , (x, y)→(1, 1), , 2x 2  y 6, , 18., , 2x  y 2, , sin xy, x 2  y2, , xy 3 cos x, , 2, , lim, , (x, y)→(0, 0), , 4, , x 2  y 2  2x  1, , 2, , (x, y)→(1, 2), , 2, , 1xy, , 19., , xy 2, , 2xy  2y, , lim, , 3xy, , lim, , 2x 2  3y 2, , (x, y)→(0, 0), , 2, , 4, , lim, , 2., , 3xy, , (x, y, z)→(0, 0, 0), , 11., , 2, , 2x 2  y 2, , (x, y, z)→(0, 0, 0), , 10., , b. f(H, W) measures the surface area of a human body as a, function of its height H and weight W., c. f(d, t) measures the fare as a function of distance d and, time t for taking a cab from O’Hare Airport to downtown, Chicago., d. f(T, P) measures the volume of a certain mass of gas as a, function of the temperature T and the pressure P., 4. Suppose f has the property that it is not defined at the point, (1, 2) but lim (x, y)→(1, 2) f(x, y)  3. Can you define f(1, 2) so, that f is continuous at (1, 2)? If so, what should the value of, f(1, 2) be?, , x 2  y2  z2, , 30., , lim, , (x, y)→(0, 0), , x 3  y3, , 28., , x 2  y2, , lim, , (x, y)→(0, 0), , sin(2x 2  2y 2), x 2  y2, , (x  y ) ln(x  y ), 2, , 2, , 2, , 2, , tan(2x 2  2y 2), tanh(3x 2  3y 2), , In Exercises 13–26, find the given limit., 13., , lim, , (x, y)→(1, 2), , 14., 15., , lim, , (x 2  2y 2), , (x, y)→(1, 1), , lim, , (x, y)→(1, 2), , In Exercises 31–40, determine where the function is continuous., 31. f(x, y) , , (2x  xy  3y  1), 2, , 2x 2  3y 3  4, 3  xy, , 16., , lim, , (x, y)→(1, 3), , x  2y 2, (x  1)(y  1), , 2xy, 2x  3y  1, , 32. f(x, y) , , x 3  xy  y 3, x 2  y2, , 33. t(x, y)  1x  y  1x  y, 34. h(x, y)  sin(2x  3y), , V Videos for selected exercises are available online at www.academic.cengage.com/login., , 35. F(x, y)  1xex>y

Page 33 :
13.3, , 2, , 48. f(x, y)  y ln x, t(t)  et, , xyz, x 2  y2  z2  4, , 49. Use the precise definition of a limit to prove that, lim (x, y)→(a, b) c  c where c is a constant., , 38. t(x, y, z)  1x  cos 1y  z, 39. h(x, y, z)  x ln(yz  1), 40. F(x, y, z)  x tan, , 1059, , 47. f(x, y)  x tan y, t(t)  cos t, , 36. G(x, y)  ln(2x  y), 37. f(x, y, z) , , Partial Derivatives, , 50. Use the precise definition of a limit to prove that, lim (x, y)→(a, b) y  b., , y, z, , 51. Use the precise definition of a limit to prove that, , 41. Let, sin xy, f(x, y)  • xy, 1, , lim, , if xy  0, if xy  0, , a. Determine all the points where f is continuous., , (x, y)→(0, 0), , 3xy 3, x 2  y2, , 0, , 52. Use the precise definition of a limit to prove that if, lim (x, y)→(a, b) f(x, y)  L and c is a constant, then, lim (x, y)→(a, b) cf(x, y)  cL., , cas b. Plot the graph of f. Does the graph give a visual confir-, , mation of your conclusion in part (a)?, 42. Let, x,  y if x  0, f(x, y)  • sin x, 1y, if x  0, a. Determine all the points where f is continuous., cas b. Plot the graph of f. Does the graph give a visual confir-, , mation of your conclusion in part (a)?, In Exercises 43–48, find h(x, y)  t( f(x, y)), and determine, where h is continuous., 43. f(x, y)  x  xy  y , t(t)  t cos t  sin t, 2, , 2, , 44. f(x, y)  x  xy  xy  y , t(t)  te, 3, , 2, , 45. f(x, y)  2x  y, t(t) , , 3, , 53. If lim (x, y)→(a, b) f(x, y)  L, then, lim (x, y)→(a, b) along C f(x, y)  L, where C is any path leading, to (a, b) ., 54. If lim (x, y)→(a, b) f(x, y)  L and f is defined at (a, b), then, f(a, b)  L., 55. If f(x, y)  t(x)h(y) , where t and h are continuous at a and, b, respectively, then f is continuous at (a, b) ., 56. If f(1, 3)  4, then lim (x, y)→(1, 3) f(x, y)  4., 57. If f is continuous at (3, 1) and f(3, 1)  2, then, lim (x, y)→(3, 1) f(x, y)  2., 58. If f is continuous at (a, b) and t is continuous at f(a, b) ,, then lim (x, y)→(a, b) t( f(x, y))  t( f(a, b)) ., , t2, t1, , 46. f(x, y)  x  2y  3, t(t)  1t , , t, , In Exercises 53–58, determine whether the statement is true or, false. If it is true, explain why it is true. If it is false, give an, example to show why it is false., , 1, t, , 13.3 Partial Derivatives, Partial Derivatives of Functions of Two Variables, For a function of one variable x, there is no ambiguity when we speak of the rate of, change of f(x) with respect to x. The situation becomes more complicated, however,, when we study the rate of change of a function of two or more variables. For example, for the function of two variables defined by the equation z  f(x, y), both the independent variables x and y may be allowed to vary in some arbitrary fashion, thus making it unclear what we mean by the phrase “the rate of change of z with respect to x, and y.”, One way of getting around this difficulty is to hold one variable constant and consider the rate of change of f with respect to the other variable. This approach might be, familiar to anyone who has used the expression “everything else being equal” while, debating the merits of a complicated issue.

Page 34 :
1060, , Chapter 13 Functions of Several Variables, , Specifically, suppose that (a, b) is a point in the domain of f. Fix y  b. Then the, function that is defined by z  f(x, b) is a function of the single variable x. Its graph, is the curve C formed by the intersection of the vertical plane y  b and the surface, z  f(x, y) (Figure 1)., Therefore, the quantity, , z, T, C, (a, b, f(a, b)), z  f (x, y), , f(a  h, b)  f(a, b), h→0, h, lim, , 0, yb, , y, (a, b), , x, , FIGURE 1, f(a  h, b)  f(a, b), lim, measures, h→0, h, the slope of T and the rate of change, of f(x, y) in the x-direction when, x  a and y  b., z, , (a, b, f (a, b)), , if it exists, measures both the slope of the tangent line T to the curve C at the point, (a, b, f(a, b)) as well as the rate of change of f(x, y) with respect to x (in the xdirection) with y held constant when x  a and y  b., Similarly, the quantity, f(a, b  h)  f(a, b), h→0, h, lim, , xa, , T, , 0, , x, , DEFINITIONS Partial Derivatives of a Function of Two Variables, Let z  f(x, y). Then the partial derivative of f with respect to x is, , y, (a, b), , FIGURE 2, f(a, b  h)  f(a, b), lim, measures, h→0, h, the slope of T and the rate of change, of f(x, y) in the y-direction when, x  a and y  b., , (2), , if it exists, measures the slope of the tangent line T to the curve C (formed by the intersection of the vertical plane x  a and the surface z  f(x, y) at (a, b, f(a, b)), and the, rate of change of f(x, y) with respect to y (in the y-direction) with x held constant when, x  a and y  b (Figure 2)., In expressions (1) and (2) the point (a, b) is fixed but otherwise arbitrary. Therefore, we may replace (a, b) by (x, y), leading to the following definitions., , z  f (x, y), C, , (1), , f, f(x  h, y)  f(x, y),  lim, x h→0, h, and the partial derivative of f with respect to y is, f, f(x, y  h)  f(x, y),  lim, y h→0, h, provided that each limit exists., , Computing Partial Derivatives, The partial derivatives of f can be calculated by using the following rules., , Computing Partial Derivatives, To compute f> x, treat y as a constant and differentiate in the usual manner, with respect to x (an operation that we denote by > x)., To compute f> y, treat x as a constant and differentiate in the usual manner with respect to y (an operation that we denote by > y).

Page 35 :
13.3, , Partial Derivatives, , 1061, , f, f, and, if f(x, y)  2x 2y 3  3xy 2  2x 2  3y 2  1., x, y, , EXAMPLE 1 Find, , Solution To compute f> x, we think of the variable y as a constant and differentiate with respect to x. Let’s write, f(x, y)  2x 2y 3  3xy 2  2x 2  3y 2  1, where the variable y to be treated as a constant is shown in color. Then, f,  4xy 3  3y 2  4x, x, To compute f> y, we think of the variable x as a constant and differentiate with, respect to y. In this case,, f(x, y)  2x 2y 3  3xy 2  2x 2  3y 2  1, and, f,  6x 2y 2  6xy  6y, y, Before looking at more examples, let’s introduce some alternative notations for the, partial derivatives of a function. If z  f(x, y), then, x, , f(x, y) , , f,  fx  z x, x, , and, , y, , f(x, y) , , f,  fy  z y, y, , EXAMPLE 2 Find fx and fy if f(x, y)  x cos xy 2., Solution To compute fx, we think of the variable y as a constant and differentiate with, respect to x. Thus,, f(x, y)  x cos xy 2, and, fx , , x, , (x cos xy 2)  x, ,  x(sin xy 2), , x, , x, , (cos xy 2)  (cos xy 2), , (xy 2)  cos xy 2, , x, , (x), , Use the Product Rule., , Use the Chain Rule on the first term., ,  xy 2 sin xy 2  cos xy 2, Next, to compute fy, we treat x as a constant and differentiate with respect to y., Thus,, f(x, y)  x cos xy 2, and, fy , , y, , (x cos xy 2)  x, ,  x(sin xy 2), , y, , y, , (cos xy 2)  (cos xy 2), , (xy 2)  0  2x 2y sin xy 2, , y, , (x)

Page 36 :
1062, , Chapter 13 Functions of Several Variables, , EXAMPLE 3 Let f(x, y)  4  2x 2  y 2. Find the slope of the tangent line at the, , point (1, 1, 1) on the curve formed by the intersection of the surface z  f(x, y) and, a. the plane y  1, , b. the plane x  1, , Solution, a. The slope of the tangent line at any point on the curve formed by the intersection, of the plane y  1 and the surface z  4  2x 2  y 2 is given by, f, , (4  2x 2  y 2)  4x, x, x, In particular, the slope of the required tangent line is, f, `,  4(1)  4, x (1, 1), b. The slope of the tangent line at any point on the curve formed by the intersection, of the plane x  1 and the surface z  4  2x 2  y 2 is given by, f, , (4  2x 2  y 2)  2y, y, y, In particular, the slope of the required tangent line is, f, `,  2(1)  2, y (1, 1), (See Figure 3.), z, , z, T, T, C, x1, , y1, , C, (1, 1, 1), , (1, 1, 1), , 0, , 0, y, , y, , x, , x, , (a) The slope of the tangent line is −4., , FIGURE 3, , (b) The slope of the tangent line is −2., , EXAMPLE 4 Electrostatic Potential Figure 4 shows a crescent-shaped region, R that lies inside the disk D1  {(x, y) 冟 (x  2)2  y 2 4} and outside the disk, D2  {(x, y) 冟 (x  1)2  y 2 1}. Suppose that the electrostatic potential along the, inner circle is kept at 50 volts and the electrostatic potential along the outer circle is, kept at 100 volts. Then the electrostatic potential at any point (x, y) in R is given by, U(x, y)  150 , volts., , 200x, x 2  y2

Page 37 :
13.3, y, , 1063, , a. Compute Ux (x, y) and Uy(x, y)., b. Compute Ux (3, 1) and Uy (3, 1) and interpret your results., U  50, D2, , 0, , Partial Derivatives, , U  100, D1, , 2, , 4, , x, , Solution, a. Ux (x, y) , , x, , c150 , , 200x, x y, 2, , (x 2  y 2), , x, , , FIGURE 4, The electrostatic potential inside the, crescent-shaped region is U(x, y)., , , Uy (x, y) , , 2, , d, , 200x, , x x  y2, 2, , (200x)  200x, , x, , b, , (x 2  y 2), , (x 2  y 2)2, 200(x 2  y 2)  200x(2x), , y, , (x 2  y 2)2, c150 , ,  200x, , y, , 200x, x y, 2, , 2, , d, ,  200x(x 2  y 2) 2(2y) , 200(9  1), (9  1)2, , , , 200(x 2  y 2), (x 2  y 2)2, , 200x, a, b, y x 2  y2, , (x 2  y 2) 1, ,  200x(1)(x 2  y 2) 2, , b. Ux (3, 1) , , a, ,  16, , y, , (x 2  y 2), , 400xy, (x  y 2) 2, , and, , 2, , Uy(3, 1) , , 400(3)(1), (9  1)2, ,  12, , This tells us that the rate of change of the electrostatic potential at the point (3, 1), in the x-direction is 16 volts per unit change in x with y held fixed at 1, and the, rate of change of the electrostatic potential at the point (3, 1) in the y-direction, is 12 volts per unit change in y with x held fixed at 3., , EXAMPLE 5 A Production Function The production function of a certain country is, given by, f(x, y)  20x 2>3y 1>3, billion dollars, when x billion dollars of labor and y billion dollars of capital are spent., a. Compute fx (x, y) and fy (x, y)., b. Compute fx (125, 27) and fy(125, 27) , and interpret your results., c. Should the government encourage capital investment rather than investment in, labor to increase the country’s productivity?, Solution, a. fx (x, y) , fy (x, y) , , 2, 40 y 1>3, (20x 2>3y 1>3)  (20) a x 1>3 b (y 1>3) , a b, x, 3, 3 x, 1, 20 x 2>3, (20x 2>3y 1>3)  (20x 2>3) a y 2>3 b , a b, y, 3, 3 y

Page 38 :
1064, , Chapter 13 Functions of Several Variables, , b. fx (125, 27) , , 40 27 1>3 40 3, a, b , a b8, 3 125, 3 5, , This says that the production is increasing at the rate of $8 billion per billion dollar increase in labor expenditure when the labor expenditure stands at $125 billion (capital expenditure held constant at $27 billion)., Next,, fy (125, 27) , , 20 125 2>3 20 25, 14, a, b , a b  18, 3 27, 3 9, 27, , This tells us that production is increasing at the rate of approximately $18.5 billion per billion dollar increase in capital outlay when the capital expenditure, stands at $27 billion (with labor expenditure held constant at $125 billion)., c. Yes. Since a unit increase in capital expenditure results in a greater increase in, production than a unit increase in labor, the government should encourage spending on capital rather than on labor., Sometimes we have available only the contour map of a function f. In such instances, we can use the contour map to help us estimate the partial derivatives of f at a specified point, as the following example shows., , EXAMPLE 6 Figure 5 shows the contour map of a function f. Use it to estimate, fx (3, 1) and fy(3, 1) ., y, , 8, 1, , 2, 0, , 6, , 10, , 2, 12, , 4, 0, , 2, , 4 14, , 6, , x, , 16, , FIGURE 5, A contour map of f, , Solution To estimate fx (3, 1), we start at the point (3, 1), where the value of f at (3, 1), can be read off from the contour map: f(3, 1)  8. Then we proceed along the positive x-axis until we arrive at the point on the next level curve whose location is approximately (3.8, 1). Using the definition of the partial derivative, we find, fx (3, 1) ⬇, , f(3.8, 1)  f(3, 1), 10  8, ,  2.5, 3.8  3, 0.8, , Similarly, starting at the point (3, 1) and moving along the positive y-axis, we find, fy (3, 1) ⬇, , f(3, 3)  f(3, 1), 68, ,  1, 31, 2

Page 39 :
13.3, , Partial Derivatives, , 1065, , Implicit Differentiation, EXAMPLE 7 Suppose z is a differentiable function of x and y that is defined implicitly by x 2  y 3  z  2yz 2  5. Find z> x and z> y., Solution, , Differentiating the given equation implicitly with respect to x, we find, x, , (x 2  y 3  z  2yz 2) , , 2x , , x, , (5), , z, z,  2ya2z b  0, x, x, , Remember that y is, treated as a constant., , z, (4yz  1)  2x  0, x, and, z, 2x, , x, 1  4yz, Next, differentiating the given equation with respect to y, we obtain, y, 3y 2 , , (x 2  y 3  z  2yz 2) , , (5), , z, z,  2ya2z b  2z 2  0, y, y, , 3y 2 , and, , y, , z, (1  4yz)  2z 2  0, y, 3y 2  2z 2, z, , y, 1  4yz, , Partial Derivatives of Functions of More Than Two Variables, The partial derivatives of a function of more than two variables are defined in much, the same way as the partial derivatives of a function of two variables. For example,, suppose that f is a function of three variables defined by w  f(x, y, z). Then the partial derivative of f with respect to x is defined as, f, f(x  h, y, z)  f(x, y, z), w, ,  lim, x, x h→0, h, where y and z are held fixed, provided that the limit exists. The other two partial derivatives, f> y and f> z, are defined in a similar manner., , Finding the Partial Derivative of a Function of More Than Two Variables, To find the partial derivative of a function of more than two variables with respect, to a certain variable, say x, we treat all the other variables as if they are constants and differentiate with respect to x in the usual manner.

Page 40 :
1066, , Chapter 13 Functions of Several Variables, , EXAMPLE 8 Find, a. fx if f(x, y, z)  x 2y  y 2z  xz, , b. h w if h(x, y, z, w) , , xw 2, y  sin zw, , Solution, a. To find fx, we treat y and z as constants and differentiate f with respect to x to, obtain, fx , , x, , (x 2y  y 2z  xz)  2xy  z, , b. To find h w, we treat x, y, and z as constants and differentiate h with respect to w,, obtaining, xw 2, b, w y  sin zw, a, , hw , , (y  sin zw), , w, , , , , , , (xw 2)  xw 2, , w, , (y  sin zw), Use the Quotient Rule., , (y  sin zw)2, (y  sin zw)(2xw)  xw 2 c0  cos zw ⴢ, , w, , (zw)d, Use the Chain Rule., , (y  sin zw)2, 2xw(y  sin zw)  xw 2z cos zw, (y  sin zw)2, , xw(2y  2 sin zw  wz cos zw), , , , (y  sin zw)2, , Higher-Order Derivatives, Consider the function z  f(x, y) of two variables. Each of the partial derivatives f> x, and f> y are functions of x and y. Therefore, we may take the partial derivatives of, these functions to obtain the four second-order partial derivatives, 2, , f, , x, , 2, , , , x, , a, , 2, , f, b,, x, , f, , y x, , , , y, , a, , f, b,, x, , 2, , f, , x y, , , , x, , a, , f, b,, y, , 2, , and, , f, , y, , 2, , , , y, , a, , f, b, y, , (See Figure 6.), , ∂, ∂x, , ∂f, ∂x, , f, ∂, ∂y, , FIGURE 6, The differential operators are shown, on the limbs of the tree diagram., , ∂f, ∂y, , ∂, ∂x, , ∂2f, ∂ ∂f, = 2, ∂x ∂x, ∂x, , ∂, ∂y, , ∂ ∂f, ∂2f, =, ∂y ∂x, ∂y ∂x, , (, (, (, (, , ∂, ∂x, , ), ), ), ), , ∂ ∂f, ∂2f, =, ∂x ∂y, ∂x ∂y, ∂2f, ∂ ∂f, = 2, ∂y ∂y, ∂y, , ∂, ∂y, , Before we turn to an example, let’s introduce some additional notation for the, second-order partial derivatives of f :, 2, , f, , x, , 2, , 2, ,  fxx, , f, , y x, , 2, ,  fxy, , f, , x y, , 2, ,  fyx, , f, , y2, ,  fyy

Page 41 :
13.3, , Partial Derivatives, , 1067, , Note the order in which the derivatives are taken: Using the notation 2f>( y x), we, differentiate first with respect to x—the independent variable that appears first when, read from right to left. In the notation fxy we also differentiate first with respect to x—, the independent variable that appears first when read from left to right. The derivatives, fxy and fyx are called mixed partial derivatives., Note If f is defined by the equation z  f(x, y), then the four partial derivatives of f, are also written, z xx, , z xy z yx, , and, , z yy, , EXAMPLE 9 Find the second-order partial derivatives of f(x, y)  2xy 2  3x 2  xy 3., Solution, , We first compute the first-order partial derivatives, fx , , Historical Biography, , x, , (2xy 2  3x 2  xy 3)  2y 2  6x  y 3, , SPL/Photo Researchers, Inc., , and, fy , , y, , (2xy 2  3x 2  xy 3)  4xy  3xy 2, , Then differentiating each of these functions, we obtain, fxx , , ALEXIS CLAUDE CLAIRAUT, (1713–1765), Alexis Claude Clairaut was one of twenty, children born to his mother but the only to, survive to adulthood. His father, a mathematics teacher in Paris, educated his son, at home with very high standards: Alexis, was taught to read using Euclid’s Elements., As a result of both nature and nurture,, Clairaut turned out to be a very precocious, mathematician. He studied calculus by the, age of 10 and wrote an original mathematical paper at 13. At 18 he published his first, paper; he also became the youngest member ever admitted to the prestigious Academie des Sciences. Clairaut excelled in, many areas of mathematics, including, geometry, calculus, and celestial mechanics. He was the first to prove the prediction, by Isaac Newton and the astronomer Christiaan Huygens that the earth is an oblate, ellipsoid. Clairaut also accurately predicted, the return of Halley’s comet in 1759, a prediction that made him famous. Clairaut, developed the notation for partial derivatives that we still use today, and he was, the first to prove that mixed second-order, partial derivatives of a function at a point, are equal if the derivatives are continuous, at that point., , fxy , fyx , fyy , , x, y, x, y, , fx , fx , fy , fy , , x, y, x, y, , (2y 2  6x  y 3)  6, (2y 2  6x  y 3)  4y  3y 2, (4xy  3xy 2)  4y  3y 2, (4xy  3xy 2)  4x  6xy, , Notice that the mixed derivatives fxy and fyx in Example 9 are equal. The following, theorem, which we state without proof, gives the conditions under which this is true., , THEOREM 1 Clairaut’s Theorem, If f(x, y) and its partial derivatives fx, fy, fxy, and fyx are continuous on an open, region R, then, fxy(x, y)  fyx (x, y), for all (x, y) in R., , A function u of two variables x and y is called a harmonic function if u xx  u yy  0, for all (x, y) in the domain of u. Harmonic functions are used in the study of heat conduction, fluid flow, and potential theory. The partial differential equation u xx  u yy  0, is called Laplace’s equation, named for Pierre Laplace (1749–1827).

Page 43 :
13.3, , 13.3, , 1069, , CONCEPT QUESTIONS, , 1. a. Define the partial derivatives of a function of two variables, x and y, with respect to x and with respect to y., b. Give a geometric and a physical interpretation of fx(x, y) ., 2. Let f be a function of x and y. Describe a procedure for finding fx and fy., , 13.3, , Partial Derivatives, , 3. Suppose F(x, y, z)  0 defines x implicitly as a function of y, and z; that is, x  f(y, z). Describe a procedure for finding, x> z. Illustrate with an example of your choice., 4. If f is a function of x and y, give a condition that will guarantee that fxy(x, y)  fyx(x, y) for all (x, y) in some open region., , EXERCISES, , 1. Let f(x, y)  x 2  2y 2., a. Find fx(2, 1) and fy (2, 1)., b. Interpret the numbers in part (a) as slopes., c. Interpret the numbers in part (a) as rates of change., 2. Let f(x, y)  9  x 2  xy  2y 2., a. Find fx(1, 2) and fy (1, 2)., b. Interpret the numbers in part (a) as slopes., c. Interpret the numbers in part (a) as rates of change., , (b), , 2, 2.0, 1.5, 1.0, 0.5, 0.0, 2, , 3. Determine the sign of f> x and f> y at the points P, Q,, and R on the graph of the function f shown in the figure., , 1, , 0, , z, , x, , 1, , 0, y, , (c), z, 1, , 2, , 1 2, , 0, , 1, , 2, , 1 2, , Q, , P, , 2, R, , z, 0, 2, , y, 2, , x, , 4. The graphs of a function f and its partial derivatives fx and fy, are labeled (a), (b), and (c). Identify the graphs of f, fx, and, fy, and give a reason for your answer., (a), 2, , 1, , 0 1, , 2, , 0, y, , 1, , 2, , 5. The figure below shows the contour map of the function T, (measured in degrees Fahrenheit) giving the temperature at each, point (x, y) on an 8 in. 5 in. rectangular metal plate. Use it, to estimate the rate of change of the temperature at the point, (3, 2) in the positive x-direction and in the positive y-direction., y (in.), 5, , 0.5, z, 0, , 4, 75, , 80, , 3, , 105, , 110, , 0.5, 2, , x, , 1, , 100 95, , 85, , 90, , 2, x, , 1, , 0, y, , 1, , 2, , V Videos for selected exercises are available online at www.academic.cengage.com/login., , 1, 0, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , x (in.)

Page 44 :
1070, , Chapter 13 Functions of Several Variables, In Exercises 46–49, show that the mixed partial derivatives fxy, and fyx are equal., , In Exercises 6–29, find the first partial derivatives of the, function., 7. f(x, y)  2x 2  3xy  y 2, , 6. f(x, y)  3x  4y  2, 8. z  2x 3  3x 2y 3  xy 2  2x, 9. z  x 1y, , 10. f(x, y)  (2x 2  y 3) 4, , 46. f(x, y)  x 2  2x 2y  y 3, , 47. f(x, y)  x sin2 y  y 2 cos x, , 48. f(x, y)  e2x cos 3y, , 49. f(x, y)  tan1 (x 2  y 3), , In Exercises 50–53, show that the mixed partial derivatives fxyz ,, fyxz , and fzyx are equal., , 11. t(r, s)  1r  s 2, , 12. h(u, √)  ln(u 2  √2), , 13. f(x, y)  xey>x, , 14. f(x, y)  ex cos y  ey sin x, , 50. f(x, y, z)  x 2y 3  y 2z 3, , 16. f(x, y)  23  2x  y, , 51. f(x, y, z)  29  x 2  2y 2  z 2, , 1, , 15. z  tan (x  y ), 17. t(u, √) , , 2, , 2, , 2, , 52. f(x, y, z)  ln(x  2y  3z), , u√, u 2  √3, , 18. f(x, y)  sinh xy, , x, 19. t(x, y)  x cosh, y, , 20. z  ln(e  y ), , 21. f(x, y)  y, , x, , 53. f(x, y, z)  ex cos yz, , 2, , 2, , y, , 22. f(x, y) , , 2, , 54. The figure shows the contour map of a function f. Use it to, determine the sign of (a) fx, (b) fy, (c) fxx, (d) fxy, and (e) fyy, at the point P., , x, y, , 冮 cos t dt, , 23. f(x, y) , , x, , 冮 te, , t, , dt, , y, , x, , 2, , 24. f(x, y, z)  2x 3  3xy  2yz  z 2, 25. t(x, y, z)  1xyz, , 26. f(u, √, w)  ue  √e  we, , 27. u  xey>z  z 2, , 28. u  x sin, , u, , √, , u, , y, xz, , P, , 4, 6, 8, 10, , 29. f(r, s, t)  rs ln st, 0, , x, , In Exercises 30–33, use implicit differentiation to find z> x and, z> y., 31. xey  yex  ez  10, , 30. x 2y  xz  yz 2  8, , 32. 2 cos(x  2y)  sin yz  1  0, 33. ln(x 2  z 2)  yz 3  2x 2  10, , 55. u  et sin, , In Exercises 34–39, find the second partial derivatives of the, function., 34. f(x, y)  x  2x y  y  3x, 4, , 2 3, , 4, , x, c, , 2k2t, , 56. u  ec, , In Exercises 57 and 58, show that the function satisfies the, one-dimensional wave equation u tt  c2u xx., 58. u  sin(kct) sin(kx), , 36. z  xe2y  ye2x, 37. w  cos(2u  √)  sin(2u  √), 39. h(x, y)  tan1, , y, x, , In Exercises 59–64, show that the function satisfies the, two-dimensional Laplace’s equation u xx  u yy  0., , 40. f(x, y)  x 3  y 3  3x 2y 2  2x  3y  4;, 42. f(x, y, z)  ln(x 2  y 2  z 2);, , fyxz, , 3, , 43. z  x cos y  y sin x;, 3, , 44. p  eu√w;, , z, x y x, , p, u w √, , 45. h(x, y, z)  e cos(y  2z); h zzy, x, , fxyx, , fxxx, , x, , 59. u  3x 2y  y 3, , 60. u , , 61. u  ln2x 2  y 2, , 62. u  ex cos y  ey cos x, , In Exercises 40–45, find the indicated partial derivative., 41. f(x, y)  x 4  2x 2y 2  xy 3  2y 4;, , cos kx, , 57. u  cos(x  ct)  2 sin(x  ct), , 35. t(x, y)  x 3y 2  xy 3  2x  3y  1, , 38. z  2x 2  y 2, , In Exercises 55 and 56, show that the function satisfies the, one-dimensional heat equation u t  c2u xx., , 63. u  tan1, , x 2  y2, , y, x, , 64. u  cosh y sin x  sinh y cos x, In Exercises 65 and 66, show that the function satisfies the, three-dimensional Laplace’s equation u xx  u yy  u zz  0., 65. u  x 2  3xy  2y 2  3z 2  4xyz, 66. u  (x 2  y 2  z 2)1>2

Page 45 :
13.3, y, 67. Show that the function z  2x 2  y 2 tan1 satisfies the, x, z, z, equation x, y,  z., x, y, y, 68. Show that the function u  20x 2 cos satisfies the equation, x, u, u, y,  2u., x, x, y, , a. Compute Tx 1 12, 12 2 , and interpret your result., b. Find the rate of change of the temperature at the point, P 1 12, 12 2 in the y-direction., y, , T  100, , 69. According to the ideal gas law, the volume V (in liters), of an ideal gas is related to its pressure P (in pascals) and, temperature T (in kelvins) by the formula, V, , kT, P, , where k is a constant. Compute V> T and V> P if, k  8.314, T  300, and P  125, and interpret your, results., 70. Refer to Exercise 69. Show that, T, P, V, ⴢ, ⴢ,  1, T, P, V, 71. The total resistance R (in ohms) of three resistors with, resistances of R1, R2, and R3 ohms connected in parallel, is given by the formula, 1, 1, 1, 1, , , , R, R1, R2, R3, Find R> R1 and interpret your result., 72. The height of a hill (in feet) is given by, h(x, y)  20(16  4x 2  3y 2  2xy  28x  18y), where x is the distance (in miles) east and y the distance, (in miles) north of Bolton. If you are at a point on the hill, 1 mile north and 1 mile east of Bolton, what is the rate of, change of the height of the hill (a) in a northerly direction, and (b) in an easterly direction?, 73. Profit Versus Inventory and Floor Space The monthly profit (in, dollars) of the Barker Department Store depends on its level, of inventory x (in thousands of dollars) and the floor space y, (in thousands of square feet) available for display of its merchandise, as given by the equation, P(x, y)  0.02x 2  15y 2  xy  39x  25y  15,000, Find P> x and P> y when x  5000 and y  200, and, interpret your result., 74. Steady-State Temperature Consider the upper half-disk, H  {(x, y) 冟 x 2  y 2 1, y  0} (see the figure). If the, temperature at points on the upper boundary is kept at, 100°C and the temperature at points on the lower boundary, is kept at 50°C, then the steady-state temperature at any, point (x, y) inside the half-disk is given by, T(x, y)  100 , , 1  x 2  y2, 100, tan1, p, 2y, , 1071, , Partial Derivatives, , 1 T  50, , 0, , x, , 1, , 75. Electric Potential A charge Q (in coulombs) located at the origin of a three-dimensional coordinate system produces an, electric potential V (in volts) given by, V(x, y, z) , , kQ, 2x  y 2  z 2, 2, , where k is a positive constant and x, y, and z are measured, in meters. Find the rate of change of the potential at the, point P(1, 2, 3) in the x-direction., 76. Surface Area of a Human The formula, S  0.007184W 0.425H 0.725, gives the surface area S of a human body (in square meters), in terms of its weight W (in kilograms) and its height H (in, centimeters). Compute S> W and S> H when W  70, and H  180, and interpret your results., 77. Arson Study A study of arson for profit conducted for a, certain city found that the number of suspicious fires is, approximated by the formula, N(x, y) , , 12021000  0.03x 2y, (5  0.2y)2, 0 x, , 150,, , 5, , y, , 35, , where x denotes the number of persons per census tract and, y denotes the level of reinvestment in conventional mortgages by the city’s ten largest banks measured in cents per, dollars deposited. Compute N> x and N> y when, x  100 and y  20, and interpret your results., 78. Production Functions The productivity of a Central American, country is given by the function, f(x, y)  20x 3>4y 1>4, when x units of labor and y units of capital are used., a. What are the marginal productivity of labor and the, marginal productivity of capital when the amounts, expended on labor and capital are 256 units and, 16 units, respectively?, b. Should the government encourage capital investment, rather than increased expenditure on labor at this time, to increase the country’s productivity?

Page 46 :
1072, , Chapter 13 Functions of Several Variables, , 79. Wind Chill Factor A formula that meteorologists use to, calculate the wind chill temperature (the temperature that, you would feel in still air that is the same as the actual, temperature when the presence of wind is taken into consideration) is, T  f(t, s)  35.74  0.6125t  35.75s 0.16  0.4275ts 0.16, s1, where t is the air temperature in degrees Fahrenheit and s is, the wind speed in mph., a. What is the wind chill temperature when the actual air, temperature is 32°F and the wind speed is 20 mph?, b. What is the rate of change of the wind chill temperature, with respect to the wind speed if the temperature is 32°F, and the wind speed is 20 mph?, 80. Wind Chill Factor The wind chill temperature is the temperature that you would feel in still air that is the same as the, actual temperature when the presence of wind is taken into, consideration. The following table gives the wind chill temperature T  f(t, s) in degrees Fahrenheit in terms of the, actual air temperature t in degrees Fahrenheit and the wind, speed s in mph., , Wind speed (mph), , Actual air, temperature (ºF), , t, , s, , 30, , 10, , 15, , 20, , 25, , 30, , 35, , 40, , 21.2, , 19.0, , 17.4, , 16.0, , 14.9, , 13.9, , 13.0, , 32, , 23.7, , 21.6, , 20.0, , 18.7, , 17.6, , 16.6, , 15.8, , 34, , 26.2, , 24.2, , 22.6, , 21.4, , 20.3, , 19.4, , 18.6, , 36, , 28.7, , 26.7, , 25.2, , 24.0, , 23.0, , 22.2, , 21.4, , 38, , 31.2, , 29.3, , 27.9, , 26.7, , 25.7, , 24.9, , 24.2, , 40, , 33.6, , 31.8, , 20.5, , 29.4, , 28.5, , 27.7, , 82. a. Use the result of Exercise 81 to find fx 1 1, p2 2 if, f(x, y)  x 2 cos xy., b. Verify the result of part (a) by evaluating fx(x, y) at 1 1, p2 2 ., cas In Exercises 83 and 84, use the result of Exercise 81 and a cal-, , culator or computer to find the partial derivative., 83. fx (2, 1) if f(x, y)  ln 1 exy  cos2x 2  y 2 2, sin pxy, , 1 1  2x 2  y 3 2 3>2, , 84. fy (2, 1) if f(x, y) , , 85. Cobb-Douglas Production Function Show that the Cobb-Douglas, production function P  kx ay 1a, where 0  a  1, satisfies the equation, x, , P, P, y, P, x, y, , Note: This equation is called Euler’s equation., , 86. Let S be the surface with equation z  f(x, y), where f has, continuous first-order partial derivatives and P(x 0, y0, z 0) is, a point on S (see the figure). Let C1 and C2 be the curves, obtained by the intersection of the surface S with the planes, x  x 0 and y  y0, respectively. Let T1 and T2 be the tangent, lines to the curves C1 and C2 at P. Then the tangent plane to, the surface S at the point P is the plane that contains both, tangent lines T1 and T2., z, T1, v1, , n, , v2, , P, , T2, S, , 26.9, , C2, , C1, 0, , y0, , x0, , a. Estimate the rate of change of the wind chill temperature, T with respect to the actual air temperature when the, wind speed is constant at 25 mph and the actual air, temperature is 34°F., Hint: Show that it is given by, , f(36, 25)  f(34, 25), T, (34, 25) ⬇, t, 2, b. Estimate the rate of change of the wind chill temperature, T with respect to the wind speed when the actual air, temperature is constant at 34°F and the wind speed is, 25 mph., Source: National Weather Service., , 81. Let f be a function of two variables., a. Put t(x)  f(x, b) , and use the definition of the derivative of a function of one variable to show that, fx (a, b)  t¿(a) ., b. Put h(y)  f(a, y), and show that fy (a, b)  h¿(b)., , x, , (x0, y0), , y, , a. Show that the vectors v1  i  fx (x 0, y0)k and, v2  j  fy(x 0, y0)k are parallel to T1 and T2, respectively., b. Using the result of part (a), find a vector n that is normal, to both v1 and v2., c. Use the result of part (b) to show that an equation of the, tangent plane to S at P is, z  z 0  fx (x 0, y0)(x  x 0)  fy(x 0, y0)(y  y0), 87. Use the result of Exercise 86 to find an equation of the tangent plane to the paraboloid z  x 2  14 y 2 at the point, (1, 2, 2) ., 88. Engine Efficiency The efficiency of an internal combustion, engine is given by, E  a1 , , √ 0.4, b, V

Page 47 :
13.4, where V and √ are the respective maximum and minimum, volumes of air in each cylinder., a. Show that E> V  0, and interpret your result., b. Show that E> √  0, and interpret your result., 89. A semi-infinite strip has faces that are insulated. If the edges, x  0 and x  p of the strip are kept at temperature zero, and the base of the strip is kept at temperature 1, then the, steady-state temperature (that is, the temperature after a long, time) is given by, T(x, y) , Find, , T, x, , 1 p2 , 1 2 and, , T, y, , 1073, , a. Find fx(x, y) and fy (x, y) for (x, y)  (0, 0)., b. Use the definition of partial derivatives to find fx(0, 0), and fy(0, 0) ., c. Show that fxy(0, 0)  1 and fyx(0, 0)  1., d. Does the result of part (c) contradict Theorem 1? Explain., 91. Does there exist a function f of two variables x and y, with continuous second-order partial derivatives, such that fx(x, y)  e2x (2 cos xy  y sin xy) and, fy (x, y)  ye2x sin xy? Explain., 92. Show that if a function f of two variables x and y has continuous third-order partial derivatives, then fxyx  fyxx  fxxy., , 2, sin x, tan1, p, sinh y, , 1 p2 , 1 2 , and interpret your results., , y, , In Exercises 93–96, determine whether the statement is true or, false. If it is true, explain why it is true. If it is false, give an, example to show why it is false., 93. If z  f(x, y) has a partial derivative with respect to x at the, point (a, b) , then, , T0, , T0, , π, , x, , 90. Let, xy(x 2  y 2), x 2  y2, 0, , f(x, b)  f(a, b), f, (a, b)  lim, xa, x, x→a, 94. If f> y (a, b)  0, then the tangent line to the curve, formed by the intersection of the plane x  a and the, surface z  f(x, y) at the point (a, b, f(a, b)) is horizontal;, that is, it is parallel to the xy-plane., , T1, 0, , f(x, y)  •, , Differentials, , if (x, y)  (0, 0), if (x, y)  (0, 0), , 95. If fxx (x, y) is defined for all x and y and fxx (a, b)  0 for, all x in the interval (a, b), then the curve C formed by the, intersection of the plane y  b and the surface z  f(x, y), is concave downward on (a, b)., 96. If f(x, y)  ln xy, then fxy(x, y)  fyx (x, y) for all (x, y) in, D  {(x, y) 冟 xy  0}., , 13.4 Differentials, Increments, Recall that if f is a function of one variable defined by y  f(x), then the increment in, y is defined to be, ⌬y  f(x  ⌬x)  f(x), where ⌬x is an increment in x (Figure 1a). The increment of a function of two or more, variables is defined in an analogous manner. For example, if z is a function of two variables defined by z  f(x, y), then the increment in z produced by increments of ⌬x, and ⌬y in the independent variables x and y, respectively, is defined to be, ⌬z  f(x  ⌬x, y  ⌬y)  f(x, y), (See Figure 1b.), , (1)

Page 48 :
1074, , Chapter 13 Functions of Several Variables, z, (x  Δ x, y  Δy, f (x  Δ x, y  Δy)), y  f (x), , y, , z  f (x, y), Δ z  f (x  Δ x, y  Δy)  f(x, y), (x, y, f (x, y)), , (x  Δ x, y  Δy), Δy  f (x  Δ x)  f (x), (x, y), , 0, y, , 0, , x, , x  Δx, , x, x, , (a) The increment Δy is the change in y as x, changes from x to x  Δ x., , (x  Δ x, y  Δy), , (x, y), , (x, y  Δy), , (b) The increment Δ z is the change in z as x, changes from x to x  Δ x and y changes, from y to y  Δy., , FIGURE 1, , EXAMPLE 1 Let z  f(x, y)  2x 2  xy. Find ⌬z. Then use your result to find the, change in z if (x, y) changes from (1, 1) to (0.98, 1.03)., Solution, , Using Equation (1), we obtain, , ⌬z  f(x  ⌬x, y  ⌬y)  f(x, y),  [2(x  ⌬x)2  (x  ⌬x)(y  ⌬y)]  (2x 2  xy),  2x 2  4x ⌬x  2(⌬x)2  xy  x ⌬y  y ⌬x  ⌬x ⌬y  2x 2  xy,  (4x  y) ⌬x  x ⌬y  2(⌬x)2  ⌬x ⌬y, Next, to find the increment in z if (x, y) changes from (1, 1) to (0.98, 1.03), we note, that ⌬x  0.98  1  0.02 and ⌬y  1.03  1  0.03. Therefore, using the result, obtained earlier with x  1, y  1, ⌬x  0.02, and ⌬y  0.03, we obtain, ⌬z  [4(1)  1](0.02)  (1)(0.03)  2(0.02)2  (0.02)(0.03),  0.0886, You can verify the correctness of this result by computing the quantity, f(0.98, 1.03)  f(1, 1)., , The Total Differential, Recall from Section 2.9 that if f is a function of one variable defined by y  f(x), then, the differential of f at x is defined by, dy  f ¿(x) dx, where dx  ⌬x is the differential in x. Furthermore,, ⌬y ⬇ dy, if ⌬x is small (see Figure 2)., , (2)

Page 49 :
13.4, , Differentials, , 1075, , z, (x  Δ x, y  Δy, f(x  Δ x, y  Δy)), y, , y  f (x), , z  f(x, y), dz, , T, dy, , Δz, , Tangent plane, Δy, 0, y, x, , 0, , x, , x  Δx, , (x, y), , x, , (a) Relationship between dy and Δy, , (x  Δ x, y  Δy), , (b) Relationship between dz and Δz. The tangent, plane is the analog of the tangent line T in the, one-variable case., , FIGURE 2, , For an analog of this result for a function of two variables, we begin with the following definition., , DEFINITION Differentials, Let z  f(x, y), and let ⌬x and ⌬y be increments of x and y, respectively. The, differentials dx and dy of the independent variables x and y are, dx  ⌬x, , and, , dy  ⌬y, , The differential dz, or total differential, of the dependent variable z is, dz , , f, f, dx , dy  fx (x, y) dx  fy(x, y) dy, x, y, , Later in this section, we will show that, ⌬z  dz  e1 ⌬x  e2 ⌬y, where e1 and e2 are functions of ⌬x and ⌬y that approach 0 as ⌬x and ⌬y approach 0., This implies that, ⌬z ⬇ dz, , (3), , if both ⌬x and ⌬y are small., Figure 2b shows the geometric relationship between ⌬z and dz. Observe that as x, changes from x to x  ⌬x and y changes from y to y  ⌬y, ⌬z measures the change, in the height of the graph of f, whereas dz measures the change in the height of the, tangent plane.*, , *For now, we will rely on our intuitive definition of the tangent plane. We will define the tangent plane in, Section 13.7.

Page 50 :
1076, , Chapter 13 Functions of Several Variables, , EXAMPLE 2 Let z  f(x, y)  2x 2  xy., a. Find the differential dz., b. Compute the value of dz if (x, y) changes from (1, 1) to (0.98, 1.03) , and compare your result with the value of ⌬z obtained in Example 1., Solution, f, f, dx , dy  (4x  y) dx  x dy, x, y, b. Here x  1, y  1, dx  ⌬x  0.02, and dy  ⌬y  0.03. Therefore,, a. dz , , dz  [4(1)  1](0.02)  1(0.03)  0.09, The value of ⌬z obtained in Example 1 was 0.0886, so dz is a good approximation of ⌬z in this case. Observe that it is easier to compute dz than to compute ⌬z., , EXAMPLE 3 A storage tank has the shape of a right circular cylinder. Suppose that, the radius and height of the tank are measured at 1.5 ft and 5 ft, respectively, with a, possible error of 0.05 ft and 0.1 ft, respectively. Use differentials to estimate the maximum error in calculating the capacity of the tank., Solution The capacity (volume) of the tank is V  pr 2h. The error in calculating the, capacity of the tank is given by, ⌬V ⬇ dV , , V, V, dr , dh  2prh dr  pr 2 dh, r, h, , Since the errors in the measurement of r and h are at most 0.05 ft and 0.1 ft, respectively, we have dr  0.05 and dh  0.1. Therefore, taking r  1.5, h  5, dr  0.05,, and dh  0.1, we obtain, dV  2prh dr  pr 2 dh, ⬇ 2p(1.5)(5)(0.05)  p(1.5)2 (0.1)  0.975p, Thus, the maximum error in calculating the volume of the storage tank is approximately 0.975p, or 3.1, ft3., , EXAMPLE 4 The Error in Computing the Range of a Projectile If a projectile is fired, with an angle of elevation u and initial speed of √ ft/sec, then its range (in feet) is, R, , √2 sin 2u, t, , where t is the constant of acceleration due to gravity. (See Figure 3.) Suppose that a, projectile is launched with an initial speed of 2000 ft/sec at an angle of elevation of, p>12 radians and that the maximum percentage errors in the measurement of √ and u, are 0.5% and 1%, respectively., a. Estimate the maximum error in the computation of the range of the projectile., b. Find the maximum percentage error in computing the range of this projectile.

Page 51 :
13.4, , FIGURE 3, We want to find the range R of a, projectile fired with an angle of, elevation u and initial speed of √ ft/sec., , Differentials, , 1077, , q, R, , Solution, a. The error in the computation of R is, ⌬R ⬇ dR , , R, R, 2√ sin 2u, 2√2 cos 2u, d√ , du , d√ , du, t, t, √, u, , The maximum error in the computation of √ is (0.005)(2000) or 10 ft/sec; that is,, 冟 d√ 冟 10. Also, the maximum error in the computation of u is (0.01)(p>12) radians. In other words, 冟 du 冟 0.01(p>12). Therefore, the maximum error in computing the range of the projectile is approximately, 冟 ⌬R 冟 ⬇ 冟 dR 冟, , 2√2 cos 2u, 2√ sin 2u, 冟 d√ 冟 , 冟 du 冟, t, t, , p, p, 2(2000) sin a b, 2(2000)2 cos a b, 6, 6, 0.01p, , (10) , a, b, 32, 32, 12, ⬇ 1192, or approximately 1192 ft., b. Using √  2000 and u  p>12, we find the range of the projectile to be, p, (2000)2 sin a b, 6, √ sin 2u, R, ,  62,500, t, 32, 2, , Therefore, the maximum percentage error in computing the range of the projectile is, 100 `, , ⌬R, 1192, ` ⬇ 100a, b, R, 62,500, , or approximately 1.91%., , Error in Approximating ⌬z by dz, The following theorem tells us that dz gives a good approximation of ⌬z if ⌬x and ⌬y, are small, provided that both fx and fy are continuous., , THEOREM 1, Let f be a function defined on an open region R. Suppose that the points (x, y), and (x  ⌬x, y  ⌬y) are in R and that fx and fy are continuous at (x, y). Then, ⌬z  fx(x, y) ⌬x  fy (x, y) ⌬y  e1 ⌬x  e2 ⌬y, where e1 and e2 are functions of ⌬x and ⌬y such that, lim, , (⌬x, ⌬y)→(0, 0), , e1  0, , and, , lim, , (⌬x, ⌬y)→(0, 0), , e2  0

Page 52 :
1078, , Chapter 13 Functions of Several Variables, , PROOF Fix x and y. By adding and subtracting f(x  ⌬x, y) to ⌬z, we have, ⌬z  f(x  ⌬x, y  ⌬y)  f(x, y),  [ f(x  ⌬x, y)  f(x, y)]  [ f(x  ⌬x, y  ⌬y)  f(x  ⌬x, y)],  ⌬z 1  ⌬z 2, where ⌬z 1 is the change in z as (x, y) changes from (x, y) to (x  ⌬x, y) and ⌬z 2 is the, change in z as (x, y) changes from (x  ⌬x, y) to (x  ⌬x, y  ⌬y). (See Figure 4a.), z, (x  Δ x, y  Δy, f(x  Δ x, y  Δy)), (x  Δ x, y, f(x  Δ x, y)), y, Δ z2, C(x  Δ x, y  Δy), , (x, y, f(x, y)), , Δ z1, , 0, A(x, y), x, , FIGURE 4, , B(x  Δ x, y), , (x  Δ x, y1), A(x, y) (x1, y), , y, C(x  Δ x, y  Δy), , (a) Δ z1  f (x  Δ x, y)  f (x, y) and, Δ z2  f (x  Δ x, y  Δy)  f(x  Δ x, y), , B(x  Δ x, y), , 0, , x, , (b) The points A, B, and C shown in the, xy-plane., , On the interval between A and B, y is constant, so the function t defined by, t(t)  f(t, y) for x t x  ⌬x is a function of one variable. (See Figure 4b.) Therefore, by the Mean Value Theorem, there exists a point (x 1, y) with x  x 1  x  ⌬x, such that, t(x  ⌬x)  t(x)  t¿(x 1) ⌬x, Since t¿(x 1)  fx(x 1, y) , we have, ⌬z 1  f(x  ⌬x, y)  f(x, y)  t(x  ⌬x)  t(x),  t¿(x 1) ⌬x  fx (x 1, y) ⌬x, , x  x 1  x  ⌬x, , Next, on the interval between B and C, both x and ⌬x are constant, so the function, h defined by h(t)  f(x  ⌬x, t) for y t y  ⌬y is a function of one variable. (See, Figure 4b.) Therefore, by the Mean Value Theorem there exists a point (x  ⌬x, y1), with y  y1  y  ⌬y such that, h(y  ⌬y)  h(y)  h¿(y1) ⌬y, Since h¿(y1)  fy (x  ⌬x, y1) , we have, ⌬z 2  f(x  ⌬x, y  ⌬y)  f(x  ⌬x, y),  h(y  ⌬y)  h(y)  h¿(y1) ⌬y  fy (x  ⌬x, y1) ⌬y, Therefore,, ⌬z  ⌬z 1  ⌬z 2,  fx (x 1, y) ⌬x  fy (x  ⌬x, y1) ⌬y

Page 53 :
13.4, , Differentials, , 1079, , Adding and subtracting fx(x, y) ⌬x  fy (x, y) ⌬y to the right-hand side of the previous, equation and rearranging terms, we obtain, ⌬z  fx (x, y) ⌬x  fy (x, y) ⌬y  [ fx (x 1, y)  fx(x, y)] ⌬x  [ fy (x  ⌬x, y1)  fy(x, y)] ⌬y,  fx (x, y) ⌬x  fy (x, y) ⌬y  e1 ⌬x  e2 ⌬y, where, e1  fx (x 1, y)  fx(x, y), and, e2  fy (x  ⌬x, y1)  fy (x, y), Observe that as (⌬x, ⌬y) → (0, 0), x 1 → x and y1 → y. Therefore, the continuity of fx, and fy implies that, lim, , (⌬x, ⌬y)→(0, 0), , e1  0, , and, , lim, , (⌬x, ⌬y)→(0, 0), , e2  0, , and this proves the result., Note, , Observe that the conclusion of Theorem 1 can be written as, ⌬z  dz  e1 ⌬x  e2 ⌬y, , Therefore, if ⌬x and ⌬y are both small, then, ⌬z  dz  (small number)(small number)  (small number)(small number), and this quantity is a very small number, which accounts for the closeness of the, approximation. Compare this with the case of a function of one variable discussed in, Section 2.9., , Differentiability of a Function of Two Variables, The conclusion of Theorem 1 can be written as, ⌬z  dz  e1 ⌬x  e2 ⌬y, , (4), , where e1 → 0 and e2 → 0 as (⌬x, ⌬y) → (0, 0). We define a function of two variables, to be differentiable if z  f(x, y) satisfies Equation (4)., , DEFINITION Differentiability of a Function of Two Variables, Let z  f(x, y). The function f is differentiable at (a, b) if ⌬z can be expressed, in the form, ⌬z  fx (a, b) ⌬x  fy (a, b) ⌬y  e1 ⌬x  e2 ⌬y, where e1 → 0 and e2 → 0 as (⌬x, ⌬y) → (0, 0). The function f is differentiable, in a region R if it is differentiable at each point of R., , EXAMPLE 5 Show that the function f defined by f(x, y)  2x 2  xy is differentiable, in the plane., Solution Write z  f(x, y)  2x 2  xy, and let (x, y) be any point in the plane. Then, using the result of Example 1, we have, ⌬z  (4x  y) ⌬x  x ⌬y  2(⌬x)2  ⌬x ⌬y

Page 54 :
1080, , Chapter 13 Functions of Several Variables, , Since fx  4x  y and fy  x, we can write, ⌬z  fx ⌬x  fy ⌬y  e1 ⌬x  e2 ⌬y, where e1  2 ⌬x and e2  ⌬x. Since e1 → 0 and e2 → 0 as (⌬x, ⌬y) → (0, 0), it follows that f is differentiable at (x, y). But (x, y) is any point in the plane, so f is differentiable in the plane., The next theorem, which is an immediate consequence of Theorem 1, guarantees, when a function of two variables is differentiable., , THEOREM 2 Criterion for Differentiability, Let f be a function of the variables x and y. If fx and fy exist and are continuous, on an open region R, then f is differentiable in R., , For the function f(x, y)  2x 2  xy of Example 5, we have fx (x, y)  4x  y and, fy (x, y)  x, both of which are continuous everywhere. Therefore, by Theorem 2 we, conclude that f is differentiable in the plane, as demonstrated earlier., , !, , Remember that the mere existence of the partial derivatives fx and fy of a function, f at a point (x, y) is not enough to guarantee the differentiability of f at (x, y). (See, Exercise 43.), , Differentiability and Continuity, Just as a differentiable function of one variable is continuous, the following theorem, shows that a differentiable function of two variables is also continuous., , THEOREM 3 Differentiable Functions Are Continuous, Let f be a function of two variables. If f is differentiable at (a, b) , then f is continuous at (a, b)., , PROOF Using the result of Theorem 1, we have, ⌬z  f(a  ⌬x, b  ⌬y)  f(a, b),  fx (a, b) ⌬x  fy (a, b) ⌬y  e1 ⌬x  e2 ⌬y, Writing x  a  ⌬x and y  b  ⌬y, we have, f(x, y)  f(a, b)  [ fx (a, b)  e1](x  a)  [ fy (a, b)  e2](y  b), Noting that e1 → 0 and e2 → 0 as (⌬x, ⌬y) → (0, 0), we see that, f(x, y)  f(a, b) → 0 as, , (⌬x, ⌬y) → (0, 0), , Equivalently,, lim, , (x, y)→(a, b), , Therefore, f is continuous at (a, b)., , f(x, y)  f(a, b)

Page 55 :
13.4, , Differentials, , 1081, , Functions of Three or More Variables, The notions of differentiability and the differential of functions of more than two variables are similar to those of functions of two variables. For example, suppose that f is, a function of three variables that is defined by w  f(x, y, z). Then the increment ⌬w, of w corresponding to increments of ⌬x, ⌬y, and ⌬z of x, y, and z, respectively, is, ⌬w  f(x  ⌬x, y  ⌬y, z  ⌬z)  f(x, y, z), The function f is differentiable at (x, y, z) if ⌬w can be written in the form, ⌬w  fx(x, y, z) ⌬x  fy (x, y, z) ⌬y  fz(x, y, z) ⌬z  e1 ⌬x  e2 ⌬y  e3 ⌬z, where e1, e2, and e3 are functions of ⌬x, ⌬y, and ⌬z that approach zero as, (⌬x, ⌬y, ⌬z) → (0, 0, 0)., The differential dw of the dependent variable w is defined to be, dw , , w, w, w, dx , dy , dz, x, y, z, , where dx  ⌬x, dy  ⌬y, and dz  ⌬z are the differentials of the independent variables, x, y, and z. If f has continuous partial derivatives and dx, dy, and dz are all small,, then ⌬w ⬇ dw., , EXAMPLE 6 Maximum Error in Calculating Centrifugal Force A centrifuge is a, machine designed for the specific purpose of subjecting materials to a sustained centrifugal force. The magnitude of a centrifugal force F in dynes is given by, F  f(M, S, R) , , p2S 2MR, 900, , where S is in revolutions per minute (rpm), M is the mass in grams, and R is the radius, in centimeters. If the maximum percentage errors in the measurement of M, S, and R, are 0.1%, 0.4%, and 0.2%, respectively, use differentials to estimate the maximum percentage error in calculating F., Solution, , The error in calculating F is ⌬F, and, ⌬F ⬇ dF , , , F, F, F, dM , dS , dR, M, S, R, p2S 2R, 2p2SMR, p2S 2M, dM , dS , dR, 900, 900, 900, , Therefore,, ⌬F, dF, dM, dS, dR, ⬇, , 2, , F, F, M, S, R, and, `, , ⌬F, dF, `⬇`, `, F, F, , `, , dM, dS, dR, `  2` `  `, `, M, S, R, , Since, `, , dM, `, M, , 0.001,, , `, , dS, `, S, , 0.004,, , and, , `, , dR, `, R, , 0.002

Page 56 :
1082, , Chapter 13 Functions of Several Variables, , we have, `, , dF, `, F, , 0.001  2(0.004)  0.002  0.011, , Thus, the maximum percentage error in calculating the centrifugal force is approximately 1.1%., , 13.4, , CONCEPT QUESTIONS, , 1. If z  f(x, y), what is the differential of x? The differential, of y? What is the total differential of z?, 2. Let z  f(x, y). What is the relationship between the actual, change ⌬z, when x changes from x to x  ⌬x and y changes, from y to y  ⌬y, and the total differential dz of f at (x, y)?, 3. a. What does it mean for a function f of two variables x and, y to be differentiable at (a, b)? To be differentiable in a, region R?, , 13.4, , EXERCISES, , 1. Let z  2x 2  3y 2, and suppose that (x, y) changes from, (2, 1) to (2.01, 0.98)., a. Compute ⌬z., b. Compute dz., c. Compare the values of ⌬z and dz., 2. Let z  x  2xy  3y , and suppose that (x, y) changes, from (2, 1) to (1.97, 1.02) ., a. Compute ⌬z., b. Compute dz., c. Compare the values of ⌬z and dz., 2, , 2, , In Exercises 3–20, find the differential of the function., 4. z  x 4  2x 2y 2  3xy 2  y 3, , 3. z  3x 2y 3, 5. z , , xy, xy, , 6. w , , 7. z  (2x 2y  3y 3)3, 9. w  ye, , b. Give a condition that guarantees that a function f of two, variables x and y is differentiable in an open region R., c. If a function f of two variables x and y is differentiable, at (a, b), what can you say about the continuity of f at, (a, b)?, , x2 y2, , xy, 1x, , 2, , 8. z  22x 2  3y 2, 10. z  ln(2x  3y), , 11. w  x ln(x  y ), , 12. z  x 2 sin 2y, , 13. z  e2x cos 3y, , y, 14. w  tan1 a b, x, , 15. w  x 2  xy  z 2, , 16. w  2x 2  xy  z 2, , 17. w  x 2eyz, , 18. w  ex sin(2y  3z), , 19. w  x 2ey  y ln z, , 20. w  x cosh yz, , 2, , 2, , 2, , 2, , x, 22. f(x, y)  12x  3y  ;, y, (2.96, 1.02) ., , 23. f(x, y, z)  ln(2x  y)  e2xz;, (2, 3, 0) to (2.01, 2.97, 0.04)., 24. f(x, y, z)  x 2y cos pz;, (0.98, 2.97, 2.01)., , (x, y, z) changes from, , (x, y, z) changes from (1, 3, 2) to, , 25. The dimensions of a closed rectangular box are measured as, 30 in., 40 in., and 60 in., with a maximum error of 0.2 in. in, each measurement. Use differentials to estimate the maximum error in calculating the volume of the box., 26. Use differentials to estimate the maximum error in calculating the surface area of the box of Exercise 25., 27. A piece of land is triangular in shape. Two of its sides are, measured as 80 and 100 ft, and the included angle is measured as p>3 rad. If the sides are measured with a maximum, error of 0.3 ft and the angle is measured with a maximum, error of p>180 rad, what is the approximate maximum error, in the calculated area of the land?, 28. Production Functions The productivity of a certain country is, given by the function, f(x, y)  30x 4>5y 1>5, when x units of labor and y units of capital are utilized., What is the approximate change in the number of units produced if the amount expended on labor is decreased from, 243 to 240 units and the amount expended on capital is, increased from 32 units to 35 units?, , In Exercises 21–24, use differentials to approximate the, change in f due to the indicated change in the independent, variables., 21. f(x, y)  x 4  3x 2y 2  y 3  2y  4;, (2, 2) to (1.98, 2.01) ., , (x, y) changes from (3, 1) to, , (x, y) changes from, , 29. The pressure P (in pascals), the volume V (in liters), and the, temperature T (in kelvins) of an ideal gas are related by the, , V Videos for selected exercises are available online at www.academic.cengage.com/login.

Page 57 :
13.4, equation PV  8.314T. Use differentials to find the approximate change in the pressure of the gas if its volume increases from 20 L to 20.2 L and its temperature decreases, from 300 K to 295 K., 30. Consider the ideal gas law equation PV  8.314T of Exercise 29. If T and P are measured with maximum errors of, 0.6% and 0.4%, respectively, determine the maximum percentage error in calculating the value of V., 31. Surface Area of Humans The surface area S of humans is, related to their weight W and height H by the formula, S  0.1091W 0.425H 0.725. If W and H are measured with maximum errors of 3% and 2%, respectively, find the approximate maximum percentage error in the measurement of S., 32. Specific Gravity The specific gravity of an object with density, greater than that of water can be determined by using the, formula, S, , A, AW, , where A and W are the weights of the object in air and in, water, respectively. If the measurements of an object are, A  2.2 lb and W  1.8 lb with maximum errors of 0.02 lb, and 0.04 lb, respectively, find the approximate maximum, error in calculating S., , pPR4, 8kL, , where L is the length of the arteriole in centimeters, R is the, radius in centimeters, P is the difference in pressure between, the two ends of the arteriole in dyne-sec/cm2, and k is the, viscosity of blood in dyne-sec/cm2. Find the approximate, maximum percentage error in measuring the flow of blood if, an error of at most 1% is made in measuring the length of, the arteriole and an error of at most 2% is made in measuring its radius. Assume that P and k are constant., 34. The figure below shows two long, parallel wires that are at, a distance of d m apart, carrying currents of I1 and I2 amps., It can be shown that the force of attraction per unit length, between the two wires as a result of magnetic fields generated by the currents is given by, f, , 36. Error in Calculating the Power of a Battery Suppose the source, of current in an electric circuit is a battery. Then the power, output P (in watts) obtained if the circuit has a resistance, of R ohms is given by, P, , E 2R, (R  r)2, , where E is the electromotive force (EMF) in volts and r is the, internal resistance of the battery. Estimate the maximum percentage error in calculating the power if an EMF of 12 volts, is applied in a circuit with a resistance of 100 ohms, the internal resistance of the battery is 5 ohms, and the possible maximum percentage errors in measuring E, R, and r are 2%, 3%,, and 1%, respectively., 37. Error in Measuring the Resistance of a Circuit The total resistance, R (in ohms) of three resistors with resistances of R1, R2, and, R3 ohms connected in parallel is given by, 1, 1, 1, 1, , , , R, R1, R2, R3, If R1, R2, and R3 are measured as 20, 30, and 50 ohms,, respectively, with a maximum error of 0.5 in each measurement, estimate the maximum error in the calculated value, of R., 38. A container with a constant cross section of A ft2 is filled, with water to a height of h ft. The water is then allowed to, flow out through an orifice of cross section a in.2 located at, the base of the container. It can be shown that the time (in, seconds) that it takes to empty the tank is given by, T  f(A, a, h) , , A 2h, aB t, , where t is the constant of acceleration. Suppose that the, measurements of A, a, and h are 5 ft2, 2 in.2, and 16 ft with, errors of 0.05 ft2, 0.04 in.2, and 0.2 ft, respectively. Find, the error in computing T. (Take t to be 32 ft/sec2.), , m0 I1I2, a, b, 2p D, , A ft 2, , teslas per meter, where m0 (4p 107 N/amp2) is a constant called the permeability of free space. Use differentials, to find the approximate percentage change in f if I1 increases, by 2%, I2 decreases by 2%, and D decreases by 5%., , h ft, , I1, d, I2, , 1083, , 35. Error in Measuring the Period of a Pendulum The period T of a, simple pendulum executing small oscillations is given by, T  2p2L>t, where L is the length of the pendulum and t, is the constant of acceleration due to gravity. If T is computed by using L  4 ft and t  32 ft/sec2, find the approximate percentage error in T if the true values for L and t are, 4.05 ft and 32.2 ft/sec2., , 33. Flow of Blood The flow of blood through an arteriole measured in cm3/sec is given by, F, , Differentials, , a in.2

Page 58 :
1084, , Chapter 13 Functions of Several Variables, , 39. Suspension Bridge Cables The supports of a cable of a suspension bridge are at the same level and at a distance of L ft, apart. The supports are a feet higher than the lowest point of, the cable (see the figure). If the weight of the cable is negligible and the bridge has a uniform weight of W lb/ft, then, the tension (in lb) in the cable at its lowest point is given by, H, , WL2, 8a, , that is located 400 ft above the target. If the initial speed of, the projectile, the angle of elevation of the cannon, and the, height of the site above the target are measured with maximum possible percentage errors of 0.05%, 0.02%, and 0.5%,, respectively, find the maximum error in computing the time, of flight of the projectile. (Take t to be 32 ft/sec2.), In Exercises 41 and 42, show that the function is differentiable, in the plane. (See Example 5.), , If W, L, and a are measured with possible maximum errors, of 1%, 2%, and 2%, respectively, determine the maximum, percentage error in calculating H., , 41. f(x, y)  x 2  y 2, 42. f(x, y)  2xy  y 2, 43. Let f be defined by, , a, , xy, , f(x, y)  • x 2  y 2, 0, , if (x, y)  0, if (x, y)  (0, 0), , Show that fx(0, 0) and fy (0, 0) both exist but that f is not differentiable at (0, 0)., , L, , Hint: Use the result of Theorem 3., , 40. Flight of a Projectile A projectile is fired with a muzzle velocity of √ ft/sec at an angle a radians above the horizontal. If, the launch site is located at a height of h ft above the target, (see the figure below), then the time of the flight of the projectile in seconds is given by, √ sin a  2(√ sin a)  2th, 2, , T, , t, , In Exercises 44–47, determine whether the statement is true or, false. If it is true, explain why it is true. If it is false, give an, example to show why it is false., 44. If z  f(x, y) and dz  0 for all x and y and for all differentials dx and dy, then fx(x, y)  0 and fy(x, y)  0 for all x, and y., 45. If f(x, y) is differentiable at (a, b) , then, f(a, b)  lim (x, y)→(a, b) f(x, y) ., 46. If F(x, y)  f(x)  t(y), where f and t are differentiable, in the interval (a, b), then F is differentiable on, R  {(x, y) 冟 a  x  b, a  y  b}., 47. The function, , h, , x 2  y2, 1, is differentiable everywhere., f(x, y)  e, , Suppose that the projectile is fired with an initial speed of, 800 ft/sec at an angle of elevation of p>4 radians from a site, , if (x, y)  (0, 0), if (x, y)  (0, 0), , 13.5 The Chain Rule, The Chain Rule for Functions Involving One Independent Variable, dy, dx, y, , dx, dt, x, dy, dy dx, , dt, dx dt, , FIGURE 1, To find dy>dt, compute dy>dx, ( y depends on x), compute dx>dt, (x depends on t), and then multiply, the two quantities together., , t, , In this section we extend the Chain Rule to functions of two or more variables. First,, let’s recall the Chain Rule for functions of one variable: If y is a differentiable function of x and x is a differentiable function of t (so that y is a function of t), then, dy, dy dx, , dt, dx dt, This rule is easily recalled by using the diagram shown in Figure 1., We begin by looking at the Chain Rule for the case in which a variable w depends, on two intermediate variables x and y, which in turn depend on a third variable t (so, w is a function of one independent variable t).

Page 59 :
13.5, , The Chain Rule, , 1085, , THEOREM 1 The Chain Rule for Functions Involving, One Independent Variable, Let w  f(x, y), where f is a differentiable function of x and y. If x  t(t) and, y  h(t), where t and h are differentiable functions of t, then w is a differentiable function of t, and, dw, w dx, w dy, , , dt, x dt, y dt, , Note Observe that the derivative of w with respect to t is written with an ordinary d, (d) rather than a curly d ( ) , since w is a function of the single variable t., , PROOF Let t change from t to t  ⌬t. This produces a change, ⌬x  t(t  ⌬t)  t(t), in x from x to x  ⌬x and a change, ⌬y  h(t  ⌬t)  h(t), in y from y to y  ⌬y. Since t and h are differentiable, they are continuous at t, so, both ⌬x and ⌬y approach zero as ⌬t approaches zero., Next, observe that the changes of ⌬x in x and ⌬y in y in turn produce a change ⌬w, in w from w to w  ⌬w. Since f is differentiable, we have, ⌬w , , w, w, ⌬x , ⌬y  e1 ⌬x  e2 ⌬y, x, y, , where e1 → 0 and e2 → 0 as (⌬x, ⌬y) → (0, 0). Dividing both sides of this equation, by ⌬t, we have, ⌬y, ⌬w, w ⌬x, w ⌬y, ⌬x, , ,  e1,  e2, ⌬t, x ⌬t, y ⌬t, ⌬t, ⌬t, Letting ⌬t → 0, we have, dw, ⌬w,  lim, dt, ⌬t→0 ⌬t, , , ⌬y, ⌬y, w, ⌬x, w, ⌬x, lim, , lim,  lim e1 lim,  lim e2 lim, x ⌬t→0 ⌬t, y ⌬t→0 ⌬t, ⌬t→0, ⌬t→0 ⌬t, ⌬t→0, ⌬t→0 ⌬t, , , , dy, w dx, w dy, dx, , 0ⴢ, 0ⴢ, x dt, y dt, dt, dt, , , , w dx, w dy, , x dt, y dt, , The tree diagram in Figure 2 will help you recall this version of the Chain Rule., There are two “limbs” on this tree leading from w to t. To find dw>dt, multiply the partial derivatives along each limb, and then add the products of these partial derivatives., ∂w, ∂x, , x, , dx, dt, , w, , FIGURE 2, w depends on t via x and y., , t, ∂w, ∂y, , y, , dy, dt

Page 60 :
Chapter 13 Functions of Several Variables, , EXAMPLE 1 Let w  x 2y  xy 3, where x  cos t and y  et. Find dw>dt and its, , value when t  0., , Solution Observe that w is a function of x and y and that both these variables are, functions of t. Thus, we have the situation depicted in the schematic in Figure 2. Using, the Chain Rule, we have, dw, w dx, w dy, , , dt, x dt, y dt,  (2xy  y 3)(sin t)  (x 2  3xy 2)et,  y(y 2  2x)sin t  x(x  3y 2)et, To find the value of dw>dt when t  0, we first observe that if t  0, then x  cos 0  1, and y  e0  1. So, dw, `,  0  1(1  3)e0  2, dt t0, The Chain Rule in Theorem 1 can be extended to the case involving a function of, any finite number of intermediate variables. For example, if w  f(x 1, x 2, p , x n) , where, f is a differentiable function of x 1, x 2, p , x n and x 1  f1 (t), x 2  f2(t), p , x n  fn (t),, where f1, f2, p , fn are differentiable functions of t, then, dw, w dx 1, w dx 2, w dx n, , ,  p , dt, x 1 dt, x 2 dt, x n dt, This is easier to recall if you look at Figure 3, which shows the dependency of the variables involved: Multiply the derivatives along each limb leading from w to t, and add, the products of these derivatives., , ∂w, ∂x1, , FIGURE 3, w depends on t via x 1, x 2, p , x n., , ∂w, ∂x2 x, 2, , dx2, dt, , ..., , ..., , w, , x1, , dx1, dt, , ∂w, ∂xn, , t, , ..., , 1086, , xn, , dxn, dt, , EXAMPLE 2 Tracking a Missile Cruiser Figure 4 depicts an AWACS (Airborne Warning and Control System) aircraft tracking a missile cruiser. The flight path of the plane, is described by the parametric equations, x  20 cos 12t,, , y  20 sin 12t,, , z3, , and the course of the missile cruiser is given by, x  30  20t,, , y  40  10t 2,, , z0, , where 0 t 1, and x, y, and z are measured in miles and t in hours. How fast is the, distance between the AWACS plane and the missile cruiser changing when t  0?

Page 62 :
1088, , Chapter 13 Functions of Several Variables, , Therefore, when t  0,, dD,  (0.24)(0)  (0.24)(20)  (0.97)(240)  (0.97)(0), dt,  228, that is, the distance between the AWACS aircraft and the missile cruiser is decreasing, at the rate of 228 mph at that instant of time., , The Chain Rule for Functions Involving Two Independent Variables, We now look at the Chain Rule for the case in which a variable w depends on two, intermediate variables x and y, each of which in turn depends on two variables u and, √ (so that w is a function of two independent variables u and √). More specifically, we, have the following theorem., , THEOREM 2 The Chain Rule for Functions Involving, Two Independent Variables, Let w  f(x, y), where f is a differentiable function of x and y. Suppose that, x  t(u, √) and y  h(u, √) and the partial derivatives t> u, t> √, h> u, and, h> √ exist. Then, , ∂w, ∂x, , ∂x, ∂u, , u, , ∂x, ∂√, , √, , ∂y, ∂u, , u, , w, w x, w y, , , u, x u, y u, and, , x, , w, w x, w y, , , √, x √, y √, , w, ∂w, ∂y, , PROOF For w> u we think of √ as a constant, so t and h are differentiable functions, , y, ∂y, ∂√, , √, , FIGURE 6, w depends on u and √ via x and y., , of u. Then the result follows from Theorem 1. The expression w> √ is derived in a, similar manner., The tree diagram shown in Figure 6 will help you to recall the Chain Rule given, in Theorem 2., To obtain w> u, observe that w is connected to u by two “limbs,” one from w to, u via x and the other from w to u via y. Multiply the partial derivatives along each of, these limbs, and add the product of these partial derivatives together to get w> u. The, expression for w> √ is found in a similar manner., , EXAMPLE 3 Let w  2x 2y, where x  u 2  √2 and y  u 2  √2. Find w> u and, w> √., Solution Observe that w is a function of x and y and that both of these variables are, functions of u and √. Thus, we have the situation depicted in Figure 6. Using the Chain, Rule (Theorem 2), we have, w, w x, w y, , , u, x u, y u,  4xy(2u)  2x 2(2u)  4xu(2y  x)

Page 63 :
13.5, , The Chain Rule, , 1089, , and, w, w x, w y, , , √, x √, y √,  4xy(2√)  2x 2(2√)  4x√(2y  x), , The General Chain Rule, The Chain Rule in Theorem 2 can be extended to the case involving any finite, number of intermediate variables and any finite number of independent variables., For example, if w  f(x 1, x 2, p , x n) , where f is a differentiable function of n intermediate variables, x 1, x 2, p , x n, and x 1  f1(t 1, t 2, p , t m) , x 2  f2 (t 1, t 2, p , t m), p ,, x n  fn (t 1, t 2, p , t m) , where f1, f2, p , fn are differentiable functions of m variables,, t 1, t 2, p , t m, then, w, w x1, w x2 p, w xn, , ,  , t1, x 1 t1, x 2 t1, x n t1, w, w x1, w x2 p, w xn, , ,  , t2, x 1 t2, x 2 t2, x n t2, o, w, w x1, w x2 p, w xn, , ,  , tm, x 1 tm, x 2 tm, x n tm, (See Figure 7.), , ∂w, ∂x1, ∂w, ∂x2, , x1, , ∂x1 t1, ∂t2, t2, ∂x1, ∂tm, tm, t1, ..., , ∂x1, ∂t1, , x2, , ..., , ∂w, ∂xn, , ..., , t2, , w, , tm, t1, t2, ..., , xn, , tm, , FIGURE 7, w depends on t 1, t 2, p , t m via x 1, x 2, p , x n., , EXAMPLE 4 Let w  x 2y  y 2z 3, where x  r cos s, y  r sin s, and z  res. Find, , the value of w> s when r  1 and s  0., , Solution Observe that w is a function of x, y, and z, which in turn are functions of r, and s (Figure 8).

Page 64 :
1090, , Chapter 13 Functions of Several Variables, ∂x, ∂r, ∂w, ∂x, , w, , ∂w, ∂y, , r, , Multiplying the partial derivatives on the limbs that connect w to s on the tree diagram, and adding the products of these derivatives, we obtain, , ∂x, ∂s, , x, , s, , ∂y, ∂r, , ∂w, ∂z, , When r  1 and s  0, we have x  1, y  0, and z  1, so, w,  2(1)(0)(0)  (1)(1)  3(0)(1)(1)  1, s, , s, , ∂z, ∂r, , r, , ∂z, ∂s, , z, ,  2xy(r sin s)  (x 2  2yz 3)(r cos s)  3y 2z 2(res), , r, , ∂y, ∂s, , y, , w, w x, w y, w z, , , , s, x s, y s, z s, , EXAMPLE 5 If w  f(x 2  y 2, y 2  x 2) and f is differentiable, show that w satisfies, s, , the equation, , FIGURE 8, w depends on r and s via x, y, and z., x, u, , y, , w, w, x, 0, x, y, , Solution Introduce the intermediate variables u  x 2  y 2 and √  y 2  x 2. Then, w  t(x, y)  f(u, √) (Figure 9)., Using the Chain Rule, we have, , y, , w, w u, w √, w, w, , , , (2x) , (2x), x, u x, √ x, u, √, , w, x, , and, , √, , w, w u, w √, w, w, , , , (2y) , (2y), y, u y, √ y, u, √, , y, , FIGURE 9, w depends on x and y via the, intermediate variables u and √., , Therefore,, y, , r, , EXAMPLE 6 Let w  f(x, y), where f has continuous second-order partial derivatives, and let x  r 2  s 2 and y  2rs. Find 2w> r 2., Solution, , x, s, w, , w, w, w, w, w, w, x,  a2xy,  2xy, b  a2xy,  2xy, b0, x, y, u, √, u, √, , We begin by calculating w> r. Using the Chain Rule, we have, w, w x, w y, w, w, , , , (2r) , (2s), r, x r, y r, x, y, , r, , (See Figure 10.) Next, we apply the Product Rule to w> r to obtain, y, , 2, , w, , s, , FIGURE 10, w depends on r and s via the, intermediate variables x and y., , r, , 2, , , 2, , r, , a2r, , w, w,  2s, b, x, y, , w, w, w,  2r, a b  2s, a b, x, r, x, r, y, , (1), , To compute the partial derivatives appearing in the last two terms of Equation (1), we, observe that since w is a function of r and s via the intermediate variables x and y, the, same is true of w> x and w> y (Figure 11).

Page 65 :
13.5, r, s, , r, , r, , a, , y, w, w, x, w, b, a b, , a b, x, x, x, r, y, x, r, 2, , y, , , , s, , w, , x, , 2, , 2, , w, (2s), y x, , (2r) , , and, , r, x, s, , ∂w, ∂y, , 1091, , Using the Chain Rule once again, we have, , x, ∂w, ∂x, , The Chain Rule, , r, , r, , a, , y, w, w, x, w, b, a b, , a b, y, x, y, r, y, y, r, 2, , , , y, s, , FIGURE 11, Both w> x and w> y depend, on r and s via the intermediate, variables x and y., , w, (2r) , x y, , 2, , w, , y2, , (2s), , Substituting these expressions into Equation (1) and observing that fxy  fyx because, they are continuous, we have, 2, , w, , r2, , 2, , 2, 2, 2, 2, w, w, w, w, w,  2r a2r, , 2s, b, , 2s, a2r, , 2s, b, 2, x, y, x, x, y, x, y2, , 2, , 2, 2, 2, w, w, w, w, 2,  4r 2, , 8rs, , 4s, 2, x, x y, x, y2, , Implicit Differentiation, The Chain Rule for a function of several variables can be used to find the derivative, of a function implicitly. We will consider two situations., First, suppose that the equation F(x, y)  0, where F is a differentiable function,, defines a differentiable function f of x via the equation y  f(x). If we differentiate, both sides of w  F(x, y)  0 with respect to x, we obtain, w, F, F dy, , , 0, x, x, y dx, x, , (see Figure 12) which implies that, , w, , y, , x, , FIGURE 12, Tree diagram showing dependency of w, on x directly and via y, , F, dy, Fx, x, , , dx, F, Fy, y, , if Fy  0, , Let’s summarize this result., , THEOREM 3 Implicit Differentiation: One Independent Variable, Suppose that the equation F(x, y)  0, where F is differentiable, defines y implicitly as a differentiable function of x. Then, dy, Fx (x, y), , dx, Fy (x, y), , if Fy (x, y)  0, , (2)

Page 66 :
1092, , Chapter 13 Functions of Several Variables, , EXAMPLE 7 Find, Solution, , dy, if x 3  xy  y 2  4., dx, , The given equation can be rewritten as, F(x, y)  x 3  xy  y 2  4  0, , Then Equation (2) immediately gives, dy, 3x 2  y, Fx,  , dx, Fy, x  2y, As a second application of the Chain Rule to implicit differentiation, suppose that, the equation F(x, y, z)  0, where F is a differentiable function, defines a differentiable, function f of x and y via the equation z  f(x, y). Differentiating both sides of, w  F(x, y, z)  0 with respect to x, we obtain, w, F, F z, z, , ,  Fx  Fz, 0, x, x, z x, x, x, w, , (see Figure 13) which gives, Fx, z, , x, Fz, , y, x, z, y, , FIGURE 13, w depends on x and y directly and via z., , provided that Fz  0., Similarly, we see that, Fy, z, , y, Fz, , if Fz  0, , THEOREM 4 Implicit Differentiation: Two Independent Variables, Suppose the equation F(x, y, z)  0, where F is differentiable, defines z implicitly as a differentiable function of x and y. Then, Fx (x, y, z), z, , x, Fz(x, y, z), , EXAMPLE 8 Find, Solution, , and, , Fy (x, y, z), z, , y, Fz(x, y, z), , if Fz(x, y, z)  0 (3), , z, z, and, if 2x 2z  3xy 2  yz  8  0., x, y, , Here, F(x, y, z)  2x 2z  3xy 2  yz  8  0, and Equation (3) gives, Fx (x, y, z), 4xz  3y 2, 3y 2  4xz, z, , , , x, Fz (x, y, z), 2x 2  y, 2x 2  y, , and, Fy (x, y, z), 6xy  z, 6xy  z, z, ,  2,  2, y, Fz (x, y, z), 2x  y, 2x  y

Page 67 :
13.5, , 13.5, , 1093, , CONCEPT QUESTIONS, , 1. Suppose that w  f(x, y), x  t(t), and y  h(t), where f, t,, and h are differentiable functions. Write an expression for, dw>dt. Illustrate with a tree diagram., 2. Suppose that w  f(x, y), x  t(u, √), and y  h(u, √) ,, where f, t, and h are differentiable functions. Write an, expression for w> √. Illustrate with a tree diagram., 3. Suppose that w  f(x 1, x 2, p , x n) , x 1  f1 (t 1, t 2, p , t m) ,, x 2  f2 (t 1, t 2, p , t m) p , x n  fn (t 1, t 2, p , t m) , where f, f1,, , 13.5, , f2, p , fn are differentiable functions. Write an expression for, w> t i, where 1 i m. Illustrate with a tree diagram., 4. a. Suppose that F(x, y)  0 defines y implicitly as a function of x and F is differentiable. Write an expression for, dy>dx. Illustrate with a tree diagram., b. Suppose that F(x, y, z)  0 defines z implicitly as a function of x and y and F is differentiable. Write an expression for z> x. Illustrate with a tree diagram., , EXERCISES, In Exercises 19–26, use the Chain Rule to find the indicated, derivative., , In Exercises 1–8, use the Chain Rule to find dw>dt., 1. w  x 2  y 2,, , x  t 2  1,, , 2. w  2x 2  2y 2,, , x  1t,, , 3. w  r cos s  s sin r,, 4. w  ln(x  y 2),, 5. w  2x 3y 2z,, 6. w  peqr,, , The Chain Rule, , y  t3  t, , r  e2t,, , x  tan t,, , 19. w  x 2  xy  y 2  z 3,, dw, z  cos 2t;, dt, , y  12t  1, s  t 3  2t, , y  sec t, , x  t, y  cos t, z  t sin t, , p  1t,, , q  sin 2t,, , r, , t, , t 1, z, 7. w  tan1 xz  , x  t, y  t 2, z  sinh t, y, 1, 8. w  x2y 2  z 2, x  , y  et cos t, z  et sin t, t, 2, , In Exercises 9–14, use the Chain Rule to find w> u and w> √., 9. w  x 3  y 3,, , x  u 2  √2,, , 10. w  sin xy, x  (u  √) ,, 3, , 11. w  ex cos y,, , y  1√, , x  ln(u 2  √2),, , 12. w  x ln y  2y,, 13. w  x tan1 yz,, , y  2u√, , 14. w  x cosh y  y sinh z,, u, z, √, , y  e2√,, x  u 2  √2,, , 20. z  x1y  1x, x  2s  t, y  s 2  7t;, z, if s  4 and t  1, t, x, du, `, 21. u  2, , x  sec 2t, y  tan t;, dt t0, x  y2, u, 22. w , , u  x  2y  3z, √  x cos p(y  z);, 2u 2  √2, w, w, and, if x  0, y  1, and z  1, x, z, u, s, u, 23. u  x csc yz, x  rs, y  s 2t, z  2 ;, and, s, t, t, , 25. w , y  ln(u  1),, , In Exercises 15–18, write the Chain Rule for finding the indicated, derivative with the aid of a tree diagram., , x 2y, 2, , , x  rest, y  sert, z  erst;, , z, if r  1, s  2, and t  0, , z  √ cos u, , xy, , x  r cos s, y  r sin s,, xz, w, w, and, r, t, , 26. w , , 27. Given the system, e, , 15. w  f(r, s, u, √), r  t(t), s  h(t), u  p(t),, dw, √  q(t);, dt, w, 16. w  f(x, y), x  t(u, √, t), y  h(u, √, t);, √, 17. w  f(x, y, z), x  t(r, s, t), y  h(r, s, t),, w, z  p(r, s, t);, t, 18. w  f(x, y), x  t(u, √, r, s), y  h(u, √, r, s);, , y  e t,, , 24. w  cos(2x  3y), x  r 2st, y  s 2tu;, , y  1u√, , x  ln u, y  ue√, x  1u,, , x  2t,, , x  u 2  √2, y  u 2  √2, , find u> x, u> y, √> x and √> y., 28. Given the system, •, w, r, , V Videos for selected exercises are available online at www.academic.cengage.com/login., , 1 2, (u  √2), 2, y  u√, x, , find u> x, u> y, √> x and √> y., , w, w, and, r, u, w, w, and, r, t, z  s tan t;

Page 68 :
1094, , Chapter 13 Functions of Several Variables, , In Exercises 29–32, use Equation (2) to find dy>dx., 29. x  2xy  y  4, , 30. x  2x y  3xy  x  5, , 31. 2x 2  3 1xy  2y  4, , 32. x sec y  y cos x  1, , 3, , 3, , 4, , 2 2, , In Exercises 33–36, use Equation (3) to find z> x and z> y., , 44. The Doppler Effect Suppose a sound with frequency f is emitted by an object moving along a straight line with speed u, and a listener is traveling along the same line in the opposite, direction with speed √. Then the frequency F heard by the, listener is given by, Fa, , 33. x 2  xy  x 2z  yz 2  0, 34. x 2  y 2  z 2  xy  yz  xz  1, 35. xey  yexz  x 2ex>y  10, 36. ln(x 2  y 2)  x ln z  cos(xyz)  0, 37. Find dy>dx if x 3  y 3  3axy  0, a  0., 38. The radius of a right circular cylinder is increasing at the, rate of 0.1 cm/sec while its height is decreasing at the rate, of 0.2 cm/sec. Find the rate at which the volume of the, cylinder is changing when its radius is 60 cm and its height, is 130 cm., 39. The radius of a right circular cone is decreasing at the rate, of 0.2 in./min while its height is increasing at the rate of, 0.1 in./min. Find the rate at which the area of its lateral, surface is changing when its radius is 10 in. and its height, is 18 in., 40. The pressure P (in pascals), the volume V (in liters), and the, temperature T (in kelvins) of 1 mole of an ideal gas are, related by the equation PV  8.314T. Find the rate at which, the pressure of the gas is changing when its volume is 20 L, and is increasing at the rate of 0.2 L/sec and its temperature, is 300 K and is increasing at the rate of 0.3 K/sec., 41. Car A is approaching an intersection from the north, and car, B is approaching the same intersection from the east. At a, certain instant of time car A is 0.4 mile from the intersection, and approaching it at 45 mph, while car B is 0.3 mile from, the intersection and approaching it at a speed of 30 mph., How fast is the distance between the two cars changing?, 42. The position of boat A at time t is given by the parametric, equations, x 1  5t,, , y1  5t, , and the position of boat B at time t is given by, x 2  5t,, , y2  2t  t 2, , where 0  t  15, and x 1, y1, x 2, y2 are measured in feet, and t is measured in seconds. How fast is the distance, between the two boats changing when t  10?, 43. The total resistance R (in ohms) of n resistors with resistances R1, R2, p , Rn ohms connected in parallel is given by, the formula, , c√, bf, cu, , where c is the speed of sound in still air—about 1100 ft/sec., (This phenomenon is called the Doppler effect.) Suppose, that a railroad train is traveling at 100 ft/sec in still air, and accelerating at the rate of 3 ft/sec2 and that a note, emitted by the locomotive whistle is 500 Hz. If a passenger, is on a train that is moving at 50 ft/sec in the direction, opposite to that of the first train and accelerating at the rate, of 5 ft/sec2, how fast is the frequency of the note he hears, changing?, 45. Rate of Change in Temperature The temperature at a point, (x, y, z) is given by, T(x, y, z) , , 60, 1  x  y2  z2, 2, , where T is measured in degrees Fahrenheit and x, y, and, z are measured in feet. Suppose the position of a flying, insect is, r(t)  2t i  t 2j  t 3k, , 0, , t, , 5, , where t is measured in seconds and the distance is measured, in feet. Find the rate of change in temperature that the insect, experiences at t  2., 46. If z  f(x, y), where x  r cos u and y  r sin u, show, that, a, , z 2, z 2, z 2, 1, z 2, b a b a b  2a b, x, y, r, u, r, , 47. If u  f(x, y), where x  er cos u and y  er sin u, show, that, a, , u 2, u 2, u 2, u 2, b  a b  e2r c a b  a b d, x, y, r, u, , 48. If u  f(x, y), where x  er cos u and y  er sin u, show, that, 2, , u, , x, , 2, , 2, , , , u, , y, , 2, ,  e2r c, , 2, , u, , r, , 2, , 2, , , , u, , u2, , d, , 49. If z  f(x, y), where x  u  √ and y  √  u, show, that, z, z, , 0, u, √, , 1, 1, 1, 1, , p, , R, R1, R2, Rn, , 50. If z  f(x, y), where x  u  √ and y  u  √, show that, , Show that, R, R, a b, Rk, Rk, , 2, , a, , z 2, z 2, z, b a b , x, y, u, , z, √

Page 69 :
13.5, 51. If z  f(x  at)  t(x  at), show that z satisfies the wave, equation, 2, 2, z, z, 2, , a, t2, x2, Hint: Let u  x  at and √  x  at., , z, z, b  xa b  0, x, y, , 61. Show that the functions u  ln2x 2  y 2 and, y, √  tan1 a b satisfy the Cauchy-Riemann equations, x, (see Exercise 60)., , fx (a, b)(x  a)  fy (a, b)(y  b)  0, , Hint: Let u  x 2  y 2., , b. Find an equation of the tangent line to the ellipse, , 53. If z  f(u, √), where u  t(x, y) and √  h(x, y), show that, 2, , 2, , z, , x2, , , , a, , z, , u2, , 2, , 2, , u √, u, z, z, b a, , b, x, √ u, u √, x x, z, , , , √2, , a, , 2, , 2, , 2, , √, z u, z √, b , , x, u x2, √ x2, , Assume that all second-order partial derivatives are continuous., 54. If z  f(u, √), where u  t(x, y) and √  h(x, y), show that, z, , y x, , 2, , , , z, , u, , 2, , 2, 2, u u, z u √, z u √, , , x y, √ u x y, u √ y x, 2, , , , z √ √, z 2u, z 2√, , , 2, u y x, √ y x, √ x y, , Assume that all second-order partial derivatives are continuous., 55. A function f is homogeneous of degree n if f(tx, ty) , t nf(x, y) for every t, where n is an integer. Show that if, f is homogeneous of degree n, then, x, , y2, x2, , 1, 4, 9, , 2, , 2, , 2, , 1095, , 62. a. Let P(a, b) be a point on the curve defined by the equation f(x, y)  0. Show that if the curve has a tangent line, at P(a, b), then an equation of the tangent line can be, written in the form, , 52. If z  f(x 2  y 2), show that, ya, , The Chain Rule, , f, f, y,  nf, x, y, , at the point 1 1, 313, 2 2., , 63. a. Use implicit differentiation to find an expression for, d 2y>dx 2 given the implicit equation f(x, y)  0. (Assume, that f has continuous second partial derivatives.), b. Use the result of part (a) to find d 2y>dx 2 if, x 3  y 3  3xy  0. What is its domain?, 64. Course Taken by a Yacht The following figure depicts a bird’seye view of the course taken by a yacht during an outing., The pier is located at the origin and the course is described, by the equation, x 3  y 3  9xy  0, , x  0, y  0, , where x and y are measured in miles. When the yacht was at, the point (2, 4), it was sailing in an easterly direction at the, rate of 16 mph. How fast was it moving in the northerly, direction at that instant of time?, y (mi), , Hint: Differentiate both sides of the given equation with respect to t., , In Exercises 56–59, find the degree of homogeneity of f and show, that f satisfies the equation, x, , f, f, y,  nf, x, y, , See Exercise 55., 0, , 56. f(x, y)  2x 3  4x 2y  y 3, 57. f(x, y) , , xy 2, 2x 2  y 2, , y, 58. f(x, y)  tan1 a b, x, , 59. f(x, y)  ex>y, 60. Suppose that the functions u  f(x, y) and √  t(x, y) satisfy, the Cauchy-Riemann equations, u, √, , x, y, , and, , √, u, , y, x, , If x  r cos u and y  r sin u, show that u and √ satisfy, u, 1 √, , r u, r, , and, , √, 1 u, , r u, r, , the polar coordinate form of the Cauchy-Riemann equations., , x (mi), , In Exercises 65–66, determine whether the statement is true or, false. If it is true, explain why it is true. If it is false, give an, example to show why it is false., 65. If F(x, y)  0, where F is differentiable, then, Fy (x, y), dx, , dy, Fx (x, y), provided that Fx (x, y)  0., 66. If z  cos xy for x  0 and y  0 and xy  np, where n is, an integer, then, z, 1, , x, x> z, provided that x> z  0.

Page 70 :
Concept Review, , CHAPTER, , 13, , 1141, , REVIEW, , CONCEPT REVIEW, In Exercises 1–17, fill in the blanks., 1. a. A function f of two variables, x and y, is a, that assigns to each ordered pair, in the domain, of f, exactly one real number f(x, y)., b. The number z  f(x, y) is called a, variable,, and x and y are, variables. The totality of the, numbers z is called the, of the function f., c. The graph of f is the set S , ., 2. a. The curves in the xy-plane with equation f(x, y)  k,, where k is a constant in the range of f, are called the, of f., b. A level surface of a function f of three variables is the, graph of the equation, , where k is a constant in, the range of, ., 3. lim (x, y)→(a, b) f(x, y)  L means there exists a number, such that f(x, y) can be made as close to, as we please by restricting (x, y) to be sufficiently close to, ., 4. If f(x, y) approaches L 1 as (x, y) approaches (a, b) along one, path, and f(x, y) approaches L 2 as (x, y) approaches (a, b), along another path with L 1  L 2, then lim (x, y)→(a, b) f(x, y), exist., 5. a. f(x, y) is continuous at (a, b) if lim (x, y)→(a, b) f(x, y) , ., b. f(x, y) is continuous on a region R if f is continuous at, every point (x, y) in, ., 6. a. A polynomial function is continuous, ; a rational, function is continuous at all points in its, ., b. If f is continuous at (a, b) and t is continuous at f(a, b),, then the composite function h  t ⴰ f is continuous at, ., 7. a. The partial derivative of f(x, y) with respect to x is, if the limit exists. The partial derivative, ( f> x)(a, b) gives the slope of the tangent line to, the curve obtained by the intersection of the plane, and the graph of z  f(x, y) at, ;, it also measures the rate of change of f(x, y) in the, -direction with y held, at, ., b. To compute f> x where f is a function of x and y, treat, as a constant and differentiate with respect to, in the usual manner., 8. If f(x, y) and its partial derivatives fx, fy, fxy, and fyx are continuous on an open region R, then fxy(x, y) , for, all (x, y) in R., , 9. a. The total differential dz of z  f(x, y) is dz , ., b. If ⌬z  f(x  ⌬x, y  ⌬y)  f(x, y), then ⌬z ⬇, ., c. ⌬z  fx(x, y) ⌬x  fy(x, y) ⌬y  e1 ⌬x  e2 ⌬y, where, and, e1 and e2 are functions of, such that lim (⌬x, ⌬y)→(0, 0) e1 , and, ., lim (⌬x, ⌬y)→(0, 0) e2 , d. The function z  f(x, y) is differentiable at (a, b) if ⌬z, can be expressed in the form ⌬z , , where, and, as (⌬x, ⌬y) →, ., 10. a. If f is a function of x and y, and fx and fy are continuous, on an open region R, then f is, in R., b. If f is differentiable at (a, b), then f is, at (a, b)., 11. a. If w  f(x, y), x  t(t), and y  h(t), then under suitable, conditions the Chain Rule gives dw>dt , ., b. If w  f(x, y), x  t(u, √), and y  h(u, √), then, ., w> u , c. If F(x, y)  0, where F is differentiable, then dy>dx , , provided that, ., d. If F(x, y, z)  0, where F is differentiable, and F defines, z implicitly as a function of x and y, then z> x , and z> y , , provided that, ., 12. a. If f is a function of x and y and u  u 1i  u 2 j is a unit, vector, then the directional derivative of f in the direction, of u is Du f(x, y) , if the limit exists., b. The directional derivative Du f(a, b) measures the rate of, change of f at, in the direction of, ., c. If f is differentiable, then Du f(x, y) , ., d. The gradient of f(x, y) is §f(x, y) , ., e. In terms of the gradient, Du f(x, y) , ., 13. a. The maximum value of Du f(x, y) is, occurs when u has the same direction as, b. The minimum value of Du f(x, y) is, occurs when u has the direction of, , , and this, ., , and this, ., , 14. a. §f is, to the level curve f(x, y)  c at P., b. §F is, to the level surface F(x, y, z)  0 at P., c. The tangent plane to the surface F(x, y, z)  0 at the, point P(a, b, c) is, ; the normal line passing, through P(a, b, c) has symmetric equations, ., 15. a. If f(x, y) f(a, b) for all points in an open disk containing (a, b), then f has a, at (a, b)., b. If f(x, y)  f(a, b) for all points in the domain of f, then f, has an, at (a, b) .

Page 71 :
1142, , Chapter 13 Functions of Several Variables, , c. If f is defined on an open region R containing the point, (a, b), then (a, b) is a critical point of f if (1) fx and/or fy, at (a, b) or (2) both, ., fx (a, b) and fy(a, b) equal, d. If f has a relative extremum at (a, b), then (a, b) must be, a, of f., e. To determine whether a critical point of f does give rise, to a relative extremum, we use the, ., 16. a. If f is continuous on a closed, bounded set D in the, plane, then f has an absolute maximum value, at some point, in D, and f has an absolute minimum value, at some point, in D., b. To find the absolute extreme values of f on a closed,, bounded set D, (1) find the values of f at the, in D, (2) find the extreme, values of f on the, of D. Then the largest and, smallest values found in (1) and (2) give the, value of f and the, value of f on D., , 17. a. If f(x, y) has an extremum at a point (a, b) lying on the, curve with equation t(x, y)  c, then the extremum is, called a, extremum., b. If f has an extremum at (a, b) subject to the constraint, , where l is a real, t(x, y)  c, then §f(a, b) , number called a Lagrange, ., c. To find the maximum and minimum values of f subject, to the constraint t(x, y)  c, we solve the system of, equations §f(x, y) , and t(x, y)  c for x, y,, and l. We then evaluate, at each of the, found in the last step. The largest, value yields the constrained, of f, and the, smallest value yields the constrained, of f., , REVIEW EXERCISES, In Exercises 17–22, find the first partial derivatives of the, function., , In Exercises 1–4, find and sketch the domain of the function., 29  x  y, 2, , 1. f(x, y) , , 2, , x 2  y2, , 2. f(x, y) , , ln(x  2y  4), yx, , 17. f(x, y)  2x 2y  1x, , 3. f(x, y)  sin1 x  tan1 y, , 19. f(r, s)  re, , 4. f(x, y)  ln(xy  1), , 21. f(x, y, z) , , In Exercises 5 and 6, sketch the graph of the function., 5. f(x, y)  4  x 2  y 2, , 2, , 9. f(x, y)  ex, , 1xy  4, (x, y)→(0, 0) 2y  3, x yx, 2, , 13., , lim, , (x, y)→(1, 0), , 12., , lim, , 26. f(u, √, w)  ue√ sin w, , (x, y)→(0, 0), , 27. If u  2x 2  y 2  z 2, show that, , x y, , x  3y, 4, , 3, , 4, , 14., , lim, , (x, y, z)→(0, 0, 0), , ln(x  y), (x 2  y 2) 3>2, , 16. f(x, y, z)  ex>y cos z  1x  y, , 2, , u, , x  2y  3z, 2, , 2, , x 2  y2  z2, , In Exercises 15 and 16, determine where the function is, continuous., 15. f(x, y) , , s, 22. f(r, s, t)  r cos st  s sina b, t, , z2  x 2, , 25. f(x, y, z)  x 2yz 3, , 2 2, , 1x  1y, , x 2  y2, , 24. f(x, y)  exy cos(2x  3y), , 10. f(x, y)  ln xy, , lim, , 20. f(u, √)  e2u cos(u 2  √2), , 23. f(x, y)  x 4  2x 2y 3  y 2  2, , In Exercises 11–14, find the limit or show that it does not exist., 11., , x 2  y2, , In Exercises 23–26, find the second partial derivatives of the, function., , 8. f(x, y)  y 2  x 2, , y2, , (r2s2), , xy 2, , 6. f(x, y)  21  x 2  y 2, , In Exercises 7–10, sketch several level curves for the function., 7. f(x, y)  x 2  2y, , 18. f(x, y) , , 2, , x, , 2, , 2, , , , u, , y, , 2, , 2, , , , u, , z2, , , , 2, u, , 28. Show that the function u  et cos(x>c) satisfies the onedimensional heat equation u t  c2u xx., In Exercises 29 and 30, show that the function satisfies Laplace’s, equation u xx  u yy  u zz  0., 29. u  2z 2  x 2  y 2, , 30. u  z tan1, , y, x

Page 72 :
Review Exercises, 31. Find dz if z  x 2 tan1 y 3., 32. Use differentials to approximate the change in, f(x, y)  x 2  3xy  y 2 if (x, y) changes from (2, 1) to, (1.9, 0.8)., 33. Use differentials to approximate (2.01)22(1.98)2  (3.02)3., 34. Estimating Changes in Profit The total daily profit function (in, dollars) of Weston Publishing Company realized in publishing and selling its English language dictionaries is given by, P(x, y)  0.0005x 2  0.003y 2  0.002xy  14x  12y  200, where x denotes the number of deluxe copies and y denotes, the number of standard copies published and sold daily., Currently, the number of deluxe and standard copies of the, dictionaries published and sold daily are 1000 and 1700,, respectively. Determine the approximate daily change in the, total daily profit if the number of deluxe copies is increased, to 1050 and the number of standard copies is decreased to, 1650 per day., 35. Does a function f such that §f  yi  xj exist? Explain., 36. According to Ohm’s Law, R  V>I, where R is the resistance in ohms, V is the electromotive force in volts, and I, is the current in amperes. If the errors in the measurements, made in a certain experiment in V and I are 2% and 1%,, respectively, use differentials to estimate the maximum, percentage error in the calculated value of R., 37. Let z  x 2y  1y, where x  e2t and y  cos t. Use the, Chain Rule to find dz>dt., 38. Let w  e cos y  y sin e , where x  u  √ and, y  1u√. Use the Chain Rule to find w> u and w> √., x, , x, , 2, , 2, , 39. Use partial differentiation to find dy>dx if, x 3  3x 2y  2xy 2  2y 3  9., 40. Find z> x and z> y if x z  yz  cos xz., 3 2, , 41. f(x, y)  2x 2  y 2; P(1, 2), 42. f(x, y)  ex tan y; P 1 0, p4 2, , 43. f(x, y, z)  xy 2  yz 2  zx 2;, 44. f(x, y, z)  x ln y  y ln z;, , P(2, 1, 3), P(2, 1, 1), , In Exercises 45–48, find the directional derivative of the function, f at the point P in the indicated direction., 45. f(x, y)  x 3y 2  xy 3; P(2, 1), in the direction of, v  3i  4j., 46. f(x, y)  e, to Q(3, 1) ., , x2, , cos y; P 1 0,, , 47. f(x, y, z)  x2y  z ;, v  i  2j  2k., 2, , 2, , p, 2, , 48. f(x, y, z)  x 2 ln y  xy 2ez; P(2, 1, 0) in the direction of, v  具3, 1, 2典., 49. Find the direction in which f(x, y)  1x  xy 2 increases, most rapidly at the point (4, 1) . What is the maximum rate, of increase?, 50. Find the direction in which f(x, y, z)  xeyz decreases most, rapidly at the point (4, 3, 0) . What is the greatest rate of, decrease?, In Exercises 51–54, find equations for the tangent plane and the, normal line to the surface with the equation at the given point., 51. 2x 2  4y 2  9z 2  27; P(1, 2, 1), 52. x 2  2y 2  3z 2  19; P(2, 3, 1), 53. z  x 2  3xy 2; P(3, 1, 18), 54. z  xey; P(1, 0, 1), , 55. Let f(x, y)  x 2  y 2, and let t(x, y)  y 2>x 2., , cas a. Plot several level curves of f and t using the same view-, , ing window., b. Show analytically that each level curve of f intersects all, level curves of t at right angles., 56. Show that if §f(x 0, y0)  0, then an equation of the tangent, line to the level curve f(x, y)  f(x 0, y0) at the point (x 0, y0), is, fx(x 0, y0)(x  x 0)  fy (x 0, y0)(y  y0)  0, In Exercises 57–60, find the relative extrema and saddle points, of the function., 57. f(x, y)  x 2  xy  y 2  5x  8y  5, 58. f(x, y)  8x 3  6xy  y 3, 59. f(x, y)  x 3  3xy  y 2, , 3, , In Exercises 41–44, find the gradient of the function f at the, indicated point., , 2 , in the direction from P(1, 3), , P(2, 3, 4) in the direction of, , 1143, , 60. f(x, y) , , 4, 2,   xy, x, y, , In Exercises 61 and 62, find the absolute extrema of the function, on the set D., 61. f(x, y)  x 2  xy 2  y 3;, D  {(x, y) 冟 1 x 1, 0, , y, , 2}, , 62. f(x, y)  (x  3y )e ; D  {(x, y) 冟 x 2  y 2, 2, , x, , 2, , 9}, , In Exercises 63–66, use the method of Lagrange multipliers to, find the extrema of the function f subject to the constraints., 63. f(x, y)  xy 2;, 64. f(x, y) , , x 2  y2  4, , 1, 1,  ;, x, y, , 1, x2, , , , 1, y2, , 65. f(x, y, z)  xy  yz  xz;, , x  2y  3z  1, , 66. f(x, y, z)  3x  2y  z ;, 2x  y  z  2, 2, , 2, , 9, , 2, , x  y  z  1,

Page 73 :
1144, , Chapter 13 Functions of Several Variables, , 67. Let f(x, y)  Ax 2  Bxy  Cy 2  Dx  Ey  F. Show that, if f has a relative maximum or a relative minimum at a point, (x 0, y0), then x 0 and y0 must satisfy the system of equations, , Show that the isothermal curves T(x, y)  k are arcs of circles that pass through the points x  1. Sketch the isothermal curve corresponding to a temperature of 75°C., , 2Ax  By  D  0, , y, , Bx  2Cy  E  0, simultaneously., 68. Let f(x, y)  x 2  2Bxy  y 2, where B  0. For what value, of B does f have a relative minimum at (0, 0)? A saddle, point at (0, 0)? Are there any values of B such that f has a, relative maximum at (0, 0)?, , 1, , x, , 2, , 2, , y, x, , 4, 25, , In Exercises 71 and 72, determine whether the statement is true, or false. If it is true, explain why it is true. If it is false, give an, example to show why it is false., , that is closest to the point (3, 0, 0)., 70. Isothermal Curves Consider the upper half-disk, H  {(x, y) 冟 x 2  y 2 1, y  0} (see the figure). If the, temperature at points on the upper boundary is kept at, 100°C and the temperature at points on the lower boundary, is kept at 50°C, then the steady-state temperature T(x, y) at, any point inside the disk is given by, T(x, y)  100 , , 0, , 1, , T(x, y)  50, , 69. Find the point on the paraboloid, z, , T(x, y)  100, , 71. The directional derivative of f(x, y) at the point (a, b) in the, positive x-direction is fx (a, b) ., 72. If we know the gradient of f(x, y, z) at the point P(a, b, c),, then we can compute the directional derivative of f in any, direction at P., , 1  x 2  y2, 100, tan1, p, 2y, , CHALLENGE PROBLEMS, 1. Find and sketch the domain of, , 4. Consider the quadratic polynomial function, , f(x, y)  236  4x  9y, 2, , 2, ,  ln(x 2  2x  y 2), , , 1, 24x  16x  4y 2  15, 2, , 2. Describe the domain of, H(x, y, z)  1x  a  1b  x  1y  c, , f(x, y)  Ax 2  2Bxy  Cy 2  2Dx  2Ey  F, Find conditions on the coefficients of f such that f has (a) a, relative maximum and (b) a relative minimum. What are the, coordinates of the point in terms of the coefficients of f ?, 5. Let a, b, and c denote the sides of a triangle of area A, and, let a, b, and g denote the angles opposite them. If, A  f(a, b, c), show that, f, 1,  R cos a, a, 2, ,  1d  y  1z  e  1f  z, where a, , b, c, , d, and e, , f., , 3. Suppose f has continuous second partial derivatives, in x and y. Then the second-order directional derivative, of f in the direction of the unit vector u  u 1i  u 2 j is, defined to be, , where R is the radius of the circumscribing circle., , ∫, , R, , a, , c, , D 2u f(x, y)  Du (Du f ), a. Find an expression in terms of the partial derivatives of f, for D 2u f., b. Find D 2u f(1, 0) if f(x, y)  xy 2  exy and u has the same, direction as v  2i  3j., , ©, , å, b

Page 74 :
Challenge Problems, 6. Linda has 24 feet of fencing with which to enclose a triangular flower garden. What should the lengths of the sides of, the garden be if the area is to be as large as possible?, , 10. Consider the problem of determining the maximum and, the minimum distances from the point (x 0, y0, z 0) to the, ellipsoid, , Hint: Heron’s formula states that the area of a triangle with sides a,, b, and c is given by A  1s(s  a)(s  b)(s  c), where, s  12 (a  b  c) is the semiperimeter., , 7. Find the directional derivative at 1 1, 2, p4 2 of the function, f(x, y, z)  x 2  y cos z in the direction of increasing t, along the curve in three-dimensional space described by, the position vector r(t)  具t, t 2, t 3典 at r(1)., , x2, a, , 2, , , , y2, b, , 2, , z2, , , , c2, , 1, , a. Show that the solutions are, x, , a 2x 0, a2  l, , y, , ,, , b 2y0, b2  l, , z, , ,, , c 2z 0, c2  l, , where l satisfies, , 8. Let, xy(x  y ), 2, , f(x, y)  •, , a 2x 20, , 2, , (x, y)  (0, 0), , x 2  y2, 0, , (a  l), 2, , (x, y)  (0, 0), , Use the definition of partial derivatives to show that, fxy(0, 0)  1 and fyx (0, 0)  1., 2, , 9. Show that Laplace’s equation, , u, 2, , 2, , , , u, , x, y, cylindrical coordinates takes the form, 2, , u, , r, , 2, , , , 2, , 2, , , , u, , z2, , 2, 2, 1, 1, u, u, u,  2ⴢ 2 20, ⴢ, r, r, r, u, z, ,  0 in, , 2, , , , b 2y 20, (b  l), 2, , 2, , , , c2z 20, (c  l)2, 2, , 1, , b. Use the result of part (a) to solve the problem with, a  2, b  3, c  1, and (x 0, y0, z 0)  (3, 2, 4)., , 1145

Page 75 :
MODULE 2, Textbook : Calculus: Soo T Tan Brooks/Cole, Cengage Learning (2010) ISBN 0-534-46579-X), Sections 13.6-13.9

Page 76 :
1096, , Chapter 13 Functions of Several Variables, , 13.6 Directional Derivatives and Gradient Vectors, To study the heat conduction properties of a certain material, heat is applied to one, corner of a thin rectangular sheet of that material. Suppose that the heated corner of, the sheet is located at the origin of the xy-coordinate plane, as shown in Figure 1 and, that the temperature at any point (x, y) on the sheet is given by T  f(x, y)., y, , (x, y), , FIGURE 1, The temperature at the point, (x, y) is T  f(x, y)., , x, , 0, , From our previous work we can find the rate at which the temperature is changing, at the point (x, y) in the x-direction by computing f> x. Similarly, f> y gives the, rate of change of T in the y-direction. But how fast does the temperature change if we, move in a direction other than those just mentioned?, In this section we will attempt to answer questions of this nature. More generally,, we will be interested in the problem of finding the rate of change of a function f in a, specified direction., , The Directional Derivative, Let’s look at the problem from an intuitive point of view. Suppose that f is a function, defined by the equation z  f(x, y), and let P(a, b) be a point in the domain D of f., Furthermore, let u be a unit (position) vector having a specified direction. Then the, vertical plane containing the line L passing through P(a, b) and having the same direction as u will intersect the surface z  f(x, y) along a curve C (Figure 2). Intuitively,, we see that the rate of change of z at the point P(a, b) with respect to the distance, measured along L is given by the slope of the tangent line T to the curve C at the point, P¿(a, b, f(a, b))., z, P'(a, b, f (a, b)), , z  f (x, y), , FIGURE 2, The rate of change of z at P(a, b), with respect to the distance measured, along L is given by the slope of T., , P(a, b), , 0, u, , T, C, y, , x, , L

Page 77 :
13.6, y, , u, u2  sin ¨, ¨, 0, , x, , u1  cos ¨, , FIGURE 3, Any direction in the plane can be, specified in terms of a unit vector u., , L, , ⌬y  hu 2,, , h  2(⌬x)2  (⌬y)2, , and, , So the point Q can be expressed as Q(a  hu 1, b  hu 2). Therefore, the slope of the, secant line S passing through the points P¿ and Q¿ (see Figure 5) is given by, , Q(a  Îx, b  Îy), , f(a  hu 1, b  hu 2)  f(a, b), ⌬z, , h, h, , Îy, P(a, b), , (1), , Observe that Equation (1) also gives the average rate of change of z  f(x, y) from, P(a, b) to Q(a  ⌬x, b  ⌬y)  Q(a  hu 1, b  hu 2) in the direction of u., , Îx, u, 0, , 1097, , Let’s find the slope of T. First, observe that u may be specified by writing, u  u 1i  u 2 j for appropriate components u 1 and u 2. Equivalently, we may specify u, by giving the angle u that it makes with the positive x-axis, in which case u 1  cos u, and u 2  sin u (Figure 3)., Next, let Q(a  ⌬x, b  ⌬y) be any point distinct from P(a, b) lying on the line L, passing through P and having, the same direction as u (Figure 4)., !, Since the vector PQ is parallel to u, it must be a scalar multiple of u. In other, words, there exists a nonzero number h such that, !, PQ  hu  hu 1i  hu 2 j, !, But PQ is also given by ⌬xi  ⌬yj, and therefore,, ⌬x  hu 1,, , y, , Directional Derivatives and Gradient Vectors, , x, , z, , FIGURE 4, The point Q(a  ⌬x, b  ⌬y) lies on L, and is distinct from P(a, b)., , P'(a, b, f (a, b)), , Q'(a  hu1, b  hu2, f (a  hu1, b  hu2)), , z  f (x, y), , 0, , FIGURE 5, The secant line S passes through the, points P¿ and Q¿ on the curve C., , P(a, b), , u, Q(a  hu1, b  hu2), , T, C, S, , y, , x, , If we let h approach zero in Equation (1), we see that the slope of the secant line, S approaches the slope of the tangent line at P¿. Also, the average rate of change of z, approaches the (instantaneous) rate of change of z at (a, b) in the direction of u. This, limit, whenever it exists, is called the directional derivative of f at (a, b) in the direction of u. Since the point P(a, b) is arbitrary, we can replace it by P(x, y) and define, the directional derivative of f at any point as follows., , DEFINITION Directional Derivative, Let f be a function of x and y and let u  u 1i  u 2 j be a unit vector. Then the, directional derivative of f at (x, y) in the direction of u is, Du f(x, y)  lim, , h→0, , if this limit exists., , f(x  hu 1, y  hu 2)  f(x, y), h, , (2)

Page 78 :
1098, , Chapter 13 Functions of Several Variables, , Historical Biography, , Note, , If u  i (u 1  1 and u 2  0) , then Equation (2) gives, f(x  h, y)  f(x, y),  fx(x, y), h→0, h, , SPL/Photo Researchers, Inc., , Di f(x, y)  lim, , That is, the directional derivative of f in the x-direction is the partial derivative of f in, the x-direction, as expected. Similarly, you can show that Dj f(x, y)  fy (x, y)., , ADRIEN-MARIE LEGENDRE, (1752–1833), Born to a wealthy family, Adrien-Marie, Legendre was able to study at the College, Mazarin in Paris, where he received, instruction from highly regarded mathematicians of the time. In 1782 Legendre, won a prize offered by the Berlin Academy, to “determine the curve described by cannonballs and bombs” and to “give rules for, obtaining the ranges corresponding to different initial velocities and to different, angles of projection.” This work was, noticed by the famous mathematicians, Pierre Lagrange and Simon Laplace, which, led to the beginning of Legendre’s research, career. Legendre went on to produce, important results in celestial mechanics,, number theory, and the theory of elliptic, functions. In 1794 Legendre published Elements de geometrie, which was a simplification of Euclid’s Elements. This book, became the standard textbook on elementary geometry for the next 100 years in, both Europe and the United States., Legendre was strongly committed to, Euclidean geometry and refused to accept, non-Euclidean geometries. For nearly thirty, years, he attempted to prove Euclid’s parallel postulate. Legendre’s research met, many obstacles, including the French Revolution, Laplace’s jealousy, and Legendre’s, arguments with Carl Friedrich Gauss over, priority. In 1824 Legendre refused to vote, for the government’s candidate for the, Institut National des Sciences et des Arts;, as a result his government pension was, stopped, and he died in poverty in 1833., , The following theorem helps us to compute the directional derivatives of functions without appealing directly to the definition of the directional derivative. More, specifically, it gives the directional derivative of f in terms of its partial derivatives, fx and fy., , THEOREM 1 If f is a differentiable function of x and y, then f has a directional, derivative in the direction of any unit vector u  u 1i  u 2 j and, Du f(x, y)  fx (x, y)u 1  fy(x, y)u 2, , (3), , PROOF Fix the point (a, b). Then the function t defined by, t(h)  f(a  hu 1, b  hu 2), is a function of the single variable h. By the definition of the derivative,, t¿(0)  lim, , t(h)  t(0), h, , h→0, , f(a  hu 1, b  hu 2)  f(a, b), h→0, h, ,  lim, ,  Du f(a, b), Next, observe that t may be written as t(h)  f(x, y) where x  a  hu 1 and, y  b  hu 2. Therefore, by the Chain Rule we have, t¿(h) , , f dx, f dy, ,  fx (x, y)u 1  fy (x, y)u 2, x dh, y dh, , In particular, when h  0, we have x  a, y  b, so, t¿(0)  fx (a, b)u 1  fy (a, b)u 2, Comparing this expression for t¿(0) with the one obtained earlier, we conclude that, Du f(a, b)  fx (a, b)u 1  fy (a, b)u 2, Finally, since (a, b) is arbitrary, we may replace it by (x, y) and the result follows., , EXAMPLE 1 Find the directional derivative of f(x, y)  4  2x 2  y 2 at the point, (1, 1) in the direction of the unit vector u that makes an angle of p>3 radians with the, positive x-axis., Solution, , Here, p, p, 1, 13, j, u  cosa b i  sina b j  i , 2, 3, 3, 2

Page 79 :
13.6, , so u 1  12 and u 2 , , z, , 13, 2 ., , Directional Derivatives and Gradient Vectors, , 1099, , Using Equation (3), we find that, , Du f(x, y)  fx (x, y)u 1  fy(x, y)u 2, C, , 1, 13,  (4x) a b  (2y) a, b  (2x  13y), 2, 2, , z  4  2x2  y2, , In particular,, (1, 1, 1), , Du f(1, 1)  (2  13) ⬇ 3.732, (See Figure 6.), , u, y, x, , 3, , T, , FIGURE 6, The slope of the tangent line to the, curve C at (1, 1, 1) is ⬇ 3.732., , EXAMPLE 2 Find the directional derivative of f(x, y)  ex cos 2y at the point 1 0, p4 2, , in the direction of v  2i  3j., Solution, , The unit vector u that has the same direction as v is, u, , v, 2, 3, , i, j, 冟v冟, 113, 113, , Using Equation (3) with u 1  2> 113 and u 2  3> 113, we have, Du f(x, y)  fx (x, y)u 1  fy(x, y)u 2,  (ex cos 2y) a, , 2, 3, b  (2ex sin 2y) a, b, 113, 113, , In particular,, Du f a0,, , p, p, 2, p, 3, 6, 6113, b  ae0 cos b a, b  2(e0) asin b a, b, , 4, 2, 2, 13, 113, 113, 113, , The Gradient of a Function of Two Variables, The directional derivative Du f(x, y) can be written as the dot product of the unit, vector, u  u 1i  u 2 j, and the vector, fx (x, y)i  fy(x, y)j, Thus,, Du f(x, y)  (u 1i  u 2 j) ⴢ [fx (x, y)i  fy (x, y)j]  fx(x, y)u 1  fy (x, y)u 2, The vector fx (x, y)i  fy (x, y)j plays an important role in many other computations and, is given a special name., , DEFINITION Gradient of a Function of Two Variables, Let f be a function of two variables x and y. The gradient of f is the vector, function, §f(x, y)  fx(x, y)i  fy (x, y)j

Page 80 :
1100, , Chapter 13 Functions of Several Variables, , Notes, 1. §f is read “del f.”, 2. §f(x, y) is sometimes written grad f(x, y)., , EXAMPLE 3 Find the gradient of f(x, y)  x sin y  y ln x at the point (e, p)., Solution, , Since, fx (x, y)  sin y , , y, and fy (x, y)  x cos y  ln x, x, , we have, §f(x, y)  fx (x, y)i  fy (x, y)j, y,  asin y  b i  (x cos y  ln x)j, x, So the gradient of f at (e, p) is, §f(e, p)  asin p , , , p, b i  (e cos p  ln e)j, e, , p, i  (1  e)j, e, , Theorem 1 can be rewritten in terms of the gradient of f as follows., y, , ◊ f (a, b), , THEOREM 2, If f is a differentiable function of x and y, then f has a directional derivative in, the direction of any unit vector u, and, , u, , Du f(x, y)  §f(x, y) ⴢ u, , (a, b), , (4), , Du f(a, b), u, 0, , x, , FIGURE 7, The directional derivative of f at (a, b), in the direction of u is the scalar, component of the gradient of f at (a, b), along u., , To give a geometric interpretation of Equation (4), suppose that (a, b) is a fixed, point in the xy-plane. Then, Du f(a, b)  §f(a, b) ⴢ u , , §f(a, b) ⴢ u, 冟u冟, , since 冟 u 冟  1, , so by Equation (6) of Section 11.3 we see that Du f(a, b) can be viewed as the scalar, component of §f(a, b) along u (Figure 7)., , EXAMPLE 4 Let f(x, y)  x 2  2xy., a. Find the gradient of f at the point (1, 2)., b. Use the result of (a) to find the directional derivative of f at (1, 2) in the direction from P(1, 2) to Q(2, 3)., Solution, a. The gradient of f at any point (x, y) is, §f(x, y)  (2x  2y)i  2xj, b. The gradient of f at the point (1, 2) is, §f(1, 2)  (2  4)i  2j  6i  2j

Page 81 :
13.6, , Directional Derivatives and Gradient Vectors, , 1101, , !, The desired direction is given by the direction, of the vector PQ  3i  j. A unit, !, vector that has the same direction as PQ is, u, , 3, 1, i, j, 110, 110, , Using Equation (4), we obtain, Du f(1, 2)  §f(1, 2) ⴢ u,  (6i  2j) ⴢ a, , 3, 1, i, jb, 110, 110, , 2, 16, 18, , , ⬇ 5.1, 110, 110, 110, , , , or a change in f of 5.1 per unit change in the direction of the vector u. The gradient vector §f(1, 2), the unit vector u, and the geometrical interpretation of, Du f(1, 2) are shown in Figure 8., y, , Q(2, 3), , P(1, 2), 1, 2 1 0, 1, , u, 1 Du f(1, 2), , 7, , x, , u, (1, 2), , FIGURE 8, Du f(1, 2) viewed as the scalar, component of §f(1, 2) along u., , ◊f (1, 2)  6i  2j, , Properties of the Gradient, The following theorem gives some important properties of the gradient of a function., , THEOREM 3 Properties of the Gradient, Suppose f is differentiable at the point (x, y)., 1. If §f(x, y)  0, then Du f(x, y)  0 for every u., 2. The maximum value of Du f(x, y) is 冟 §f(x, y) 冟, and this occurs when u has, the same direction as §f(x, y)., 3. The minimum value of Du f(x, y) is 冟 §f(x, y) 冟, and this occurs when u, has the direction of  §f(x, y)., , PROOF Suppose §f(x, y)  0. Then for any u  u i i  u 2 j, we have, Du f(x, y)  §f(x, y) ⴢ u  (0i  0j) ⴢ (u 1i  u 2 j)  0, Next, if §f(x, y)  0, then, Du f(x, y)  §f(x, y) ⴢ u  冟 §f(x, y) 冟冟 u 冟 cos u  冟 §f(x, y) 冟 cos u, where u is the angle between §f(x, y) and u. Since the maximum value of cos u is 1, and this occurs when u  0, we see that the maximum value of Du f(x, y) is 冟 §f(x, y) 冟

Page 82 :
1102, , Chapter 13 Functions of Several Variables, , and this occurs when both §f(x, y) and u have the same direction. Similarly, Property, (3) is proved by observing that cos u has a minimum value of 1 when u  p., Notes, 1. Property (2) of Theorem 3 tells us that f increases most rapidly in the direction of, §f(x, y). This direction is called the direction of steepest ascent., 2. Property (3) of Theorem 3 says that f decreases most rapidly in the direction of, §f(x, y) . This direction is called the direction of steepest descent., , EXAMPLE 5 Quickest Descent Suppose a hill is described mathematically by using, the model z  f(x, y)  300  0.01x 2  0.005y 2, where x, y, and z are measured in, feet. If you are at the point (50, 100, 225) on the hill, in what direction should you aim, your toboggan if you want to achieve the quickest descent? What is the maximum rate, of decrease of the height of the hill at this point?, Solution, , The gradient of the “height” function is, §f(x, y)  fx (x, y)i  fy (x, y)j  0.02xi  0.01yj, , z, , Therefore, the direction of greatest increase in z when you are at the point (50, 100, 225), is given by the direction of, , (50, 100, 225), , §f(50, 100)  i  j, So by pointing the toboggan in the direction of the vector,  §f(50, 100)  (i  j)  i  j, , 0, (50, 100), y, x, , you will achieve the quickest descent., The maximum rate of decrease of the height of the hill at the point (50, 100, 225), is, , Direction of  f(50, 100), , FIGURE 9, The direction of greatest descent is in, the direction of  §f(50, 100) ., , 冟 §f(50, 100) 冟  冟 i  j 冟  12, or approximately 1.41 ft/ft. The graph of f and the direction of greatest descent are, shown in Figure 9., , EXAMPLE 6 Path of a Heat-Seeking Object A heat-seeking object is located at the, point (2, 3) on a metal plate whose temperature at a point (x, y) is T(x, y) , 30  8x 2  2y 2. Find the path of the object if it moves continuously in the direction, of maximum increase in temperature at each point., Solution, , Let the path of the object be described by the position function, r(t)  x(t)i  y(t)j, , where, r(0)  2i  3j, Since the object moves in the direction of maximum increase in temperature, its velocity vector at time t has the same direction as the gradient of T at time t. Therefore,, there exists a scalar function of t, k, such that v(t)  k §T(x, y). But, v(t)  r¿(t) , , dy, dx, i, j, dt, dt

Page 83 :
13.6, , Directional Derivatives and Gradient Vectors, , 1103, , and §T  16x i  4yj. So we have, dy, dx, i, j  16kx i  4ky j, dt, dt, , y, , or, equivalently, the system, dx,  16kx, dt, , dy,  4ky, dt, , Therefore, dy, dy, 4ky, dt, , , dx, dx, 16kx, dt, , x, , x, , y, ◊ f (x0, y0), , 2y 4, 81, , The path of the heat-seeking object is shown in Figure 10., , P(x0, y0), f (x, y)  k, (level curve), , FIGURE 11, §f(x 0, y0) is perpendicular to the level, curve f(x, y)  k at P(x 0, y0)., , dy, y, , dx, 4x, , This is a first-order separable differential equation. The solution of this equation is, x  Cy 4, where C is a constant. (See Section 8.1.) Using the initial condition y(2)  3,, we have 2  C(34), or C  2>(81). So, , FIGURE 10, The path of the heat-seeking object, , 0, , or, , x, , In Figure 10 observe that at each point where the path intersects a level curve that, is part of the contour map of T, the gradient vector §T is perpendicular to the level, curve at that point. To see why this makes sense, refer to Figure 11, where we show, the level curve f(x, y)  k of a function f for some k and a point P(x 0, y0) lying on the, curve. If we move away from P(x 0, y0) along the level curve then the values of f remain, constant (at k). It seems reasonable to conjecture that by moving away in a direction, that is perpendicular to the tangent line to the level curve at P(x 0, y0), f will increase, at the fastest rate. But this direction is given by the direction of §f(x 0, y0). This will, be demonstrated in Section 13.7., , Functions of Three Variables, The definitions of the directional derivative and the gradient of a function of three or, more variables are similar to those for a function of two variables. Also, the algebraic, results that are obtained for the case of a function of two variables carry over to the, higher-dimensional case and are summarized in the following theorem., , THEOREM 4 Directional Derivative and Gradient of a Function, of Three Variables, Let f be a differentiable function of x, y, and z, and let u  u 1i  u 2 j  u 3k be, a unit vector. The directional derivative of f in the direction of u is given by, Du f(x, y, z)  fx (x, y, z)u 1  fy(x, y, z)u 2  fz(x, y, z)u 3, The gradient of f is, §f(x, y, z)  fx (x, y, z)i  fy (x, y, z)j  fz (x, y, z)k, We also write, Du f(x, y, z)  §f(x, y, z) ⴢ u

Page 84 :
1104, , Chapter 13 Functions of Several Variables, , The properties of the gradient given in Theorem 3 for a function of two variables, are also valid for a function of three or more variables. For example, the direction, of greatest increase of f coincides with that of the gradient of f and has magnitude, 冟 §f(x, y, z) 冟., , EXAMPLE 7 Electric Potential Suppose a point charge Q (in coulombs) is located, at the origin of a three-dimensional coordinate system. This charge produces an electric potential V (in volts) given by, V(x, y, z) , , kQ, 2x 2  y 2  z 2, , where k is a positive constant and x, y, and z are measured in meters., a. Find the rate of change of the potential at the point P(1, 2, 3) in the direction of, the vector v  2i  j  2k., b. In which direction does the potential increase most rapidly at P, and what is the, rate of increase?, Solution, a. We begin by computing the gradient of V. Since, Vx , , x, , , , [kQ(x 2  y 2  z 2) 1>2]  kQ 1 12 2 (x 2  y 2  z 2)3>2 (2x), kQx, , (x  y 2  z 2) 3>2, 2, , and by symmetry, Vy  , , kQy, 2, , Vz  , , and, , (x  y  z ), 2, , 2 3>2, , kQz, (x  y 2  z 2)3>2, 2, , we obtain, §V(x, y, z)  Vx i  Vy j  Vz k, , , kQ, (x  y 2  z 2) 3>2, 2, , (xi  y j  zk), , In particular,, §V(1, 2, 3)  , , kQ, 143>2, , (i  2j  3k), , A unit vector u that has the same direction as v  2i  j  2k is, u  13 (2i  j  2k), By Theorem 4 the rate of change of V at P(1, 2, 3) in the direction of v is, DuV(1, 2, 3)  §V(1, 2, 3) ⴢ u  , , , kQ, 14, , 3>2, , (i  2j  3k) ⴢ, , (2i  j  2k), 3, , kQ, kQ, 114kQ, (2  2  6) , , 294, (3)(14) 114, 21114, , In other words, the potential is increasing at the rate of 114kQ>294 volts/m.

Page 85 :
13.6, , Directional Derivatives and Gradient Vectors, , 1105, , b. The maximum rate of change of V occurs in the direction of the gradient of V,, that is, in the direction of the vector (i  2j  3k). Observe that this vector, points toward the origin from P(1, 2, 3). The maximum rate of change of V at, P(1, 2, 3) is given by, 冟 §V(1, 2, 3) 冟  ` , , , kQ, 143>2, , kQ, 3>2, , 14, , (i  2j  3k) `, , 11  4  9 , , kQ, 14, , or kQ>14 volts/m., , 13.6, , CONCEPT QUESTIONS, , 1. a. Let f be a function of x and y, and let u  u 1i  u 2 j be, a unit vector. Define the directional derivative of f in the, direction of u. Why is it necessary to use a unit vector to, indicate the direction?, b. If f is a differentiable function of x, y, and z and, u  u 1 i  u 2 j  u 3k is a unit vector, express, Du f(x, y, z) in terms of the partial derivatives of f and the, components of u., 2. a. What is the gradient of a function f(x, y) of two variables, x and y?, , 13.6, , b. What is the gradient of a function f(x, y, z) of three variables x, y, and z?, c. If f is a differentiable function of x and y and u is a unit, vector, write Du f(x, y) in terms of f and u., d. If f is a differentiable function of x, y, and z and u is a, unit vector, write Du f(x, y, z) in terms of f and u., 3. a. If f is a differentiable at (x, y), what can you say about, Du f(x, y) if §f(x, y)  0?, b. What is the maximum (minimum) value of Du f(x, y),, and when does it occur?, , EXERCISES, , In Exercises 1–4, find the directional derivative of the function f, at the point P in the direction of the unit vector that makes the, angle u with the positive x-axis., p, 1. f(x, y)  x 3  2x 2  y 3; P(1, 2), u , 6, 3p, 2, 2, 2. f(x, y)  2y  x ; P(4, 5), u , 4, p, 3. f(x, y)  (x  1)ey; P(3, 0) , u , 2, p, 4. f(x, y)  sin xy; P(1, 0), u  , 4, , In Exercises 11–28, find the directional derivative of the function, f at the point P in the direction of the vector v., , In Exercises 5–10, find the gradient of f at the point P., , 16. f(x, y)  xexy;, , 5. f(x, y)  2x  3xy  3y  4; P(2, 1), 6. f(x, y) , , 1, x 2  y2, , ;, , P(1, 2), , 7. f(x, y)  x sin y  y cos x;, xy, 8. f(x, y, z) , ;, xz, 9. f(x, y, z)  xeyz;, , P(1, 2, 3), , P(1, 0, 2), , 10. f(x, y, z)  ln(x  y  z );, 2, , P 1 p4 , p2 2, , 2, , 2, , P(1, 1, 1), , 11. f(x, y)  x 3  x 2y 2  xy  y 2; P(1, 1), v  i  2j, 12. f(x, y)  x 3  y 3;, y, 13. f(x, y)  ;, x, , P(2, 1), v , v  i, , P(3, 1),, , 14. f(x, y)  2x 2  y 2  1;, xy, 15. f(x, y) , ;, xy, , P(2, 0),, , 17. f(x, y)  x sin y;, y, 18. f(x, y)  tan1 ;, x, 19. f(x, y, z)  x 2y 3z 4;, , v  3i  4j, , P(2, 2),, , v  i  3j, , P(2, 1),, , 2, , 1, (i  j), 12, , v  2i  j, , P 1 1, p4 2 ,, , v  2i  3j, vij, , P(1, 1),, , P(3, 2, 1),, , vijk, , 20. f(x, y, z)  x  2xy  2yz ;, v  i  2j  2k, , P(2, 1, 1),, , 21. f(x, y, z)  1xyz;, , v  2i  4j  4k, , 2, , 2, , 3, , P(4, 2, 2),, , 22. f(x, y, z)  2xy  6y z ;, , V Videos for selected exercises are available online at www.academic.cengage.com/login., , 2, , 2 2, , P(2, 3, 1),, , v  2i  k

Page 86 :
1106, , Chapter 13 Functions of Several Variables, , 23. f(x, y, z)  x 2eyz;, , P(2, 3, 0), v  i  2j  3k, , 24. f(x, y, z)  ln(x  y 2  z 2) ;, v  3i  2j  k, 2, , 25. f(x, y, z)  x 2y cos 2z;, , P(1, 2, 1),, , P 1 1, 2, p4 2 ,, , 26. f(x, y, z)  ex (2 cos y  3 sin z);, v  2i  j  3k, , S, , vijk, , P 1 1, p6 , p6 2 ,, , 400, 300, , y, 27. f(x, y, z)  x tan1 a b ;, z, , P(3, 2, 2),, , 28. f(x, y, z)  x 2 sin1 yz;, , 1, P(2, 1, 0), v , (i  j  k), 13, , v  i  2j  k, , In Exercises 29–32, find the directional derivative of the function, f at the point P in the direction from P to the point Q., , P, 200, , Note: This path is called the path of steepest ascent., , 43. Path of Steepest Descent The figure shows a topographical, map of a 620-ft hill with contours at 100-ft intervals., , 29. f(x, y)  x 3  y 3; P(1, 2), Q(2, 5), 30. f(x, y)  xey;, , P(2, 0),, , Q(1, 2), , p, 31. f(x, y, z)  x sin(2y  3z); P 1 1, p4 , 12, 2,, , 32. f(x, y, z) , , 600, 500, , xy, ; P(2, 1, 1) , Q(3, 2, 2), yz, , N, A, , Q 1 3, p2 , p4 2, , W, , E, S, , C, D, 620 600 500 400, , 300, , 200, , B, , In Exercises 33–36, find a vector giving the direction in which, the function f increases most rapidly at the point P. What is the, maximum rate of increase?, 33. f(x, y)  22x  3y 2; P(3, 2), 34. f(x, y)  e2x cos y; P 1 0, p4 2, , 35. f(x, y, z)  x 3  2xz  2yz 2  z 3; P(1, 3, 2), 36. f(x, y, z)  ln(x 2  2y 2  3z 2) ; P(1, 2, 1), In Exercises 37–40, find a vector giving the direction in which, the function f decreases most rapidly at the point P. What is the, maximum rate of decrease?, 37. f(x, y)  tan1 (2x  y); P(0, 0), 38. f(x, y)  xey ; P(1, 0), 2, , 39. f(x, y, z) , , y, x,  ;, y, z, , P(1, 1, 2), , 40. f(x, y, z)  1xy cos z;, , P 1 4, 1, p4 2, , 41. The height of a hill (in feet) is given by, h(x, y)  20(16  4x 2  3y 2  2xy  28x  18y), where x is the distance (in miles) east and y the distance (in, miles) north of Bolton. In what direction is the slope of the, hill steepest at the point 1 mile north and 1 mile east of, Bolton? What is the steepest slope at that point?, 42. Path of Steepest Ascent The following figure shows the contour map of a hill with its summit denoted by S. Draw the, curve from P to S that is associated with the path you will, take to reach the summit by ascending the direction of the, greatest increase in altitude., Hint: Study Figure 10., , a. If you start from A and proceed in a southwesterly direction, will you be ascending, descending, or neither, ascending nor descending? What if you start from B?, b. If you start from C and proceed in a westerly direction,, will you be ascending, descending, or neither ascending, nor descending?, c. If you start from D, in what direction should you proceed, to have the steepest ascent?, d. If you want to climb to the summit of the hill using the, gentlest ascent, would you start from the east or the west?, 44. Steady-State Temperature Consider the upper half-disk, H  {(x, y) 冟 x 2  y 2 1, y  0} (see the figure). If the, temperature at points on the upper boundary is kept at, 100°C and the temperature at points on the lower boundary, is kept at 50°C, then the steady-state temperature at any, point (x, y) inside the half-disk is given by, T(x, y)  100 , , 1  x 2  y2, 100, tan1, p, 2y, , y, , T  100, , 1 T  50, , 0, , 1, , x

Page 87 :
13.6, Find the rate of change of the temperature at the point, P 1 12, 12 2 in the direction of the vector v  2i  3j., , 45. Steady-State Temperature Consider the upper half-disk, H  {(x, y) 冟 x 2  y 2 1, y  0} (see the figure). If the, temperature at points on the upper boundary is kept at, 100°C and the temperature at points on the lower boundary, is kept at 0°C, then the steady-state temperature at any point, (x, y) inside the half-disk is given by, T(x, y) , , 2y, 200, tan1, p, 1  x 2  y2, , 1, a. Find the gradient of T at the point 1 17, 4 , 4 2 , and interpret, your result., b. Sketch the isothermal curve of T passing through the, 1, point 1 17, 4 , 4 2 and the gradient vector §T at that point on, the same coordinate system., , y, , Directional Derivatives and Gradient Vectors, , 50. Cobb-Douglas Production Function The output of a finished product is given by the production function, f(x, y)  100x 0.6y 0.4, where x stands for the number of units of labor and y stands, for the number of units of capital. Currently, the amount, being spent on labor is 500 units, and the amount being, spent on capital is 250 units. If the manufacturer wishes to, expand production by injecting an additional 10 units into, labor, how much more should be put into capital to maximize the increase in output?, 51. The figure shows the contour map of a function f of two, variables x and y. Use it to estimate the directional derivative of f at P0 in the indicated direction., y, 20, 4, , 25, 30, , P0, , 35, , 3, P1, , 2, , T  100, , 1107, , 40, , 1, 0, 1, , T0, , 0, , 1, , x, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , x, , Hint: If u is a unit vector having indicated direction, then, , 46. The temperature at a point P(x, y, z) of a solid ball, of radius 4 with center at the origin is given by, T(x, y, z)  xy  yz  xz. Find the direction in which, T is increasing most rapidly at P(1, 1, 2)., 47. Let T(x, y, z) represent the temperature at a point P(x, y, z), of a region R in space. If the isotherms of T are concentric, spheres, show that the temperature gradient §T points either, toward or away from the center of the spheres., Hint: Recall that the isotherms of T are the sets on which T is constant., , 48. Suppose the temperature at the point (x, y) on a thin sheet of, metal is given by, T(x, y) , , 1, , 100(1  3x  2y), 1  2x 2  3y 2, , degrees Fahrenheit. In what direction will the temperature be, increasing most rapidly at the point (1, 2)? In what direction, will it be decreasing most rapidly?, 49. The temperature (in degrees Fahrenheit) at a point (x, y) on, a metal plate is, T(x, y)  90  6x 2  2y 2, An insect located at the point (1, 1) crawls in the direction, in which the temperature drops most rapidly., a. Find the path of the insect., b. Sketch a few level curves of T and the path found in, part (a)., , Du f(P0) ⬇, , f(P1)  f(P0), d(P0, P1), , when f(Pi) is the value of f at Pi (i  0, 1) and d(P0, P1) is the distance between P0 and P1., , 52. A rectangular metal plate of dimensions 8 in. 4 in. is, placed on a rectangular coordinate system with one corner at, the origin and the longer side along the positive x-axis. The, figure shows the contour map of the function f describing, the temperature of the plate in degrees Fahrenheit. Use the, contour map to estimate the rate of change of the temperature at the point (3, 1) in the direction from the point (3, 1), toward the point (5, 4)., y (in.), 4, , (5, 4), , 3, , 60, , 2, (3, 1), , 1, , 90 85, 0, , 1, , 2, , 3, , 4, , 65, , 75 70, , 80, 5, , 6, , 7, , 8 x (in.), , Hint: See Exercise 51., , 53. Suppose that f is differentiable, and suppose that the directional derivative of f at the origin attains a maximum value, of 5 in the direction of the vector from the origin to the, point (3, 4). Find §f(0, 0) .

Page 88 :
1108, , Chapter 13 Functions of Several Variables, , 54. Find unit vector(s) u  具u 1, u 2典 such that the directional, derivative of f(x, y)  x 2  exy at the point (1, 0) in the, direction of u has value 1., cas In Exercises 55 and 56, (a) Plot several level curves of each, , pair of functions f and t using the same viewing window, and, (b) Show analytically that each level curve of f intersects all, level curves of t at right angles., 55. f(x, y)  x 2  y 2,, , t(x, y)  xy, , 56. f(x, y)  ex cos y,, , t(x, y)  ex sin y, , 57. Let f(x, y)  x 2  y 2, and let t(x, y)  x 2  y 2. Find the, direction in which f increases most rapidly and the direction, in which t increases most rapidly at (0, 0). Is Theorem 3, applicable here?, , 13.7, , In Exercises 58–62, determine whether the statement is true or, false. If it is true, explain why it is true. If it is false, give an, example to show why it is false., 58. If f is differentiable at each point, then the directional derivative of f exists in all directions., 59. If f is differentiable at each point, then the value of the, directional derivative at any point in a given direction, depends only on the direction and the partial derivatives, fx and fy at that point., 60. If fx (a, b)  0 and fy (a, b)  0, then §f(a, b)  0., 61. The maximum value of Du f(x, y) is 2f 2x (x, y)  f 2y (x, y)., 62. If §f is known, then we can determine f completely., , Tangent Planes and Normal Lines, One compelling reason for studying the tangent line to a curve is that the curve may be, approximated by its tangent line near a point of tangency (Figure 1). Answers to questions about the curve near a point of tangency may be obtained indirectly by analyzing, the tangent line, a relatively simple task, rather than by studying the curve itself. As you, might recall, both the approximation of the change in a function using its differential, and Newton’s method for finding the zeros of a function are based on this observation., Our motivation for studying tangent planes to a surface in space is the same as that, for studying tangent lines to a curve: Near a point of tangency, a surface may be approximated by its tangent plane (Figure 1). We will show later that approximating the change, in z  f(x, y) using the differential is tantamount to approximating this change by the, change in z on the tangent plane., z, y, y  f (x), T, z  f (x, y), 0, , FIGURE 1, Near a point of tangency, the tangent, line approximates the curve, and the, tangent plane approximates the surface., , y, 0, (a) The tangent line to a curve, , x, , x, (b) The tangent plane to a surface, , Geometric Interpretation of the Gradient, We begin by looking at the geometric interpretation of the gradient of a function. This, vector will play a central role in our effort to find the tangent plane to a surface., Suppose that the temperature T at any point (x, y) in the plane is given by the function f ; that is, T  f(x, y). Then the level curve f(x, y)  c, where c is a constant, gives, the set of points in the plane that have temperature c (Figure 2). Recall that such a, curve is called an isothermal curve.

Page 89 :
13.7, y, , C, P, , (a, b), f (x, y)  c, , 0, , x, , FIGURE 2, The level curve C defined by, f(x, y)  c is an isothermal curve., , Tangent Planes and Normal Lines, , 1109, , If we are at the point P(a, b), in what direction should we move if we want to experience the greatest increase in temperature? Since the temperature remains constant if, we move along C, it seems reasonable to conjecture that proceeding in the direction, perpendicular to the tangent line to C at P will result in the greatest increase in temperature. But as we saw in Section 13.6, the function f (and hence the temperature), increases most rapidly in the direction given by its gradient §f(a, b). These observations suggest that the gradient §f(a, b) is perpendicular to the tangent line to the level, curve f(x, y)  c at P. That this is indeed the case can be demonstrated as follows:, Suppose that the curve C is represented by the vector function, r(t)  t(t)i  h(t)j, where t and h are differentiable functions, a  t(t 0) and b  h(t 0), and t 0 lies in the, parameter interval (Figure 3). Since the point (x, y)  (t(t), h(t)) lies on C, we have, f(t(t), h(t))  c, for all t in the parameter interval., y, ◊f (a, b), C, , P(a, b), r(t0), f (x, y)  c, , r(t), , FIGURE 3, The curve C may be represented by, r(t)  xi  yj  t(t)i  h(t)j., , t0, , t, , x, , 0, , Differentiating both sides of this equation with respect to t and using the Chain, Rule for a function of two variables, we obtain, f dx, f dy, , 0, x dt, y dt, Recalling that, §f(x, y) , , f, f, i, j, x, y, , and, , rⴕ(t) , , dy, dx, i, j, dt, dt, , we can write this last equation in the form, §f(x, y) ⴢ rⴕ(t)  0, In particular, when t  t 0, we have, §f(a, b) ⴢ rⴕ(t 0)  0, Thus, if rⴕ(t 0)  0, the vector §f(a, b) is orthogonal to the tangent vector rⴕ(t 0) at, P(a, b). Loosely speaking, we have demonstrated the following:, §f is orthogonal to the level curve f(x, y)  c at P., , See Figure 3., , EXAMPLE 1 Let f(x, y)  x 2  y 2. Find the level curve of f passing through the point, (5, 3). Also, find the gradient of f at that point, and make a sketch of both the level, curve and the gradient vector.

Page 90 :
1110, , Chapter 13 Functions of Several Variables, y, , 4, 4 0, 4, , Solution Since f(5, 3)  25  9  16, the required level curve is the hyperbola, x 2  y 2  16. The gradient of f at any point (x, y) is, , Tangent, line, , §f(x, y)  2x i  2y j, , (5, 3), x, , 4, , and, in particular, the gradient of f at (5, 3) is, §f(5, 3)  10i  6j, , ◊ f (5, 3), , The level curve and §f(5, 3) are shown in Figure 4., FIGURE 4, The gradient §f(5, 3) is orthogonal to, the level curve x 2  y 2  16 at (5, 3)., , EXAMPLE 2 Refer to Example 1. Find equations of the normal line and the tangent, line to the curve x 2  y 2  16 at the point (5, 3)., Solution We think of the curve x 2  y 2  16 as the level curve f(x, y)  k of the, function f(x, y)  x 2  y 2 for k  16. From Example 1 we see that §f(5, 3) , 10i  6j. Since this gradient is normal to the curve x 2  y 2  16 at (5, 3) (see Figure 4), we see that the slope of the required normal line is, m1  , , 6, 3, , 10, 5, , Therefore, an equation of the normal line is, 3, y  3   (x  5), 5, , 3, y x6, 5, , or, , The slope of the required tangent line is, m2  , , 1, 5, , m1, 3, , so an equation of the tangent line is, y3, , 5, (x  5), 3, , or, , y, , 5, 16, x, 3, 3, , Next, suppose that F(x, y, z)  k is the level surface S of a differentiable function, F defined by T  F(x, y, z). You may think of the function F as giving the temperature at any point (x, y, z) in space and interpret the following argument in terms of this, application., Suppose that P(a, b, c) is a point on S and let C be a smooth curve on S passing, through P. Then C can be described by the vector function, r(t)  f(t)i  t(t)j  h(t)k, where f(t 0)  a, t(t 0)  b, h(t 0)  c, and t 0 is a point in the parameter interval (Figure 5)., z, F(a, b, c), r'(t0), , S, , P(a, b, c), C, , FIGURE 5, The curve C is described by, r(t)  f(t)i  t(t)j  h(t)k with, P(a, b, c) corresponding to t 0., , t0, , 0, , t, , y, x

Page 91 :
13.7, , Tangent Planes and Normal Lines, , 1111, , Since the point (x, y, z)  ( f(t), t(t), h(t)) lies on S, we have, F( f(t), t(t), h(t))  k, for all t in the parameter interval. If r is differentiable, then we can use the Chain Rule, to differentiate both sides of this equation to obtain, F dx, F dy, F dz, , , 0, x dt, y dt, z dt, z, , This is the same as, [Fx (x, y, z)i  Fy(x, y, z)j  Fz(x, y, z)k] ⴢ c, , F(a, b, c), , dy, dx, dz, i, j, kd  0, dt, dt, dt, , r'(t), , or, in an even more abbreviated form,, , P, , §F(x, y, z) ⴢ rⴕ(t)  0, , C, , In particular, at t  t 0 we have, , 0, , §F(a, b, c) ⴢ rⴕ(t 0)  0, y, , x, , FIGURE 6, The gradient §F(a, b, c) is orthogonal, to the tangent vector of every curve, on S passing through P(a, b, c)., , This shows that if rⴕ(t 0)  0, then the gradient vector §F(a, b, c) is orthogonal to, the tangent vector rⴕ(t 0) to C at P (Figure 6). Since this argument holds for any differentiable curve passing through P(a, b, c) on S, we have shown that §F(a, b, c) is, orthogonal to the tangent vector of every curve on S passing through P. Thus, loosely, speaking, we have demonstrated the following result., , §F is orthogonal to the level surface F(x, y, z)  0 at P., , Note Interpreting the function F as giving the temperature at any point (x, y, z) in, space as was suggested earlier, we see that the level surface F(x, y, z)  k gives all, points (x, y, z) in space whose temperature is k. The result that was just derived simply states that if you are at any point on this surface, then moving away from it in a, direction of §F (perpendicular to the surface at that point) will result in the greatest, increase in temperature., z, , F(0, 3, 4), , 5, , EXAMPLE 3 Let F(x, y, z)  x 2  y 2  z 2. Find the level surface that contains the, point (0, 3, 4). Also, find the gradient of F at that point, and make a sketch of both the, level surface and the gradient vector., , 5, , y, , 5, x, , FIGURE 7, The gradient §F(0, 3, 4) is orthogonal, to the level surface x 2  y 2  z 2  25, at (0, 3, 4)., , Solution Since F(0, 3, 4)  0  9  16  25, the required level surface is the sphere, x 2  y 2  z 2  25 with center at the origin and radius 5. The gradient of F at any, point (x, y, z) is, §F(x, y, z)  2xi  2yj  2zk, so the gradient of F at (0, 3, 4) is, §F(0, 3, 4)  6j  8k, The level surface and §F(0, 3, 4) are sketched in Figure 7.

Page 92 :
1112, , Chapter 13 Functions of Several Variables, , Tangent Planes and Normal Lines, We are now in a position to define a tangent plane to a surface in space. But before, doing so, let’s digress a little to talk about the representation of surfaces in space. Up, to now, we have assumed that a surface in space is described by a function f with, explicit representation z  f(x, y)., Another way of describing a surface in space is via a function that is represented, implicitly by the equation, F(x, y, z)  0, , (1), , Here, F is a function of the three variables x, y, and z described by the equation, w  F(x, y, z). Thus, we can think of Equation (1) as representing a level surface, of F., For a surface S that is given explicitly by z  f(x, y), we define, F(x, y, z)  z  f(x, y), This shows that we can also view S as the level surface of F given by Equation (1)., For example, the surface described by z  x 2  2y 2  1 can be viewed as the level, surface of F defined by F(x, y, z)  0, where F(x, y, z)  z  x 2  2y 2  1., To define a tangent plane, let S be a surface described by F(x, y, z)  0, and let, P(a, b, c) be a point on S. Then, as we saw earlier, the gradient §F(a, b, c) at P is, orthogonal to the tangent vector of every curve on S passing through P (Figure 8). This, suggests that we define the tangent plane to S at P to be the plane passing through P, and containing all these tangent vectors. Equivalently, the tangent plane should have, §F(a, b, c) as a normal vector., z, F(a, b, c), S, P, F(x, y, z), , FIGURE 8, The tangent plane to S at P contains, the tangent vectors to all curves, on S passing through P., , t0, , t, , 0, y, x, , DEFINITIONS Tangent Plane and Normal Line, Let P(a, b, c) be a point on the surface S described by F(x, y, z)  0, where F, is differentiable at P, and suppose that §F(a, b, c)  0. Then the tangent plane, to S at P is the plane that passes through P and has normal vector §F(a, b, c)., The normal line to S at P is the line that passes through P and has the same, direction as §F(a, b, c)., , Using Equation (4) from Section 11.5, we see that an equation of the tangent plane, is, Fx(a, b, c)(x  a)  Fy(a, b, c)(y  b)  Fz (a, b, c)(z  c)  0, , (2)

Page 93 :
13.7, , Tangent Planes and Normal Lines, , 1113, , and using Equation (2) of Section 11.5, we see that the equations of the normal line, (in symmetric form) are, yb, xa, zc, , , Fx (a, b, c), Fy(a, b, c), Fz(a, b, c), , (3), , EXAMPLE 4 Find equations of the tangent plane and normal line to the ellipsoid with, equation 4x 2  y 2  4z 2  16 at the point (1, 2, 12)., Solution The given equation can be written in the form F(x, y, z)  0, where, F(x, y, z)  4x 2  y 2  4z 2  16. The partial derivatives of F are, Fx (x, y, z)  8x,, , z, , Fy(x, y, z)  2y,, , and, , Fz(x, y, z)  8z, , In particular, at the point (1, 2, 12), , L, , Fx(1, 2, 12)  8,, T, , Fy (1, 2, 12)  4,, , and, , Fz(1, 2, 12)  812, , Then, using Equation (2), we find that an equation of the tangent plane to the ellipsoid, at (1, 2, 12) is, 8(x  1)  4(y  2)  812(z  12)  0, y, , x, , FIGURE 9, The tangent plane and normal line to, the ellipsoid 4x 2  y 2  4z 2  16, at (1, 2, 12)., , or 2x  y  212z  8. Next, using Equation (3), we obtain the following parametric equations of the normal line:, y2, x1, z  12, , , 8, 4, 812, , or, , x1, z  12, y2, 2, 212, , The tangent plane and normal line are shown in Figure 9., , EXAMPLE 5 Find equations of the tangent plane and normal line to the graph of the, function f defined by f(x, y)  4x 2  y 2  2 at the point where x  1 and y  1., Solution, , Here, the surface is defined by, z  f(x, y)  4x 2  y 2  2, , z, , and we recognize it to be a paraboloid. This equation can be rewritten in the form, 12, , F(x, y, z)  z  f(x, y)  0, where F(x, y, z)  z  4x  y 2  2. The partial derivatives of F are, , 1100, , 2, , Fx (x, y, z)  8x,, 7 (1, 1, 7), , Fy (x, y, z)  2y,, , and, , Fz(x, y, z)  1, , If x  1 and y  1, then z  f(1, 1)  4  1  2  7. At the point (1, 1, 7) we have, Fx(1, 1, 7)  8,, , T, , Fy (1, 1, 7)  2,, , and, , Fz(1, 1, 7)  1, , Then, using Equation (2), we find an equation of the tangent plane to the paraboloid, at (1, 1, 7) to be, , L, , 8(x  1)  2(y  1)  1(z  7)  0, 1, , 1 2, 3, , y, , x, , FIGURE 10, The tangent plane and normal line to, the paraboloid z  4x 2  y 2  2 at, (1, 1, 7)., , or 8x  2y  z  3. Next, using Equation (3), we find that the parametric equations, of the normal line at (1, 1, 7) are, y1, x1, z7, , , 8, 2, 1, The tangent plane and normal line are shown in Figure 10., , Copyright 2009 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.

Page 94 :
1114, , Chapter 13 Functions of Several Variables, , Using the Tangent Plane of f to Approximate, the Surface z ⴝ f(x, y), We conclude this section by showing that in approximating the change ⌬z in, z  f(x, y) as (x, y) changes from (a, b) to (a  ⌬x, b  ⌬y) by the differential, dz  fx(a, b) ⌬x  fy (a, b) ⌬y, we are in effect using the tangent plane of f near P(a, b), to approximate the surface z  f(x, y) near P(a, b)., We begin by finding an expression for the tangent plane to the surface z  f(x, y), at (a, b). Writing z  f(x, y) in the form F(x, y, z)  z  f(x, y)  0 we see that, Fx (a, b, c)  fx (a, b),, , Fy (a, b, c)  fy(a, b),, , Fz(a, b, c)  1, , and, , Using Equation (2), we find that the required equation is, fx (a, b)(x  a)  fy (a, b)(y  b)  (z  c)  0, or, z  f(a, b)  fx(a, b) ⌬x  fy (a, b) ⌬y, , c  f(a, b), , (4), , But the expression on the right is just the differential of f at (a, b). So Equation (4), implies that dz  z  f(a, b); that is, dz represents the change in height of the tangent, plane (Figure 11)., z, (a  Δ x, b  Δy, f(a  Δ x, b  Δy)), z  f (x, y), , dz, , Δz, , (a, b, f (a, b)), Tangent plane, 0, (a  Δ x, b), y, , FIGURE 11, The relationship between ⌬z and dz, , x, , (a  Δ x, b  Δy), , (a, b), , By Theorem 1 of Section 13.4 we have, ⌬z  fx (a, b) ⌬x  fy (a, b) ⌬y  e1 ⌬x  e2 ⌬y, or, ⌬z  dz  e1 ⌬x  e2 ⌬y, where e1 and e2 are functions of ⌬x and ⌬y that approach 0 as ⌬x and ⌬y approach 0., Therefore, as was pointed out in Section 13.4, ⌬z ⬇ dz if ⌬x and ⌬y are small. Recalling the meaning of ⌬z (see Figure 11), we see that we are using the tangent plane at, (a, b) to approximate the surface z  f(x, y) when (x, y) is close to (a, b)., , 13.7, , CONCEPT QUESTIONS, , 1. a. Consider the level curve f(x, y)  c, where f is differentiable and c is a constant. What can you say about §f at, a point P on the level curve? Illustrate with a figure., b. Repeat part (a) for a level surface F(x, y, z)  c., , 2. a. Define the tangent plane to a surface S described by, F(x, y, z)  0 at the point P(a, b, c) on S. Illustrate with, a figure and give an equation of the tangent plane., b. Repeat part (a) for the normal line to S at P.

Page 95 :
13.7, , 13.7, , 1. f(x, y)  y 2  x 2; P(1, 2), 2. f(x, y)  4x 2  y 2;, 3. f(x, y)  x 2  y;, , P 1 13, 2 , 12, , P(1, 3), , y, 29. z  tan1 a b ;, x, , Pa1, 1,, , p, b, 4, , 30. z  x cos y  0;, , Pa2,, , p, , 1b, 3, , 31. sin xy  3z  3;, , P(0, 3, 1), , 32. e (cos y  1)  2z  2;, x, , 4. f(x, y)  2x  3y; P(3, 4), , x2, a, , 1 313, 2 , 22, , y2, x2, ,  1;, 9, 16, , 6. x 4  x 2  y 2  0;, 4, , 1 12, 134 2, , xx 0, , 2, , 8. 2x  y  exy  2;, , 2, , In Exercises 9–14, sketch (a) the level surface of the function F, that passes through the point P and (b) the gradient of F at P., 9. F(x, y, z)  x 2  y 2  z 2;, , 2, , a2, , P(2, 3, 1), 2, , P 1 2, 12, 0 2, , 15. x 2  4y 2  9z 2  17;, , P(2, 1, 1), , 16. 2x 2  y 2  3z 2  2;, , P(2, 3, 1), , 17. x  2y  4z  4;, , P(4, 2, 1), , 2, , 2, , 2, , 19. xy  yz  xz  11;, 20. xyz  4;, , 22. z  y  2x ;, 2, , 2, , 25. z  xey;, , a, , P(2, 0, 2), , 26. z  e sin py;, 27. z  ln(xy  1);, x, 28. z  ln ;, y, , P(0, 1, 0), P(3, 0, 0), , P(2, 2, 0), , 1, , , , yy0, b2, , , , zz 0, c2, , 1, , , , yy0, b2, , , , zz 0, c2, , 1, , 2, , , , 2yy0, b2, ,  c(z  z 0), , 37. Find the points on the sphere x 2  y 2  z 2  14 at which, the tangent plane is parallel to the plane x  2y  3z  12., , P(2, 4, 8), , 3, , x, , c2, , at the point (x 0, y0, z 0) can be written as, 2xx 0, , P(1, 2, 25), , 24. x  xy  z  2x  6  0;, 2, , P(1, 0, 2), , P(1, 2, 3), , 23. xz 2  yx 2  y 2  2x  3y  6  0;, 3, , z2, , at the point (x 0, y0, z 0), and express it in a form similar to, that of Exercise 34., , P(2, 1, 2), , 21. z  9x 2  4y 2;, , , , 36. Show that an equation of the tangent plane to the elliptic, paraboloid, y2, x2,  2  cz, 2, a, b, , 18. x  y  z  2xy  4xz  x  y  12;, 2, , b, , 2, , 35. Find an equation of the tangent plane to the hyperboloid of, two sheets, y2, x2, z2,  2 21, 2, a, b, c, , P(1, 3, 2), , In Exercises 15–32, find equations for the tangent plane and, the normal line to the surface with the equation at the given, point., , 2, , 2, , xx 0, , P(0, 2, 4), , 13. F(x, y, z)  x  y  z ;, , 2, , y2, , at the point (x 0, y0, z 0) can be written as, , P(1, 1, 2), , 12. F(x, y, z)  2x  3y  z;, 14. F(x, y, z)  xy;, , , , 34. Show that an equation of the tangent plane to the hyperboloid, y2, x2, z2, , , 1, a2, b2, c2, , P(1, 2, 2), , 10. F(x, y, z)  z  x 2  y 2;, , 2, , a, , ( 15, 1), , (1, 1), , 11. F(x, y, z)  x 2  y 2;, , 2, , at the point (x 0, y0, z 0) can be written as, , 7. x  2x y  y  9x  9y  0;, 2 2, , P(0, 0, 2), , 33. Show that an equation of the tangent plane to the ellipsoid, , In Exercises 5–8, find equations of the normal and tangent lines, to the curve at the given point., , 4, , 1115, , EXERCISES, , In Exercises 1–4, sketch (a) the level curve of the function f that, passes through the point P and (b) the gradient of f at P., , 5., , Tangent Planes and Normal Lines, , P(2, 1, 3), , P(1, 2, 1), , 38. Find the points on the hyperboloid of two sheets, x 2  2y 2  z 2  4 at which the tangent plane is parallel, to the plane 2x  2y  4z  1., 39. Find the points on the hyperboloid of one sheet, 2x 2  y 2  z 2  1 at which the normal line is parallel to, the line passing through the points (1, 1, 2) and (3, 3, 3)., 40. Find the points on the surface x 2  4y 2  3z 2  2xy  16, at which the tangent plane is horizontal., , V Videos for selected exercises are available online at www.academic.cengage.com/login.

Page 96 :
1116, , Chapter 13 Functions of Several Variables, , 41. Two surfaces are tangent to each other at a point P if and, only if they have a common tangent plane at that point., Show that the elliptic paraboloid 2x 2  y 2  z  5  0, and the sphere x 2  y 2  z 2  6x  8y  z  17  0 are, tangent to each other at the point (1, 2, 1)., 42. Two surfaces are orthogonal to each other at a point of, intersection P if and only if their normal lines at P are, orthogonal. Show that the sphere x 2  y 2  z 2  17  0, and the elliptic paraboloid 2x 2  y  2z 2  2  0 are, orthogonal to each other at the point (1, 4, 0)., , parametric equations of the tangent line to C at the point, 1 122, 122, 1 2 ., In Exercises 46–49, determine whether the statement is true or, false. If it is true, explain why it is true. If it is false, give an, example to show why it is false., 46. The tangent line at a point P on the level curve f(x, y)  c, is orthogonal to §f at P., 47. The line with equations, y1, x2, z, , , 4, 6, 2, , 43. Show that any line that is tangent to the ellipse, (x 2>a 2)  (y 2>b 2)  1 has the equation, , is perpendicular to the plane with equation 2x  3y  z  4., , (b cos u)x  (a sin u)y  ab, where u lies in the interval [0, 2p)., 44. Suppose that two surfaces F(x, y, z)  0 and G(x, y, z)  0, intersect along a curve C and that P(x 0, y0, z 0) is a point on, C. Show that the vector §F(x 0, y0, z 0), §G(x 0, y0, z 0) is, parallel to the tangent line to C at P. Illustrate with a sketch., 45. Refer to Exercise 44. Let C be the intersection of the sphere, x 2  y 2  z 2  2 and the paraboloid z  x 2  y 2. Find the, , 48. If an equation of the tangent plane at the point P0(x 0, y0, z 0), on the surface described by F(x, y, z)  0 is, ax  by  cz  d, then §F(x 0, y0, z 0)  k具a, b, c典 for some, scalar k., 49. The vector equation of the normal line passing through the, point P0 (x 0, y0, z 0) on the surface with equation, F(x, y, z)  0 is r(t)  具x 0, y0, z 0典  t §F(x 0, y0, z 0)., , 13.8 Extrema of Functions of Two Variables, Relative and Absolute Extrema, In Chapter 3 we saw that the solution of a problem often reduces to finding the extreme, values of a function of one variable. A similar situation arises when we solve problems, involving a function of two or more variables., For example, suppose that the Scandi Company manufactures computer desks in, both assembled and unassembled versions. Then its weekly profit P is a function of, the number of assembled units, x, and the number of unassembled units, y, sold per, week; that is, P  f(x, y). A question of paramount importance to the manufacturer is:, How many assembled desks and how many unassembled desks should the company, manufacture per week to maximize its weekly profit? Mathematically, the problem is, solved by finding the values of x and y that will make f(x, y) a maximum., In this section and the next section we will focus our attention on finding the extrema, of a function of two variables. As in the case of a function of one variable, we distinguish between the relative (or local) extrema and the absolute extrema of a function of, two variables., , DEFINITION Relative Extrema of a Function of Two Variables, Let f be a function defined on a region R containing the point (a, b). Then f has, a relative maximum at (a, b) if f(x, y) f(a, b) for all points (x, y) in an open, disk containing (a, b). The number f(a, b) is called a relative maximum value., Similarly, f has a relative minimum at (a, b) with relative minimum value, f(a, b) if f(x, y)  f(a, b) for all points (x, y) in an open disk containing (a, b).

Page 97 :
13.8, , Extrema of Functions of Two Variables, , 1117, , Loosely speaking, f has a relative maximum at (a, b) if the point (a, b, f(a, b)) is, the highest point on the graph of f when compared to all nearby points. A similar interpretation holds for a relative minimum., If the inequalities in this last definition hold for all points (x, y) in the domain of, f, then f has an absolute maximum (absolute minimum) at (a, b) with absolute maximum value (absolute minimum value) f(a, b). Figure 1 shows the graph of a function defined on a domain D with relative maxima at (a, b) and (e, t) and a relative minimum at (c, d). The absolute maximum of f occurs at (e, t), and the absolute minimum, of f occurs at (h, i)., z, , Relative, maximum, , Absolute maximum, (also a relative, maximum), , Absolute, minimum, Relative, minimum (a, b), , FIGURE 1, The relative and absolute extrema of, the function f over the domain D, , x, , (h, i), , y, , (c, d) (e, t), , D, , Critical Points—Candidates for Relative Extrema, Figure 2a shows the graph of a function f with a relative maximum at a point (a, b), lying inside the domain of f. As you can see, the tangent plane to the surface z  f(x, y), at the point (a, b, f(a, b)) is horizontal. This means that all the directional derivatives, of f at (a, b), if they exist, must be zero. In particular, fx(a, b)  0 and fy (a, b)  0., Next, Figure 2b shows the graph of a function with a relative maximum at a point, (a, b) . Note that both fx (a, b) and fy (a, b) do not exist because the surface z  f(x, y), has a point (a, b, f(a, b)) that looks like a jagged mountain peak., z, (a, b, f (a, b)), fy(a, b)  0, fx(a, b)  0, , z, (a, b, f (a, b)), , FIGURE 2, At a relative extremum of f, either, fx  fy  0 or one or both partial, derivatives do not exist., , 0, , 0, x, , (a, b), , (a) fx  fy  0, , y, , x, , (a, b), , y, , (b) fx and fy do not exist., , You are encouraged to draw similar graphs of functions having relative minima at, points lying inside the domain of the functions. All of these points are critical points, of a function of two variables., , Copyright 2009 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.

Page 98 :
1118, , Chapter 13 Functions of Several Variables, , DEFINITION Critical Points of a Function, Let f be defined on an open region R containing the point (a, b). We call (a, b), a critical point of f if, a. fx and/or fy do not exist at (a, b) or, b. both fx (a, b)  0 and fy (a, b)  0., , The next theorem tells us that the relative extremum of a function f defined on an, open region can occur only at a critical point of f., , THEOREM 1 The Critical Points of f Are Candidates for Relative Extrema, If f has a relative extremum (relative maximum or relative minimum) at a point, (a, b) in the domain of f, then (a, b) must be a critical point of f., , PROOF If either fx or fy does not exist at (a, b), then (a, b) is a critical point of f., , So suppose that both fx (a, b) and fy(a, b) exist. Let t(x)  f(x, b). If f has a relative, extremum at (a, b), then t has a relative extremum at a, so by Theorem 1 of Section 3.1,, t¿(a)  0. But, t¿(a)  lim, , h→0, , f(a  h, b)  f(a, b),  fx(a, b), h, , Therefore, fx (a, b)  0. Similarly, by considering the function f(y)  f(a, y), we obtain, fy (a, b)  0. Thus, (a, b) is a critical point of f., , EXAMPLE 1 Let f(x, y)  x 2  y 2  4x  6y  17. Find the critical point of f, and, , z, , show that f has a relative minimum at that point., z  f (x, y), , 3, , 1, , x, , 0, , 1, , 2, (2, 3), , and, , fy(x, y)  2y  6  2(y  3), , Observe that both fx and fy are continuous for all values of x and y. Setting fx and fy, equal to zero, we find that x  2 and y  3, so (2, 3) is the only critical point of f., Next, to show that f has a relative minimum at this point, we complete the squares in, x and y and write f(x, y) in the form, , (2, 3, 4), , 2, , 1, , To find the critical point of f, we compute, , fx (x, y)  2x  4  2(x  2), , 4, , 2, , Solution, , f(x, y)  (x  2)2  (y  3)2  4, , 3, y, , FIGURE 3, The function f has a relative minimum, at (2, 3)., , Notice that (x  2)2  0 and (y  3)2  0, so f(x, y)  4 for all (x, y) in the domain, of f. Therefore, f(2, 3)  4 is a relative minimum value of f. In fact, we have shown, that 4 is the absolute minimum value of f. The graph of f shown in Figure 3 confirms, this result., , EXAMPLE 2 Let f(x, y)  3  2x 2  y 2. Show that (0, 0) is the only critical point, , of f and that f(0, 0)  3 is a relative maximum value of f., Solution, , The partial derivatives of f are, fx (x, y)  , , x, 2x  y, 2, , 2, , and, , fy(x, y)  , , y, 2x  y 2, 2

Page 99 :
13.8, z, , 3 (0, 0, 3), z3, , x2  y2, , (0, 0), 3, , 3, , y, , x, , FIGURE 4, The function f has a relative maximum, at (0, 0)., , Extrema of Functions of Two Variables, , 1119, , Since both fx (x, y) and fy (x, y) are undefined at (0, 0), we conclude that (0, 0) is a critical point of f. Also, fx (x, y) and fy(x, y) are not both equal to zero at any point. This, tells us that (0, 0) is the only critical point of f. Finally, since 2x 2  y 2  0 for all, values of x and y, we see that f(x, y) 3 for all points (x, y). We conclude that, f(0, 0)  3 is a relative (indeed, the absolute) maximum of f. The graph of f shown in, Figure 4 confirms this result., As in the case of a function of one variable, a critical point of a function of two, variables is only a candidate for a relative extremum of the function. A critical point, need not give rise to a relative extremum, as the following example shows., , EXAMPLE 3 Show that the point (0, 0) is a critical point of f(x, y)  y 2  x 2 but, that it does not give rise to a relative extremum of f., Solution, , The partial derivatives of f,, fx (x, y)  2x, , and, , fy (x, y)  2y, , are continuous everywhere. Since fx and fy are both equal to zero at (0, 0), we conclude, that (0, 0) is a critical point of f and that it is the only candidate for a relative extremum, of f. But notice that for points on the x-axis we have y  0, so f(x, y)  x 2  0 if, x  0; and for points on the y-axis we have x  0, so f(x, y)  y 2  0 if y  0. Therefore, every open disk containing (0, 0) has points where f takes on positive values as, well as points where f takes on negative values. This shows that f(0, 0)  0 cannot be, a relative extremum of f. The graph of f is shown in Figure 5. The point (0, 0, 0) is, called a saddle point., , z, , FIGURE 5, The point (0, 0) is a critical point, of f(x, y)  y 2  x 2, but it does not, give rise to a relative extremum of f., , y, x, , fx (0, 0)  0, , fy(0, 0)  0, , As we have observed, the critical point (0, 0) in Example 3 does not yield a relative, maximum or minimum. In general, a critical point of a differentiable function of two, variables that does not give rise to a relative extremum is called a saddle point. A saddle point is the analog of an inflection point for the case of a function of one variable., , The Second Derivative Test for Relative Extrema, In Examples 1 and 3 we were able to determine, either by inspection or with the help, of simple algebraic manipulations, whether f did or did not possess a relative extremum, at a critical point. For more complicated functions, the following test may be used., This test is the analog of the Second Derivative Test for a function of one variable. Its, proof will be omitted.

Page 100 :
1120, , Chapter 13 Functions of Several Variables, , THEOREM 2 The Second Derivative Test for a Function of Two Variables, Suppose that f has continuous second-order partial derivatives on an open region, containing a critical point (a, b) of f. Let, D(x, y)  fxx(x, y)fyy (x, y)  f 2xy (x, y), a. If D(a, b), value., b. If D(a, b), c. If D(a, b), d. If D(a, b), ,  0 and fxx(a, b)  0, then f(a, b) is a relative maximum,  0 and fxx(a, b)  0, then f(a, b) is a relative minimum value.,  0, then (a, b, f(a, b)) is a saddle point.,  0, then the test is inconclusive., , EXAMPLE 4 Find the relative extrema of f(x, y)  x 3  y 2  2xy  7x  8y  2., Solution, , First, we find the critical points of f. Since, fx (x, y)  3x 2  2y  7, , and, , fy(x, y)  2y  2x  8, , are both continuous for all values of x and y, the only critical points of f, if any, are, found by solving the system of equations fx (x, y)  0 and fy (x, y)  0, that is, by, solving, 3x 2  2y  7  0, , and, , 2y  2x  8  0, , From the second equation we obtain y  x  4, which upon substitution into the first, equation yields, 3x 2  2x  1  0, , or, , (3x  1)(x  1)  0, , 13 or, 11, 3 and, , Therefore, x , x  1. Substituting each of these values of x into the expression, for y gives y , y  5, respectively. Therefore, the critical points of f are 1 13, 113 2, and (1, 5)., Next, we use the Second Derivative Test to determine the nature of each of these, critical points. We begin by computing fxx(x, y)  6x, fyy(x, y)  2, fxy(x, y)  2, and, D(x, y)  fxx (x, y)fyy(x, y)  f 2xy(x, y),  (6x)(2)  (2)2  4(3x  1), , To test the point 1 13, 113 2 , we compute, , D 1 13, 113 2  4(1  1)  8  0, , from which we deduce that 1 13, 113 2 gives rise to the saddle point 1 13, 113, 373, 27 2 of f., Next, to test the critical point (1, 5), we compute, D(1, 5)  4(3  1)  8  0, which indicates that (1, 5) gives a relative extremum of f. Since, fxx(1, 5)  6(1)  6  0, we see that (1, 5) yields a relative minimum of f. Its value is, f(1, 5)  (1)3  (5)2  2(1)(5)  7(1)  8(5)  2,  15, The graph and contour map of f are shown in Figures 6a and 6b.

Page 101 :
13.8, 2, , 4, , Extrema of Functions of Two Variables, , 1121, , y, , 6, 7, , 6, , 0, Saddle, point, , 15, , 5, , 14.3, , 10, 4, , 1, , Relative, minimum, , 2, , 12.7, , 3, , 0, 1, , 2, 2, , 2, , FIGURE 6, , 13.4, , (a) The graph of f (x, y)  x 3  y2  2xy  7x  8y  2, , 1, , 0, , 1, , 2, , x, , (b) The contour plot of f, , EXAMPLE 5 Priority Mail Regulations Postal regulations specify that the combined, length and girth of a package sent by priority mail may not exceed 108 in. Find the, dimensions of a rectangular package with the greatest possible volume satisfying these, regulations., Solution Let the length, width, and height of the package be x, y, and z inches respectively, as shown in Figure 7. Then the volume of the package is V  xyz. Observe that, the combined length and girth of the package is (x  2y  2z) inches. Clearly, we, should let this quantity be as large as possible, that is, we should let, , z, , x  2y  2z  108, , x, y, , FIGURE 7, The combined length and girth of, the package is x  2y  2z inches., , With the help of this equation we can express V as a function of two variables. For, example, solving the equation for x in terms of y and z, we obtain, x  108  2y  2z, which, upon substitution into the expression for V, gives, V  f(y, z)  (108  2y  2z)yz  108yz  2y 2z  2yz 2, To find the critical points of f, we set, fy  108z  4yz  2z 2  2z(54  2y  z)  0, and, fz  108y  2y 2  4yz  2y(54  y  2z)  0, Since y and z are both nonzero (otherwise, V would be zero), we are led to the system, 54  2y  z  0, 54  y  2z  0, Multiplying the second equation by 2 gives 108  2y  4z  0. Then subtracting this, equation from the first equation gives 54  3z  0, or z  18. Substituting this value, of z into either equation in the system then yields y  18. Therefore, the only critical, point of f is (18, 18) .

Page 102 :
1122, , Chapter 13 Functions of Several Variables, , We could use the Second Derivative Test to show that the point (18, 18) gives, a relative maximum of V, or, as in this situation, we can simply argue from physical, considerations that V must attain an absolute maximum at (18, 18). Finally, from the, equation, x  108  2y  2z, found earlier, we see that when y  z  18,, x  108  2(18)  2(18)  36, Therefore, the dimensions of the required package are 18 in., , 18 in., , 36 in., , Finding the Absolute Extremum Values, of a Continuous Function on a Closed Set, Recall that if f is a continuous function of one variable on a closed interval [a, b], then, the Extreme Value Theorem guarantees that f has an absolute maximum value and an, absolute minimum value. The analog of this theorem for a function of two variables, follows., , THEOREM 3 The Extreme Value Theorem for Functions of Two Variables, If f is continuous on a closed, bounded set D in the plane, then f attains an, absolute maximum value f(a, b) at some point (a, b) in D and an absolute minimum value f(c, d) at some point (c, d) in D., , The following procedure for finding the extreme values of a function of two variables is the analog of the one for finding the extreme values of a function of one variable discussed in Section 3.1., Finding the Absolute Extremum Values of f on a Closed, Bounded Set D, 1. Find the values of f at the critical points of f in D., 2. Find the extreme values of f on the boundary of D., 3. The absolute maximum value of f and the absolute minimum value of f are, precisely the largest and the smallest numbers found in Steps 1 and 2., , The justification for this procedure is similar to that for a function of one variable, on a closed interval [a, b]: If an absolute extremum of f occurs in the interior of D,, then it must also be a relative extremum of f, and hence it must occur at a critical point, of f. Otherwise, the absolute extremum of f must occur at a boundary point of D., , EXAMPLE 6 Find the absolute maximum and the absolute minimum values of the, function f(x, y)  2x 2  y 2  4x  2y  3 on the rectangle, D  {(x, y) 冟 0, , x, , 3, 0, , y, , 2}, , Solution Since f is a polynomial, it is continuous on the closed, bounded set D. Therefore, Theorem 3 guarantees the existence of an absolute maximum and an absolute, minimum value of f on D.

Page 103 :
13.8, Step 1, , y, , L3, , 2, L4, , Step 2, L2, , 1, , 0, , 1, , L1, , 2, , 1123, , To find the critical point of f in D, we set fx  4x  4  0 and, fy  2y  2  0. Solving this system of equations gives (1, 1) as, the only critical point of f. The value of f at this point is f(1, 1)  0., Next, we look for extreme values of f on the boundary of D. We can think, of this boundary as being made up of four line segments L 1, L 2, L 3, and L 4,, as shown in Figure 8., On L 1: Here y  0, so we have, f(x, 0)  2x 2  4x  3, , x, , 3, , Extrema of Functions of Two Variables, , 0, , x, , 3, , To find the extreme values of the continuous function f(x, 0) of one variable on, the closed bounded interval [0, 3], we use the method of Section 3.1. Setting, , FIGURE 8, The boundary of D consists of the four, line segments L 1, L 2, L 3, and L 4., , f ¿(x, 0)  4x  4  0, gives x  1 as the only critical number of f(x, 0) in (0, 3). Evaluating f(x, 0), at x  1, as well as at the endpoints of the interval [0, 3], gives f(0, 0)  3,, f(1, 0)  1, and f(3, 0)  9. Thus, f has the absolute minimum value of 1, and the absolute maximum value of 9 on L 1., On L 2: Here x  3, so we have, f(3, y)  y 2  2y  9, , 0, , y, , 2, , Setting f ¿(3, y)  2y  2  0 yields y  1 as the only critical number of, f(3, y) in (0, 2). Evaluating f(3, y) at the endpoints of [0, 2] and at the critical number y  1 gives f(3, 0)  9, f(3, 1)  8, and f(3, 2)  9. We see, that f has the absolute minimum value of 8 and the absolute maximum value, of 9 on L 2., On L 3: Here y  2, so we have, f(x, 2)  2x 2  4x  3, , 0, , x, , 3, , Setting f ¿(x, 2)  4x  4  0 gives x  1 as the only critical number of, f(x, 2) in (0, 3). Since f(0, 2)  3, f(1, 2)  1, and f(3, 2)  9, we see that, f(x, 2) has the absolute minimum value of 1 and the absolute maximum, value of 9 on L 3., On L 4: Here x  0, so we have, f(0, y)  y 2  2y  3, , Step 3, , 0, , y, , 2, , Setting f ¿(0, y)  2y  2  0 gives y  1 as the only critical number of, f(0, y) in (0, 2). Since f(0, 0)  3, f(0, 1)  2, and f(0, 2)  3, we see that, f(0, y) has the absolute minimum value of 2 and the absolute maximum, value of 3 on L 4., Table 1 summarizes the results of our computations. Comparing the value of f, obtained at the various points, we conclude that the absolute minimum value of, f on D is 0 attained at the critical point (1, 1) of f and that the absolute maximum value of f on D is 9 attained at the boundary points (3, 0) and (3, 2)., , TABLE 1, Critical, point, , Boundary, point on L 1, , Boundary, point on L 2, , Boundary, point on L 3, , Boundary, point on L 4, , (x, y), , (1, 1), , (1, 0), , (3, 0), , (3, 1), , (3, 0), , (3, 2), , (1, 2), , (3, 2), , (0, 1), , (0, 0), , (0, 2), , Extreme, value: f(x, y), , 0, , 1, , 9, , 8, , 9, , 9, , 1, , 9, , 2, , 3, , 3

Page 104 :
1124, , Chapter 13 Functions of Several Variables, , 13.8, , CONCEPT QUESTIONS, , 1. a. What does it mean when one says that f has a relative, maximum (relative minimum) at a point (a, b) in the, domain of f ? What is f(a, b) called in each case?, b. What does it mean when one says that f has an absolute, maximum (absolute minimum) at (a, b) ? What is f(a, b), called in each case?, 2. a. What is a critical point of a function f(x, y) ?, b. What role does a critical point of a function play in the, determination of the relative extrema of the function?, c. State the Second Derivative Test for a function of two, variables., , 13.8, , 3. a. What can you say about the existence of a maximum, value and a minimum value of a continuous function of, two variables defined on a closed, bounded set on the, plane?, b. Describe a strategy for finding the absolute extreme values of a continuous function on a closed, bounded set in, the plane., , EXERCISES, , In Exercises 1–22, find and classify the relative extrema and, saddle points of the function., 1. f(x, y)  x 2  y 2  2x  4y, , In Exercises 23 and 24, (a) use the graph and the contour map, of f to estimate the relative extrema and saddle point(s) of f, and, (b) verify your guess analytically., 23. f(x, y)  x 3  3xy 2  y 4, , 2. f(x, y)  2x 2  y 2  6x  2y  1, 3. f(x, y)  x 2  3y 2  4x  6y  8, 4. f(x, y)  2x 2  3y 2  6x  4y  6, 5. f(x, y)  x  3xy  3y, 2, , z, , y, , 2.0, , 2, , 6. f(x, y)  x 2  3xy  2y 2  1, 7. f(x, y)  2x 2  y 2  2xy  8x  2y  2, , 0, , x, , 8. f(x, y)  x 2  3y 2  6xy  2x  4y, 9. f(x, y)  x 2  2y 2  x 2y  3, , y, , 10. f(x, y)  x  y  2xy  1, 2, , 2, , 2, , x, , 11. f(x, y)  x 2  5y 2  x 2y  2y 3, , z, , 13. f(x, y)  x 2  6x  x1y  y, 14. f(x, y)  xy(3  x  y), x y  2y  4x, xy, , x, y, , 4y, , x  0,, x  0,, , 0, 0, , x  0,, , 22. f(x, y)  sin x  sin y,, , 0, , y, , 1.5, , 2p, , y, , 20. f(x, y)  xex sin y, x  0, 0, 21. f(x, y)  ex cos y,, , x, , 1.0, , 2y2, , 17. f(x, y)  ex, , 19. f(x, y)  x sin y,, , 0.5, 0, , 2, , 18. f(x, y)  ex sin y,, , 1.0, , 0.5, , x y 1, 2, , y, , 1.5, , 2 2, , 16. f(x, y)  , , 1.0 2.0, , 24. f(x, y)  3xy 2  x 3, , 12. f(x, y)  x 3  3xy  y 3  3, , 15. f(x, y) , , 2.0, 2.0 1.0 0, , 1.0, , 2p, y, , 2p, , 0, , y, , 2p, , x, , 2p, 0, , y, , 2p, , V Videos for selected exercises are available online at www.academic.cengage.com/login., , 0, , 1.0

Page 105 :
13.8, cas In Exercises 25–28, plot the graph and the contour map of f, and, , use them to estimate the relative maximum and minimum values, and the saddle point(s) of f. Then find these values and saddle, point(s) analytically., 1, 25. f(x, y)  x 4  2x 3  4xy  y 2, 2, 2, 4,  8, x, y, , 28. f(x, y)  6xy  2x  3y, 2, , 3, , 4, , to find the critical points of f correct to three decimal places., Then use these results to find the relative extrema of f. Plot, the graph of f., , 49. Find the dimensions of the rectangular box of maximum, volume with faces parallel to the coordinate planes that can, be inscribed in the ellipsoid, , 29. f(x, y)  2x 4  8x 2  y 2  4x  2y  5, 30. f(x, y)  2  2x 2  5xy  2y  y 4, , y2, x2, z2, , , 1, 4, 9, 16, , 31. f(x, y)  x 4  y 4  2x 2y  x 2  y  2, 32. f(x, y)  x 4  2x 2  x  y 2  ey, In Exercises 33–40, find the absolute extrema of the function on, the set D., , 34. f(x, y)  x  xy  y ;, D  {(x, y) 冟 2 x 2, 1, , 45. Find three positive real numbers whose sum is 500 and, whose product is as large as possible., , 48. An open rectangular box having a volume of 108 in3. is to, be constructed from cardboard. Find the dimensions of such, a box if the amount of cardboard used in its construction is, to be minimized., , cas In Exercises 29–32, use a graphing calculator or computer, , 2, , 44. Find the points on the surface xy 2z  4 that are closest to, the origin. What is the shortest distance from the origin to, the surface?, , 47. Find the dimensions of a closed rectangular box of maximum volume that can be constructed from 48 ft2 of cardboard., , (“Monkey Saddle”), , 33. f(x, y)  2x  3y  6;, D  {(x, y) 冟 0 x 2, 2, , 1125, , 46. Find the dimensions of an open rectangular box of maximum volume that can be constructed from 48 ft2 of cardboard., , 26. f(x, y)  (x 2  y 2)ey, 27. f(x, y)  xy , , Extrema of Functions of Two Variables, , 50. Solve the problem posed in Exercise 49 for the general case, of an ellipsoid with equation, x2, a2, , y, , 3}, , , , y2, b2, , , , z2, c2, , 1, , where a, b, and c are positive real numbers., , 2, , 35. f(x, y)  3x  4y  12; D is the closed triangular region, with vertices (0, 0), (3, 0), and (3, 4)., , 51. Find the dimensions of the rectangular box of maximum, volume lying in the first octant with three of its faces lying, in the coordinate planes and one vertex lying in the plane, 2x  3y  z  6. What is the volume of such a box?, , 36. f(x, y)  3x 2  2xy  y 2; D is the closed triangular region, with vertices (2, 1), (1, 1), and (1, 2)., , 52. Solve the problem posed in Exercise 51 for the general case, of a plane with equation, , 37. f(x, y)  xy  x 2; D is the region bounded by the parabola, y  x 2 and the line y  4., , y, x, z,   1, a, c, b, , y, , 1}, , 38. f(x, y)  4x 2  y 2; D is the region bounded by the, parabola y  4  x 2 and the x-axis., 39. f(x, y)  x 2  4y 2  3x  1;, , D  {(x, y) 冟 x 2  y 2, , 40. f(x, y)  4x 2  y 2  2x  y;, , D  {(x, y) 冟 4x 2  y 2, , where a, b, and c are positive real numbers., 4}, 1}, , 41. Find the shortest distance from the origin to the plane, x  2y  z  4., Hint: The square of the distance from the origin to any point (x, y, z), on the plane is d 2  x 2  y 2  z 2  x 2  y 2  (4  x  2y)2., Minimize d 2  f(x, y)  x 2  y 2  (4  x  2y)2., , 42. Find the point on the plane x  2y  z  5 that is closest to, the point (2, 3, 1)., Hint: Study the hint of Exercise 41., , 43. Find the points on the surface z 2  xy  x  4y  21 that, are closest to the origin. What is the shortest distance from, the origin to the surface?, , 53. An open rectangular box is to have a volume of 12 ft3. If, the material for its base costs three times as much (per, square foot) as the material for its sides, what are the, dimensions of the box that can be constructed at a, minimum cost?, 54. A closed rectangular box is to have a volume of 16 ft3. If, the material for its base costs twice as much (per square, foot) as the material for its top and sides, find the dimensions of the box that can be constructed at a minimum, cost., 55. Locating a Radio Station The following figure shows the locations of three neighboring communities. The operators of a, newly proposed radio station have decided that the site, P(x, y) for the station should be chosen so that the sum of

Page 106 :
1126, , Chapter 13 Functions of Several Variables, , the squares of the distances from the site to each community, is minimized. Find the location of the proposed radio station., , One criterion for determining the straight line, that “best” fits the data calls for minimizing the, n, , y, , sum of the squares of the deviations a d 2k, where, , C(4, 24), , k1, , dk  yk  (mx k  b)  yk  mx k  b. This sum is, a function of m and b; that is,, , 20, , n, , P(x, y), , t(m, b)  a (yk  mx k  b)2, k1, , 10, , B(20, 8), , and it is minimized with respect to the variables m and b by, solving the system comprising the equations tm (a, b)  0, and tb(a, b)  0 for m and b. Show that this leads to the, system, , A(2, 4), 0, , 10, , x, , 20, , n, , 56. Parcel Post Regulations Postal regulations specify that a parcel, sent by parcel post may have a combined length and girth of, no more than 130 inches. Find the dimensions of a cylindrical package of greatest volume that can be sent through the, mail. What is the volume of such a package?, Hint: The length plus the girth is 2pr  l., , r, , n, , a a x k bm  nb  a yk, k1, , k1, , n, , n, , n, , k1, , k1, , k1, , a a x 2k bm  a a x k bb  a x kyk, This method of determining the equation y  mx  b is, called the method of least squares, and the line with equation y  mx  b is called a least squares or regression, line., 58. Use the method of least squares (Exercise 57) to find the, straight line y  mx  b that best fits the data points (1, 3),, (2, 5), (3, 5), (4, 7), and (5, 8). Plot the scatter diagram, and, sketch the graph of the regression line on the same set of, axes., , l, , 57. Suppose a relationship exists between two quantities x and y, and that we have obtained the following data relating y to x:, x, , x1, , x2, , p, , xn, , y, , y1, , y2, , p, , yn, , 59. Information Security Software Sales Refer to Exercise 57. As, online attacks persist, spending on information security software continues to rise. The following table gives the forecast, for the worldwide sales (in billions of dollars) of information security software through 2007 (x  0 corresponds to, 2002)., Year, x, , The figure below shows the points (x 1, y1), (x 2, y2), p ,, (x n, yn) plotted in the xy-plane. (This figure is called a, scatter diagram.) If the data points are scattered about a, straight line, as in this illustration, then it is reasonable to, describe the relationship between x and y in terms of a, linear equation y  mx  b., , Spending, y, , 0, , 1, , 2, , 3, , 4, , 5, , 6.8, , 8.3, , 9.8, , 11.3, , 12.8, , 14.9, , a. Find an equation of the least-squares line for these data., b. Use the result of part (a) to forecast the spending on, information security software in 2010, assuming that this, trend continues., Source: International Data Corp., , y, (xn, yn), , y  mx  b, , (xk, yk), dk, (x2, y2), , (xk, mxk  b), , (x1, y1), , 60. Male Life Expectancy At 65 Refer to Exercise 57. The projections of male life expectancy at age 65 in the United States, are summarized in the following table (x  0 corresponds to, 2000)., Year, x, , x, , Years beyond 65, y, , 0, , 10, , 20, , 30, , 40, , 50, , 15.9 16.8 17.6 18.5 19.3 20.3

Page 107 :
13.9, , Lagrange Multipliers, , 1127, , a. Find an equation of the least-squares line for these data., b. Use the result of (a) to estimate the life expectancy at 65, of a male in 2040. How does this result compare with the, given data for that year?, c. Use the result of (a) to estimate the life expectancy at 65, of a male in 2030., , 63. Let f(x, y)  3x 2  6x  4y 2  4y  3., a. Show that f has no minimum value., cas b. Find the maximum and minimum values of f in the, region D  {(x, y) 冟 x 2  y 2 1}., , Source: U.S. Census Bureau., , 64. Let f(x, y)  Ax 2  2Bxy  Cy 2, where B 2  4AC  0., Find conditions in terms of A, B, and C such that f has a, relative minimum at (0, 0); a relative maximum at (0, 0);, and a saddle point at (0, 0)., , 61. Operations Management Consulting Spending Refer to Exercise 57. The following table gives the projected operations, management consulting spending (in billions of dollars), from 2005 through 2010. Here, x  5 corresponds to 2005., Year, x, , 5, , 6, , 7, , 8, , 9, , 10, , Spending, y, , 40, , 43.2, , 47.4, , 50.5, , 53.7, , 56.8, , a. Find an equation of the least-squares line for these data., b. Use the results of part (a) to estimate the average rate of, change of operations management consulting spending, from 2005 through 2010., c. Use the results of part (a) to estimate the amount of, spending on operations management consulting in 2011,, assuming that the trend continues., Source: Kennedy Information., , 62. Let f(x, y)  x 2  y 2  2xy  2., a. Show that f has no maximum or minimum values., cas b. Find the maximum and minimum values of f in the, region D  {(x, y) 冟 x 2  4y 2 4}., , Hint: On the boundary of D, let x  cos t, y  sin t for, 0, , t, , 2p., , In Exercises 65–68, determine whether the statement is true or, false. If it is true, explain why it is true. If it is false, give an, example to show why it is false., 65. If f(x, y) has a relative maximum at (a, b), then fx (a, b)  0, and fy(a, b)  0., 66. Let h(x, y)  f(x)  t(y), where f and t have second-order, derivatives. If the graph of f is concave upward on (⬁, ⬁), and the graph of t is concave downward on (⬁, ⬁), then h, cannot have a relative maximum or a relative minimum at, any point., 67. If §f(a, b)  0 , then f has a relative extremum at (a, b)., 68. If f(x, y) has continuous second-order partial derivatives and, fxx(x, y)  fyy(x, y)  0 and fxy (x, y)  0 for all (x, y), then f, cannot have a relative extremum., , Hint: On the boundary of D, let x  2 cos t, y  sin t for, 0, , t, , 2p., , 13.9 Lagrange Multipliers, Constrained Maxima and Minima, Many practical optimization problems involve maximizing or minimizing an objective, function subject to one or more constraints, or side conditions. In Example 5 of Section 13.8 we discussed the problem of maximizing the (volume) function, , y, , a, , V  f(x, y, z)  xyz, , y, , subject to the constraint, x, , FIGURE 1, We want to find the core of largest, surface area that can be inserted into, a coil of radius a., , t(x, y, z)  x  2y  2z  108, In this case the constraint expresses the condition that the combined length plus girth, of a package is 108 in. (the maximum allowed by postal regulations)., As another example, consider a problem encountered in the construction of an AC, transformer. Here, we are required to find the cross-shaped iron core of largest surface, area that can be inserted into a coil of radius a (Figure 1). In terms of x and y we see, that the surface area of the iron core is, S  4xy  4y(x  y)  8xy  4y 2

Page 108 :
1128, , Chapter 13 Functions of Several Variables, , Next, observe that x and y must satisfy the equation x 2  y 2  a 2. Therefore, the problem is equivalent to one of maximizing the objective function, f(x, y)  8xy  4y 2, subject to the constraint, t(x, y)  x 2  y 2  a 2, We will complete the solution of this problem in Example 2., Figure 2a shows the graph of a function f defined by the equation z  f(x, y)., Observe that f has an absolute minimum at (0, 0) and an absolute minimum value of, 0. However, if the independent variables x and y are subjected to a constraint of the, form t(x, y)  k, then the points (x, y, z) that satisfy both z  f(x, y) and t(x, y)  k, lie on the curve C, the intersection of the surface z  f(x, y) and the cylinder t(x, y)  k, (Figure 2b). From the figure you can see that the absolute minimum of f subject to the, constraint t(x, y)  k occurs at the point (a, b). Furthermore, f has the constrained, absolute minimum value f(a, b) rather than the unconstrained absolute minimum value, of 0 at (0, 0)., z, , z, z  f (x, y), , z  f (x, y), , C, , (a, b, f(a, b)), , FIGURE 2, The function f has an unconstrained, minimum value of 0, but it has a, constrained minimum value of f(a, b), when subjected to the constraint, t(x, y)  k., , (0, 0), g(x, y)  k, , y, x, , (a, b), , y, , x, , (a) f is not subject to any constraints., , (b) f is subject to a constraint., , The problem that we discussed at the beginning of this section (maximizing the, volume of a box subject to a given constraint) was first solved in Section 13.8. Recall, the method of solution that we used:, First, we solved the constraint equation, t(x, y, z)  x  2y  2z  108, for x in terms of y and z. We then substituted this expression for x into the equation, V  f(x, y, z)  xyz, thereby obtaining an expression for V involving the variables y and z and satisfying the, constraint equation. Next, we found the maximum of V by treating V as an unconstrained function of y and z., The major drawback of this method is that it relies on our ability to solve the constraint equation t(x, y)  k for one variable explicitly in terms of the other (or, t(x, y, z)  k for one variable explicitly in terms of the other two variables in the case, of a constraint involving three variables). This might not always be possible or convenient. Moreover, even when we are able to solve the constraint equation t(x, y)  k, for y explicitly in terms of x, the resulting function of one variable that is obtained by, substituting this expression for y into the objective function f(x, y) might turn out to, be unnecessarily complicated.

Page 109 :
13.9, , Lagrange Multipliers, , 1129, , The Method of Lagrange Multipliers, We will now consider a method, called the method of Lagrange multipliers (named, after the French mathematician Joseph Lagrange, 1736–1813), which obviates the need, to solve the constraint equation for one variable in terms of the other variables. To see, how this method works, let’s reexamine the problem of finding the absolute minimum, of the objective function f subject to the constraint t(x, y)  k that we considered earlier. Figure 3a shows the level curves of f drawn in the xyz-coordinate system. These, level curves are reproduced in the xy-plane in Figure 3b., , z, z  f (x, y), , z  c1, z  c2, , g(x, y)  k, (a, b), , z  f (a, b), , C, zc, z  c2 1, z  f (a, b), zc, z  c4 3, , y, , (a, b, f (a, b)), , x, z  c3, , g(x, y)  k, , (a, b), , y, , z  c4, , x, , FIGURE 3, , (a) The level curves of f in the xyz-plane, , (b) The level curves of f in the xy-plane, , Observe that the level curves of f with equations f(x, y)  c, where c  f(a, b),, have no points in common with the graph of the constraint equation t(x, y)  k (for, example, the level curves f(x, y)  c1 and f(x, y)  c2 shown in Figure 3). Thus, points, lying on these curves are not candidates for the constrained minimum of f., On the other hand, the level curves of f with equation f(x, y)  c, where, c  f(a, b), do intersect the graph of the constraint equation t(x, y)  k (such as the, level curves of f(x, y)  c3 and f(x, y)  c4). These points of intersection are candidates for the constrained minimum of f., Finally, observe that the larger c is for c  f(a, b), the larger the value f(x, y) is for, (x, y) lying on the level curve t(x, y)  k. This observation suggests that we can find, the constrained minimum of f by choosing the smallest value of c so that the level curve, f(x, y)  c still intersects the curve t(x, y)  k. At such a point (a, b) the level curve, of f just touches the graph of the constraint equation t(x, y)  k. That is, the two curves, have a common tangent at (a, b) (see Figure 3b). Equivalently, their normal lines at, this point coincide. Putting it yet another way, the gradient vectors §f(a, b) and §t(a, b), have the same direction, so §f(a, b)  l§t(a, b) for some scalar l (lambda)., A similar result holds for the problem of maximizing or minimizing a function f of, three variables defined by w  f(x, y, z) and subject to the constraint t(x, y, z)  k., In this situation, f has a constrained maximum or constrained minimum at a point, (a, b, c) where the level surface f(x, y, z)  f(a, b, c) is tangent to the level surface, t(x, y, z)  k. But this means that the normals of these surfaces, and therefore their, gradient vectors, at the point (a, b, c) must be parallel to each other. Thus, there is a, scalar l such that §f(a, b, c)  l§t(a, b, c)., These geometric arguments suggest the following theorem.

Page 110 :
1130, , Chapter 13 Functions of Several Variables, , THEOREM 1 Lagrange’s Theorem, Let f and t have continuous first partial derivatives in some region D in the plane., If f has an extremum at a point (a, b) on the smooth constraint curve t(x, y)  c, lying in D and §t(a, b)  0, then there is a real number l such that, §f(a, b)  l§t(a, b), , The number l in Theorem 1 is called a Lagrange multiplier., , PROOF Suppose that the smooth curve C described by t(x, y)  c is represented by, the vector function, r(t)  x(t)i  y(t)j,, , rⴕ(t)  0, , where x¿ and y¿ are continuous on an open interval I (Figure 4). Then the values assumed, by f on C are given by, h(t)  f(x(t), y(t)), for t in I. Suppose that f has an extreme value at (a, b). If t 0 is the point in I corresponding to the point (a, b) , then h has an extreme value at t 0. Therefore, h¿(t 0)  0. Using, the Chain Rule, we have, h¿(t 0)  fx (x(t 0), y(t 0))x¿(t 0)  fy(x(t 0), y(t 0))y¿(t 0),  fx (a, b)x¿(t 0)  fy (a, b)y¿(t 0),  §f(a, b) ⴢ rⴕ(t 0)  0, This shows that §f(a, b) is orthogonal to rⴕ(t 0) . But as we demonstrated in Section 13.7, §t(a, b) is orthogonal to rⴕ(t 0). Therefore, the gradient vectors §f(a, b) and, §t(a, b) are parallel, so there is a number l such that §f(a, b)  l §t(a, b)., y, r(t0), , ◊ g(a, b), , (a, b), , r(t), t0, (a) The parameter interval I, , FIGURE 4, , t, , C, , g(x, y)  c, , 0, , x, , (b) The smooth curve C is represented by the, vector function r(t)., , The proof of Lagrange’s Theorem for functions of three variables is similar to that, for functions of two variables. In the case involving three variables, level surfaces rather, than level curves are involved. Lagrange’s Theorem leads to the following procedure, for finding the constrained extremum values of functions. We state it for the case of, functions of three variables.

Page 111 :
13.9, , Lagrange Multipliers, , 1131, , The Method of Lagrange Multipliers, Suppose f and t have continuous first partial derivatives. To find the maximum, and minimum values of f subject to the constraint t(x, y, z)  k (assuming that, these extreme values exist and that §t  0 on t(x, y, z)  k):, 1. Solve the equations, §f(x, y, z)  l§t(x, y, z), , t(x, y, z)  k, , and, , for x, y, z, and l., 2. Evaluate f at each solution point found in Step 1. The largest value yields, the constrained maximum of f, and the smallest value yields the constrained minimum of f., , Note, , Since, §f(x, y, z)  fx (x, y, z)i  fy (x, y, z)j  fz (x, y, z)k, , and, §t(x, y, z)  tx (x, y, z)i  ty (x, y, z)j  tz (x, y, z)k, we see, by equating like components, that the vector equation, §f(x, y, z)  l §t(x, y, z), is equivalent to the three scalar equations, fx (x, y, z)  ltx(x, y, z) ,, , fy (x, y, z)  lty(x, y, z) ,, , and, , fz(x, y, z)  ltz(x, y, z), , These scalar equations together with the constraint equation t(x, y, z)  k give a system of four equations to be solved for the four unknowns x, y, z, and l., , EXAMPLE 1 Find the maximum and minimum values of the function f(x, y) , x 2  2y subject to x 2  y 2  9., Solution, , The constraint equation is t(x, y)  x 2  y 2  9. Since, §f(x, y)  2x i  2j, , and, , §t(x, y)  2x i  2yj, , the equation §f(x, y)  l§t(x, y) becomes, 2xi  2j  l(2x i  2yj)  2lx i  2ly j, Equating like components and rewriting the constraint equation lead to the following system of three equations in the three variables x, y, and l:, 2x  2lx, , (1a), , 2  2ly, , (1b), , x y 9, 2, , 2, , From Equation (1a) we have, 2x(1  l)  0, , (1c)

Page 112 :
1132, , Chapter 13 Functions of Several Variables, , so x  0, or l  1. If x  0, then Equation (1c) gives y  3. If l  1, then, Equation (1b) gives y  1, which upon substitution into Equation (1c) yields, x  212. Therefore, f has possible extreme values at the points (0, 3), (0, 3),, (212, 1), and (2 12, 1). Evaluating f at each of these points gives, f(0, 3)  6,, , f(0, 3)  6,, , f(2 12, 1)  10,, , and, , f(2 12, 1)  10, , We conclude that the maximum value of f on the circle x  y  9 is 10, attained at, the points (212, 1) and (2 12, 1), and that the minimum value of f on the circle is 6, attained at the point (0, 3) ., Figure 5 shows the graph of the constraint equation x 2  y 2  9 and some level, curves of the objective function f. Observe that the extreme values of f are attained at, the points where the level curves of f are tangent to the graph of the constraint equation., 2, , 2, , f (x, y)  x2  2y  0, , f (x, y)  x2  2y  6 y, (0, 3), , f (x, y)  x2  2y  10, , FIGURE 5, The extreme values of f occur at, the points where the level curves, of f are tangent to the graph of, the constraint equation (the circle)., , 0, , (2 √ 2, 1), , (2 √ 2, 1), , x, , x2  y2  9, , EXAMPLE 2 Complete the solution to the problem posed at the beginning of this, section: Find the cross-shaped iron core of largest surface area that can be inserted into, a coil of radius a (Figure 6)., , y, , a, , y, x, , Solution Recall that the problem reduces to one of finding the largest value of the objective function f(x, y)  8xy  4y 2 subject to the constraint t(x, y)  x 2  y 2  a 2., Since, §f(x, y)  8y i  (8x  8y)j, , §t(x, y)  2xi  2y j, , and, , the equation §f(x, y)  l§t(x, y) becomes, 8y i  (8x  8y)j  l(2x i  2yj)  2lxi  2ly j, FIGURE 6, A cross-shaped iron core of largest, surface area is to be inserted into the, coil., , Equating like components and rewriting the constraint equation, we get the following, system of three equations in the three variables x, y, and l:, 8y  2lx, , (2a), , 8x  8y  2ly, , (2b), , x y a, , (2c), , 2, , 2, , 2, , Solving Equation (2a) for y, we obtain y  lx. Substituting this expression for y into, Equation (2b) gives, 1, 4, , 8x  2lx , , 1 2, lx, 2

Page 113 :
13.9, , Lagrange Multipliers, , 1133, , or, x(l2  4l  16)  0, Observe that x  0; otherwise, Equation (2a) implies that y  0, so Equation (2c), becomes 0  a 2, which is impossible. So we have l2  4l  16  0. Using the quadratic formula, we obtain, l, , 4, , 116  64,  2, 2, , 215, , Observe that l must be positive; otherwise, Equation (2a) implies that x or y must be, negative. So we choose l  2  215 ⬇ 2.4721. Next, substituting y  14 lx into, Equation (2c) gives, x2 , , 1 2 2, l x  a2, 16, , x 2 a1 , x2a, , l2, b  a2, 16, , l2  16, b  a2, 16, , or, x, , y, , 4a, 2l  16, 2, , ⬇, , 4, 2(2.4721) 2  16, , a, , Recall that l ⬇ 2.4721., , ⬇ 0.8507a, (0.8507a, 0.5258a), a, , a, , x, , FIGURE 7, The maximum value of f occurs at the, point where the level curve of f is, tangent to the level curve of the, constraint equation., , Finally,, y, , 1, 1, lx ⬇ (2.4721)(0.8507a) ⬇ 0.5258a, 4, 4, , Therefore, the core will have the largest surface area if x ⬇ 0.8507a and y ⬇ 0.5258a,, where a is the radius of the coil., Figure 7 shows the graph of the constraint equation x 2  y 2  a 2 (the circle of, radius a centered at the origin) and several level curves of the objective function f., Once again, observe that the maximum value of f, f(0.8507a, 0.5258a) ⬇ 2.4725a 2,, occurs at the point (0.8507a, 0.5258a) , where the level curve of f is tangent to the graph, of the constraint equation., , EXAMPLE 3 Find the dimensions of a rectangular package having the greatest possible volume and satisfying the postal regulation that specifies that the combined length, and girth of the package may not exceed 108 inches. (See Example 5 in Section 13.8.), Solution Recall that to solve this problem, we need to find the largest value of the volume function f(x, y, z)  xyz subject to the constraint t(x, y, z)  x  2y  2z  108., To solve this problem using the method of Lagrange multipliers, observe that, §f(x, y, z)  yz i  xz j  xy k, , and, , §t(x, y, z)  i  2j  2k, , so the equation §f(x, y, z)  l§t(x, y, z) becomes, yz i  xz j  xy k  l(i  2j  2k)

Page 114 :
1134, , Chapter 13 Functions of Several Variables, , Equating components and rewriting the constraint equation give the following system, of four equations in the four variables x, y, z, and l:, yz  l, , (3a), , xz  2l, , (3b), , xy  2l, , (3c), , x  2y  2z  108, , (3d), , Substituting Equation (3a) into Equation (3b) yields, xz  2yz, , z(x  2y)  0, , or, , Since z  0, we have x  2y. Next, substituting Equation (3a) into Equation (3c) gives, xy  2yz, , y(x  2z)  0, , or, , Since y  0, we have x  2z. Equating the two expressions for x just obtained gives, 2y  2z, , or, , yz, , Finally, substituting the expressions for x and y into Equation (3d) gives, 2z  2z  2z  108, , or, , z  18, , So y  18 and x  2(18)  36. Therefore, the dimensions of the package are, 18 in. 18 in. 36 in., as was obtained before., , Interpreting Our Results, Geometrically, this problem is one of finding the point on the plane x  2y  2z  108, at which f(x, y, z)  xyz has the largest value. The point (36, 18, 18) is precisely the, point at which the level surface xyz  f(36, 18, 18)  11,664 is tangent to the plane, x  2y  2z  108., , EXAMPLE 4 Find the dimensions of the open rectangular box of maximum volume, that can be constructed from a rectangular piece of cardboard having an area of 48 ft2., What is the volume of the box?, Solution Let the length, width, and height of the box (in feet) be x, y, and z, as shown, in Figure 8. Then the volume of the box is V  xyz. The area of the bottom of the box, plus the area of the four sides is, z, , xy  2xz  2yz, x, , square feet, and this is equal to the area of the cardboard; that is,, , y, , FIGURE 8, An open rectangular box of maximum, volume is to be constructed from a, piece of cardboard. What are the, dimensions of the box?, , xy  2xz  2yz  48, Thus, the problem is one of maximizing the objective function, f(x, y, z)  xyz, subject to the constraint, t(x, y, z)  xy  2xz  2yz  48, Since, §f(x, y, z)  yz i  xz j  xy k

Page 116 :
1136, , Chapter 13 Functions of Several Variables, , EXAMPLE 5 Find the absolute extreme values of f(x, y)  2x 2  y 2  2y  1 sub-, , ject to the constraint x 2  y 2, , 4., , Solution The inequality x 2  y 2 4 defines the disk D, which is a closed, bounded, set with boundary given by the circle x 2  y 2  4. So, following the procedure given, in Section 13.8, we first find the critical number(s) of f inside D. Setting, fx (x, y)  4x  0, fy (x, y)  2y  2  2(y  1)  0, simultaneously gives (0, 1) as the only critical point of f in D., Next, we find the critical numbers of f on the boundary of D using the method of, Lagrange multipliers. Writing t(x, y)  x 2  y 2  4, we have, §f(x, y)  4xi  2(y  1)j, , §t(x, y)  2xi  2yj, , and, , The equation §f(x, y)  lt(x, y) and the constraint equation give the system, 4x  2lx, , (8a), , 2(y  1)  2ly, , (8b), , x y 4, , (8c), , 2, , 2, , Equation (8a) gives, 2x(l  2)  0, that is, x  0 or l  2. If x  0, then Equation (8c) gives y , then Equation (8b) gives, 2(y  1)  4y, , 2. Next, if l  2,, , y  1, , or, , in which case x  13. So f has the critical points (0, 2), (0, 2), ( 13, 1) and, ( 13, 1) on the boundary of D., Finally, we construct the following table., (x, y), , f(x, y) ⴝ 2x 2 ⴙ y 2 ⴚ 2y ⴙ 1, , (0, 1), ( 13, 1), ( 13, 1), (0, 2), (0, 2), , 0, 10, 10, 9, 1, , From the table we see that f has an absolute minimum value of 0 attained at (0, 1) and, an absolute maximum value of 10 attained at (13, 1) and ( 13, 1)., , Optimizing a Function Subject to Two Constraints, Some applications involve maximizing or minimizing an objective function f subject, to two or more constraints. Consider, for example, the problem of finding the extreme, values of f(x, y, z) subject to the two constraints, t(x, y, z)  k, , and, , h(x, y, z)  l, , It can be shown that if f has an extremum at (a, b, c) subject to these constraints, then, there are real numbers (Lagrange multipliers) l and m such that, §f(a, b, c)  l§t(a, b, c)  m §h(a, b, c), , (9)

Page 117 :
13.9, h(x, y, z)  l, , z, , f, , ¬ g, P, μ h, , Lagrange Multipliers, , 1137, , Geometrically, we are looking for the extreme values of f(x, y, z) on the curve of, intersection of the level surfaces t(x, y, z)  k and h(x, y, z)  l. Condition (9) is a, statement that at an extremum point (a, b, c), the gradient of f must lie in the plane, determined by the gradient of t and the gradient of h. (See Figure 9.) The vector equation (9) is equivalent to three scalar equations. When combined with the two constraint, equations, this leads to a system of five equations that can be solved for the five, unknowns x, y, z, l, and m., , C, 0, , x, , g(x, y, z)  k, , EXAMPLE 6 Find the maximum and minimum values of the function f(x, y, z) , 3x  2y  4z subject to the constraints x  y  2z  1 and x 2  y 2  4., , y, , FIGURE 9, If f has an extreme value at P(a, b, c),, then §f(a, b, c)  l §t(a, b, c) , m §h(a, b, c)., , Solution, , Write the constraint equations in the form, t(x, y, z)  x  y  2z  1, , h(x, y, z)  x 2  y 2  4, , and, , Then the equation §f(x, y, z)  l §t(x, y, z)  m §h(x, y, z) becomes, 3i  2j  4k  l(i  j  2k)  m(2x i  2yj),  (l  2mx)i  (l  2my)j  2lk, Equating like components and rewriting the constraint equations lead to the following, system of five equations in the five variables, x, y, z, l, and m:, 3, , l  2mx, , (10a), , 2  l  2my, , (10b), , 4  2l, , (10c), , x  y  2z  1, , (10d), , x 2  y2  4, , (10e), , From Equation (10c) we have l  2. Next, substituting this value of l into Equations, (10a) and (10b) gives, 3  2  2mx, , or, , 1  2mx, , (11a), , and, 2  2  2my, , 4  2my, , or, , (11b), , Solving Equations (11a) and (11b) for x and y gives x  1>(2m) and y  2>m. Substituting these values of x and y into Equation (10e) yields, a, , 1 2, 2 2, b a b 4, m, 2m, 1  16  16m2, , Therefore, m , we have, , 117>4, so x , z, , , m2 , , or, , 2> 117 and y , , 8> 117. Using Equation (10d),, , 1, 1, 2, (1  x  y)  a1 , 2, 2, 117, 1, a1, 2, , 6, b, 117, , 17, 16, , 8, b, 117

Page 118 :
1138, , Chapter 13 Functions of Several Variables, , The value of f at the point, 3a, , 2, 8, 3, , 117, , 12  117, 1 117, 2 is, , 2, 8, 1, 3, 34, b  2a, b  4a , b2,  2(1  117), 2, 117, 117, 117, 117, , 2, 8, and the value of f at the point 1 117, , 117, , 12 , , 3a, , 3, 117, , 2 is, , 2, 8, 1, 3, 34, b  2a, b  4a , b2,  2(1  117), 2, 117, 117, 117, 117, , Therefore, the maximum value of f is 2(1  117) , and the minimum value of f is, 2(1  117) ., , 13.9, , CONCEPT QUESTIONS, y, , 1. What is a constrained maximum (minimum) value problem?, Illustrate with examples., 2. Describe the method of Lagrange multipliers for finding the, extrema of f(x, y) subject to the constraint t(x, y)  c. State, the method for the case in which f and t are functions of, three variables., 3. The figure at the right shows the contour map of a function f, and the curve of the equation t(x, y)  4. Use the figure to, obtain estimates of the maximum and minimum values of f, subject to the constraint t(x, y)  4., , k  6, , g(x, y)  4, 4, 2, , 4, , 0, , 2, , x, , 4, , 2, , 2, , k6, k4, k2, , 4, , 13.9, , k  4, k  2, , EXERCISES, , In Exercises 1–4, use the method of Lagrange multipliers to find, the extrema of the function f subject to the given constraint., Sketch the graph of the constraint equation and several level, curves of f. Include the level curves that touch the graph of the, constraint equation at the points where the extrema occur., 1. f(x, y)  3x  4y;, , x 2  y2  1, , 2. f(x, y)  x 2  y 2;, , 2x  4y  5, , 3. f(x, y)  x 2  y 2;, , xy  1, , 4. f(x, y)  xy;, , 12. f(x, y, z)  x  y  z;, , x 2  y2  z2  1, , 13. f(x, y, z)  x  2y  2z; x 2  2y 2  4z 2  1, 14. f(x, y, z)  x 2  y 2  z 2;, 15. f(x, y, z)  xyz;, , x 2  2y 2 , , 16. f(x, y, z)  xy  xz;, , x 2  y2  z2  8, , 17. f(x, y, z)  2x  y; x  y  z  1,, 18. f(x, y, z)  x  y  z;, 19. f(x, y, z)  yz  xz;, , 6. f(x, y)  x 2  y 2;, , 20. f(x, y, z)  x 2  y 2  z 2;, x  2y  3z  4, , x 2  y2  1, , x 2  4y 2  1, 4x 2  9y 2  36, , 9. f(x, y)  x  xy  y 2; x 2  y 2  8, 2, , 10. f(x, y)  x 2  y 2;, , x 4  y4  1, , xz  1,, , xz2, , y z 1, 2, , 2, , 2x  y  z  2,, , In Exercises 21–22, use the method of Lagrange multipliers to find, the extrema of the function subject to the inequality constraint., 21. f(x, y)  3x 2  2y 2  2x  1; x 2  y 2, , 11. f(x, y, z)  x  2y  z; x  4y  z  0, 2, , y2  z2  9, , x 2  y 2  1,, , 5. f(x, y)  xy; 2x  3y  6, , 8. f(x, y)  8x  9y;, , 1 2, z 6, 2, , In Exercises 17–20, use the method of Lagrange multipliers to, find the extrema of the function subject to the given constraints., , x  y2  4, 2, , In Exercises 5–16, use the method of Lagrange multipliers to, find the extrema of the function f subject to the given constraint., , 7. f(x, y)  xy;, , yx1, , 2, , 22. f(x, y)  x y; 4x  y, , V Videos for selected exercises are available online at www.academic.cengage.com/login., , 2, , 2, , 2, , 4, , 9

Page 119 :
13.9, 23. Find the point on the plane x  2y  z  4 that is closest to, the origin., 24. Find the maximum and minimum distances from the origin, to the curve 5x 2  6xy  5y 2  10  0., 25. Find the point on the plane x  2y  z  5 that is closest to, the point (2, 3, 1)., 26. Find the points on the surface z 2  xy  x  4y  21 that, are closest to the origin. What is the shortest distance from, the origin to the surface?, 27. Find the points on the surface xy 2z  4 that are closest to, the origin. What is the shortest distance from the origin to, the surface?, 28. Find three positive real numbers whose sum is 500 and, whose product is as large as possible., 29. Find the dimensions of a closed rectangular box of maximum, volume that can be constructed from 48 ft2 of cardboard., 30. Find the dimensions of an open rectangular box of maximum, volume that can be constructed from 12 ft2 of cardboard., 31. An open rectangular box having a volume of 108 in3. is to, be constructed from cardboard. Find the dimensions of such, a box if the amount of cardboard used in its construction is, to be minimized., 32. Find the dimensions of the rectangular box of maximum, volume with faces parallel to the coordinate planes that can, be inscribed in the ellipsoid, y2, x2, z2, , , 1, 4, 9, 16, 33. Solve the problem posed in Exercise 32 for the general case, of an ellipsoid with equation, x2, a2, , , , y2, b2, , , , z2, c2, , 1, , where a, b, and c are positive real numbers., 34. Find the dimensions of the rectangular box of maximum, volume lying in the first octant with three of its faces lying, in the coordinate planes and one vertex lying in the plane, 2x  3y  z  6. What is the volume of the box?, 35. Solve the problem posed in Exercise 34 for the general case, of a plane with equation, y, x, z,   1, a, c, b, where a, b, and c are positive real numbers., 3, , 36. An open rectangular box is to have a volume of 12 ft . If the, material for its base costs three times as much (per square, foot) as the material for its sides, what are the dimensions of, the box that can be constructed at the minimum cost?, 37. A rectangular box is to have a volume of 16 ft3. If the material for its base costs twice as much (per square foot) as the, material for its top and sides, find the dimensions of the box, that can be constructed at the minimum cost., , Lagrange Multipliers, , 1139, , 38. Maximizing Profit The total daily profit (in dollars) realized by, Weston Publishing in publishing and selling its dictionaries, is given by the profit function, P(x, y)  0.005x 2  0.003y 2  0.002xy  14x  12y  200, where x stands for the number of deluxe editions and y, denotes the number of standard editions sold daily. Weston’s, management decides that publication of these dictionaries, should be restricted to a total of exactly 400 copies per day., How many deluxe copies and how many standard copies, should be published each day to maximize Weston’s daily, profit?, 39. Cobb-Douglas Production Function Suppose x units of labor and, y units of capital are required to produce, f(x, y)  100x 3>4y 1>4, units of a certain product. If each unit of labor costs $200, and each unit of capital costs $300 and a total of $60,000 is, available for production, determine how many units of labor, and how many units of capital should be used to maximize, production., 40. a. Find the distance between the point P(x 1, y1) and the, line ax  by  c  0 using the method of Lagrange, multipliers., b. Use the result of part (a) to find the distance between the, point (2, 1) and the line 2x  3y  6  0., 41. Let f(x, y)  x  y and t(x, y)  x  x 5  y., a. Use the method of Lagrange multipliers to find the, point(s) where f may have a relative maximum or relative, minimum subject to the constraint t(x, y)  1., b. Plot the graph of t and the level curves of f(x, y)  k, for k  2, 1, 0, 1, 2, using the viewing window, [4, 4] [4, 4]. Then use this to explain why the, point(s) found in part (a) does not give rise to a relative, maximum or a relative minimum of f, c. Verify the observation made in part (b) analytically., 42. Let f(x, y)  x 2  y 2, and let t(x, y)  x  y., a. Show that f has no maximum or minimum values when, subjected to the constraint t(x, y)  1., b. What happens when you try to use the method of, Lagrange multipliers to find the extrema of f subject to, t(x, y)  1? Does this contradict Theorem 1?, 43. Find the point on the line of intersection of the planes, x  2y  3z  9 and 2x  3y  z  4 that is closest to the, origin., 44. Find the shortest distance from the origin to the curve with, equation y  (x  1)3>2. Explain why the method of, Lagrange multipliers fails to give the solution., 45. a. Find the maximum distance from the origin to the, Folium of Descartes, x 3  y 3  3axy  0, where a  0,, x  0 and y  0, using symmetry., b. Verify the result of part (a), using the method of, Lagrange multipliers.

Page 120 :
1140, , Chapter 13 Functions of Several Variables, , In Exercises 46 and 47, use the fact that a vector in n-space has, the form √  具 √1, √2, p , √n典 and the gradient of a function of n, variables, f(x 1, x 2, p , x n), is defined by §f  具 fx1, fx2, p , fxn 典., Also, assume that Theorem 1 holds for the n-dimensional case., , b. Use the result of part (a) to show that if x 1, x 2, p , x n are, positive numbers, then, n, 1x 1 x 2 p x n, , 46. a. Find the maximum value of, , x1  x2  p  xn, n, , This shows that the geometric mean of n positive numbers cannot exceed the arithmetic mean of the numbers., , f(x 1, x 2, p , x n, y1, y2, p , yn), n, ,  a x iyi  x 1y1  x 2y2  p  x nyn, i1, , subject to the constraints, n, 2, 2, 2, 2, p, a xi  x1  x2   xn  1, , 49. Snell’s Law of Refraction According to Fermat’s Principle in, optics, the path POQ taken by a ray of light (see the figure, below) in traveling across the plane separating two optical, media is such that the time taken is minimal. Using this, principle, derive Snell’s Law of Refraction:, , i1, , √1, √2, , sin u1, sin u2, , and, n, 2, 2, 2, 2, p, a yi  y1  y2   yn  1, i1, , where u1 is the angle of incidence, u2 is the angle of refraction, and √1 and √2 are the speeds of light in the two media., , b. Use the result of part (a) to show that if a1, a2, p , an,, b1, b2, p , bn are any numbers, then, n, , n, 2, , ai, , Hint: Put x i , , 2, , a ai B a b i, B i1, i1, , a aibi, i1, , bi, , and yi , , n, 2, , ¨1, , a, , a bi, B i1, , ¨2, , Note: This inequality is called the Cauchy-Schwarz Inequality., (Compare this with Exercise 9 in the Challenge Problems for, Chapter 4.), , 47. a. Let p and q be positive numbers satisfying, (1>p)  (1>q)  1. Find the minimum value of, q, , y, xp, , p, q, , x  0,, , Medium I, O, , ., , n, 2, , a ai, B i1, , f(x, y) , , P, , n, , y0, , subject to the constraint xy  c, where c is a constant., b. Use the result of part (a) to show that if x and y are positive numbers, then, yq, xp, ,  xy, p, q, where p  0 and q  0 and (1>p)  (1>q)  1., 48. a. Let x 1, x 2, p , x n be positive numbers. Find the maximum value of, n, f(x 1, x 2, p , x n)  1x 1 x 2 p x n, , subject to the constraint x 1  x 2  p  x n  c, where, c is a constant., , k, , b, , Medium II, , Q, , Hint: Show that the time taken by the ray of light in traveling from, P to Q is, t, , a, b, , √1 cos u1, √2 cos u2, , Then minimize t  f(cos u1, cos u2) subject to a tan u1 , b tan u2  k, where k is a constant., , In Exercises 50–52, determine whether the statement is true or, false. If it is true, explain why it is true. If it is false, give an, example to show why it is false., 50. Suppose f and t have continuous first partial derivatives, in some region D in the plane. If f has an extremum at a, point (a, b) subject to the constraint t(x, y)  c, then there, exists a constant l such that (a, b) is a critical point of, F  f  lt; that is, Fx (a, b)  0, Fy(a, b)  0, and, Fl(a, b)  0., 51. If (a, b) gives rise to a (constrained) extremum of f subject, to the constraint t(x, y)  0, then (a, b) also gives rise to an, unconstrained extremum of f., 52. If (a, b) gives rise to a (constrained) extremum of f subject to the constraint t(x, y)  0, then fx(a, b)  0 and, fy (a, b)  0 simultaneously.

Page 121 :
Concept Review, , CHAPTER, , 13, , 1141, , REVIEW, , CONCEPT REVIEW, In Exercises 1–17, fill in the blanks., 1. a. A function f of two variables, x and y, is a, that assigns to each ordered pair, in the domain, of f, exactly one real number f(x, y)., b. The number z  f(x, y) is called a, variable,, and x and y are, variables. The totality of the, numbers z is called the, of the function f., c. The graph of f is the set S , ., 2. a. The curves in the xy-plane with equation f(x, y)  k,, where k is a constant in the range of f, are called the, of f., b. A level surface of a function f of three variables is the, graph of the equation, , where k is a constant in, the range of, ., 3. lim (x, y)→(a, b) f(x, y)  L means there exists a number, such that f(x, y) can be made as close to, as we please by restricting (x, y) to be sufficiently close to, ., 4. If f(x, y) approaches L 1 as (x, y) approaches (a, b) along one, path, and f(x, y) approaches L 2 as (x, y) approaches (a, b), along another path with L 1  L 2, then lim (x, y)→(a, b) f(x, y), exist., 5. a. f(x, y) is continuous at (a, b) if lim (x, y)→(a, b) f(x, y) , ., b. f(x, y) is continuous on a region R if f is continuous at, every point (x, y) in, ., 6. a. A polynomial function is continuous, ; a rational, function is continuous at all points in its, ., b. If f is continuous at (a, b) and t is continuous at f(a, b),, then the composite function h  t ⴰ f is continuous at, ., 7. a. The partial derivative of f(x, y) with respect to x is, if the limit exists. The partial derivative, ( f> x)(a, b) gives the slope of the tangent line to, the curve obtained by the intersection of the plane, and the graph of z  f(x, y) at, ;, it also measures the rate of change of f(x, y) in the, -direction with y held, at, ., b. To compute f> x where f is a function of x and y, treat, as a constant and differentiate with respect to, in the usual manner., 8. If f(x, y) and its partial derivatives fx, fy, fxy, and fyx are continuous on an open region R, then fxy(x, y) , for, all (x, y) in R., , 9. a. The total differential dz of z  f(x, y) is dz , ., b. If ⌬z  f(x  ⌬x, y  ⌬y)  f(x, y), then ⌬z ⬇, ., c. ⌬z  fx(x, y) ⌬x  fy(x, y) ⌬y  e1 ⌬x  e2 ⌬y, where, and, e1 and e2 are functions of, such that lim (⌬x, ⌬y)→(0, 0) e1 , and, ., lim (⌬x, ⌬y)→(0, 0) e2 , d. The function z  f(x, y) is differentiable at (a, b) if ⌬z, can be expressed in the form ⌬z , , where, and, as (⌬x, ⌬y) →, ., 10. a. If f is a function of x and y, and fx and fy are continuous, on an open region R, then f is, in R., b. If f is differentiable at (a, b), then f is, at (a, b)., 11. a. If w  f(x, y), x  t(t), and y  h(t), then under suitable, conditions the Chain Rule gives dw>dt , ., b. If w  f(x, y), x  t(u, √), and y  h(u, √), then, ., w> u , c. If F(x, y)  0, where F is differentiable, then dy>dx , , provided that, ., d. If F(x, y, z)  0, where F is differentiable, and F defines, z implicitly as a function of x and y, then z> x , and z> y , , provided that, ., 12. a. If f is a function of x and y and u  u 1i  u 2 j is a unit, vector, then the directional derivative of f in the direction, of u is Du f(x, y) , if the limit exists., b. The directional derivative Du f(a, b) measures the rate of, change of f at, in the direction of, ., c. If f is differentiable, then Du f(x, y) , ., d. The gradient of f(x, y) is §f(x, y) , ., e. In terms of the gradient, Du f(x, y) , ., 13. a. The maximum value of Du f(x, y) is, occurs when u has the same direction as, b. The minimum value of Du f(x, y) is, occurs when u has the direction of, , , and this, ., , and this, ., , 14. a. §f is, to the level curve f(x, y)  c at P., b. §F is, to the level surface F(x, y, z)  0 at P., c. The tangent plane to the surface F(x, y, z)  0 at the, point P(a, b, c) is, ; the normal line passing, through P(a, b, c) has symmetric equations, ., 15. a. If f(x, y) f(a, b) for all points in an open disk containing (a, b), then f has a, at (a, b)., b. If f(x, y)  f(a, b) for all points in the domain of f, then f, has an, at (a, b) .

Page 122 :
1142, , Chapter 13 Functions of Several Variables, , c. If f is defined on an open region R containing the point, (a, b), then (a, b) is a critical point of f if (1) fx and/or fy, at (a, b) or (2) both, ., fx (a, b) and fy(a, b) equal, d. If f has a relative extremum at (a, b), then (a, b) must be, a, of f., e. To determine whether a critical point of f does give rise, to a relative extremum, we use the, ., 16. a. If f is continuous on a closed, bounded set D in the, plane, then f has an absolute maximum value, at some point, in D, and f has an absolute minimum value, at some point, in D., b. To find the absolute extreme values of f on a closed,, bounded set D, (1) find the values of f at the, in D, (2) find the extreme, values of f on the, of D. Then the largest and, smallest values found in (1) and (2) give the, value of f and the, value of f on D., , 17. a. If f(x, y) has an extremum at a point (a, b) lying on the, curve with equation t(x, y)  c, then the extremum is, called a, extremum., b. If f has an extremum at (a, b) subject to the constraint, , where l is a real, t(x, y)  c, then §f(a, b) , number called a Lagrange, ., c. To find the maximum and minimum values of f subject, to the constraint t(x, y)  c, we solve the system of, equations §f(x, y) , and t(x, y)  c for x, y,, and l. We then evaluate, at each of the, found in the last step. The largest, value yields the constrained, of f, and the, smallest value yields the constrained, of f., , REVIEW EXERCISES, In Exercises 17–22, find the first partial derivatives of the, function., , In Exercises 1–4, find and sketch the domain of the function., 29  x  y, 2, , 1. f(x, y) , , 2, , x 2  y2, , 2. f(x, y) , , ln(x  2y  4), yx, , 17. f(x, y)  2x 2y  1x, , 3. f(x, y)  sin1 x  tan1 y, , 19. f(r, s)  re, , 4. f(x, y)  ln(xy  1), , 21. f(x, y, z) , , In Exercises 5 and 6, sketch the graph of the function., 5. f(x, y)  4  x 2  y 2, , 2, , 9. f(x, y)  ex, , 1xy  4, (x, y)→(0, 0) 2y  3, x yx, 2, , 13., , lim, , (x, y)→(1, 0), , 12., , lim, , 26. f(u, √, w)  ue√ sin w, , (x, y)→(0, 0), , 27. If u  2x 2  y 2  z 2, show that, , x y, , x  3y, 4, , 3, , 4, , 14., , lim, , (x, y, z)→(0, 0, 0), , ln(x  y), (x 2  y 2) 3>2, , 16. f(x, y, z)  ex>y cos z  1x  y, , 2, , u, , x  2y  3z, 2, , 2, , x 2  y2  z2, , In Exercises 15 and 16, determine where the function is, continuous., 15. f(x, y) , , s, 22. f(r, s, t)  r cos st  s sina b, t, , z2  x 2, , 25. f(x, y, z)  x 2yz 3, , 2 2, , 1x  1y, , x 2  y2, , 24. f(x, y)  exy cos(2x  3y), , 10. f(x, y)  ln xy, , lim, , 20. f(u, √)  e2u cos(u 2  √2), , 23. f(x, y)  x 4  2x 2y 3  y 2  2, , In Exercises 11–14, find the limit or show that it does not exist., 11., , x 2  y2, , In Exercises 23–26, find the second partial derivatives of the, function., , 8. f(x, y)  y 2  x 2, , y2, , (r2s2), , xy 2, , 6. f(x, y)  21  x 2  y 2, , In Exercises 7–10, sketch several level curves for the function., 7. f(x, y)  x 2  2y, , 18. f(x, y) , , 2, , x, , 2, , 2, , , , u, , y, , 2, , 2, , , , u, , z2, , , , 2, u, , 28. Show that the function u  et cos(x>c) satisfies the onedimensional heat equation u t  c2u xx., In Exercises 29 and 30, show that the function satisfies Laplace’s, equation u xx  u yy  u zz  0., 29. u  2z 2  x 2  y 2, , 30. u  z tan1, , y, x

Page 123 :
Review Exercises, 31. Find dz if z  x 2 tan1 y 3., 32. Use differentials to approximate the change in, f(x, y)  x 2  3xy  y 2 if (x, y) changes from (2, 1) to, (1.9, 0.8)., 33. Use differentials to approximate (2.01)22(1.98)2  (3.02)3., 34. Estimating Changes in Profit The total daily profit function (in, dollars) of Weston Publishing Company realized in publishing and selling its English language dictionaries is given by, P(x, y)  0.0005x 2  0.003y 2  0.002xy  14x  12y  200, where x denotes the number of deluxe copies and y denotes, the number of standard copies published and sold daily., Currently, the number of deluxe and standard copies of the, dictionaries published and sold daily are 1000 and 1700,, respectively. Determine the approximate daily change in the, total daily profit if the number of deluxe copies is increased, to 1050 and the number of standard copies is decreased to, 1650 per day., 35. Does a function f such that §f  yi  xj exist? Explain., 36. According to Ohm’s Law, R  V>I, where R is the resistance in ohms, V is the electromotive force in volts, and I, is the current in amperes. If the errors in the measurements, made in a certain experiment in V and I are 2% and 1%,, respectively, use differentials to estimate the maximum, percentage error in the calculated value of R., 37. Let z  x 2y  1y, where x  e2t and y  cos t. Use the, Chain Rule to find dz>dt., 38. Let w  e cos y  y sin e , where x  u  √ and, y  1u√. Use the Chain Rule to find w> u and w> √., x, , x, , 2, , 2, , 39. Use partial differentiation to find dy>dx if, x 3  3x 2y  2xy 2  2y 3  9., 40. Find z> x and z> y if x z  yz  cos xz., 3 2, , 41. f(x, y)  2x 2  y 2; P(1, 2), 42. f(x, y)  ex tan y; P 1 0, p4 2, , 43. f(x, y, z)  xy 2  yz 2  zx 2;, 44. f(x, y, z)  x ln y  y ln z;, , P(2, 1, 3), P(2, 1, 1), , In Exercises 45–48, find the directional derivative of the function, f at the point P in the indicated direction., 45. f(x, y)  x 3y 2  xy 3; P(2, 1), in the direction of, v  3i  4j., 46. f(x, y)  e, to Q(3, 1) ., , x2, , cos y; P 1 0,, , 47. f(x, y, z)  x2y  z ;, v  i  2j  2k., 2, , 2, , p, 2, , 48. f(x, y, z)  x 2 ln y  xy 2ez; P(2, 1, 0) in the direction of, v  具3, 1, 2典., 49. Find the direction in which f(x, y)  1x  xy 2 increases, most rapidly at the point (4, 1) . What is the maximum rate, of increase?, 50. Find the direction in which f(x, y, z)  xeyz decreases most, rapidly at the point (4, 3, 0) . What is the greatest rate of, decrease?, In Exercises 51–54, find equations for the tangent plane and the, normal line to the surface with the equation at the given point., 51. 2x 2  4y 2  9z 2  27; P(1, 2, 1), 52. x 2  2y 2  3z 2  19; P(2, 3, 1), 53. z  x 2  3xy 2; P(3, 1, 18), 54. z  xey; P(1, 0, 1), , 55. Let f(x, y)  x 2  y 2, and let t(x, y)  y 2>x 2., , cas a. Plot several level curves of f and t using the same view-, , ing window., b. Show analytically that each level curve of f intersects all, level curves of t at right angles., 56. Show that if §f(x 0, y0)  0, then an equation of the tangent, line to the level curve f(x, y)  f(x 0, y0) at the point (x 0, y0), is, fx(x 0, y0)(x  x 0)  fy (x 0, y0)(y  y0)  0, In Exercises 57–60, find the relative extrema and saddle points, of the function., 57. f(x, y)  x 2  xy  y 2  5x  8y  5, 58. f(x, y)  8x 3  6xy  y 3, 59. f(x, y)  x 3  3xy  y 2, , 3, , In Exercises 41–44, find the gradient of the function f at the, indicated point., , 2 , in the direction from P(1, 3), , P(2, 3, 4) in the direction of, , 1143, , 60. f(x, y) , , 4, 2,   xy, x, y, , In Exercises 61 and 62, find the absolute extrema of the function, on the set D., 61. f(x, y)  x 2  xy 2  y 3;, D  {(x, y) 冟 1 x 1, 0, , y, , 2}, , 62. f(x, y)  (x  3y )e ; D  {(x, y) 冟 x 2  y 2, 2, , x, , 2, , 9}, , In Exercises 63–66, use the method of Lagrange multipliers to, find the extrema of the function f subject to the constraints., 63. f(x, y)  xy 2;, 64. f(x, y) , , x 2  y2  4, , 1, 1,  ;, x, y, , 1, x2, , , , 1, y2, , 65. f(x, y, z)  xy  yz  xz;, , x  2y  3z  1, , 66. f(x, y, z)  3x  2y  z ;, 2x  y  z  2, 2, , 2, , 9, , 2, , x  y  z  1,

Page 124 :
1144, , Chapter 13 Functions of Several Variables, , 67. Let f(x, y)  Ax 2  Bxy  Cy 2  Dx  Ey  F. Show that, if f has a relative maximum or a relative minimum at a point, (x 0, y0), then x 0 and y0 must satisfy the system of equations, , Show that the isothermal curves T(x, y)  k are arcs of circles that pass through the points x  1. Sketch the isothermal curve corresponding to a temperature of 75°C., , 2Ax  By  D  0, , y, , Bx  2Cy  E  0, simultaneously., 68. Let f(x, y)  x 2  2Bxy  y 2, where B  0. For what value, of B does f have a relative minimum at (0, 0)? A saddle, point at (0, 0)? Are there any values of B such that f has a, relative maximum at (0, 0)?, , 1, , x, , 2, , 2, , y, x, , 4, 25, , In Exercises 71 and 72, determine whether the statement is true, or false. If it is true, explain why it is true. If it is false, give an, example to show why it is false., , that is closest to the point (3, 0, 0)., 70. Isothermal Curves Consider the upper half-disk, H  {(x, y) 冟 x 2  y 2 1, y  0} (see the figure). If the, temperature at points on the upper boundary is kept at, 100°C and the temperature at points on the lower boundary, is kept at 50°C, then the steady-state temperature T(x, y) at, any point inside the disk is given by, T(x, y)  100 , , 0, , 1, , T(x, y)  50, , 69. Find the point on the paraboloid, z, , T(x, y)  100, , 71. The directional derivative of f(x, y) at the point (a, b) in the, positive x-direction is fx (a, b) ., 72. If we know the gradient of f(x, y, z) at the point P(a, b, c),, then we can compute the directional derivative of f in any, direction at P., , 1  x 2  y2, 100, tan1, p, 2y, , CHALLENGE PROBLEMS, 1. Find and sketch the domain of, , 4. Consider the quadratic polynomial function, , f(x, y)  236  4x  9y, 2, , 2, ,  ln(x 2  2x  y 2), , , 1, 24x  16x  4y 2  15, 2, , 2. Describe the domain of, H(x, y, z)  1x  a  1b  x  1y  c, , f(x, y)  Ax 2  2Bxy  Cy 2  2Dx  2Ey  F, Find conditions on the coefficients of f such that f has (a) a, relative maximum and (b) a relative minimum. What are the, coordinates of the point in terms of the coefficients of f ?, 5. Let a, b, and c denote the sides of a triangle of area A, and, let a, b, and g denote the angles opposite them. If, A  f(a, b, c), show that, f, 1,  R cos a, a, 2, ,  1d  y  1z  e  1f  z, where a, , b, c, , d, and e, , f., , 3. Suppose f has continuous second partial derivatives, in x and y. Then the second-order directional derivative, of f in the direction of the unit vector u  u 1i  u 2 j is, defined to be, , where R is the radius of the circumscribing circle., , ∫, , R, , a, , c, , D 2u f(x, y)  Du (Du f ), a. Find an expression in terms of the partial derivatives of f, for D 2u f., b. Find D 2u f(1, 0) if f(x, y)  xy 2  exy and u has the same, direction as v  2i  3j., , ©, , å, b

Page 125 :
Challenge Problems, 6. Linda has 24 feet of fencing with which to enclose a triangular flower garden. What should the lengths of the sides of, the garden be if the area is to be as large as possible?, , 10. Consider the problem of determining the maximum and, the minimum distances from the point (x 0, y0, z 0) to the, ellipsoid, , Hint: Heron’s formula states that the area of a triangle with sides a,, b, and c is given by A  1s(s  a)(s  b)(s  c), where, s  12 (a  b  c) is the semiperimeter., , 7. Find the directional derivative at 1 1, 2, p4 2 of the function, f(x, y, z)  x 2  y cos z in the direction of increasing t, along the curve in three-dimensional space described by, the position vector r(t)  具t, t 2, t 3典 at r(1)., , x2, a, , 2, , , , y2, b, , 2, , z2, , , , c2, , 1, , a. Show that the solutions are, x, , a 2x 0, a2  l, , y, , ,, , b 2y0, b2  l, , z, , ,, , c 2z 0, c2  l, , where l satisfies, , 8. Let, xy(x  y ), 2, , f(x, y)  •, , a 2x 20, , 2, , (x, y)  (0, 0), , x 2  y2, 0, , (a  l), 2, , (x, y)  (0, 0), , Use the definition of partial derivatives to show that, fxy(0, 0)  1 and fyx (0, 0)  1., 2, , 9. Show that Laplace’s equation, , u, 2, , 2, , , , u, , x, y, cylindrical coordinates takes the form, 2, , u, , r, , 2, , , , 2, , 2, , , , u, , z2, , 2, 2, 1, 1, u, u, u,  2ⴢ 2 20, ⴢ, r, r, r, u, z, ,  0 in, , 2, , , , b 2y 20, (b  l), 2, , 2, , , , c2z 20, (c  l)2, 2, , 1, , b. Use the result of part (a) to solve the problem with, a  2, b  3, c  1, and (x 0, y0, z 0)  (3, 2, 4)., , 1145

Page 126 :
MODULE 3, Textbook : Calculus: Soo T Tan Brooks/Cole, Cengage Learning (2010) ISBN 0-534-46579-X), Sections 14.1-14.8

Page 127 :
1148, , Chapter 14 Multiple Integrals, , 14.1, , Double Integrals, An Introductory Example, Suppose a piece of straight, thin wire of length (b  a) is placed on the x-axis of a, coordinate system, as shown in Figure 1. Further suppose that the wire has linear mass, density given by f(x) at x for a  x  b, where f is a nonnegative continuous function, on [a, b]. Let P  {x 0, x 1, x 2, p , x n}, where a  x 0 and b  x n, be a regular partition of [a, b]. Then the continuity of f tells us that, f(x) ⬇ f(ck) for each x in the kth, subinterval [x k1, x k], where ck is an evaluation point in [x k1, x k], provided n is large, enough. Therefore, the mass of the piece of wire lying on [x k1, x k] is, , y, y  f(x), , ⌬m k ⬇ f(ck) ⌬x, 0, , a, , b, , ⌬x , , x, , ba, n, , This leads to the definition of the mass of the wire as, , FIGURE 1, The mass of a straight wire of length, (b  a) is given by 兰ab f(x) dx, where, f(x) is the density of the wire at any, point x for a  x  b., , n, , b, , n, , m  lim a ⌬m k  lim a f(ck) ⌬x , n→⬁, n→⬁, k1, , k1, , 冮 f(x) dx, a, , Thus, the mass of the curve has the same numerical value as that of the area under the, graph of the (nonnegative) density function f shown in Figure 1., Now let’s consider a thin rectangular plate occupying the region, R  {(x, y) 冟 a  x  b, c  y  d}, (See Figure 2.) If the plate is homogeneous (having a constant mass density of k g/cm2),, then its mass is given by, m  k(b  a)(d  c), mass density ⴢ area, y, d, R, c, R, , a, , 0, , FIGURE 2, , (a) A thin rectangular plate, , b, , x, , (b) The plate placed in the xy-plane, , Observe that m has the same numerical value as that of the volume of the rectangular, box bounded above by the graph of the constant function f(x, y)  k and below by R., (See Figure 3a.) Next, instead of being constant, suppose that the mass density of the, z, , z, , z  f (x, y), , zk, , S, , FIGURE 3, The mass of the plate R is numerically, equal to that of the volume of the solid, region lying directly above R and, below the surface z  f(x, y)., , R, , y, , R, , x, , x, , (a), , (b), , y

Page 128 :
14.1, , Double Integrals, , 1149, , plate is given by the mass density function f. Then it seems reasonable to conjecture, that the mass of the plate is given by the “volume” of the solid region S lying directly, above R and below the graph of z  f(x, y). (See Figure 3b.) We will show in Section 14.4 that this is indeed the case., , Volume of a Solid Between a Surface and a Rectangle, We will now show that the volume of a solid S can be defined as a limit of Riemann, sums. Suppose that f is a nonnegative continuous function* of two variables that is, defined on a rectangle, R  [a, b]  [c, d]  {(x, y) 冟 a  x  b, c  y  d}, and suppose that f(x, y)  0 on R. Let, a  x 0  x 1  p  x i1  x i  p  x m  b, be a regular partition of the interval [a, b] into m subintervals of length ⌬x , (b  a)>m, and let, c  y0  y1  p  yj1  yj  p  yn  d, be a regular partition of the interval [c, d] into n subintervals of length ⌬y  (d  c)>n., The grid comprising segments of the vertical lines x  x i for 0  i  m and the horizontal lines y  yj for 0  j  n partition R into N  mn subrectangles R11, R12, p , Rij,, p , Rmn, where Rij  [x i1, x i]  [yj1, yj]  {(x, y) 冟 x i1  x  x i, yj1  y  yj} as, shown in Figure 4. The area of each subrectangle is ⌬A  ⌬x ⌬y. This partition is, called a regular partition of R., y, d  yn, yn1, yj, yj1, y2, y1, c  y0, , FIGURE 4, A partition P  {Rij} of R, , 0, , (xij*, y*ij), R1n, , R2n, , ..., , Rin, , ., ., ., , ., ., ., , ., ., ., , ., ., ., , ., ., ., , ., ., ., , R1j, , R2j, , ..., , Rij, , ..., , Rmj, , ., ., ., , ., ., ., , ., ., ., , ., ., ., , ., ., ., , ., ., ., , R12, , R22, , ..., , Ri2, , ..., , Rm2, , R11, , R21, , ..., , Ri1, , ..., , Rm1, , a  x0 x1, , x2, , ..., , xi1, , Rmn, , xi, , y, , . . . xm1 xm  b, , x, , x, , The partition P  {R11, R12, p , Rij, p , Rmn} divides the solid S between the graph, of z  f(x, y) and R into N  mn solids; the solid Sij is bounded below by Rij and, bounded above by the part of the surface z  f(x, y) that lies directly above Rij. (See, Figure 5.), , *As in the case of the integral of a function of one variable, these assumptions will simplify the discussion.

Page 129 :
1150, , Chapter 14 Multiple Integrals, z, , z, , Sij, , S, , y, , FIGURE 5, , x, , x, , (a) The solid S is the union of N  mn, solids (shown here with m  3, n  4), , (b) A typical solid Sij, , Rij, , Let (x *, ij , y *, ij ) be an evaluation point in Rij. Then the parallelepiped with base Rij,, *, height f(x *, ,, y, ij, ij ) , and volume, , (x ij*, yij*, f (xij*, yij*)), , f(x *, ij , y *, ij ) ⌬A, , Sij, , f (xij*, yij*), , gives an approximation of the volume of Sij. (See Figure 6.), Therefore, the volume V of S is approximated by the volume of the sum of N  mn, parallelepipeds; that is,, m, , (xij*, yij*), , y, , n, , V ⬇ a a f(x *, ij , y *, ij ) ⌬A, , FIGURE 6, The volume of Sij is approximated, by the volume of the parallelepiped, with base Rij and height f(x *, ij , y *, ij ) ., , (1), , i1 j1, , Rij, , If we take m and n to be larger and larger, then, intuitively, we can expect the approximation (1) to improve. This suggests the following definition., , DEFINITION Volume Under the Graph of z ⴝ f(x, y), Let f be defined on the rectangle R and suppose f(x, y)  0 on R. Then the volume V of the solid S that lies directly above R and below the surface z  f(x, y), is, m, , V  lim, , n, , ij , y *, ij ) ⌬A, a a f(x *, , m, n→⬁ i1 j1, , (2), , if the limit exists., , Because of the assumption that f be continuous, it can be shown that the limit in, Equation (2) always exists regardless of how the evaluation points (x *, ij , y *, ij ) in Rij, for, 1  i  m and 1  j  n, are chosen., , EXAMPLE 1 Approximate the volume V of the solid lying under the graph of the, elliptic paraboloid z  8  2x 2  y 2 and above the rectangle R  {(x, y) 冟 0  x  1,, 0  y  2}. Use the partition P of R that is obtained by dividing R into four subrectangles with the lines x  12 and y  1, and choose the evaluation point (x *, ij , y *, ij ) to be, the upper right-hand corner of Rij. (See Figure 7.)

Page 130 :
14.1, y, , ( ), , 2, R12, , R22, (1, 1), , R11, , 8, , (1, 2), , 1, , ( , 1), 1_, 2, , R21, , R11, R21, , R12, R22, , 1, , x, , 1, , 2, , y, , x, , (a) The region R is divided into, four subrectangles, , Solution, , 1151, , z, 1_, 2, 2, , FIGURE 7, , Double Integrals, , (b) The solid lying under the graph of, z  8  2x 2  y2 and above R, , Here,, ⌬x , , 10, 1, , 2, 2, , 20, 1, 2, , ⌬y , , and, , so ⌬A  1 12 2 (1)  12. Also, x 0  0, x 1  12, and x 2  1, and y0  0, y1  1, and, 1, 1, y2  2. Taking (x *, 11, y *, 11)  (x 1, y1)  1 2 , 1 2 , (x *, 12, y *, 12)  (x 1, y2)  1 2 , 2 2 ,, (x *, 21, y *, 21)  (x 2, y1)  (1, 1) , and (x *, 22, y *, 22)  (x 2, y2)  (1, 2) , we have, 2, , 2, , V ⬇ a a f(x *, ij , y *, ij ) ⌬A  f(x *, 11, y *, 11) ⌬A, , f(x *, 12, y *, 12) ⌬A, , f(x *, 21, y *, 21) ⌬A, , f(x *, 22, y *, 22) ⌬A, , i1 j1, , 1,  f a , 1b ⌬A, 2, a, , 13 1, ba b, 2, 2, , 1, f a , 2b ⌬A, 2, 7 1, a ba b, 2 2, , f(1, 1) ⌬A, , 1, (5) a b, 2, , f(1, 2) ⌬A, , 1, 17, (2) a b , 2, 2, , The approximations to the volume in Example 1 get better and better as m and n, increase, as shown in Figure 8., , FIGURE 8, The approximation of V using the sum of the volumes of 16 parallelepipeds in, (a), 64 parallelepipeds in (b), and 256 parallelepipeds in (c)., , Note Suppose that the mass density of a rectangular plate R  {(x, y) 冟 0  x  1,, 0  y  2} is f(x, y)  8  2x 2  y 2 g/cm2. Then the result of Example 1 tells us, that the mass of the plate is approximately 172 g.

Page 131 :
1152, , Chapter 14 Multiple Integrals, , The Double Integral Over a Rectangular Region, Thus far, we have assumed that f(x, y)  0 on the rectangle R. This condition was, imposed so that we could give a simple geometric interpretation for the limit in Equation (2). In the general situation we have the following., , DEFINITION Riemann Sum, Let f be a continuous function of two variables defined on a rectangle R, and let, P  {Rij} be a regular partition of R. A Riemann sum of f over R with respect, to the partition P is a sum of the form, m, , n, , ij , y *, ij ) ⌬A, a a f(x *, , (3), , i1 j1, , where (x *, ij , y *, ij ) is an evaluation point in Rij., , DEFINITION Double Integral of f Over a Rectangle R, Let f be a continuous function of two variables defined on a rectangle R. The, double integral of f over R is, , 冮冮, , m, , f(x, y) dA  lim, , R, , n, , ij , y *, ij ) ⌬A, a a f(x *, , m, n→⬁ i1 j1, , (4), , if this limit exists for all choices of the evaluation point (x *, ij , y *, ij ) in Rij., , Notes, 1. If the double integral of f over R exists, then f is said to be integrable over R. It, can be shown, although we will not do so here, that if f is continuous on R, then f, is integrable over R., 2. If f is integrable, then the Riemann sum (3) is an approximation of the double, integral (4)., 3. If f(x, y)  0 on R, then 兰兰R f(x, y) dA gives the volume of the solid lying, directly above R and below the surface z  f(x, y)., 4. We use the (double) integral sign to denote the limit, whenever it exists, because, it is related to the definite integral, as you will see in Section 14.2., y, , EXAMPLE 2 Find an approximation for 兰兰R (x  4y) dA, where R  {(x, y) 冟 0 , , x  2, 0  y  1}, using the Riemann sum of f(x, y)  x  4y over R with, m  n  2 and taking the evaluation point (x *, ij , y *, ij ) to be the center of Rij., , 1, R12, , R22, , R11, , R21, , 1, 2, , 0, , 1, 2, , 1, , Solution, 3, 2, , FIGURE 9, The partition, P  {R11, R12, R21, R22} of R, , 2, , x, , Here,, ⌬x , , 20, 1, 2, , ⌬y , , 10, 1, , 2, 2, , and x 0  0, x 1  1, x 2  2, y0  0, y1  12, y2  1. The partition P is shown in Figure 9. Using Equation (3) with f(x, y)  x  4y, ⌬A  ⌬x ⌬y  (1) 1 12 2  12, and

Page 132 :
14.1, , (x *, 11, y *, 11) , obtain, , 1 12, 14 2 ,, , (x *, 12, y *, 12) , 2, , 1 12, 34 2 ,, , (x *, 21, y *, 21) , , 1 32, 14 2 ,, , Double Integrals, , and (x *, 22, y *, 22) , , 1 32, 34 2 ,, , 1153, we, , 2, , 冮冮 f(x, y) dA ⬇ a a f(x *, y*) ⌬A, ij, , R, ,  f(x *, 11, y *, 11) ⌬A, 1 1 1,  fa , b, 2 4 2, 1 1,  a b a b, 2 2, , y, , R, , ij, , i1 j1, , f(x *, 12, y *, 12) ⌬A, 1 3 1, fa , b, 2 4 2, , f(x *, 21, y *, 21) ⌬A, , 3 1 1, fa , b, 2 4 2, , 5 1, a b a b, 2 2, , 1 1, a ba b, 2 2, , f(x *, 22, y *, 22) ⌬A, , 3 3 1, fa , b, 2 4 2, 3 1, a b a b  2, 2 2, , (xij*, y*ij), , Rij, , Double Integrals Over General Regions, D, x, , 0, , FIGURE 10, A partition Q of R, , Next we will extend the definition of the double integral to more general functions and, regions. Suppose that f is a bounded function defined on a bounded plane region D. If, you like, you can think of f as a mass density function for a thin plate occupying a, nonrectangular region D (in which case f(x, y)  0 on D) and of what follows as a way, of finding the mass of the plate. Since D is bounded, it can be enclosed in a rectangle, R. Let Q be a regular partition of R into subrectangles R11, R12, p , Rij, p , Rmn. (See, Figure 10.), Let’s define the function, fD(x, y)  e, , z, , f(x, y), 0, , if (x, y) is in D, if (x, y) is in R but not in D, , Note that fD takes on the same value as f if (x, y) is in D, but it takes on the value zero, if (x, y) lies outside D. (See Figure 11.), Now let (x ij*, y ij*) be an evaluation point in the subrectangle Rij of Q for 1  i  m, and 1  j  n. Then the sum, , Graph of fD, , m, , n, , ij , y *, ij ) ⌬A, a a fD(x *, , y, , D, , i1 j1, , x, , FIGURE 11, fD (x, y)  f(x, y) if (x, y) lies in D, but, fD (x, y)  0 if (x, y) lies outside D., , is a Riemann sum of f over D with respect to the partition Q. Taking the limit of these, sums as m, n → ⬁ gives the double integral of f over D. Thus,, , 冮冮 f(x, y) dA , D, , m, , lim, , n, , ij , y *, ij ) ⌬A, a a fD(x *, , m, n→⬁ i1 j1, , (5), , if the limit exists. Again, it can be shown that if f is continuous, then the limit (5), always exists regardless of how the evaluation points (x *, ij , y *, ij ) in Rij are chosen., , y, 3, R, , x, , Notes, 1. If f(x, y)  0 on D, then 兰兰D f(x, y) dA gives the volume of the solid lying, directly above D and below the surface z  f(x, y)., 2. If r(x, y)  0 on D, where r is a mass density function, then 兰兰D r(x, y) dA gives, the mass of the thin plate occupying the plane region D in the xy-plane. This will, be demonstrated in Section 14.4., , FIGURE 12, fD (x, y)  f(x, y) if (x, y) lies in D, but, fD (x, y)  0 if (x, y) lies outside D., , EXAMPLE 3 Find an approximation for 兰兰D (x 2y) dA, where D is the region, shown in Figure 12, using the Riemann sum of f(x, y)  x 2y over D with respect, to the partition Q obtained by dividing the rectangle {(x, y) 冟 0  x  2, 0  y  3}, , 2, , D, , 1, , 0, , 1, 2, , 1, , 3, 2, , 2

Page 133 :
1154, , Chapter 14 Multiple Integrals, , into 12 subrectangles by taking m  4 and n  3 and taking the evaluation points to, be the center of Rij., Solution, , Here,, ⌬A  (⌬x)(⌬y)  a, , 20 30, 1, ba, b, 4, 3, 2, , Next, define, fD (x, y)  e, , f(x, y), 0, , if (x, y) is in D, if (x, y) is not in D, , Then, , 冮冮, , 4, , (x, , 3, , 2y) dA ⬇ a a fD(x *, ij , y *, ij ) ⌬A, , f(x, y)  x, , 2y, , i1 j1, , D, , 1 1,  cfD a , b, 4 2, , 1 3, fD a , b, 4 2, , 5 1, fD a , b, 4 2, , 5 3, fD a , b, 4 2, , 1, 1 3,  cf a , b, 2, 4 2, , , 1 1, ec, 2 4, , 3 1, fa , b, 4 2, , 3, 2a b d, 2, c, , 5, 4, , 1 5, fD a , b, 4 2, , c, , 3, 2a b d, 2, , 5 5, fD a , b, 4 2, 3 3, fa , b, 4 2, 1, 2a b d, 2, , 3, 4, c, , 3 1, fD a , b, 4 2, , 7, 4, , 7 1, fD a , b, 4 2, 3 5, fa , b, 4 2, , c, , 3 3, fD a , b, 4 2, , 3, 4, , 3 5, fD a , b, 4 2, , 7 3, fD a , b, 4 2, 5 1, a , b, 4 2, , 3, 2a b d, 2, , c, , 3, 4, , 7 5, fD a , b d ⌬A, 4 2, , 5 3, fa , b, 4 2, 5, 2a b d, 2, , c, , 7 1, f a , bd, 4 2, 5, 4, , 1, 2a b d, 2, , 1, 2a b d f  11.875, 2, , Properties of Double Integrals, Double integrals have many of the properties that single integrals enjoy. We list some, of them in the following theorem, the proof of which will be omitted., , THEOREM 1 Properties of the Definite Integral, Let f and t be defined on a suitably restricted region D, so that both 兰兰D f(x, y) dA, and 兰兰D t(x, y) dA exist, and let c be a constant. Then, 1., , 冮冮 cf(x, y) dA  c冮冮 f(x, y) dA, D, , 2., , 冮冮[f(x, y), , D, , t(x, y)] dA , , D, , 3. If f(x, y)  0 on D, then, , 冮冮 f(x, y) dA 冮冮 t(x, y) dA, D, , D, , 冮冮 f(x, y) dA  0, D, , 4. If f(x, y)  t(x, y) on D, then, , 冮冮 f(x, y) dA  冮冮 t(x, y) dA, D, , D

Page 134 :
14.1, y, , Double Integrals, , 1155, , 5. If D  D1 傼 D2, where D1 and D2 are two nonoverlapping subregions, with the possible exception of their common boundaries, then, D2, , D, , 冮冮 f(x, y) dA  冮冮 f(x, y) dA 冮冮 f(x, y) dA, , D1, , D, , FIGURE 13, D  D1 傼 D2 where D1 傽 D2 , , 14.1, , ., , CONCEPT QUESTIONS, , 1. Let f(x, y)  x 2y., a. Complete the table of values for f(x, y) in the following, table., y, x, , D2, , (See Figure 13.), , x, , 0, , D1, , 3, 2, , 1, , 2, , 5, 2, , 3, , 7, 2, , 4, , 0, 1, 4, 1, 2, 3, 4, , b. Use the table of values from part (a) to estimate the volume of the solid lying under the graph of z  x 2y, and above the rectangular region R  [0, 1]  [1, 4], using a regular partition with m  2 and n  3 and, choosing the evaluation point (x *, ij , y *, ij ) to be the lower, left-hand corner of Rij., c. Repeat part (b), this time choosing the evaluation point, (x *, ij , y *, ij ) to be the center of Rij., 2. a. Let f be a continuous function defined on the rectangular, region [a, b]  [c, d]. Define 兰兰R f(x, y) dA., b. Suppose that f(x, y)  k, where k is a constant. Find, 兰兰R k dA for R  [a, b]  [c, d] using your definition, from part (a)., , 1, , 14.1, , EXERCISES, , In Exercises 1–4, find an approximation for the volume V, of the solid lying under the graph of the elliptic paraboloid, z  8  2x 2  y 2 and above the rectangular region, R  {(x, y) 冟 0  x  1, 0  y  2}. Use a regular partition, P of R with m  n  2, and choose the evaluation point, (x *, ij , y *, ij ) as indicated in each exercise., 1. The lower left-hand corner of Rij, 2. The upper left-hand corner of Rij, 3. The lower right-hand corner of Rij, 4. The center of Rij, n, In Exercises 5–8, find the Riemann sum 兺m, i1 兺j1 f(x *, ij , y *, ij ) ⌬A, of f over the region R with respect to the regular partition P with, the indicated values of m and n., , 5. f(x, y)  2x 3y; R  [0, 1]  [0, 3]; m  2, n  3;, (x ij*, y *, ij ) is the lower left-hand corner of Rij, , V Videos for selected exercises are available online at www.academic.cengage.com/login., , 6. f(x, y)  x 2  2y; R  [1, 5]  [1, 3]; m  4, n  2;, (x *, ij , y *, ij ) is the upper right-hand corner of Rij, 7. f(x, y)  x 2 2y 2; R  [1, 3]  [0, 4];, (x *, ij , y *, ij ) is the center of Rij, 8. f(x, y)  2xy; R  [1, 1]  [2, 2];, (x *, ij , y *, ij ) is the center of Rij, , m  4, n  4;, , m  4, n  4;, , 9. The figure on the following page shows a region D enclosed, by a rectangular region R and a partition Q of R into subrectangles with m  5 and n  3. Suppose that f is continuous, on D and the values of f at the evaluation points of Q that, lie in D are as shown in the figure (next to the evaluation, points). Define, fD (x, y)  e, , f(x, y), 0, , if (x, y) is in D, if (x, y) is in R but not in D

Page 135 :
1156, , Chapter 14 Multiple Integrals, part of the room using a regular partition with m  n  3, and choosing the evaluation point (x *, ij , y *, ij ) to be the center, of Rij., , Compute 兺5i1 兺3j1 fD (x *, ij , y *, ij ) ⌬A., y, , R, , 3, , D, 2, , 1, , 3, , 2, , 3, , 1, , 0, , 2, , y (ft), , 4, 9, , 3, , 74, , (x, y), 1, , 2, , 0, , 1, , 1, , 2, , 6, , 1, , 3, , 76, 4, , x, , 5, , 78, , 3, 82, , 10. Volume of Water in a Pond The following figure depicts a pond, that is 40 ft long and 20 ft wide. The depth of the pond is, measured at the center of each subrectangle in the imaginary, partition of the rectangle that is superimposed over the aerial, view of the pond. These measurements (in feet) are shown, in the figure. Estimate the volume of water in the pond., Hint: See Exercise 9., R, , 20, , 2, , 10, , 4, , 2, , 5, , 6, , 5, , 3, , 14., , 2, , 1, , 6, , 8, , 8, , 6, , 4, , 3, , 1, , 3, , 2, , 0, , 1, , 2, , 10, , 5, , 15, , 2, , 15., , 9, , x (ft), , 冮冮 2 dA, where R  [1, 3]  [2, 5], , 20, , 25, , 30, , 35, , y, , 冮冮 2x dA, where R  {(x, y) 冟 0  x  2, 0  y  1}, 冮冮 (6  2y) dA, where R  {(x, y) 冟 0  x  4, 0  y  2}, R, , 40, , 11. The figure shows the contour map of a function f on, the set R  {(x, y) 冟 0  x  2, 0  y  2}. Estimate, 兰兰R f(x, y) dA using a Riemann sum with m  n  2 and, choosing the evalution point (x *, ij , y *, ij ) to be the center of Rij., , 冮冮 29  x, , 2, ,  y 2 dA, where, , R, , R  {(x, y) 冟 x 2, , y 2  9, x  0, y  0}, , In Exercises 17 and 18, the double integral gives the volume of a, solid. Describe the solid., 17., , 冮冮 (4  x ) dA, where R  {(x, y) 冟 0  y  x, 0  x  2}, 2, , R, , 2, 40, , 50, , 18., , 3, , 1, , R  {(x, y) 冟 2x, , 1, , 3y  12, x  0, y  0}, , In Exercises 19 and 20, the expression is the limit of a Riemann, sum of a function f over a rectangle R. Write this expression as a, double integral over R., , 20, 10, 1, , 冮冮 a3  2 x  4 yb dA, where, R, , 30, , 0, , 6, , R, , 16., 0, , 3, , R, , 1, , 6, , 86, , In Exercises 13–16, find the double integral by interpreting it as, the volume of a solid., , 3, , 0, 15, , 0, , 13., , D, 3, , 90, , 2, , x, , m, , 19., , lim, , n, , ij, a a (3  2x *, , m, n→⬁ i1 j1, m, , n, , y*, ij ) ⌬A,, , R  [1, 2]  [1, 3], , 12. Room Temperature The figure represents a part of a room with, 2, 2, 20. lim a a 2(x *, R  [0, 1]  [0, 2], 2(y *, ij ), ij ) ⌬A,, a fireplace located at the origin. The curves shown are the, m, n→⬁ i1 j1, level curves of the temperature function T, and are called, isothermals because the temperature is the same at all points cas In Exercises 21 and 22, use a computer algebra system (CAS) to, on an isothermal. Estimate the average temperature in this, obtain an approximate value of the double integral using a regu-

Page 136 :
14.2, lar partition with the given value of m and n and choosing the, evaluation point (x *, ij , y *, ij ) to be the center of Rij., 21., , 冮冮 21, , 1157, , In Exercises 27–30, determine whether the statement is true or, false. If it is true, explain why. If it is false, explain why or give, an example that shows it is false., , y 2 dA, where R  [0, 1]  [0, 1];, , x2, , Iterated Integrals, , 27. If f and t are continuous on D, then, , R, , m  10, n  10, 22., , 冮冮 1, , 1, exy, , 冮冮 [2f(x, y)  3t(x, y)] dA  2冮冮 f(x, y) dA  3冮冮 t(x, y) dA, , dA, where R  [0, 1]  [1, 3];, , D, , 冮冮 [ f(x, y)t(x, y)] dA  c冮冮 f(x, y) dAd c冮冮 t(x, y) dAd, , 23. Use Property 4 (in Theorem 1) of the double integral to, show that if f and 冟 f 冟 are integrable over D, then, , D, , 冮冮 f(x, y) dA `  冮冮 冟 f(x, y) 冟 dA, D, , 冮冮, , 29., , D, , 2x 2, , R, , D, , xy, , y2, , cos(x 2, , y 2), , R  {(x, y) 冟 x 2, , 24. Use a geometric argument and Theorem 1 to show that if, f(x, y)  k, where k is a constant, then, 兰兰R k dA  k ⴢ area of R., , 1, , D, , dA  p, where, , y 2  1}, , 30. If f is nonnegative and continuous on, D  {(x, y) 冟 0  x  1, 0  y  1} and, E  5 1 (x, y 2 冟 0  x  1, 12  y  1 6 , then, , 25. Let R  {(x, y) 冟 0  x  1, 0  y  1}. Show that, 0  兰兰R ex cos y dA  1., , 冮冮 f(x, y) dA  冮冮 f(x, y) dA, , 26. Let R  C0, 12 D  C0, 12 D . Show that, 0  兰兰R sin(2x 3y) dA  14., , 14.2, , D, , 28. If f and t are continuous on D, then, , R, , m  10, n  20, , `, , D, , D, , E, , Iterated Integrals, Iterated Integrals Over Rectangular Regions, Just as it is difficult to find the value of an integral of a function of one variable directly, from its definition, the task is even harder in the case of double integrals. Fortunately,, as you will see, the value of a double integral can be found by evaluating two single, integrals., We begin by looking at the simple case in which f is a continuous function defined, on the rectangular region R  {(x, y) 冟 a  x  b, c  y  b} shown in Figure 1b., z, y, , z  f (x, y), , d, , R, (x, y), , S, y, c, , a, b, , R, , c, , d, y, , a, , x, , b, , x, , x, , FIGURE 1, , (a) The graph of f, , (b) The domain R of f, , If we fix x, then f(x, y) is a function of the single variable y for c  y  d. As such,, we can integrate the function with respect to y over the interval [c, d]. This operation

Page 137 :
1158, , Chapter 14 Multiple Integrals, , is called partial integration with respect to y and is the reverse of the operation of partial differentiation studied in Chapter 13. The result is the number, d, , 冮 f(x, y) dy, c, , that depends on the value of x in [a, b]. In other words, the rule, d, , 冮 f(x, y) dy, , A(x) , , axb, , (1), , c, , defines a function A of x on [a, b]. If we integrate the function A with respect to x over, [a, b], we obtain, , 冮, , b, , b, , A(x) dx , , a, , d, , 冮 冮 f(x, y) dyd dx, a, , c, , (2), , c, , The integral on the right-hand side of Equation (2) is usually written in the form, b, , d, , 冮 冮 f(x, y) dy dx, a, , (3), , c, , without the brackets and is called an iterated or repeated integral., Similarly, by holding y fixed and integrating the resulting function with respect to, x over [a, b], we obtain a function of y on the interval [c, d]. If this function is then, integrated with respect to y over [c, d], we obtain the iterated integral, d, , 冮冮, c, , b, , d, , f(x, y) dx dy , , a, , b, , 冮 冮 f(x, y) dx d dy, c, , c, , a, , Observe that when we evaluate an iterated integral, we work from the inside out., , EXAMPLE 1 Evaluate the iterated integrals:, 2, , a., , 冮冮, 1, , 1, , 1, , 3x 2y dx dy, , 2, , 冮 冮 3x y dy dx, 2, , b., , 0, , 0, , 1, , Solution, a. By definition,, 2, , 冮冮, 1, , 1, , 2, , 3x 2y dx dy , , 0, , 1, , 冮 冮 3x y dx d dy, 1, , c, , 2, , 0, , Now the integral inside the brackets is found by integrating with respect to x, while treating y as a constant. This gives, 1, , 冮 3x y dx  Cx yD, 2, , x1, x0, , 3, , y, , 0, , Therefore,, 2, , 冮冮, 1, , 0, , 1, , 2, , 3x 2y dx dy , , 冮 y dy, 1, , 2, 3, 1,  c y2d , 2, 2, 1, , (4)

Page 138 :
14.2, , Historical Biography, , b. Here, we first integrate with respect to y and then with respect to x, obtaining, 1, , 冮冮, , GUIDO FUBINI, , 0, , (1879–1943), Making contributions to analysis, group, theory, mathematical physics, and nonEuclidean spaces, Guido Fubini was one of, Italy’s most productive and eclectic mathematicians. He began his career with a doctoral thesis in differential geometry, but he, later made significant contributions in, analysis, differential geometry, mathematical physics, group theory, and even engineering. His technical skills and geometric, intuition allowed him to discover simpler, expressions of very difficult results. For, example, he revisited the expression of, surface integrals and demonstrated that, they could be written in terms of two simple integrals. Fubini taught at the University of Catania in Sicily, the University of, Genoa, the Politecnico in Turin, and the, University of Turin. In 1938 Fubini was, forced to retire from his chair position in, Turin after Benito Mussolini published the, Manifesto of Fascist Racism and the subsequent anti-Semitism policy forced all Jews, from positions in government, banking, and, education. Fubini, who was of Jewish, descent, was worried about his sons’, future in Italy. The Institute for Advanced, Study in Princeton made him an offer in, 1939, and, although Fubini and his family, did not wish to leave Italy, they decided to, emigrate to the United States to give their, sons better career opportunities. Fubini, was able to teach for a few years, but his, failing health overtook him, and he died of, heart problems in 1943., , 1159, , Iterated Integrals, , 2, , 1, , 3x 2y dy dx , , 1, , 0, , , , 冮, , 1, , 冮, , 1, , 2, , 1, , y2, 3, c x 2y 2 d, dx, 2, y1, , 0, , , , 2, , 冮 冮 3x y dyd dx, c, , 0, , 1, 9 2, 3, 3, x dx  c x 3 d , 2, 2, 2, 0, , Fubini’s Theorem for Rectangular Regions, Observe that the two iterated integrals in Example 1 are equal. Thus, the example seems, to suggest that the order of integration of the iterated integrals does not matter. To see, why this might be true for continuous functions, consider the special case in which f, is nonnegative. Let’s calculate the volume V of the solid S lying under the graph of, z  f(x, y) and above the rectangular region R  {(x, y) 冟 a  x  b, c  y  d}., Using the method of cross sections of Section 5.2, we see that, b, , V, , 冮 A(x) dx, a, , where A(x) is the area of the cross section of S in the plane perpendicular to the x-axis, at x. (See Figure 2a.) But from the figure, you can see that A(x) is the area under the, graph C of the function defined by t(y)  f(x, y) for c  y  d, where x is fixed. So, A(x) , , 冮, , d, , d, , t(y) dy , , c, , 冮 f(x, y) dy, , x fixed, , c, , Therefore,, V, , 冮, , b, , b, , d, , 冮 冮 f(x, y) dyd dx, , A(x) dx , , a, , a, , c, , c, , Similarly, using cross sections perpendicular to the y-axis (Figure 2b), you can show, that, d, , V, , b, , 冮 冮 f(x, y) dxd dy, c, , c, , a, , z, , z, , C, , z = f(x, y), , z = f(x, y), C, , S, , S, c, , c, a, b, , x, , a, , d, A(x), , y, , x, , FIGURE 2, , (a) A(x) is the area of a cross section of S in the, plane perpendicular to the x-axis., , A(y), , y, , d, y, , b, x, (b) A(y) is the area of a cross section of S in the, plane perpendicular to the y-axis.

Page 139 :
1160, , Chapter 14 Multiple Integrals, , Now, by definition,, , 冮冮 f(x, y) dA, , V, , R, , Therefore, we have shown that, b, , d, , d, , b, , 冮冮 f(x, y) dA  冮 冮 f(x, y) dy dx  冮 冮 f(x, y) dx dy, a, , R, , c, , c, , a, , This discussion suggests the following theorem, which is named after the Italian, mathematician Guido Fubini (1879–1943). Its proof lies outside the scope of this book, and will be omitted., , THEOREM 1 Fubini’s Theorem for Rectangular Regions, Let f be continuous over the rectangle R  {(x, y) 冟 a  x  b, c  y  d}. Then, , 冮冮, , b, , f(x, y) dA , , 冮冮, a, , R, , d, , d, , f(x, y) dy dx , , c, , b, , 冮 冮 f(x, y) dx dy, c, , a, , Fubini’s Theorem provides us with a practical method for finding double integrals, by expressing them in terms of iterated integrals that we can evaluate by integrating, with respect to one variable at a time. It also states that the order in which the integration is carried out does not matter, an important option, as you will see later on. Finally,, observe that Fubini’s Theorem holds for any continuous function; f(x, y) may assume, negative as well as positive values on R., , EXAMPLE 2 Evaluate 兰兰R(1  2xy 2) dA, where, R  {(x, y) 冟 0  x  2, 1  y  1}, Solution, , Using Fubini’s Theorem, we obtain, 1, , 2, , 冮冮 (1  2xy ) dA  冮 冮 (1  2xy ) dx dy, 2, , 2, , 1, , R, , , , 冮, , 1, , 冮, , 1, , , , 0, , Cx  x 2y 2 D x0 dy, x2, , 1, , 1, , (2  4y 2) dy  c2y , , 4,  a2  b  a2, 3, , 4 3 1, y d, 3, 1, , 4, 4, b, 3, 3, , We leave it for you to verify that, , 冮冮, R, , as well., , 2, , (1  2xy 2) dA , , 冮冮, 0, , 1, , 1, , (1  2xy 2) dy dx , , 4, 3

Page 140 :
14.2, , Iterated Integrals, , 1161, , EXAMPLE 3 Find the volume of the solid lying under the elliptic paraboloid z  8 , , 2x 2  y 2 and above the rectangular region R  {(x, y) 冟 0  x  1, 0  y  2}. (See, Figure 3.) Compare with Example 1 in Section 14.1., Solution, , Using Fubini’s Theorem, we see that the required volume is, V, , 冮冮, , 2, , (8  2x 2  y 2) dA , , , , 冮, , 2, , 冮, , 2, , 0, , , , 0, , 冮 冮 (8  2x, 0, , R, , c8x , a, , 1, , 2, ,  y 2) dx dy, , 0, , x1, 2 3, x  xy 2 d, dy, 3, x0, , 2, 22, 22, 1,  y 2 b dy  c y  y 3 d  12, 3, 3, 3, 0, , y, , z, , 2, , 8, z = 8 – 2x 2 – y2, , 6, 4, , R, , 2, , R, 1, , 2, , x, , 1, , y, , x, (a) The solid between the graph of z = 8 – 2x 2 – y2, and the rectangular region R, , FIGURE 3, , (b) The region R, , Iterated Integrals Over Nonrectangular Regions, Fubini’s Theorem is valid for regions that are more general than rectangular regions., More specifically, it is valid for the two types of regions that we will now describe. A, plane region R is said to be y-simple if it lies between two functions of x; that is,, R  {(x, y) 冟 a  x  b, t1(x)  y  t2(x)}, where t1 and t2 are continuous on [a, b]. (See Figure 4.), An x-simple region R is one that lies between two functions of y; that is,, R  {(x, y) 冟 c  y  d, h 1(y)  x  h 2(y)}, where h 1 and h 2 are continuous on [c, d]. (See Figure 5.), y, , y, y  g2(x), d, R, x  h1(y), , y  g1(x), , FIGURE 4, A y-simple region, , 0, , a, , R, , x  h2(y), , c, , b, , x, , FIGURE 5, An x-simple region, , 0, , x

Page 141 :
1162, , Chapter 14 Multiple Integrals, , The following theorem tells us that a double integral over a y-simple or an xsimple region can be found by evaluating an iterated integral., , THEOREM 2 Fubini’s Theorem for General Regions, Let f be continuous on a region R., 1. If R is a y-simple region, then, , 冮冮, , b, , f(x, y) dA , , 冮冮, , t2(x), , d, , h2(y), , a, , R, , f(x, y) dy dx, , t1(x), , 2. If R is an x-simple region, then, , 冮冮, , f(x, y) dA , , c, , R, , 0.0, , 0, , 1, , 0.5, , x, 1.0, 1.5, 2.0, , y, 2, , 3, , f(x, y) dx dy, , h1(y), , EXAMPLE 4 Find the volume of the solid S lying under the graph of the surface, z  x 3 4y and above the region R in the xy-plane bounded by the line y  2x and, the parabola y  x 2. (See Figure 6.), , 4, 20, z, 10, , Solution First, we make a sketch of the region R. (See Figure 7a). We see that R can, be viewed as a y-simple region; that is,, R  {(x, y) 冟 0  x  2, x 2  y  2x}, , 0, , FIGURE 6, The graph of the solid S, , 冮冮, , where t1 (x)  x 2 and t2 (x)  2x. Observe that if we integrate over a y-simple region,, we integrate with respect to y first. The appropriate limits of integration can be found, by drawing a vertical arrow as shown in Figure 7a. The arrow begins at the lower, boundary of the region described by y  t1 (x)  x 2, giving the lower limit of integration as t1 (x)  x 2, and terminates at the upper boundary of the region described by, y  t2 (x)  2x, giving the upper limit of integration as t2(x)  2x. To find the limits, for integrating with respect to x, observe that a vertical line sweeping from left to right, meets the extreme left point of R when x  0 (the lower limit of integration) and meets, the extreme right point of R when x  2 (the upper limit of integration). Using Fubini’s, Theorem for general regions, we obtain, 2, , V, , 冮冮 f(x, y) dA  冮 冮, 0, , R, , , , 冮, , 2, , 冮, , 2, , 0, , , , 0, , Cx 3y, , 2x, , (x 3, , x, , 2y 2 D yx2 dx , y2x, , 4y) dy dx, , 2, , 2, , 冮 [(2x, , 4, , 8x 2)  (x 5, , 2x 4)] dx, , 0, , 2, 8, 1, 32, (8x 2  x 5) dx  c x 3  x 6 d , 3, 6, 3, 0, , Alternative Solution, , We can view the region R as an x-simple region, R  5 (x, y) 冟 0  y  4, 2y  x  1y 6, , where h 1 (y)  y>2 and h 2 (y)  1y are obtained by solving y  2x and y  x 2 for x, in terms of y, respectively. (See Figure 7b.) If we integrate over an x-simple region, we

Page 142 :
14.2, , 1163, , y, , y, 4, , 4, , y  2x, , 3, , yx, , x  √y, , 2, , R, , 1, , x  12 y, , 3, , 2, , 2, , R, , 1, 1, , FIGURE 7, , Iterated Integrals, , x, , 2, , 1, , (a) The region R viewed as a y-simple region, , x, , 2, , (b) The region R viewed as an x-simple region, , integrate with respect to x first. A horizontal arrow starting from the left boundary of, R described by h 1(y)  y>2 and terminating at the right boundary of R described by, h 2 (y)  1y gives the lower and upper limits of integration with respect to x. The limits of integration with respect to y are found by letting a horizontal line sweep through, the region. This line meets the lowest point of R when y  0 (the lower limit of integration) and the highest point of R when y  4 (the upper limit of integration). Once, again using Fubini’s Theorem, we obtain, V, , 冮冮, 冮, , 4, , 冮, , 4, , 0, , , , 0, , 冮冮, 0, , R, , , , 1y, , 4, , f(x, y) dA , 1, c x4, 4, , (x 3, , x1y, , dy , , 4xyd, xy>2, , 7, a y 2, 4, , 4y) dx dy, , y>2, , 冮, , 4, , 0, , 4y 3>2 , , 1, c a y2, 4, , 4y 3>2 b  a, , 1 4, y, 64, , 2y 2 b d dy, , 8 5>2, 1 5 4 32, y , y d , 5, 320, 3, 0, , 1 4, 7, y b dy  c y 3, 64, 12, , as before., , EXAMPLE 5 Evaluate 兰兰R (2x  y) dA, where R is the region bounded by the parab-, , ola x  y 2 and the straight line x  y  2., , Solution The region R is shown in Figure 8. It is both y-simple and x-simple. But, observe that it is more convenient to view it as an x-simple region because the lower, boundary of R consists of two curves when viewed as a y-simple region. In fact, viewing R as a y-simple region (Figure 8a) and using Fubini’s Theorem, we find, , 冮冮, , 1, , (2x  y) dA , , 1x, , 4, , (2x  y) dy dx, , yx2, , 2, , y  √x, , 1, 2, , x, 1, y   √x, , (a) R viewed as a y-simple region, , (2x  y) dy dx, , x2, , y, , 1, 4, , 1x, , xy, , 2, , 2, , R, 3, , 冮冮, 1, , y, , 1, , FIGURE 8, , 冮冮, 0, , R, , 1x, , 2, , R, 3, , 4, , x  y2, , (b) R viewed as an x-simple region, , x

Page 143 :
1164, , Chapter 14 Multiple Integrals, 1, 0 0.5, , x, , 1, 2, , 3, 4, , 0, , y, 0.5, , On the other hand, viewing R as an x-simple region (Figure 8b), we have, 1, , 1.5, , 冮冮, , 2, 6, , 2, , (2x  y) dA , , R, , 4, , , , z, 2, 0, , , , 冮 冮, 1, , 冮, , 2, , 冮, , 2, , y 2, , y2, , (2x  y) dx dy, , Cx 2  xyD xy2 dy , xy 2, , 1, , (4, , 冮, , 2, , 1, , 5 C (y, , 2)2  y(y, , y 3  y 4) dy  c4y, , 2y, , 1, , y2, , 2) D  C y 4  y 3 D 6 dy, , 1 4 1 5 2, 243, y  y d , 4, 5, 20, 1, , which is easier to evaluate., The double integral 兰兰R (2x  y) dA gives the volume of the solid S shown in Figure 9., , FIGURE 9, The solid S, , Example 5 shows that it is sometimes easier to integrate in one order rather than, the other because of the shape of R. In certain instances the nature of the function dictates the order of integration, as the next example shows., 1, , EXAMPLE 6 Evaluate, , 冮冮, 0, , Solution, , 1, , y, , sin x, dx dy., x, , Because, , 冮, , sin x, dx, x, , cannot be expressed in terms of elementary functions, the given integral cannot be evaluated as it stands. So let’s attempt to evaluate it by reversing the order of integration., We begin by using Fubini’s Theorem to express the iterated integral as a double integral. The order of integration of the given integral suggests that, 1, , 冮冮, 0, , 1, , y, , sin x, dx dy , x, , 冮冮, , sin x, dA, x, , R, , where R  {(x, y) 冟 0  y  1, y  x  1} is viewed as an x-simple region (see Figure 10a)., , x1, , y, , xy, , 1, , R, , yx, , 1, , R, , 1, , FIGURE 10, , x1, , y, , x, , (a) R viewed as an x-simple region, , 1, (b) R viewed as a y-simple region, , x

Page 144 :
14.2, , 0.0, , that, 1, , 冮冮, , 0.8, , 0, , 0.6, 0.4, , y, , 1, , sin x, dx dy , x, , sin x, dA, x, , 1, , , , 冮冮, 0, , 0.0, , , , 0.2, 0.6 0.4, 1.0 0.8, x, , 冮, , 1, , The double integral, , 冮冮, , 冮冮, 0, , y, , 1, , x, , 0, , 1, , sin x, dy dx , x, , 冮, , 0, , 1, , c, , y sin x yx, d, dx, x, y0, , sin x dx  Ccos xD 0  cos 1, 1, , 0, , FIGURE 11, The solid S represented by the, 1 1, sin x, dx dy, double integral, x, 0 y, , 1 ⬇ 0.46, , sin x, dx dy gives the volume of the solid S shown in Figure 11., x, , CONCEPT QUESTIONS, , 1. Suppose that f is continuous on the rectangular region, R  [a, b]  [c, d]., a. Explain the difference between the iterated integrals, d, , 冮 c 冮 f(x, y) dyd dx, a, , 冮冮, R, , z, , 0.2, , b, , d, , 2. a. What is a y-simple region, and what is an x-simple, region?, b. Express 兰兰R f(x, y) dA as an iterated integral if R is a, y-simple region. As an x-simple region., c. Explain why it is sometimes advantageous to reverse the, order of integration of an iterated integral., , b, , 冮 c 冮 f(x, y) dxd dy, , and, , c, , c, , a, , b. Give a geometric interpretation of each of the iterated, integrals in part (a), where f is nonnegative., c. What does Fubini’s Theorem say about the two iterated, integrals in part (a)?, , 14.2, , EXERCISES, In Exercises 13–32, evaluate the double integral., , In Exercises 1–12, evaluate the iterated integral., 1, , 1., , 2, , 冮冮, 0, , 4, , 冮冮, 0, , 冮冮1, 0, p>2, , 6., , 冮 冮, 0, , 2xy dy dx, , 8., , 21y2, , 1, , 10., , 冮冮, , 1 x, , 2x, y, , dy dx, , 12., , 13., , ex sin y dx dy, , 14., , 15., , p, , ecos x, , 2, , 2xy 3) dA, where, , 冮冮 (x cos y, , y sin x) dA, where, , R  5 (x, y) 冟 0  x  p2 , 0  y  p4 6, R, , (x, , e2x, , 冮冮 (3x, R, , y) dy dx, , 0, , 0, , y 2) dA, where, , R  {(x, y) 冟 1  x  2, 0  y  2}, , 2xy dy dx, , 冮冮, , 冮冮, , 冮冮 (x, , R  {(x, y) 冟 0  x  1, 1  y  2}, , dy dx, , ln 2, , 21x2, , 0, , ex, , xy, , 0, , 1, , 0, , y) dx dy, , R, , x, , 冮 冮, 0, , x dx dy, , 2, , 0, 11x, , 1>2, , 0, , 1, , 11., , y) dy dx, , 冮 冮 (3x, 0, , 0, 1x, , 冮冮, 0, , 9., , 4., , p, , cos(x, , 3, , 1 0, 1 1, , 1, , p, , 7., , 2., , 4, , 冮 冮 y2x dy dx, 0, , 5., , 2y) dy dx, , 0, , 2, , 3., , 1, , 冮 冮 (x, 0, , 1165, , Viewing R as a y-simple region (Figure 10b), we find, again by Fubini’s Theorem,, , y, 0.5, 1.0, 1.0, , 14.2, , Iterated Integrals, , ln y, dy dx, y, , 16., , V Videos for selected exercises are available online at www.academic.cengage.com/login., , 冮冮 ye, R, , xy, , dA, where R  {(x, y) 冟 0  x  1, 0  y  1}

Page 145 :
1166, , 17., , Chapter 14 Multiple Integrals, , 冮冮 (x, , 2y) dA, where R  {(x, y) 冟 0  x  1, 0  y  x}, , 31., , R, , 18., , 冮冮 21  x, 冮冮 (x, , x, , dA, where R is the triangular region with vertices, , R, , 2, , (0, 0), (4, 4), and (6, 0), , dA, where R  {(x, y) 冟 0  x  1, 0  y  x}, 32., , R, , 19., , 冮冮 ye, , 冮冮 y dA, where R is the half-disk defined by the inequalities, R, , 3, , 2y) dA, where, , y 2  1 and y  0, , x2, , R, , R  {(x, y) 冟 0  x  2, x 2  y  2x}, 20., , In Exercises 33–38, find the volume of the solid shown in the, figure., , 冮冮 xy dA, where, , 33., , R  {(x, y) 冟 1  x  2, x 2  y  1, 21., , 冮冮 (1, , 2x, , 34., , z, , R, , z, z=4–x+, , 6, , x 2}, , z=6–y, , 2y) dA, where, , _1_ y, 2, , 4, , R, , R  {(x, y) 冟 0  y  1, y  x  2y}, 22., , 冮冮 (x, , 2, , y 2) dA, where, , R  {(x, y) 冟 0  y  1, y  1  x  y  1}, 23., , 36., , z, , z, 4, , R, , 4, , 冮冮 xy dA, where R  {(x, y) 冟 1  y  e, y  x  y }, 1, , y, , x, , 35., , 冮冮 x cos y dA, where, , 4, , 3, , x, , R  5 (x, y) 冟 0  y  p2 , 0  x  sin y 6, 24., , y, , 4, , 3, , R, , 2, , z = 4 – x 2 – y2, , z = 4 – x 2 – y2, , 2, , R, , 25., , 冮冮 x y dA, where R is the region bounded by the graphs of, 2, , 2, , R, , y  x, y  2x, x  1, and x  2, 26., , y, 1, , x, , 冮冮 xy dA, where R is the region bounded by the graphs of, R, , y  x , y  1, and x  0, , y, x + 2y = 2, , 2, x, , 37., , 3, , 27., , 2, , 38., , z, , 冮冮 (sin x  y) dA, where R is the region bounded by the, , 4, , z, , z = 4 – x2, y=4, , 2, z = 2e –x–y, , R, , graphs of y  cos x, y  0, x  0, and x  p>2, 28., , 冮冮 (x, , 2, , y) dA, where R is the region bounded by the, , 2, , R, , graphs of y  x 2, 29., , 冮冮 4x, , 3, , 2, x  0, x  1 and y  0, , dA, where R is the region bounded by the graphs of, , R, , y  (x  1) and y  x, 2, , 30., , 冮冮 2xy, , 2, , 3, , dA, where R is the region bounded by the graphs of, , R, , x  y 2 and x  3  2y 2, , x, , 2, 4, , y, , 2, , y, , x, , In Exercises 39–46, find the volume of the solid., 39. The solid under the plane z  4  2x  y and above, the region R  {(x, y) 冟 0  x  1, 0  y  2} lying in, the xy-plane, 40. The solid under the plane z  x 2y and above the triangular region in the xy-plane bounded by the lines y  2x,, y  0, and x  2

Page 146 :
14.2, 41. The solid under the surface z  xy and above the triangular, region in the xy-plane bounded by the lines y  2x,, y  x 6, and y  0, 42. The solid under the surface z  x 2 y and above the region, in the xy-plane bounded by the parabolas y  x 2 and, y  2  x2, 43. The solid under the paraboloid z  x 2 y 2 and above the, region in the xy-plane bounded by the line y  x and the, parabola y  x 2, , Iterated Integrals, , 1167, , 62. Suppose that f(x, y) has continuous second-order partial, derivatives. Find, , 冮冮 f, , xy (x,, , y) dA, , R, , where R  {(x, y) 冟 a  x  b, c  y  d}., 63. The following figure depicts a semicircular metal plate, whose density at the point (x, y) is (1 y) slugs/ft2. What is, the mass of the plate?, y (ft), , 44. The solid under the paraboloid z  x 2 3y 2 and above the, region in the xy-plane bounded by the graphs of y  1x,, y  0, and x  4, 45. The solid bounded by the cylinder y 2 z 2  9 and the, planes x  0, y  0, z  0, and 2x y  2, y 2  4 and the, , 46. The solid bounded by the cylinder x 2, planes z  4  y and z  0, , 2, , In Exercises 47–54, sketch the region of integration for the iterated integral, and reverse the order of integration., 1, , 47., , 冮冮, , 1x, , 1, , 3, 1, y, , 0, , 49., , 1, , 48., , f(x, y) dy dx, , 0, , 冮冮, , 0, , 50., , f(x, y) dx dy, , 52., , 冮 冮, , 冮 冮, , 4y2, , f(x, y) dx dy, , 24y2, , (3>2)y3>2, , 冮 冮, 1, 1, , f(x, y) dy dx, , 2x, , 2, , y2, 5>2, , 64. Population Density The population density (number of people, per square mile) of a coastal town is described by the function, , 2, , 2, , 0, , 51., , 冮冮, , f(x, y) , , 10,000ey, 1 0.5 冟 x 冟, , y (mi), , f(x, y) dy dx, , (ocean), (0, 0), , f(x, y) dy dx, , 0, p>4, , 5, , 1, , 54., , 冮 冮, 0, , tan x, , 1, , 2, , 冮冮e, , 2, , x2, , dx dy, , 56., , 2y, 4 2, , 冮冮, , sin y 3 dy dx, , 58., , 60., , 0, , 1, , 1y 2x, p>4, , 冮冮, , 冮冮, 0, , 冮冮, 1, , 1, , ey>x dx dy, , y>2, 2 4, , 1x, 4 2, , 0, , 冮冮, 0, , 0, , 59., , 5, , x (mi), , 2, , f(x, y) dy dx, , 0, , 57., , R, , 0, , In Exercises 55–60, evaluate the integral by reversing the order, of integration., 55., , 4  y  0, , R  {(x, y) 冟 5  x  5, 2  y  0}, , f(x, y) dx dy, , y24, 32x2, , 冮冮, , 10  x  10,, , where x and y are measured in miles. Find the population, inside the rectangular area described by, , 1 x2, e ln x, , 53., , x (ft), , 2, , tan1 y, , 3, , x cos y 2 dy dx, , x2, , dx dy, sec2 x dx dy, , 61. Suppose that f(x, y)  t(x)h(y) and let, R  {(x, y) 冟 a  x  b, c  y  d}. Show that, , 冮冮, R, , f(x, y) dA  c, , 冮, , a, , b, , t(x) dxd c, , 66. Population Density The population density (number of people, per square mile) of a certain city is given by the function, f(x, y) , , 50,000 冟 xy 冟, (x, , 2, , 20)(y 2, , 36), , where the origin (0, 0) gives the location of the government, center. Find the population inside the rectangular area, described by R  {(x, y) 冟 15  x  15, 20  y  20}., , 1, , sec2 x 21, , 65. Population Density Refer to Exercise 64. Find the average population density inside the rectangular area R., , d, , 冮 h(y) dyd, c, , cas 67. a. Plot the region R bounded by the graphs of y  cos x, , and y  x 2 x and the y-axis., b. Find the x-coordinate of the point of intersection of the, graphs of y  cos x and y  x 2 x for x 0 accurate, to three decimal places., c. Estimate 兰兰R x dA.

Page 147 :
1168, , Chapter 14 Multiple Integrals, , cas 68. a. Plot the region R bounded by the graphs of y  e2x and, , In Exercises 73–78, determine whether the statement is true or, false. If it is true, explain why. If it is false, explain why or give, an example that shows it is false., , y  x21  x 2. Then find the x-coordinates of the points, of intersection of the two graphs accurate to three decimal places., b. Estimate 兰兰R x 1>3y 2>3 dA., , 73. If f is continuous on R  [a, b]  [c, d], then, , 冮冮, , cas In Exercises 69–72, use a calculator or computer to compute the, 1, , 2 3, , 0, , 2, , 冮 冮 2x, 0, , 1, , 1, , 71., , 冮冮, , 1x, , 2, , 4x2, , 0, , 72., , xy, , d, , c, , 2, , y, , 21, , 2, , x2, , c, , b, , 冮 冮 f(x, y) dxd dy, c, , c, , a, , b, , c, , a, , a, , d, , b, , 75. If f is a nonnegative continuous function on the interval, [a, b], then the area under the graph of f on [a, b] is, 兰ab C 兰0f(x) dyD dx., , y 3 dy dx, , x2, , 1, , d, , f(x, y) dyd dx , , dy dx, , exy, , 24x2, , d, , 冮 c 冮 f(x, y) dxd dy  冮 c 冮 f(x, y) dxd dy, , 0, , 冮冮, 0, , y) dy dx, , 0, , 1, , 70., , cos(x, , c, , 74. If f is continuous on R  [a, b]  [c, d], then, , 2, , 冮冮xy, , 冮 冮, a, , R, , iterated integral accurate to four decimal places., 69., , b, , f(x, y) dA , , y2, , 76. If f is continuous on R  [0, 1]  [0, 1], then, , dy dx, , 1, , y, , 1, , x, , 冮 c 冮 f(x, y) dxd dy  冮 c 冮 f(x, y) dyd dx, 0, , 2, , 77., , 0, , 0, , 0, , 1, , 冮冮, , 1, 1 1, , x cos(y 2) dx dy  0, , 0, , 78., , 冮 冮 ( 1x, 0, , 14.3, , y)cos( 1xy) dx dy  1.2, , 0, , Double Integrals in Polar Coordinates, Polar Rectangles, Some double integrals are easier to evaluate if they are expressed in terms of polar coordinates. This is especially true when the region of integration is a polar rectangle,, R  {(r, u) 冟 a  r  b, a  u  b}, ¨∫, rb, R, ra, ∫, , ¨å, , å, , (See Figure 1.) Observe that R is a part of an annular ring with inner radius r  a and, outer radius r  b. Therefore, its area is the difference between the area of the circular sector of radius b and central angle ⌬u  b  a, and the area of the circular sector of radius a and the same central angle ⌬u. Since the area of a circular sector of, radius r and central angle u is 12 r 2u, we see that the area of R is, A, , O, , FIGURE 1, A polar rectangle is bounded by, circular arcs and rays., , , , 1 2, 1, 1, b ⌬u  a 2 ⌬u  (b 2  a 2) ⌬u, 2, 2, 2, 1, (b, 2, , where ⌬r  b  a and r  12 (b, , (1), , a)(b  a) ⌬u  r ⌬r ⌬u, a) is the average radius of the polar rectangle., , Double Integrals Over Polar Rectangles, To define a double integral over a polar rectangle R, suppose f is a continuous function on R. We start by taking a regular partition, a  r0  r1  r2  p  ri1  ri  p  rm  b

Page 148 :
14.3, , Double Integrals in Polar Coordinates, , 1169, , of [a, b] into m subintervals of equal length ⌬r  (b  a)>m, and a regular partition, a  u0  u1  u2  p  uj1  uj  p  un  b, , Rij, , R, , rb, , ∫, ra, , å, O, , FIGURE 2, A polar partition of the polar region R, with m  6 and n  6, , of [a, b] into n subintervals of equal length ⌬u  (b  a)>n. Then the circles r  ri, and the rays u  uj determine a polar partition P of R into N  mn polar rectangles, R11, R12, p , Rij, p , Rmn, where Rij  {(r, u) 冟 ri1  r  ri, uj1  u  uj}, as shown, in Figure 2. Figure 3 shows a typical polar subrectangle Rij enlarged for the sake of, clarity. The center of Rij is the point (r *, i , u*, j ) , where r *, i is the average radius of, 1, Rij, and u*, is, the, average, angle, of, R, ., In, other, words,, r*, ri) and, j, ij, i  2 (ri1, 1, u*, , (u, u, ), ., Observe, that, the, center, of, R, ,, when, expressed, in, terms, of, rectanj, j, ij, 2 j1, *, *, *, gular coordinates, takes the form (r *, cos, u, ,, r, sin, u, ), ., Also,, from, Equation, (1) we, i, j, i, j, see that the area of Rij is ⌬Ai  r *, ⌬r, ⌬u, ., Therefore,, the, Riemann, sum, of, f, over, the, i, polar partition P is, m, , Rij, , ¨  ¨j, , n, , m, , (ri*, ¨*j ), , i1 j1, , i1 j1, m, , n, ,  a a t(r *, i , u*, j ) ⌬r ⌬u, , ¨  ¨j  1, , Ψ, , n, , i cos u*, j , r*, i sin u*, j ) ⌬Ai  a a f(r *, i cos u*, j , r*, i sin u*, j ) r*, i ⌬r ⌬u, a a f(r *, , i1 j1, , where t(r, u)  rf(r cos u, r sin u). But the last sum is just a Riemann sum associated, with the double integral, , FIGURE 3, A polar subrectangle Rij and, its center (r *, i , u*, j ), , b, , b, , 冮 冮 t(r, u) dr du, a, , a, , Therefore, we have, , 冮冮, , m, , f(x, y) dA  lim, , n, , i cos u*, j , r*, i sin u*, j ) ⌬A, a a f(r *, , m, n→⬁ i1 j1, , R, , m, ,  lim, , n, , i , u*, j ) ⌬r ⌬u, a a t(r *, , m, n→⬁ i1 j1, b, , b, , , , 冮冮, a, , b, , b, , t(r, u) dr du , , a, , 冮 冮 f(r cos u, r sin u) r dr du, a, , a, , Transforming a Double Integral Over a Polar Rectangle to Polar Coordinates, Let f be continuous on a polar rectangle R  {(r, u) 冟 0  a  r  b, a  u  b},, where 0  b  a  2p. Then, b, , b, , 冮冮 f(x, y) dA  冮 冮, R, , a, , f(r cos u, r sin u) r dr du, , (2), , a, , Thus, we formally transform a double integral over a polar rectangle from rectangular, to polar coordinates by substituting, , r d¨, dr, , x  r cos u,, , d¨, r, O, , FIGURE 4, The infinitesimal polar rectangle has, “area” dA  r dr du., , y  r sin u,, , dA  r dr du, , and inserting the appropriate limits., , !, , Do not forget the factor r on the right-hand side of Equation (2). You can remember the expression for dA by making a sketch of the “infinitesimal polar rectangle”, shown in Figure 4. The polar rectangle is similar to an ordinary rectangle with sides, of length r du and dr, and therefore, it has “area” dA  (r du) dr  r dr du.

Page 149 :
1170, , Chapter 14 Multiple Integrals, , EXAMPLE 1 Evaluate 兰兰R (2x, , y, , bounded by the circles x, , 2, , Solution The region R is a polar rectangle that can also be described in terms of polar, coordinates by, , R, , 0, , 3y) dA, where R is the region in the first quadrant, y  1 and x 2 y 2  4., , 2, , 1, , 2, , R  5 (r, u) 冟 1  r  2, 0  u  p2 6, , x, , (See Figure 5.) Using Equation (2), we obtain, , 冮冮, , 3y) dA , , 冮 冮 (2r cos u, 0, , R, , FIGURE 5, The region, R  5 (r, u) 冟 1  r  2, 0  u  p2 6, , 2, , p>2, , (2x, , 2, , p>2, , , , 冮 冮, 0, , , , 冮, , p>2, , 冮, , p>2, , 0, , c, z, , (2r 2 cos u, , 3r 2 sin u) dr du, , 1, , 2, c r 3 cos u, 3, , 0, , , , 3r sin u) r dr du, , 1, , a, , 14, cos u, 3, , r2, , r 3 sin ud, , du, r1, , 7 sin ub du, , p>2, 14, 35, sin u  7 cos ud, , 3, 3, 0, , EXAMPLE 2 Find the volume of the solid S that lies below the hemisphere, z  29  x 2  y 2, above the xy-plane, and inside the cylinder x 2 y 2  1., , 3, , Solution The solid S is shown in Figure 6. It lies between the hemisphere, z  29  x 2  y 2 and the circular disk centered at the origin with radius 1., A polar representation of R is, , S, 0, , 1, , 3, , 3, x, , FIGURE 6, The solid S lies above the disk, x 2 y 2  1 and under the, hemisphere z  29  x 2  y 2., , y, , R  {(r, u) 冟 0  r  1, 0  u  2p}, Also, in polar coordinates we can write z  29  x 2  y 2  29  r 2. Therefore,, the required volume is given by, V, , 冮冮, , 2p, , f(x, y) dA , , , , 冮, , 0, , 2p, , , , 冮 冮 29  r, 0, , R, , 1, , 2, , r dr du, , 0, , r1, 1, c (9  r 2)3>2 d, du, 3, r0, , 1, (27  1612), 3, , 冮, , 0, , 2p, , du , , 2p, (27  1612), 3, , or approximately 9.16., Note You can appreciate the role played by polar coordinates in Example 2 by observing that in rectangular coordinates,, 1, , V, , 冮冮, , which is not easy to evaluate., , 21y2, , 1 21y2, , 29  x 2  y 2 dx dy

Page 150 :
14.3, , 1171, , Double Integrals in Polar Coordinates, , Double Integrals Over General Regions, R, , rb, , ∫, å, , ra, , O, , FIGURE 7, An inner polar partition of the region R, , The results obtained thus far can be extended to more general regions. If the bounded, region R is such a region, then we can transform the double integral 兰兰R f(x, y) dA into, one involving polar coordinates by expressing it as a limit of Riemann sums associated with the function, f(x, y) if (x, y) is in R, fR (x, y)  e, 0, if (x, y) is outside R, (See Figure 7.), We will not pursue the details. Instead, we will state the result for the type of region, that occurs most frequently in practice: A region is r-simple if it is bounded by the, graphs of two functions of u. The r-simple region described by, R  {(r, u) 冟 a  u  b, t1(u)  r  t2 (u)}, , ¨∫, , where t1 and t2 are continuous on [a, b], is shown in Figure 8., r  g2(¨), R, , ¨å, , Transforming a Double Integral Over a Polar Region to Polar Coordinates, Let f be continuous on a polar region of the form, R  {(r, u) 冟 a  u  b, t1(u)  r  t2 (u)}, , ∫, å, , r  g1(¨), , where 0  b  a  2p. Then, , O, , FIGURE 8, The polar region, R  {(r, u) 冟 a  u  b,, t1(u)  r  t2 (u)}. Observe, that r runs from the curve, r  t1(u) to the curve r  t2(u), as indicated by the arrow., , 冮冮, , t2(u), , b, , f(x, y) dA , , 冮冮, a, , R, , f(r cos u, r sin u) r dr du, , (3), , t1(u), , Note u-simple regions (regions that are bounded by the graphs of functions of r) will, be considered in Exercise 44., , EXAMPLE 3 Use a double integral to find the area enclosed by one loop of the threeleaved rose r  sin 3u., ¨ π, 3, , R, , ¨0, , Solution The graph of r  sin 3u is shown in Figure 9. Observe that a loop of the, rose is described by the region, R  5 (r, u) 冟 0  u  p3 , 0  r  sin 3u 6, , and may be viewed as being r-simple, where t1(u)  0 and t2(u)  sin 3u. Taking, f(x, y)  1 in Equation (3), we see that the required area is given by, A, , 冮冮, , , , 冮, , p>3, , 0, , , , 1, 2, , 冮 冮, 0, , R, , FIGURE 9, The region R viewed as an r-simple, region, , sin 3u, , p>3, , dA , , 冮, , r dr du, , 0, , rsin 3u, 1, c r 2d, du, 2, r0, , p>3, , sin2 3u du , , 0, , 1, 4, , 冮, , or approximately 0.26., , 1, 1, cu  sin 6ud, 4, 6, u0, , (1  cos 6u) du, , 0, , up>3, , , , p>3, , , , p, 12, , sin2 u , , 1  cos 2u, 2

Page 151 :
1172, , Chapter 14 Multiple Integrals, , EXAMPLE 4 Evaluate 兰兰R y dA, where R is the region in the first quadrant that is, outside the circle r  2 and inside the cardioid r  2(1 cos u)., Solution, , The required region, R  5 (r, u) 冟 0  u  p2 , 2  r  2(1, , is sketched in Figure 10 and may be viewed as being r-simple. Recalling that y  r sin u, and using Equation (3), we obtain, , 2, R, 0, , 2, r2, , cos u) 6, , 2, , 冮冮, , 冮 冮, 0, , R, , 4, , , , 冮, , p>2, , , , cos ¨), , FIGURE 10, The polar region, R  5 (r, u) 冟 0  u  p2 ,, 2  r  2(1 cos u) 6, , 8, 3, , 冮, , 0, , 冮 冮, 0, , r2(1, 1, c r 3 sin ud, 3, r2, , p>2, , C (1, , 2(1 cos u), , p>2, , r(sin u) r dr du , , 2, , 0, , 2, r  2(1, , 2(1 cos u), , p>2, , y dA , , r 2(sin u) dr du, , 2, , cos u), , du, , cos u)3 sin u  sin uD du, p>2, , 8 1,  c (1, 3 4, , cos u)4, , , , cos ud, 0, , 22, 3, , EXAMPLE 5 Find the volume of the solid that lies below the paraboloid, z  4  x 2  y 2, above the xy-plane, and inside the cylinder (x  1)2 y 2  1., Solution The solid S under consideration is shown in Figure 11a. It lies above the, disk R bounded by the circle with center (1, 0) and radius 1 shown in Figure 11b. This, unit circle has polar equation r  2 cos u, as you can verify by replacing x and y in, the rectangular equation of the circle by x  r cos u and y  r sin u. Therefore,, R  5 (r, u) 冟 p2  u  p2 , 0  r  2 cos u 6, and may be viewed as being r-simple, where t1 (u)  0 and t2 (u)  2 cos u. Using the, relationship x 2 y 2  r 2 and taking advantage of symmetry, we see that the required, volume is, V, , 冮冮, R, , 2, , 冮, , p>2, , 冮, , p>2, , 冮, , p>2, , 0, , 8, , 0, , 8, , 2 cos u, , p>2, , (4  x 2  y 2) dA , , 0, , 5,  8c u, 8, , c2r 2 , , 冮 冮, p>2, , c1, , cos 2u  a, , c, , 1, 1, cos 2u , 2, , 3, 4, , 0, , 1 4 r2 cos u, r d, du  8, 4, r0, 1, , p>2, , (2 cos2 u  cos4 u) du, , 0, , cos 2u 2, b d du, 2, cos 4u, d du, 8, , p>2, 1, 1, 5p, sin 2u , sin 4ud, , 4, 32, 2, 0, , or approximately 7.85., , 冮 冮, 0, , 冮, , 2 cos u, , p>2, , (4  r 2) r dr du  2, , cos2 u , , 1, , cos 2u, 2, , 0, , (4r  r 3) dr du

Page 152 :
14.3, , Double Integrals in Polar Coordinates, , (x – 1)2 + y2 = 1, y, , z, , R, , –2, R, , 2, , (a) The solid S, , (b) The region R is r-simple., , CONCEPT QUESTIONS, , 1. a. What is a polar rectangle? Illustrate with a sketch., b. Suppose f is continuous on a polar rectangle, R  {(r, u) 冟 a  r  b, a  u  b}, where, 0  b  a  2p. Write 兰兰R f(x, y) dA in terms of polar, coordinates., , 14.3, , 2. a. What is an r-simple region? Illustrate with a sketch., b. Suppose that f is continuous on a region of the form, R  {(r, u) 冟 a  u  b, t1 (u)  r  t2 (u)} where, 0  b  a  2p. Write 兰兰R f(x, y) dA in terms of polar, coordinates., , EXERCISES, , In Exercises 1–4, determine whether to use polar coordinates or, rectangular coordinates to evaluate the integral 兰兰R f(x, y) dA,, where f is a continuous function. Then write an expression for, the (iterated) integral., 2., , y, , p>2, , 7., 8., , R, , 3, , x, , 2 1 0, , 4., , y, , 9., 1, , 2, , 1, , √2, , x, , 2, , 0, , 2, , x, , 6., , 11., , 2, , 1, , 12., , 0, , 冮冮 xy dA, where R is the region in the first quadrant, 冮冮 2x, , 2, , y 2  4 and the lines x  0 and, , y 2 dA, where R is the region in the first quadrant, , R, , bounded by the circle x 2, y  13x, 13., , 4 sin u, , y2  9, , R, , 冮 冮 f(r cos u, r sin u) r dr du, 冮冮, , 2y) dA, where R is the region in the first quadrant, , bounded by the circle x 2, xy, , 4, , p, , 冮冮 (x, , bounded by the circle x 2, , In Exercises 5–8, sketch the region of integration associated with, the integral., , 0, , 冮冮 3y dA, where R is the disk of radius 2 centered at the, R, , R, , 1, , 5., , f(r cos u, r sin u) r dr du, , 0, , origin, 10., , y, 2, , R, , p, , f(r cos u, r sin u) r dr du, , 0, 1 cos u, , R, , x, , 1, , 0, , 212, , In Exercises 9–16, evaluate the integral by changing to polar, coordinates., , 1, , 0, , 冮 冮, 0, , 2, R, , 冮 冮, p>4, 2p, , y, , 2, , 3., , x, , x2 + y2 = 4, , –2, , x, , 1., , 2, , y, , 2, , 14.3, , r = 2 cos q, , or, , 2, , S, , FIGURE 11, , 1173, , f(r cos u, r sin u) r dr du, , 0, , V Videos for selected exercises are available online at www.academic.cengage.com/login., , 冮冮 x, R, , y2, 2, , y2, , y 2  4 and the lines y  0 and, , dA, where R is the annular region bounded by, , the circles x 2, , y 2  1 and x 2, , y2  2

Page 153 :
1174, , 14., , Chapter 14 Multiple Integrals, , 冮冮 sin(x, , 2, , y 2) dA, where R is the region in the first, , In Exercises 33–40, evaluate the integral by changing to polar, coordinates., , R, , quadrant bounded by the circles x 2, x 2 y2  9, 15., , y 2  1 and, , 2, , 33., , 2, , 冮冮 y dA, where R is the smaller of the two regions bounded, , 34., , 16., , 冮冮, , (x, , y 2  2x and the line y  x, , y) dA, where R is the region in the first quadrant, , R, , bounded by the circles x 2, , y 2  4 and x 2, , y 2  2y, , 36., , 冮 冮, 1, , 0, , 3, , x, , 冮冮, 1, , In Exercises 17–26, use polar coordinates to find the volume of, the solid region T., y , above the xy-plane,, 17. T lies below the paraboloid z  x, and inside the cylinder x 2 y 2  4., 2, , 19. T lies below the cone z  2x 2 y 2, above the xy-plane,, and inside the cylinder x 2 y 2  4., , 39., , 20. T lies below the cone z  2x 2 y 2, above the xy-plane,, inside the cylinder x 2 y 2  4, and outside the cylinder, x 2 y 2  1., , 40., , 22. T lies under the paraboloid z  x 2 y 2, above the xy-plane,, and inside the cylinder x 2 y 2  2y., 23. T is bounded by the paraboloid z  9  2x 2  2y 2 and the, plane z  1., 24. T is bounded by the paraboloids z  5x, z  12  x 2  y 2., , 2, , 5y and, , 25. T is below the sphere x 2, z  2x 2 y 2., , y2, , z 2  2 and above the cone, , 26. T is inside the sphere x 2, cylinder x 2 y 2  2y., , y2, , z 2  4 and inside the, , In Exercises 27–32, use a double integral to find the area of the, region R., , 1, , 29. R is bounded by the cardioid r  3  3 sin u., , 32. R is inside the circle r  3 sin u and outside the cardioid, r  1 sin u., , dx dy, , dy dx, , y2, 2 y2, , ex, , 冮冮, , dy dx, , 2, , 冮冮, , 22xx2, , 1, , 21x2, , cos(x 2, , y 2) dx dy, , 0, , x dy dx, , 22xx2, , 0, , 冮冮, 0, , y, tan1 a b dy dx, x, , 0, , In Exercises 41 and 42, write the sum of the double integrals as, a simple double integral using polar coordinates. Then evaluate, the resulting integral., 12, , 41., , 冮 冮, 1, , 42., , x, , 冮冮, , 24x2, , 2, , 冮 冮, , xy dy dx, , 12, , 0, , 0, , xy dy dx, , 0, , 24x2, , 2x, , 24x2, , 2, , 2, , 冮冮, , 2, , y dy dx, , 21x2, , 0, , 1, , 2x 2, , y 2 dy dx, , 0, , 43. a. Suppose that f is continuous on the region R bounded by, the lines y  x, y  x, and y  1. Show that, , 冮冮, , 3p>4, , f(x, y) dA , , R, , 冮 冮, , csc u, , f(r cos u, r sin u) r dr du, , 0, , p>4, , b. Use the result of part (a) to evaluate, 1, , 冮冮, 0, , y, , 2x 2, , y 2 dx dy, , y, , 44. A region is U-simple if it is bounded by the graphs of two, functions of r. A u-simple region is described by, R  {(r, u) 冟 a  r  b, t1(r)  u  t2 (r)}, , 30. R is bounded by the lemniscate r 2  4 cos 2u., 31. R is outside the circle r  a and inside the circle, r  2a sin u., , y2, , 24x2, , 27. R is bounded by the circle r  3 cos u., 28. R is bounded by one loop of the four-leaved rose r  cos 2u., , 2, , 21y2, , 0, , 1, x2, , 1, , 0, , 1, , 18. T lies below the paraboloid z  9  x 2  y 2, above the, xy-plane, and inside the cylinder x 2 y 2  1., , 21. T lies under the plane 3x 4y z  12, above the xyplane, and inside the cylinder x 2 y 2  2x., , 冮 冮, 2, , 38., , y 2) 3>2 dy dx, , 21y2, , 2x, , 0, , 2, , 37., , 2, , 2, , (x 2, , 0, , 1, , 35., , y 2 dy dx, , 29x2, , 冮冮, 0, , 2x 2, , 0, , 3, , R, , by the circle x 2, , 冮 冮, , 24x2, , where t1 and t2 are continuous on [a, b]. It can be shown, that if f is continuous on R, then, , 冮冮, R, , b, , f(x, y) dA , , 冮冮, a, , t 2(r), , t 1(r), , f(r cos u, r sin u) r du dr

Page 154 :
14.4 Applications of Double Integrals, 2, , 2, , d. Show that 兰0⬁ ex dx  lim a→⬁ 兰0a ex dx  1p>2, and, hence deduce the result I  12p., , y, , ¨  g2(r), , Use the results of Exercise 45 to evaluate the integrals in Exercises 46 and 47., , rb, R, , 46., ¨  g1(r), , ra, , x, , Use this formula to find the area of the smaller region, bounded by the spiral r u  1, the circles r  1 and r  2,, and the polar axis., , ⬁, 45. The integral I  兰⬁, ex >2 dx occurs in the study of probability and statistics. Show that I  12p by verifying the, following steps., a. Sketch the regions R1  {(x, y) 冟 x 2 y 2  a 2,, x  0, y  0}, R2  {(x, y) 冟 0  x  a, 0  y  a},, and R3  {(x, y) 冟 x 2 y 2  2a 2, x  0, y  0} on the, same plane. Observe that R1 lies inside R2 and that R2, lies inside R3., b. Show that, 2, , 冮冮 f(r, u) dA  4 1 1  e, p, , R1, , a2, , 2, , 2, , p, , R3, , 2a2, , 2, 2 y2, , c. By considering 兰兰 R2 f(x, y) dA, where f(x, y)  ex, and using the results of part (b), show that, p, 2, 1 1  ea 2  a, 4, , a, , 冮e, , x2, , 2, , dxb , , 0, , ⬁, , 2, , x 2ex dx, , 47., , 冮, , 0, , ⬁ x, , e, dx, 1x, , 48. Water Delivered by a Water Sprinkler A lawn sprinkler sprays, water in a circular pattern. It delivers water to a depth of, f(r)  0.1re0.1r ft/hr at a distance of r ft from the sprinkler., a. Find the total amount of water that is accumulated in an, hour in a circular region of radius 50 ft centered at the, sprinkler., b. What is the average amount of water that is delivered to, the region in part (a) in an hour?, Hint: The average value of f over a region, D, , 1, A(D), , 冮冮 f(x, y) dA, D, , where A(D), is the area of D., , In Exercises 49 and 50, determine whether the statement is true, or false. If it is true, explain why. If it is false, explain why or, give an example that shows it is false., 49. If R  {(r, u) 冟 a  u  b, 0  r  t(u)}, where, 0  b  a  2p and f(r cos u, r sin u)  1 for all (r, u) in, t, R, then 兰ab 兰0 (u) f(r cos u, r sin u) r dr du gives the area of R., , where f(r, u)  er , and that, , 冮冮 f(r, u) dA  4 1 1  e, , 冮, , 0, , 0, , 14.4, , 1175, , ,, , 50. If R is the triangular region, whose vertices in rectangular, coordinates are (0, 0), (1, 0), and (1, 1), then, 兰兰R f(x, y) dA  兰 0p>4 兰 0sec u f(r cos u, r sin u) r dr du, where, r and u are polar coordinates., , p, 2, 1 1  e2a 2, 4, , Applications of Double Integrals, z, , Mass of a Lamina, z = r (x, y), , 0, , T, R, , y, , x, , FIGURE 1, The mass of the plate R is numerically, equal to the volume of the solid T., , We mentioned in Section 14.1 that the mass of a thin rectangular plate R lying in the, xy-plane and having mass density r(x, y) at a point (x, y) in R is given by the volume, of the solid region T lying directly above R and bounded above by z  r(x, y). (See Figure 1.) We will now show that this is the case. In fact, we will demonstrate that the mass, of a lamina occupying a region R in the xy-plane and having mass density r(x, y) at a, point (x, y), where r is a nonnegative continuous function, is given by 兰兰R r(x, y) dA., The double integral also gives the volume of the solid region lying directly above R and, bounded above by the surface z  r(x, y). (See Figure 2.)

Page 155 :
1176, , Chapter 14 Multiple Integrals, z, z = r (x, y), , 0, T, , y, , x, , FIGURE 2, The mass of the lamina in part (a), is numerically equal to the, volume of the solid T in part (b)., , Sij, , y, , R, , (a), , (b), , Let S be a rectangle containing R, and let P  {S11, S12, p , Sij, p , Smn} be a regular partition of S. (See Figure 3.) Define, , (xij*, y*ij), , rR (x, y)  e, , R, , S, 0, , FIGURE 3, P  {S11, S12, p , Sij, p , Smn} is a, partition of S., , x, , r(x, y), 0, , if (x, y) is in R, if (x, y) is inside S but outside R, , Let (x *, ij , y *, ij ) be a point in Sij that also lies in R. If both m and n are large (so that the, dimensions of Sij are small), then the continuity of r implies that r(x, y) is approximately equal to r(x *, ij , y *, ij ) for all points (x, y) in Sij. Therefore, the mass of that piece, of R lying in Sij with area ⌬A is approximately, r(x *, ij , y *, ij ) ⌬A, , constant density ⴢ area, , Summing the masses of all such pieces gives an approximation of the mass of R:, m, , n, , ij , y *, ij ) ⌬A, a a r(x *, i1 j1, , We can expect the approximation to improve as both m and n get larger and larger., Therefore, it is reasonable to define the mass of the lamina as the limiting value of the, sums of this form. But each of these sums is just the Riemann sum of rR over S. This, leads to the following definition., , DEFINITION Mass of a Lamina, Suppose that a lamina occupies a region R in the plane and the mass density of, the lamina at a point (x, y) in R is r(x, y), where r is a continuous density function. Then the mass of the lamina is given by, m, , 冮冮 r(x, y) dA, , (1), , R, , Note We obtain other physical interpretations of the double integral 兰兰R f(x, y) dA by, letting f represent various types of densities. For example, if an electric charge is spread

Page 156 :
14.4 Applications of Double Integrals, , 1177, , over a plane surface R and the charge density (charge per unit area) at a point (x, y) in, R is s(x, y), then the total charge on the surface is given by, Q, , 冮冮 s(x, y) dA, , (2), , R, , For another example, suppose that the population density (number of people per unit, area) at a point (x, y) in a plane region R is d(x, y); then the total population in the, region is given by, N, , 冮冮 d(x, y) dA, , (3), , R, , EXAMPLE 1 Find the mass of a lamina occupying a triangular region R with vertices, (0, 0), (2, 0), and (0, 2) if its mass density at a point (x, y) in R is r(x, y)  x 2y., , y, , Solution The region R is shown in Figure 4. Viewing R as a y-simple region and using, Equation (1), we see that the required mass is given by, , 2, y2x, , m, , 1, R, 0, , 冮冮, , 2, , r(x, y) dA , , 2, , 1, , , , x, , 2, , 冮, , 0, , FIGURE 4, The region R is both x-simple and, y-simple. Here, we view it as y-simple., , 冮冮, 0, , R, , Cxy, , y 2 D y0, , y2x, , 2x, , (x, , dx , , 冮, , 2, , [x(2  x), , (2  x)2] dx, , 0, , 2, , , , 2y) dy dx, , 0, , 冮 (4  2x) dx  C4x  x D, , 2 2, 0, , 0, , 4, , EXAMPLE 2 Electric Charge Over a Region An electric charge is spread over a region, R lying in the first quadrant and inside the circle x 2 y 2  4. Find the total charge, on R if the charge density (measured in coulombs per square meter) at a point (x, y), in R is directly proportional to the square of the distance between the point and the, origin., Solution The region R is shown in Figure 5. The charge density function is given, by s(x, y)  k(x 2 y 2), where k is the constant of proportionality. Viewing R as a, y-simple region and using Equation (2), we see that the total charge on R is given by, , y, 2, x2, , Q, , y2  4, , 冮冮, R, , 冮冮, 0, , k(x 2, , y 2) dy dx, , 0, , or, changing to polar coordinates,, , R, , 2, , p>2, , Q, 0, , 24x2, , 2, , s(x, y) dA , , 2, , x, , FIGURE 5, The region R is both x-simple and, y-simple. Here, we view it as y-simple., , 冮 冮, 0, ,  4k, , 0, , 冮, , p>2, , 0, , or 2pk coulombs., , 2, , p>2, , (kr 2) r dr du  k, du  2pk, , 冮 冮, 0, , 0, , r 3 dr du  k, , 冮, , 0, , p>2, , r2, 1, c r 4d, du, 4, r0

Page 157 :
1178, , Chapter 14 Multiple Integrals, , y, xij*, , Moments and Center of Mass of a Lamina, , Sij, (xij* , y*ij), , R, y*ij, S, x, , 0, , FIGURE 6, The region R contained in a rectangle S, , We considered the moments and the center of mass of a homogeneous lamina in Section 5.7. Using double integrals, we can now find the moments and center of mass of, a lamina with variable density. Suppose that a lamina with continuous mass density, function r occupies a region R in the xy-plane. (See Figure 6.), Let S be a rectangle containing R, and let P  {S11, S12, p , Sij, p , Smn} be a regular partition of S. Choose (x *, ij , y *, ij ) to be any evaluation point in Sij. If m and n are, large, then the mass of the part of the lamina occupying the subrectangle Sij is approximately r(x *, ij , y *, ij ) ⌬A. Consequently, the moment of this part of the lamina with respect, to the x-axis is approximately, [r(x *, ij , y *, ij ) ⌬A]y *, ij, , mass ⴢ moment arm, , Adding up these mn moments and taking the limit of the resulting sum as m and n, approach infinity, we obtain the moment of the lamina with respect to the x-axis. A, similar argument gives the moment of the lamina about the y-axis. These formulas and, the formula for the center of mass of a lamina follow., , DEFINITION Moments and Center of Mass of a Lamina, Suppose that a lamina occupies a region R in the xy-plane and the mass density, of the lamina at a point (x, y) in R is r(x, y), where r is a continuous density, function. Then the moments of mass of the lamina with respect to the x- and, y-axes are, Mx , , 冮冮 yr(x, y) dA, , My , , and, , R, , 冮冮xr(x, y) dA, , (4a), , R, , Furthermore, the center of mass of the lamina is located at the point (x, y),, where, x, , My, m, , , , 1, m, , 冮冮 xr(x, y) dA, , y, , Mx, 1, , m, m, , R, , 冮冮 yr(x, y) dA, , (4b), , R, , where the mass of the lamina is given by, m, , 冮冮 r(x, y) dA, R, , y, 1, , Note If the density function r is constant on R, then the point (x, y) is also called the, centroid of the region R. (See Section 5.7.), , y1, , R, , 1, , 0, , 1 x, , FIGURE 7, The region R occupied by the, lamina viewed as being y-simple, , EXAMPLE 3 A lamina occupies a region R in the xy-plane bounded by the parabola, y  x 2 and the line y  1. (See Figure 7.) Find the center of mass of the lamina if its, mass density at a point (x, y) is directly proportional to the distance between the point, and the x-axis., Solution The mass density of the lamina is given by r(x, y)  ky, where k is the constant of proportionality. Since R is symmetric with respect to the y-axis and the density of the lamina is directly proportional to the distance from the x-axis, we see that

Page 158 :
14.4 Applications of Double Integrals, , 1179, , the center of mass is located on the y-axis. Thus, x  0. To find y, we view R as being, y-simple and first compute, m, , 冮冮, , 1, , r(x, y) dA , , R, , , , k, 2, , 冮 冮, 1, , 1, , x2, , ky dy dx  k, , 1, , y1, 1, c y2d, dx, yx2, 1 2, , 冮, , 1, , 1, k, 1, 4k, (1  x 4) dx  cx  x 5 d , 2, 5, 5, 1, 1, , 冮, , Then using Equation (4b), we obtain, y, , 1, , 冮冮 yr(x, y) dA  4k 冮 冮, , 1, m, , 5, , 1, , R, , , , , 5, 4, , 1, , 冮 冮, , 5, 12, , 1, , 冮, , 1, , x2, , y 2 dy dx , , 5, 4, , 1, , 1, , (1  x 6) dx , , 冮, , 1, , 1, , 1, , y(ky) dy dx, , x2, , y1, 1, c y3d, dx, 3, yx2, , 1, 5, 1, 5, cx  x 7 d , 12, 7, 7, 1, , Therefore, the center of mass of the lamina is located at 1 0, 57 2 ., , Moments of Inertia, The moments of mass of a lamina, M x and M y, are called the first moments of the, lamina with respect to the x- and y-axes. We can also consider the second moment or, moment of inertia of a lamina about an axis. We begin by recalling that the moment, of inertia of a particle of mass m with respect to an axis is defined to be, I  mr 2, m, r, Axis, , mass ⴢ the square of the distance of the moment arm, , To understand the physical significance of the moment of inertia of a particle, suppose that a particle of mass m rotates with constant angular velocity v about a stationary axis. (See Figure 8.) The velocity of the particle is √  rv, where r is the distance, of the particle from the axis. The kinetic energy of the particle is, 1 2 1, 1, m√  mr 2v2  Iv2, 2, 2, 2, , FIGURE 8, A particle of mass m rotating, about a stationary axis, , I  mr 2, , This tells us that the moment of inertia I of the particle with respect to the axis plays, the same role in rotational motion that the mass m of a particle plays in rectilinear, motion. Since the mass m is a measure of the inertia or resistance to rectilinear motion, (the larger m is, the greater the energy needed), we see that the moment of inertia I is, a measure of the resistance of the particle to rotational motion., To define the moment of inertia of a lamina occupying a region R in the xy-plane, and having mass density described by a continuous function r, we proceed as before, by enclosing R with a rectangle and partitioning the latter using a regular partition. The, moment of inertia of the piece of the lamina occupying the subrectangle Rij about the, x-axis is approximately [r(x ij*, y ij*) ⌬A](y ij*) 2, where (x ij*, y ij*) is a point in Rij. Taking, the limit of the sum of the second moments as m and n approach infinity, we obtain, the moment of inertia of the lamina with respect to the x-axis. In a similar manner, we obtain the moment of inertia of a lamina with respect to the y-axis., The formulas for these quantities and the formulas for the moment of inertia of a, lamina with respect to the origin (the sum of the moments with respect to x and with, respect to y) follow.

Page 159 :
1180, , Chapter 14 Multiple Integrals, , DEFINITION Moments of Inertia of a Lamina, The moment of inertia of a lamina with respect to the x-axis, the y-axis, and, the origin are, respectively, as follows:, m, , Ix  lim, , n, , 2, ij ) r(x *, ij , y *, ij ) ⌬A , a a (y *, , m, , Iy  lim, , 2, ij ) r(x *, ij , y *, ij ) ⌬A , a a (x *, , 冮冮(x, , R, , ij ), a a [(x *, , 2, (y *, ij ) ]r(x *, ij , y *, ij ) ⌬A, , 2, , (5c), , y 2)r(x, y) dA  Ix, , 2, , (5b), , n, , m, n→⬁ i1 j1, , , , 冮冮 x r(x, y) dA, 2, , m, n→⬁ i1 j1, , I0  lim, , (5a), , R, , n, , m, , 冮冮 y r(x, y) dA, 2, , m, n→⬁ i1 j1, , Iy, , R, , EXAMPLE 4 Find the moments of inertia with respect to the x-axis, the y-axis, and, the origin of a thin homogeneous disk of mass m and radius a, centered at the origin., Solution Since the disk is homogeneous, its density is constant and given by, r(x, y)  m>(pa)2. Using Equation (5a), we see that the moment of inertia of the disk, about the x-axis is given by, Ix , , 冮冮, , y 2r(x, y) dA , , R, , , , 2p, , m, pa 2, 2, , , , , ma, 4p, , 冮 冮, 0, , 冮, , 2p, , m, pa 2, , a, , 冮 冮 (r sin u), 0, , r 3 sin2 u dr du , , 0, 2, , sin2 u du , , 0, , ma, 8p, , r dr du, , 0, , a, , 2p, , 2, , 冮, , m, pa 2, , 冮, , 0, , 2p, , ra, 1, c r 4 sin2 ud, du, 4, r0, , 2p, , (1  cos 2u) du, , 0, , 2p, ma 2, 1, 1, cu  sin 2ud  ma 2, 8p, 2, 4, 0, , By symmetry we see that Iy  Ix  14 ma 2. Finally, using Equation (5c), we see that the, moment of inertia of the disk about the origin is given by, I0  Ix, , Axis, , Iy , , 1, ma 2, 4, , 1, 1, ma 2  ma 2, 4, 2, , Radius of Gyration of a Lamina, m, , R, , FIGURE 9, R is the radius of gyration of the, lamina with respect to the axis., , If we imagine that the mass of a lamina is concentrated at a point at a distance R from, the axis, then the moment of inertia of this “point mass” would be the same as the, moment of inertia of the lamina. (See Figure 9.) The distance R is called the radius, of gyration of the lamina with respect to the axis. Thus, if the mass of the lamina, is m and its moment of inertia with respect to the axis is I, then, mR 2  I

Page 160 :
14.4 Applications of Double Integrals, , 1181, , from which we see that, R, , I, Bm, , (6), , EXAMPLE 5 Find the radius of gyration of the disk of Example 4 with respect to the, y-axis., Solution Using the result of Example 4, we have Iy  14 ma 2. Therefore, using Equation (6), we see that the radius of gyration of the disk about the y-axis is, x, , Iy, Bm, , , , 1, 2, 4 ma, , B m, , , , 1, a, 2, , Note In Example 5 we have used the customary notation x for the radius of gyration, of a lamina with respect to the y-axis. The radius of gyration of a lamina with respect, to the x-axis is denoted by y., , 14.4, , CONCEPT QUESTIONS, , 1. A lamina occupies a region R in the plane. If the mass density of the lamina is r(x, y), write an integral giving (a) the, mass of the lamina, (b) the moments of mass of the lamina, with respect to the x- and y-axes, and (c) the center of mass, of the lamina., 2. A lamina occupies a region R in the plane., a. Write an integral giving the moment of inertia of the lamina with respect to the x-axis, the y-axis, and the origin., , 14.4, , b. Write an integral giving the moment of inertia of the, lamina with respect to a line L., Hint: Let d(x, y) denote the distance between a point (x, y) in R, and the line L., , 3. What is the radius of gyration of a lamina with respect to an, axis? Illustrate with a sketch., , EXERCISES, 9. R is the region bounded by the graphs of y  sin x, y  0,, x  0, and x  p; r(x, y)  y, , In Exercises 1–12, find the mass and the center of mass of the lamina occupying the region R and having the given mass density., 1. R is the rectangular region with vertices (0, 0), (3, 0), (3, 2),, and (0, 2); r(x, y)  y, , 10. R is the region in the first quadrant bounded by the circle, x 2 y 2  1; r(x, y)  x y, , 2. R is the rectangular region with vertices (0, 0) , (3, 0), (3, 1),, and (0, 1); r(x, y)  x 2 y 2, , 11. R is the region bounded by the circle r  2 cos u;, r(r, u)  r, , 3. R is the triangular region with vertices (0, 0), (2, 1), and, (4, 0); r(x, y)  x, , 12. R is the region bounded by the cardioid r  1, r(r, u)  3, , 4. R is the triangular region with vertices (1, 0), (1, 1), and, (0, 1); r(x, y)  x y, 5. R is the region bounded by the graphs of the equations, y  1x, y  0, and x  4; r(x, y)  xy, , cos u;, , 13. An electric charge is spread over a rectangular region, R  {(x, y) 冟 0  x  3, 0  y  1}. Find the total charge, on R if the charge density at a point (x, y) in R (measured in, coulombs per square meter) is s(x, y)  2x 2 y 3., , 7. R is the region bounded by the graphs of, y  ex, y  0, x  0, and x  1; r(x, y)  2xy, , 14. Electric Charge on a Disk An electric charge is spread over, the half-disk H described by x 2 y 2  4, y  0. Find, the total charge on H if the charge density at any point, (x, y) in H (measured in coulombs per square meter) is, s(x, y)  2x 2 y 2., , 8. R is the region bounded by the graphs of y  ln x, y  0,, and x  e; r(x, y)  y>x, , 15. Temperature of a Hot Plate An 8-in. hot plate is described by, the set S  {(x, y) 冟 x 2 y 2  16}. The temperature at the, , 6. R is the region bounded by the parabola y  4  x 2 and the, x-axis; r(x, y)  y, , V Videos for selected exercises are available online at www.academic.cengage.com/login.

Page 161 :
1182, , Chapter 14 Multiple Integrals, , point (x, y) is T(x, y)  400 cos(0.12x 2 y 2), measured, in degrees Fahrenheit. What is the average temperature of, the hot plate?, 16. Population Density of a City The population density (number of, people per square mile) of a certain city is, s(x, y)  3000e, , 25. A thin metal plate has the shape of the region R inside the, circle x 2 y 2  4, below the line y  x, to the right of the, line x  1, and above the x-axis. Its density is r(x, y)  y>x, for (x, y) in R. Find the mass of the plate., y, , (x2 y2), , where x and y are measured in miles. Find the population, within a 1-mi radius of the town hall, located at the, origin., , x1, , yx, , R, 2, , x, , In Exercises 17–20, find the moments of inertia Ix, Iy, and I0 and, the radii of gyration x and y for the lamina occupying the region, R and having uniform density r., 17. R is the rectangular region with vertices (0, 0), (a, 0), (a, b),, and (0, b)., 18. R is the triangular region with vertices (0, 0), (a, 0), and, (0, b)., 19. D is the half-disk H  {(x, y) 冟 x 2, , y  R , y  0}., 2, , 20. R is the region bounded by the ellipse, , 2, , x2, , y2, , a2, , b2, ,  1., , In Exercises 21–24, find the moments of inertia Ix, Iy, and I0 and, the radii of gyration x and y for the lamina., 21. The lamina of Exercise 1, 22. The lamina of Exercise 3, 23. The lamina of Exercise 5, 24. The lamina of Exercise 10, , 14.5, , 26. Find the rectangular coordinates of the centroid of the region, lying between the circles r  2 cos u and r  4 cos u., In Exercises 27–29, determine whether the statement is true or, false. If it is true, explain why. If it is false, explain why or give, an example that shows it is false., 27. A piece of metal is laminated from two thin sheets of metal, with mass density r1 (x, y) and r2 (x, y). If it occupies a, region R in the plane, then the mass of the laminate is, 兰兰R r1(x, y) dA 兰兰R r2(x, y) dA., 28. If the region occupied by a lamina is symmetric with respect, to both the x- and y-axes, then the center of mass of the, lamina must be located at the origin., 29. If a lamina occupies a region R in the plane, then its center, of mass must be located in R., , Surface Area, In Section 5.4 we saw that the area of a surface of revolution can be found by evaluating a simple integral. We now turn our attention to the problem of finding the area, of more general surfaces. More specifically, we will consider surfaces that are graphs, of functions of two variables. As you will see, the area of these surfaces can be found, by using double integrals., , Area of a Surface z ⴝ f(x, y), For simplicity we will consider the case in which f is defined in an open set containing a rectangular region R  [a, b]  [c, d]  {(x, y) 冟 a  x  b, c  y  d} and, f(x, y)  0 on R. Furthermore, we assume that f has continuous first-order partial derivatives in that region. We wish first to define what we mean by the area of the surface, S with equation z  f(x, y) (Figure 1) and then to find a formula that will enable us to, calculate this area., Let P be a regular partition of R into N  mn subrectangles R11, R12, p , Rmn. Corresponding to the subrectangle Rij, there is the part Sij of S (called a patch) that lies

Page 162 :
14.5, , Surface Area, , 1183, , z, Sij, z = f (x, y), , S, 0, , y, , FIGURE 1, The surface S is the graph of, z  f(x, y) for (x, y) in R., , Rij, , x, , R, , directly above Rij with area denoted by ⌬Sij. Since the subrectangles Rij are nonoverlapping except for their common boundaries, so are the patches Sij of S, so the area of, S is given by, m, , n, , A  a a ⌬Sij, (xi, yj, f (xi, yj)), , Tij, , a, , Sij, , (xi, yj), , (1), , i1 j1, , b, , Rij, , FIGURE 2, The tangent plane determined by a and, b approximates S well if Rij is small., , Next, let’s find an approximation of ⌬Sij. Let (x i, yj) be the corner of Rij closest to, the origin, and let (x i, yj, f(x i, yj)) be the point directly above it. If you refer to Figure 2, you can see that ⌬Sij is approximated by the area of ⌬Tij of the parallelogram Tij, that is part of the tangent plane to S at the point (x i, yj, f(x i, yj)) and lying directly above, Rij. To find a formula for ⌬Tij, let a and b be vectors that have initial point (x i, yj, f(x i, yj)), and lie along the sides of the approximating parallelogram. Now from Section 13.3, we see that the slopes of the tangent lines passing through (x i, yj, f(x i, yj)) and having, the directions of a and b are given by fx(x i, yj) and fy (x i, yj), respectively. Therefore,, a  ⌬xi, , fx (x i, yj) ⌬xk, , b  ⌬yj, , and, , fy(x i, yj) ⌬yk, , From Section 11.4 we have ⌬Tij  冟 a  b 冟. But, i, j, k, a  b  † ⌬x 0 fx (x i, yj) ⌬x †, 0 ⌬y fy (x i, yj) ⌬y,  fx(x i, yj) ⌬x ⌬yi  fy(x i, yj) ⌬x ⌬yj,  [fx (x i, yj)i  fy(x i, yj)j, , ⌬x ⌬yk, , k] ⌬A, , where ⌬A  ⌬x ⌬y is the area of Rij. Therefore,, ⌬Tij  冟 a  b 冟  2[ fx (x i, yj)]2, , [ fy (x i, yj)]2, , 1 ⌬A, , (2), , If we approximate ⌬Sij by ⌬Tij, then Equation (1) becomes, m, , n, , A ⬇ a a ⌬Tij, i1 j1, , Intuitively, we see that the approximation should get better and better as both m and n, get larger and larger. This suggests that we define, m, , A  lim, , n, , a a 2[ fx (x i, yj)], , m, n→⬁ i1 j1, , 2, , [ fy (x i, yj)]2, , 1 ⌬A, , Using the definition of the double integral, we obtain the following result, which is, stated for the general case in which R is not necessarily rectangular and f(x, y) is not, necessarily positive.

Page 163 :
1184, , Chapter 14 Multiple Integrals, , y, , Formula for Finding the Area of a Surface z ⴝ f(x, y), Let f be defined on a region R in the xy-plane and suppose that fx and fy are continuous. The area A of the surface z  f(x, y) is, , (1, 1), , 1, R, yx, , A, , 冮冮2[ f (x, y)], , 2, , [ fy (x, y)]2, , x, , 1 dA, , (3), , R, , 0, , 1, , x, , FIGURE 3, The region, R  {(x, y) 冟 0  x  y, 0  y  1}, viewed as an x-simple region, , EXAMPLE 1 Find the area of the part of the surface with equation z  2x, , y 2 that, lies directly above the triangular region R in the xy-plane with vertices (0, 0), (1, 1),, and (0, 1)., Solution The region R is shown in Figure 3. It is both a y-simple and an x-simple, region. Viewed as an x-simple region, R  {(x, y) 冟 0  x  y, 0  y  1}, Using Equation (3) with f(x, y)  2x, A, , 冮冮 2[ f (x, y)], , 2, , y 2, we see that the required area is, [ fy(x, y)]2, , x, , 1 dA, , R, , , , 冮冮, , 1, , 222, , 冮冮, , 1 dA , , (2y)2, , 0, , R, , , , 冮, , 1, , 0, , cx24y 2, , 1 2,  c ⴢ (4y 2, 8 3, , xy, , dy , , 5d, x0, , 冮, , y, , 24y 2, , 5 dx dy, , 0, , 1, , y24y 2, , 5 dy, , 0, , 1, , 5) 3>2 d , 0, , 1, (27  515), 12, , or approximately 1.32., , EXAMPLE 2 Find the surface area of the part of the paraboloid z  9  x 2  y 2 that, , lies above the plane z  5., , Solution The paraboloid is sketched in Figure 4a. The paraboloid intersects the plane, z  5 along the circle x 2 y 2  4. Therefore, the surface of interest lies directly above, the disk R  {(x, y) 冟 x 2 y 2  4} shown in Figure 4b. Using Equation (3) with, f(x, y)  9  x 2  y 2, we find the required area to be, A, , 冮冮 2[ f (x, y)], , 2, , x, , [ fy (x, y)]2, , R, , , , 冮冮 2(2x), , 2, , (2y)2, , R, , , , 冮冮 24x, R, , 2, , 4y 2, , 1 dA, , 1 dA, , 1 dA

Page 164 :
14.5, , 1185, , Surface Area, , z, , Historical Biography, Jaime Abecasis/Photo Researchers, Inc., , z = 9 – x2 – y2, , y, 2, x2 + y2 = 4, , z=5, R, –2, 3, , GASPARD MONGE, , 2, , 2, , 3, , 2, , x, , y, , (1746–1818), In 1789, at the beginning of the French Revolution, Gaspard Monge was one of the, best-known mathematicians in France. In, addition to doing theoretical work in, descriptive geometry, Monge applied his, skills to construction projects, general, architecture, and military applications., Before the revolution he was appointed, examiner of naval cadets. This position, took him away from his professorship in, Mézières, but he used his salary to pay, other people to fulfill his teaching duties., With this arrangement in place, in 1796, Monge embarked on a prolonged absence, from France. He traveled first to Italy,, where he became friendly with Napoleon, Bonaparte. Two years later, he joined Bonaparte’s expeditionary force to Egypt. While, in Egypt, Monge carried out many technical, and scientific tasks, including the establishment of the Institut d’Egypt in Cairo., Monge returned to Paris in 1799, where he, resumed his teaching duties and returned, to his research. He received numerous, awards for his work and accepted an, appointment as a senator for life during, Napoleon’s military dictatorship. However,, after the defeat of Napoleon in 1815, life, became difficult for Monge. He was, expelled from the Institut de France, and, his life was continually threatened. Monge, died in Paris in 1818. He is known today primarily for his application of the calculus to, the study of curvature of surfaces, and he, is considered the father of differential, geometry., , –2, , x, (a) The part of the paraboloid that lies above the plane z = 5, , (b) The disk R = {(x,y) | x 2 + y 2 < 4}, , FIGURE 4, , Changing to polar coordinates, we have, 2p, , A, , 冮 冮 24r, 0, , , , 2, , 1 r dr du, , 0, , 冮, , 2p, , 冮, , 2p, , 0, , , , 2, , 0, , 1 2, c ⴢ (4r 2, 8 3, c, , 1)3>2 d, , r2, , du, r0, , 1, 1, 1, (173>2  1)d du  2pa b (17217  1)  p(17 117  1), 12, 12, 6, , or approximately 36.2., , Area of Surfaces with Equations y ⴝ g(x, z) and x ⴝ h( y, z), Formulas for finding the area of surfaces that are graphs of y  t(x, z) and x  h(y, z), are developed in a similar manner., Formulas for Finding the Area of Surfaces in the Form y ⴝ g(x, z), and x ⴝ h( y, z)., Let t be defined on a region R in the xz-plane, and suppose that tx and tz are, continuous. The area A of the surface y  t(x, z) is, A, , 冮冮 2[t (x, z)], , 2, , x, , [tz(x, z)]2, , 1 dA, , (4), , R, , Let h be defined on a region R in the yz-plane, and suppose that h y and h z are, continuous. The area A of the surface x  h(y, z) is, A, , 冮冮 2[h (y, z)], , 2, , y, , [h z (y, z)]2, , 1 dA, , R, , These situations are depicted in Figure 5., , Copyright 2009 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part., , (5)

Page 165 :
1186, , Chapter 14 Multiple Integrals, z, , z, y = t(x, z), , R, R, , S, S, y, , y, , x, (a) The surface S has equation y = t(x, z) and, projection R onto the xz-plane., , FIGURE 5, , x = h(y, z), , x, , (b) The surface S has equation x = h(y, z) and, projection R onto the yz-plane., , EXAMPLE 3 Find the area of that part of the plane y, x, , 2, , z  2 inside the cylinder, , z  1., 2, , Solution The surface S of interest is sketched in Figure 6a. The projection of S onto, the xz-plane is the disk R  {(x, z) 冟 x 2 z 2  1} shown in Figure 6b. Using Equation (4) with t(x, z)  2  z, we see that the area of S is, A, , 冮冮 2[t (x, z)], , 2, , [tz(x, z)]2, , x, , 1 dA, , R, , , , 冮冮 20, , 2, , (1)2, , 1 dA  12, , R, , 冮冮 1 dA  12p, R, , upon observing that the area of R is p., z, z, 1, , y + z = 2 (y = 2 – z), , R, , R, , S, y, , x, , FIGURE 6, , 14.5, , (a) The surface S, , –1, , 1, , x, , –1, (b) The projection R of S onto the xz-plane, , CONCEPT QUESTIONS, , 1. Write an integral giving the area of the surface z  f(x, y), defined over a region R in the xy-plane., , 2. Write an integral giving the area of the surface x  f(y, z), defined over a region R in the yz-plane.

Page 166 :
14.5, , 14.5, , Surface Area, , 1187, , EXERCISES, , In Exercises 1–14, find the area of the surface S., 1. S is the part of the plane 2x 3y z  12 that lies above, the rectangular region R  {(x, y) 冟 0  x  2, 0  y  1}., 2. S is the part of the plane 3x 2y z  6 that lies above, the triangular region with vertices (0, 0), (1, 3) , and (0, 3) ., 3. S is the part of the surface z  12 x 2 y that lies above the, triangular region with vertices (0, 0), (1, 0), and (1, 1)., 4. S is the part of the surface z  2  x 2 y that lies above, the triangular region with vertices (0, 1), (1, 0), and (0, 1)., 5. S is the part of the paraboloid z  9  x 2  y 2 that lies, above the xy-plane., 6. S is the part of the paraboloid y  9  x  z that lies, between the planes y  0 and y  5., 2, , 7. S is the part of the sphere x, the plane z  2., , 2, , y, , 2, , 2, , z  9 that lies above, 2, , 8. S is the part of the hyperbolic paraboloid z  y 2  x 2 that, lies above the annular region A  {(x, y) 冟 1  x 2 y 2  4}., 9. S is the part of the surface x  yz that lies inside the cylinder y 2 z 2  16., 10. S is the part of the sphere x 2 y 2 z 2  9 that lies to the, right of the xz-plane and inside the cylinder x 2 z 2  4., 11. S is the part of the sphere x 2, the cone z 2  x 2 y 2., , y2, , z 2  8 that lies inside, , 12. S is the part of the hyperbolic paraboloid y  x 2  z 2 that, lies in the first octant and inside the cylinder x 2 z 2  4., 13. S is the part of the sphere x 2 y 2, the cylinder x 2  ax y 2  0., , z 2  a 2 that lies inside, , 14. S comprises the parts of the cylinder x 2, within the cylinder y 2 z 2  1., , z 2  1 that lie, , z, , 15. Let S be the part of the plane ax by cz  d lying in the, first octant whose projection onto the xy-plane is a region R., Prove that the area of S is (1>c) 2a 2 b 2 c2 A(R) ,, where A(R) is the area of R., 16. a. Let S be the part of the sphere x 2 y 2 z 2  a 2, that lies above the region R  {(x, y) 冟 x 2 y 2  b 2,, 0  b  a} in the xy-plane. Show that the area of S is, 2pa(a  2a 2  b 2) ., b. Use the result of part (a) to deduce that the area of a, sphere of radius a is 4pa 2., cas In Exercises 17–20, use a calculator or a computer to approxi-, , mate the area of the surface S, accurate to four decimal places., 17. S is the part of the paraboloid z  x 2 y 2 that lies above, the square region R  {(x, y) 冟 0  x  2, 0  y  2}., 18. S is the part of the paraboloid z  9  x 2  y 2 that lies, above the square region R  {(x, y) 冟 2  x  2,, 2  y  2}., 19. S is the part of the surface z  ex, cylinder x 2 y 2  4., , 2 y2, , 20. S is the part of the surface z  sin(x 2, the disk x 2 y 2  1., , that lies inside the, y 2) that lies above, , In Exercises 21–24, write a double integral that gives the surface, area of the part of the graph of f that lies above the region R., Do not evaluate the integral., 21. f(x, y)  3x 2y 2; R  {(x, y) 冟 1  x  1, 1  y  1}, 22. f(x, y)  x 2  3xy y 2; R is the triangular region with, vertices (0, 0) , (1, 1) , and (0, 1), 23. f(x, y) , , 1, 2x, , 3y, , ; R  {(x, y) 冟 0  x  2, 0  y  x}, , 24. f(x, y)  exy; R  {(x, y) 冟 0  x  1, 0  y  2}, x2 + z 2 = 1, , y2 + z 2 = 1, , y, x, , Hint: The figure shows the intersection of the two cylinders in the, , In Exercises 25 and 26, determine whether the statement is true, or false. If it is true, explain why. If it is false, explain why or, give an example that shows it is false., 25. If f(x, y)  24  x 2  y 2, then, 2, 2, 兰兰R 2f x f y 1 dA  8p, where, R  {(x, y) 冟 0  x 2 y 2  4}., 26. If z  f(x, y) is defined over a region R in the xy-plane, then, 2, 2, 兰兰R 2f x f y 1 dA  A(R), where A(R) denotes the area, of R. (Assume that fx and fy exist.), , first octant. Use symmetry., , V Videos for selected exercises are available online at www.academic.cengage.com/login.

Page 167 :
1188, , Chapter 14 Multiple Integrals, , 14.6, , Triple Integrals, Triple Integrals Over a Rectangular Box, Just as the mass of a piece of straight, thin wire of linear mass density d(x), where, a  x  b, is given by the single integral 兰ab d(x) dx, and the mass of a thin plate D of, mass density s(x, y) is given by the double integral 兰兰D s(x, y) dA, we will now see, that the mass of a solid object T with mass density r(x, y, z) is given by a triple integral., Let’s consider the simplest case in which the solid takes the form of a rectangular, box:, B  [a, b]  [c, d]  [p, q]  {(x, y, z) 冟 a  x  b, c  y  d, p  z  q}, Suppose that the mass density of the solid is r(x, y, z) g/m3, where r is a positive continuous function defined on B. Let, , z, , a  x 0  x 1  p  x i1  x i  p  x l  b, , Bijk, , c  y0  y1  p  yj1  yj  p  ym  d, p  z 0  z 1  p  z k1  z k  p  z n  q, 0, y, x, , FIGURE 1, A partition P  {Bijk} of B, , be regular partitions of the intervals [a, b], [c, d], and [p, q] of length ⌬x  (b  a)>l,, ⌬y  (d  c)>m, and ⌬z  (q  p)>n, respectively. The planes x  x i, for 1  i  l,, y  yj, for 1  j  m, and z  z k, for 1  k  n, parallel to the yz-, xz-, and xycoordinate planes divide the box B into N  lmn boxes B111, B112, p , Bijk, p , Blmn,, as shown in Figure 1. The volume of Bijk is ⌬V  ⌬x ⌬y ⌬z., Let (x *, ijk, y *, ijk, z *, ijk) be an arbitrary point in Bijk. If l, m, and n are large (so that the, dimensions of Bijk are small), then the continuity of r implies that r(x, y, z) does not, vary appreciably from r(x *, ijk, y *, ijk, z *, ijk) , whenever (x, y, z) is in Bijk. Therefore, we can, approximate the mass of Bijk by, r(x *, ijk, y *, ijk, z *, ijk) ⌬V, , constant mass density ⴢ volume, , where ⌬V  ⌬x ⌬y ⌬z. Adding up the masses of the N boxes, we see that the mass of, the box B is approximately, l, , m, , n, , ijk, y *, ijk, z *, ijk) ⌬V, a a a r(x *, , (1), , i1 j1 k1, , We expect the approximation to improve as l, m, and n get larger and larger. Therefore, it is reasonable to define the mass of the box B as, l, , lim, , m, , n, , ijk, y *, ijk, z *, ijk) ⌬V, a a a r(x *, , l, m, n→⬁ i1 j1 k1, , (2), , The expression in (1) is an example of a Riemann sum of a function of three, variables over a box and the corresponding limit in (2) is the triple integral of f over, B. More generally, we have the following definitions. Notice that no assumption, regarding the sign of f(x, y, z) is made in these definitions.

Page 168 :
14.6 Triple Integrals, , 1189, , DEFINITION Triple Integral of f Over a Rectangular Box B, Let f be a continuous function of three variables defined on a rectangular box, B, and let P  {Bijk} be a partition of B., 1. A Riemann sum of f over B with respect to the partition P is a sum of the, form, l, , m, , n, , ijk, y *, ijk, z *, ijk) ⌬V, a a a f(x *, i1 j1 k1, , * , y*, where (x ijk, ijk, z *, ijk) is a point in Bijk., 2. The triple integral of f over B is, , 冮冮冮, , l, , f(x, y, z) dV , , B, , m, , n, , ijk, y *, ijk, z *, ijk) ⌬V, a a a f(x *, , lim, , l, m, n→⬁ i1 j1 k1, , if the limit exists for all choices of (x *, ijk, y *, ijk, z *, ijk) in Bijk., , As in the case of double integrals, a triple integral may be found by evaluating an, appropriate iterated integral., , THEOREM 1, Let f be continuous on the rectangular box, B  {(x, y, z) 冟 a  x  b, c  y  d, p  z  q}, Then, q, , d, , b, , 冮冮冮 f(x, y, z) dV  冮 冮 冮 f(x, y, z) dx dy dz, p, , B, , c, , (3), , a, , The iterated integral in Equation (3) is evaluated by first integrating with respect, to x while holding y and z constant, then integrating with respect to y while holding z, constant, and finally integrating with respect to z. The triple integral in Equation (3), can also be expressed as any one of five other iterated integrals, each with a different, order of integration. For example, we can write, , 冮冮冮, , b, , f(x, y, z) dV , , B, , q, , d, , 冮 冮 冮 f(x, y, z) dy dz dx, a, , p, , c, , where the iterated integral is evaluated by successively integrating with respect to y,, z, and then x. (Remember, we work “from the inside out.”), , EXAMPLE 1 Evaluate 兰兰兰B (x 2y, , yz 2) dV, where, , B  {(x, y, z) 冟 1  x  1, 0  y  3, 1  z  2}

Page 169 :
1190, , Chapter 14 Multiple Integrals, , Solution We can express the given integral as one of six integrals. For example, if, we choose to integrate with respect to x, y, and z, in that order, then we obtain, , 冮冮冮, , 2, , (x 2y, , yz 2) dV , , 冮冮冮, 1, , B, , 3, , 冮冮, , 3, , 2, , 3, , 1, , , , 0, , 冮冮, 1, , , , 0, , 冮, , 2, , 1, , (x 2y, , 1, c x 3y, 3, 2, c y, 3, , 冮 (3, , xyz 2 d, , x1, , dy dz, x1, , 2yz 2 d dy dz, , 1, c y2, 3, , 2, , , , yz 2) dx dy dz, , 1, , 0, , 2, , , , 1, , y 2z 2 d, , y3, , dz, y0, , 9z 2) dz  C3z, , 1, , 3z 3 D 1  24, 2, , Triple Integrals Over General Bounded Regions in Space, z, , T, , Bijk, , y, x, , We can extend the definition of the triple integral to more general regions using the same, technique that we used for double integrals. Suppose that T is a bounded solid region, in space. Then it can be enclosed in a rectangular box B  [a, b]  [c, d]  [p, q]., Let P be a regular partition of B into N  lmn boxes with sides of length, ⌬x  (b  a)>l, ⌬y  (d  c)>m, ⌬z  (q  p)>n, and volume ⌬V  ⌬x ⌬y ⌬z., Thus, P  {B111, B112, p , Bijk, p , Blmn}. (See Figure 2.), Define, F(x, y, z)  e, , f(x, y, z), 0, , if (x, y, z) is in T, if (x, y, z) is in B but not in T, , Then a Riemann sum of f over T with respect to the partition P is given by, , FIGURE 2, The box Bijk is a typical element of the, partition of B., , l, , m, , n, , ijk, y *, ijk, z *, ijk) ⌬V, a a a F(x *, i1 j1 k1, , where (x *, ijk, y *, ijk, z *, ijk) is an arbitrary point in Bijk and ⌬V is the volume of Bijk. If we take, the limit of these sums as l, m, n approach infinity, we obtain the triple integral of f, over T. Thus,, , 冮冮冮, T, , l, , f(x, y, z) dV , , lim, , m, , n, , ijk, y *, ijk, z *, ijk) ⌬V, a a a F(x *, , l, m, n→⬁ i1 j1 k1, , provided that the limit exists for all choices of (x *, ijk, y *, ijk, z *, ijk) in T., Notes, 1. If f is continuous and the surface bounding T is “sufficiently smooth,” it can be, shown that f is integrable over T., 2. The properties of double integrals that are listed in Theorem 1, Section 14.1, with, the necessary modifications are also enjoyed by triple integrals., , Evaluating Triple Integrals Over General Regions, We will now restrict our attention to certain types of regions. A region T is z-simple, if it lies between the graphs of two continuous functions of x and y, that is, if, T  {(x, y, z) 冟 (x, y) 僆 R, k 1 (x, y)  z  k 2(x, y)}

Page 170 :
14.6 Triple Integrals, , 1191, , where R is the projection of T onto the xy-plane. (See Figure 3.) If f is continuous on, T, then, , 冮冮冮, , 冮冮 冮, c, , f(x, y, z) dV , , T, , R, , k2(x, y), , f(x, y, z) dzd dA, , (4), , k1(x, y), , z, z  k 2 (x,y), , T, z  k 1 (x,y), a, , FIGURE 3, A z-simple region T is bounded by the, surfaces z  k 1(x, y) and z  k 2(x, y)., , y, , y  t1 (x), , b, , R, , x, , y  t2 (x), , The iterated integral on the right-hand side of Equation (4) is evaluated by first integrating with respect to z while holding x and y constant. The resulting double integral, is then evaluated by using the method of Section 14.2. For example, if R is y-simple,, as shown in Figure 3, then, R  {(x, y) 冟 a  x  b, t1(x)  y  t2(x)}, in which case Equation (4) becomes, , 冮冮冮, , b, , f(x, y, z) dV , , 冮冮 冮, a, , T, , t2(x), , t1(x), , k2(x, y), , f(x, y, z) dz dy dx, , k1(x, y), , To determine the “limits of integration” with respect to z, notice that z runs from the, lower surface z  k 1(x, y) to the upper surface z  k 2(x, y) as indicated by the arrow, in Figure 3., , EXAMPLE 2 Evaluate 兰兰兰T z dV where T is the solid in the first octant bounded by, the graphs of z  1  x 2 and y  x., Solution The solid T is shown in Figure 4a. The solid is z-simple because it is bounded, below by the graph of z  k 1(x, y)  0 and above by z  k 2(x, y)  1  x 2., y, , z, z  1  x2, , 1, , (1, 1), , 1, y x, , T, 1, , R, , R, y, , x, , z 0, , y x, , (a) The solid T is z-simple., , FIGURE 4, , 0, , 1, , x, , (b) The projection of the solid T onto, R in the xy-plane is y-simple.

Page 171 :
1192, , Chapter 14 Multiple Integrals, , The projection of T onto the xy-plane is the set R that is sketched in Figure 4b., Regarding R as a y-simple region, we obtain, , 冮冮冮, , z dV , , T, , 冮冮 冮, c, , 1, , 冮冮, 0, , , , , , 1, 2, 1, 2, 1, 2, , 1, , x, , 0, , 0, , 0, , z dz dy dx, , 0, , 2, , 1 z1x, c z2d, dy dx, 2 z0, , 1, , x, , 冮 冮 (1  x ), , 2 2, , 0, , dy dx, , 0, , 1, , 冮 C (1  x ) yD, 2 2, , 0, , 冮, , 1x2, , x, , 冮冮冮, , z dzd dA , , zk1(x, y), , R, , , , zk2(x, y), , yx, y0, , dx, , 0, , 1, , 1, 1, 1 1, 1, x(1  x 2) 2 dx  c a b a b a b (1  x 2) 3 d , 2, 2 3, 12, 0, , There are two other simple regions besides the z-simple region just considered. An, x-simple region T is one that lies between the graphs of two continuous functions of, y and z. In other words, T may be described as, T  {(x, y, z) 冟 (y, z) 僆 R, k 1 (y, z)  x  k 2(y, z)}, where R is the projection of T onto the yz-plane. (See Figure 5.) Here, we have, , 冮冮冮, , f(x, y, z) dV , , T, , 冮冮 冮, c, , R, , k2(y, z), , f(x, y, z) dxd dA, , (5), , k1(y, z), , The (double) integral over the plane region R is evaluated by integrating with respect, to y or z first depending on whether R is y-simple or z-simple., z, , R, 0, y, x = k1(y, z), , T, , FIGURE 5, An x-simple region T is bounded by the, surfaces x  k 1 (y, z) and x  k 2(y, z)., , x, x = k2(y, z), , A y-simple region T lies between the graphs of two continuous functions of x, and z. In other words, T may be described as, T  {(x, y, z) 冟 (x, z) 僆 R, k 1 (x, z)  y  k 2(x, z)}, where R is the projection of T onto the xz-plane. (See Figure 6.) In this case we have, , 冮冮冮, T, , f(x, y, z) dV , , 冮冮 冮, c, , R, , k2(x, z), , k1(x, z), , f(x, y, z) dyd dA, , (6)

Page 172 :
14.6 Triple Integrals, , 1193, , z, , y = k2(x, z), , R, T, y = k1(x, z), , 0, , FIGURE 6, A y-simple region T is bounded by the, surfaces y  k 1 (x, z) and y  k 2(x, z)., , y, x, , Again depending on whether R is an x-simple or z-simple plane region, the double integration is carried out first with respect to x or z., , EXAMPLE 3 Evaluate 兰兰兰T 2x 2, inder x, , 2, , z 2 dV, where T is the region bounded by the cylz  2 and y  0., , z  1 and the planes y, 2, , Solution The solid T is shown in Figure 7a. Although T can be viewed as an xsimple or z-simple region, it is easier to view it as a y-simple region. (Try It!) In this, case we see that T is bounded to the left by the graph of the function y  k 1 (x, z)  0, and to the right by the graph of the function y  k 2 (x, z)  2  z. The projection of, T onto the xz-plane is the set R, which is sketched in Figure 7b. We have, , 冮冮冮, , 2x 2, , z 2 dV , , T, , 冮冮 冮, , k2(x, z), , 冮冮 冮, , 2z, , c, , R, , , , c, , R, , , , 2x 2, , 2x 2, , z 2 dyd dA, , 0, , 冮冮 c2x, , y2z, 2, , z 2 yd, , 冮冮 2x, , 2, , dA, y0, , R, , , , z 2 dyd dA, , k1(x, z), , z 2 (2  z) dA, , R, , z, , z, x2 + z2 = 1, , 1, , y=2–z, , R, , T, y, x, (a) The solid T is viewed as being y-simple., , FIGURE 7, , x2 + z2 = 1, , –1, , 1, , x, , –1, (b) The projection of the solid T onto R, in the xz-plane

Page 173 :
1194, , Chapter 14 Multiple Integrals, , Since R is a circular region, it is more convenient to use polar coordinates when integrating over R. So letting x  r cos u and z  r sin u, we have, , 冮冮, , 2p, , 2x 2, , z 2 (2  z) dA , , 1, , 冮 冮 r(2  r sin u) r dr du, 0, , R, , 0, , 2p, , , , 1, , 冮 冮 (2r, 0, , , , 冮, , 2p, ,  r 3 sin u) dr du, , 冮, , 2p, , r1, 2, 1, c r 3  r 4 sin ud, du, 3, 4, r0, , 0, , , , 2, , 0, , 0, , 2, 1, a  sin ub du, 3, 4, , 2, c u, 3, , 2p, 1, 4p, cos ud , 4, 3, 0, , Therefore,, , 冮冮冮 2x, , 2, , z 2 dV , , 4p, 3, , T, , Volume, Mass, Center of Mass, and Moments of Inertia, Before looking at other examples, let’s list some applications of triple integrals. Let, f(x, y, z)  1 for all points in a solid T. Then the triple integral of f over T gives the, volume V of T; that is,, V, , 冮冮冮 dV, , (7), , T, , We also have the following., , DEFINITIONS Mass, Center of Mass, and Moments of Inertia, for Solids in Space, Suppose that r(x, y, z) gives the mass density at the point (x, y, z) of a solid T., Then the mass m of T is, m, , 冮冮冮 r(x, y, z) dV, , (8), , T, , The moments of T about the three coordinate planes are, M yz , , 冮冮冮 xr(x, y, z) dV, , (9a), , 冮冮冮 yr(x, y, z) dV, , (9b), , 冮冮冮 zr(x, y, z) dV, , (9c), , T, , M xz , , T, , M xy , , T

Page 174 :
14.6 Triple Integrals, , 1195, , The center of mass of T is located at the point (x, y, z), where, x, , M yz, m, , y, , ,, , M xz, m, , ,, , z, , M xy, , (10), , m, , and the moments of inertia of T about the three coordinate axes are, Ix , , 冮冮冮 (y, , 2, , z 2)r(x, y, z) dV, , (11a), , 2, , z 2)r(x, y, z) dV, , (11b), , 2, , y 2)r(x, y, z) dV, , (11c), , T, , Iy , , 冮冮冮 (x, T, , Iz , , 冮冮冮 (x, T, , If the mass density is constant, then the center of mass of a solid is called the centroid of T., , EXAMPLE 4 Let T be the solid tetrahedron bounded by the plane x y z  1, and the three coordinate planes x  0, y  0, and z  0. Find the mass of T if the, mass density of T is directly proportional to the distance between a base of T and a, point on T., Solution The solid T is shown in Figure 8a. It is x-, y-, and z-simple. For example, it, can be viewed as being x-simple if you observe that it is bounded by the surface, x  k 1 (y, z)  0 and the surface x  k 2(y, z)  1  y  z. (Solve the equation, x y z  1 for x.), z, , z, , 1, x 1y z, , 1, T, , R, , x 0, , z 1y, R, , 1, , y, , x 1, (a) The solid T viewed as an x-simple region, , FIGURE 8, , 1, , y, , (b) The projection of the solid T onto the, yz-plane viewed as a z-simple region, , The projection of T onto the yz-plane is the set R shown in Figure 8b. Observe that, the upper boundary of R lies along the line that is the intersection of x y z  1, and the plane x  0 and hence has equation y z  1 or z  1  y. If we take, the base of T as the face of the tetrahedron lying on the xy-plane (actually, by symmetry, any face will do), then the mass density function for T is r(x, y, z)  kz,

Page 175 :
1196, , Chapter 14 Multiple Integrals, , where k is the constant of proportionality. Using Equation (8), we see that the required, mass is, m, , 冮冮冮 r(x, y, z) dV  冮冮冮 kz dV, T, , T, , 1, , k, , 1y, , 冮冮 冮, 0, , 0, , 冮冮, 0, , k, , 冮, 冮, , 1, , 0, , k, , 1y, , 0, , 1, , 0, , z dx dz dy, , View T as x-simple., , 0, , 1, , k, , 1yz, , CzxD x0, , x1yz, , 1, , dz dy  k, , 冮冮, 0, , 1y, , [(1  y)z  z 2] dz dy, , View R as z-simple., , 0, , 1, 1 z1y, c (1  y)z 2  z 3 d, dy, 2, 3 z0, 1, 1, 1, 1, k, (1  y)3 dy  kc a b a b (1  y)4 d , 6, 6, 4, 24, 0, , EXAMPLE 5 Let T be the solid that is bounded by the parabolic cylinder y  x 2 and, , the planes z  0 and y z  1. Find the center of mass of T, given that it has uniform density r(x, y, z)  1., Solution The solid T is shown in Figure 9a. It is x-, y-, and z-simple. Let’s choose to, view it as being z-simple. (You are also encouraged to solve the problem by viewing, T as x-simple and y-simple.) In this case we see that T lies between the xy-plane, z  k 1 (x, y)  0 and the plane z  k 2 (x, y)  1  y. The projection of T onto the, xy-plane is the region R shown in Figure 9b. As a first step toward finding the center, of mass of T, let’s find the mass of T. Using Equation (8), we have, m, , 冮冮冮 r(x, y, z) dV  冮冮冮 dV, T, , T, , 1, , , , 1, , 冮 冮 冮, 1, , x, , 1, , , , , 2, , 1y, , dz dy dx , , 0, , 1, , 冮 冮 (1  y) dy dx  冮, 1, , 冮, , 1, , 1, , 1, , 2, , 冮 冮 CzD, 1, , 1, , x2, , 1, , 1, a  x2, 1 2, , x, , cy , , 1 2 y1, y d, dx, 2, yx2, 1 5 1, 8, x d , 10, 15, 1, , z 1, , T, , 1, R, , R, z 0, , y, , x, , FIGURE 9, , dy dx, , y, y, , y x, , z0, , 1 4, 1, 1, x b dx  c x  x 3, 2, 2, 3, , z, , 2, , z1y, , (a) The solid T is viewed as a, z-simple region., , 1, , y  x2, 1, , x, , (b) The projection R of T onto the, xy-plane viewed as being y-simple

Page 176 :
14.6 Triple Integrals, , 1197, , By symmetry we see that x ⫽ 0. Next, using Equations (9b) and (10), we have, y⫽, , 冮冮冮 yr(x, y, z) dV ⫽ 8 冮冮冮 y dV, , 1, m, , 15, , T, , ⫽, , T, , 1, , 1, , 冮 冮冮, , 15, 8, , ⫺1, , x2, , 1, , ⫽, , 15, 8, , 冮 冮, , 15, 8, , 1, , ⫽, , ⫺1, , 冮, , 1⫺y, , y dz dy dx ⫽, , 0, , 1, , x2, , 1, , 冮 冮 C yzD, ⫺1, , x2, , z⫽1⫺y, z⫽0, , dy dx, , 1, , y⫽1, 1, 1, c y2 ⫺ y3d, dx, 3, y⫽x2, ⫺1 2, , 冮, , 15, 8, , (y ⫺ y 2) dy dx ⫽, , 1, , 15, 8, , 1, 1, 1, 15, a ⫺ x 4 ⫹ x 6 b dx ⫽ 2a b, 2, 3, 8, ⫺1 6, , 1, , 冮 a6 ⫺ 2 x, 1, , 1, , 4, , ⫹, , 0, , 1 6, x b dx, 3, , The integrand is an even function., , 1, , 15 1, 1 5, 1 7, 3, ⫽, c x⫺, x ⫹, x d ⫽, 4 6, 10, 21, 7, 0, Similarly, you can verify that, z⫽, , 冮冮冮 zr(x, y, z) dV ⫽ 8 冮冮冮 z dV, , 1, m, , 15, , T, , ⫽, , 15, 8, , Use Equation (9c)., , T, , 1, , 1, , ⫺1, , 2, , 冮 冮 冮, x, , 1⫺y, , 2, 7, , z dz dy dx ⫽, , 0, , Therefore, the center of mass of T is located at the point 1 0, 37, 27 2 ., z, , EXAMPLE 6 Find the moments of inertia about the three coordinate axes for the solid, rectangular parallelepiped of constant density k shown in Figure 10., , b, , Solution, , Using Equation (11a) with r(x, y, z) ⫽ k, we obtain, , 0, , c, , a, , Ix ⫽, , y, , x, , 冮冮冮 (y, , 2, , ⫹ z 2)k dV, , T, , FIGURE 10, The center of the solid is placed, at the origin., , ⫽, , c>2, , b>2, , a>2, , ⫺c>2, , ⫺b>2, , ⫺a>2, , 冮 冮 冮, , k(y 2 ⫹ z 2) dx dy dz, , Observe that the integrand is an even function of x, y, and z. Taking advantage of symmetry, we can write, c>2, , Ix ⫽ 8k, , b>2, , 冮 冮 冮, 0, , 0, , 冮 冮, 0, , ⫽ 4ka, , 冮, , c>2, , ⫽, , (y 2 ⫹ z 2) dy dz ⫽ 4ka, , a, , 冮, , c>2, , 0, , 3, , 冮 冮, 0, , b>2, , 0, , 0, , ⫽ 4kaa, , c>2, , (y 2 ⫹ z 2) dx dy dz ⫽ 8k, , 0, , c>2, , ⫽ 4ka, , a>2, , 2, , 0, , b>2, , C (y 2 ⫹ z 2)xD x⫽0 dy dz, , y⫽b>2, 1, c y 3 ⫹ z 2yd, dz, 3, y⫽0, , 3, , z⫽c>2, b, bz, b, b, ⫹, b dz ⫽ 4kaa z ⫹ z 3 b `, 24, 2, 24, 6, z⫽0, , b 3c, bc3, kabc 2, ⫹, b⫽, (b ⫹ c2), 48, 48, 12, , 1, m(b 2 ⫹ c2), 12, , m ⫽ kabc ⫽ mass of the solid, , x⫽a>2

Page 177 :
1198, , Chapter 14 Multiple Integrals, , Similarly, we find, Iy , , 14.6, , 1, m(a 2, 12, , c 2), , 14.6, , 1. f(x, y, z)  x y z; B  {(x, y, z) 冟 0  x  2, 0  y  1,, 0  z  3}. Integrate (a) with respect to x, y, and z, in that, order, and (b) with respect to z, y, and x, in that order., , 3. a. What is a z-simple region in space? An x-simple region?, A y-simple region?, b. Write the integral 兰兰兰T f(x, y, z) dV, where T is a zsimple region. An x-simple region. A y-simple region., , In Exercises 11–14, the figure shows the region of integration for, 兰兰兰T f(x, y, z) dV. Express the triple integral as an iterated integral in six different ways using different orders of integration., 11., , 3, , 3. f(x, y, z)  xy 2 yz 2; B  {(x, y, z) 冟 0  x  2,, 1  y  1, 0  z  3}. Integrate (a) with respect to z, y,, and x, in that order, and (b) with respect to x, y, and z, in, that order., , 4. f(x, y, z)  xy 2 cos z; B  5 (x, y, z) 冟 0  x  2, 0  y  3,, 0  z  p2 6 . Integrate (a) with respect to y, z, and x, in that, order, and (b) with respect to y, x, and z, in that order., , 冮冮冮, 0, , 7., , 0, , p>2, , 2, , 2, , 24z2, , 1, , 0, , 0, , 4, , 1, , x, , 0, , 0, , e, , 10., , x, , 1, , 1, , y 2z dx dz dy, 2 1y ex dz dx dy, 2, , 1>(xy), , 0, , 0, , 2 ln y dz dy dx, , 2, , 2, , y, , x2, , z2  4, y 3, , 3, , y, , x, , 14., , z, , z, z 1, , x, , y, , z 1, , x  y2, , 2xz dx dy dz, , z  x2, , 0, , 1, , 0, , 冮冮冮, , 冮冮冮, , z, , 4z  12, , 4, , 13., , y cos x dy dz dx, , 0, , 冮 冮冮, , z, , 3y, , x, , 11z, , 1, , 冮冮冮, , 6., , 0, , 冮 冮冮, 1, , 9., , 1, , x dz dy dx, , 0, , 0, , 8., , y, , 6x, , 2, , In Exercises 5–10, evaluate the iterated integral., x, , 12., , z, , 2. f(x, y, z)  xyz; B  {(x, y, z) 冟 1  x  1, 0  y  2,, 2  z  6}. Integrate (a) with respect to y, x, and z, in, that order, and (b) with respect to x, z, and y, in that order., , 5., , b 2), , EXERCISES, , In Exercises 1–4, evaluate the integral 兰兰兰B f(x, y, z) dV using, the indicated order of integration., , x, , 1, m(a 2, 12, , CONCEPT QUESTIONS, , 1. a. Define the Riemann sum of f over a rectangular box B., b. Define the triple integral of f over B., 2. Suppose that f is continuous on the rectangular box, B  [a, b]  [c, d]  [p, q]., a. Explain how you would evaluate 兰兰兰B f(x, y, z) dV., b. Write all iterated integrals that are associated with the, triple integral of part (a)., , 1, , Iz , , and, , x, , y, , y2, , y, x, , In Exercises 15–22, evaluate the integral 兰兰兰T f(x, y, z) dV., 15. f(x, y, z)  x; T is the tetrahedron bounded by the planes, x  0, y  0, z  0, and x y z  1, 16. f(x, y, z)  y; T is the region bounded by the planes x  0,, y  0, z  0, and 2x 3y z  6, 17. f(x, y, z)  2z; T is the region bounded by the cylinder, y  x 3 and the planes y  x, z  2x, and z  0, , V Videos for selected exercises are available online at www.academic.cengage.com/login.

Page 178 :
14.6 Triple Integrals, 18. f(x, y, z)  x 2y; T is the region bounded by the cylinder y  1x and the planes y  x, z  2x, and z  0, 19. f(x, y, z)  y; T is the region bounded by the paraboloid, y  x 2 z 2 and the plane y  4, 20. f(x, y, z)  z; T is the region bounded by the parabolic, cylinder y  x 2 and the planes y z  1 and z  0, 21. f(x, y, z)  z; T is the region bounded by the cylinder, x 2 z 2  4 and the planes x  2y, y  0, and z  0, 22. f(x, y, z)  2x 2 z 2; T is the region bounded by the, paraboloids y  x 2 z 2 and y  8  x 2  z 2, , 1199, , of the subrectangles Rijk (1  i, j, k  2) to estimate, 兰兰兰B f(x, y, z) dV., b. Find the exact value of 兰兰兰B f(x, y, z) dV., 40. Let f(x, y, z)  2x 2 y 2 z 2 and let, B  {(x, y, z) 冟 0  x  4, 0  y  2, 0  z  1}., a. Use a Riemann sum with m  n  p  2, and choose, the evaluation point (x *, ijk, y *, ijk, z *, ijk) to be the midpoint, of the subrectangles Rijk (1  i, j, k  2) to estimate, 兰兰兰B f(x, y, z) dV., cas b. Use a computer algebra system to estimate, 兰兰兰B f(x, y, z) dV accurate to four decimal places., , In Exercises 23–28, sketch the solid bounded by the graphs of, the equations, and then use a triple integral to find the volume of cas In Exercises 41 and 42, use a computer algebra system to estithe solid., mate the triple integral accurate to four decimal places., 23. 3x, , z  6, x  0, y  0, z  0, , 2y, , 24. y  2z,, , 1, , 41., , y  x , y  4, z  0, 2, , 25. x  4  y 2, x, , z  4, x  0, z  0, , y2, , 42., , In Exercises 31–34, sketch the solid whose volume is given by, the iterated integral., 1, , 1y, , 0, , 0, , 2, , 33., , 1xy, , 32., , 0, , 4y2, , 0, , y, , 2, , 0, , 1, , dz dx dy, , 34., , 1y, , 冮冮 冮, 0, , 冮冮 冮, 2 0, , 1, , dz dx dy, , 21, , 1, , 1x, , 0, , cos xy, xyz 2, , dx dy dz, , 1x2, , xeyz dz dy dx, , 0, , In Exercises 43–46, find the center of mass of the solid T having, the given mass density., , z2  4, , 30. Find the volume of the tetrahedron with vertices (0, 0, 0) ,, (1, 0, 0) , (1, 0, 1) , and (1, 1, 0) ., , 冮冮 冮, , 2, , 冮冮 冮, 0, , 29. Find the volume of the tetrahedron with vertices (0, 0, 0),, (1, 0, 0) , (0, 3, 0), and (0, 0, 2) ., , 31., , 0, , 1, , y 2, z  8  x 2  y 2, , z 2  4,, , 28. x 2, , 冮 冮冮, 1, , 26. z  1  x 2, y  x, y  2  x, z  0, 27. z  x 2, , 2, , 冮冮, 0, , 22z, , dx dz dy, , 0, , 11y, , 11y, , 冮, , y, , dz dx dy, , 0, , In Exercises 35–38, express the triple integral 兰兰兰T f(x, y, z) dV, as an iterated integral in six different ways using different orders, of integration., 35. T is the solid bounded by the planes x, x  0, y  0, and z  0., , 2y, , 3z  6,, , 36. T is the tetrahedron bounded by the planes z  0, x  0,, y  0, y  2  2z, and z  1  x., 37. T is the solid bounded by the circular cylinder x 2, and the planes z  0 and z  2., , y2  1, , 38. T is the solid bounded by the parabolic cylinder y  x and, the planes z  0 and z  4  y., 2, , 39. Let f(x, y, z)  x y z and let B  {(x, y, z) 冟 0  x  4,, 0  y  4, 0  z  4}., a. Use a Riemann sum with m  n  p  2, and choose, the evaluation point (x *, ijk, y *, ijk, z *, ijk) to be the midpoint, , 43. T is the tetrahedron bounded by the planes x  0, y  0,, z  0, and x y z  1. The mass density at a point P of, T is directly proportional to the distance between P and the, yz-plane., 44. T is the wedge bounded by the planes x  0, y  0, z  0,, z  23 y 2 and x  1. The mass density at a point P of, T is directly proportional to the distance between P and the, xy-plane., 45. T is the solid bounded by the cylinder y 2 z 2  4 and the, planes x  0 and x  3. The mass density at a point P of T, is directly proportional to the distance between P and the, yz-plane., 46. T is the solid bounded by the parabolic cylinder z  1  x 2, and the planes y z  1, y  0, and z  0. T has uniform, mass density r(x, y, z)  k, where k is a constant., In Exercises 47–50, set up, but do not evaluate, the iterated integral giving the mass of the solid T having mass density given by, the function r., 47. T is the solid bounded by the cylinder x 2 z 2  1 in the, first octant and the plane z y  1; r(x, y, z)  xy z 2, 48. T is the solid bounded by the ellipsoid, 36x 2 9y 2 4z 2  36 and the planes y  0, and z  0; r(x, y, z)  1yz, 49. T is the solid bounded by the parabolic cylinder z  1  y 2, and the planes 2x y  2, y  0, and z  0;, r(x, y, z)  2x 2 y 2 z 2

Page 179 :
1200, , Chapter 14 Multiple Integrals, , 50. T is the upper hemisphere bounded by the sphere, x 2 y 2 z 2  1 and the plane z  0;, r(x, y, z)  21 x 2 y 2, , 56. Find the average value of f(x, y, z)  x 2, tetrahedron bounded by the planes x y, y  0, and z  0., , 51. Let T be a cube bounded by the planes x  0, x  1, y  0,, y  1, z  0, and z  1. Find the moments of inertia of T, with respect to the coordinate axes if T has constant mass, density k., , 57. Find the average value of f(x, y, z)  xyz over the solid, region lying inside the spherical ball of radius 2 with center, at the origin and in the first octant., 58. Average Temperature in a Room A rectangular room can be, described by the set B  {(x, y, z) 冟 0  x  20,, 0  y  40, 0  z  9}. If the temperature (in degrees, Fahrenheit) at a point (x, y, z) in the room is given by, f(x, y, z)  60 0.2x 0.1y 0.2z, what is the average, temperature in the room?, , 52. Let T be a rectangular box bounded by the planes x  0,, x  a, y  0, y  b, z  0, and z  c. Find the moments of, inertia of T with respect to the coordinate axes if T has constant mass density k., 53. Let T be the solid bounded by the planes x y z  1,, x  0, y  0, and z  0. Find the moments of inertia of T, with respect to the x-, y-, and z-axes if T has mass density, given by r(x, y, z)  x., , 59. Find the region T that will make the value of, 2, 2, 2 1>3, 兰兰兰T (1  2x  3y  z ) dV as large as possible., 60. Find the values of a and b that will maximize, 2, 2, 2, 兰兰兰T (4  x  y  z ) dV, where, T  {(x, y, z) 冟 1  a  x 2 y 2 z 2  b  2}., , 54. Let T be the solid bounded by the cylinder y  x and the, planes y  x, z  0, and z  x. Find the moments of inertia, of T with respect to the coordinate axes if T has mass density given by r(x, y, z)  z., 2, , In Exercises 61–64, determine whether the statement is true or, false. If it is true, explain why. If it is false, explain why or give, an example that shows it is false., , The average value of a function f of three variables over a solid, region T is defined to be, 1, fav , V(T), , 61. If B  [1, 1]  [2, 2]  [3, 3], then, 2, y 2 z 2 dV 0., 兰兰兰B 2x, , 冮冮冮 f(x, y, z) dV, , 62. If T is a solid sphere of radius a centered at the origin, then, 兰兰兰T x dV  0., , T, , where V(T) is the volume of T. Use this definition in Exercises, 55–58., , 2, , 63. 12 , 64., , 3, , 4, , 冮 冮 冮 21, 1, , 55. Find the average value of f(x, y, z)  x y z over the, rectangular box T bounded by the planes x  0, x  1,, y  0, y  2, z  0, and z  3., , 14.7, , y 2 z 2 over the, z  1, x  0,, , 1, , x2, , y2, , z 2 dz dy dx  6130, , 1, , 冮冮冮 k dV , , 28pk, , where, 3, , T, , T  {(x, y, z) 冟 1  (x  1)2, and k is a constant, , (y  2) 2, , (z, , 1)2  4}, , Triple Integrals in Cylindrical and Spherical Coordinates, Just as some double integrals are easier to evaluate by using polar coordinates, we will, see that some triple integrals are easier to evaluate by using cylindrical or spherical, coordinates., , Cylindrical Coordinates, Let T be a z-simple region described by, T  {(x, y, z) 冟 (x, y) 僆 R, h 1 (x, y)  z  h 2(x, y)}, where R is the projection of T onto the xy-plane. (See Figure 1.) As we saw in Section, 14.6, if f is continuous on T, then, , 冮冮冮, T, , f(x, y, z) dV , , 冮冮 冮, c, , R, , h2(x, y), , h1(x, y), , f(x, y, z) dzd dA, , (1)

Page 180 :
14.7, , Triple Integrals in Cylindrical and Spherical Coordinates, , 1201, , z, z = h2(x, y), , T, , z = h1(x, y), 0, r = t1(¨), , R, , ¨=b, r = t2(¨), , x, , FIGURE 1, T viewed as a z-simple region, , ¨=a, , y, , Now suppose that the region R can be described in polar coordinates by, R  {(r, u) 冟 a  u  b, t1(u)  r  t2 (u)}, Then, since x  r cos u, y  r sin u, and z  z in cylindrical coordinates, we use Equation (2) in Section 14.3 to obtain the following formula., Triple Integral in Cylindrical Coordinates, , 冮冮冮, , b, , f(x, y, z) dV , , 冮冮 冮, a, , T, , t 2(u), , t 1(u), , h2(r cos u, r sin u), , f(r cos u, r sin u, z) r dz dr du (2), , h1(r cos u, r sin u), , Note As an aid to remembering Equation (2), observe that the element of volume in, cylindrical coordinates is dV  r dz dr du, as is suggested by Figure 2., z, r d¨, , dz, , dr, , d¨, r, , FIGURE 2, The element of volume in cylindrical, coordinates is dV  r dz dr du., , y, , x, , EXAMPLE 1 A solid T is bounded by the cone z  2x 2, , y 2 and the plane z  2., (See Figure 3.) The mass density at any point of the solid is proportional to the distance between the axis of the cone and the point. Find the mass of T., Solution, , The solid T is described by, , T  5 (x, y, z) 冟 (x, y) 僆 R, 2x 2, , y2  z  2 6

Page 181 :
1202, , Chapter 14 Multiple Integrals, , where R  {(x, y) 冟 0  x 2, , z, , y 2  4}. In cylindrical coordinates,, , T  {(r, u, z) 冟 0  u  2p, 0  r  2, r  z  2}, and, T, , R, x, , R  {(r, u) 冟 0  u  2p, 0  r  2}, , z=2, , Since the density of the solid at (x, y, z) is proportional to the distance from the z-axis, to the point in question, we see that the density function is, , y, 2, , r(x, y, z)  k2x 2, , 2, , x +y =4, , y 2  kr, , where k is the constant of proportionality. Therefore, if we use Equation (8) in Section 14.6, the mass of T is, , FIGURE 3, The arrow runs from the lower surface, z  h 1 (x, y)  2x 2 y 2 to the, upper surface z  h 2(x, y)  2 of T., , m, , 冮冮冮 r(x, y, z) dV  冮冮冮 k2x, T, , 2, , 冮 冮冮, 0, , 0, , (kr) r dz dr du, , 2, , 冮 冮 Cr zD, 2, , 0, , k, , 2, , r, , 2p, , k, , y 2 dV, , T, , 2p, , , , 2, , 0, , 2p, , 冮, , 0, , 2p, , z2, , dr du  k, , zr, , 2, , 冮 冮 (2r, 0, , 2, ,  r 3) dr du, , 0, , r2, 2, 1, 4, c r 3  r 4d, du  k, 3, 4, 3, r0, , 冮, , 2p, , du , , 0, , 8, pk, 3, , EXAMPLE 2 Find the centroid of a homogeneous solid hemisphere of radius a., z, , z = √a2 – x2 – y2, , a, , where, , T, , a, , R, , The solid T is shown in Figure 4. In rectangular coordinates we can write, T  5 (x, y, z) 冟 (x, y) 僆 R, 0  z  2a 2  x 2  y 2 6, R  {(x, y) 冟 0  x 2, , y 2  a 2}, , In cylindrical coordinates we have, , a, y, 2, , x, , Solution, , 2, , x +y =a, , FIGURE 4, A homogeneous solid hemisphere of, radius a, , 2, , and, , T  5 (r, u, z) 冟 0  u  2p, 0  r  a, 0  z  2a 2  r 2 6, R  {(r, u) 冟 0  u  2p, 0  r  a}, , By symmetry the centroid lies on the z-axis. Therefore, it suffices to find z  M xy>V,, where V, the volume of T, is 12 ⴢ 43 pa 3, or 23 pa 3. Using Equation (9c) in Section 14.6,, with r(x, y, z)  1, we obtain, 2p, , M xy , , 冮冮冮 z dV  冮 冮 冮, 0, , T, , 2p, , , , 冮 冮, 0, , , , 1, 2, , a, , 0, , 冮, , 0, , 2p, , 2a2 r2, , a, , 0, , 2, , 1 z2a, c z2d, 2 z0, , z r dz dr du, , 0, , r2, , r dr du , , 1, 2, , 2p, , a, , 冮 冮 (a, 0, , 2, ,  r 2) r dr du, , 0, , ra, , 1, 1, c a 2r 2  r 4 d, du, 2, 4, r0, , 1 1,  a a4b, 2 4, , 冮, , 0, , 2p, , du , , 1 4, 1, a (2p)  pa 4, 8, 4, , Copyright 2009 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.

Page 182 :
14.7, , Triple Integrals in Cylindrical and Spherical Coordinates, , 1203, , Therefore,, z, , M xy, V, , , , pa 4, 3, 3, ⴢ,  a, 4, 8, 2pa 3, , so the centroid is located at the point 1 0, 0, 3a8 2 ., , Spherical Coordinates, When the region of integration is bounded by portions of spheres and cones, a triple, integral is generally easier to evaluate if it is expressed in terms of spherical coordinates. Recall from Section 11.7 that the relationship between spherical coordinates, r, f, u and rectangular coordinates x, y, z is given by, x  r sin f cos u,, z, , ®, , z  r cos f, , (3), , (See Figure 5.), To see the role played by spherical coordinates in integration, let’s consider the, simplest case in which the region of integration is a spherical wedge (the analog of a, rectangular box), , P( ®, ƒ, ¨ ), or P(x, y, z), , ƒ, , T  {(r, f, u) 冟 a  r  b, c  f  d, a  u  b}, , O, ¨, , y  r sin f sin u,, , y, , x, , FIGURE 5, The point P has representation (r, f, u), in spherical coordinates and (x, y, z) in, rectangular coordinates., , where a  0, 0  d  c  p, and 0  b  a  2p. To integrate over such a region,, let, a  r0  r1  p  ri1  ri  p  rl  b, c  f0  f1  p  fj1  fj  p  fm  d, a  u0  u1  p  uk1  uk  p  un  b, be regular partitions of the intervals [a, b], [c, d], and [a, b], respectively, where, ⌬r  (b  a)>l, ⌬f  (d  c)>m and ⌬u  (b  a)>n. The concentric spheres ri,, where 1  i  l, half-cones f  fj, where 1  j  m, and the half-planes u  uk,, where 1  k  n, divide the spherical wedge T into N  lmn spherical wedges, T111, T112, p , Tlmn. A typical wedge Tijk comprising the spherical partition P  {Tijk} is, shown in Figure 6., z, , ® = ® i+1, , ƒ = ƒj, , ® = ®i, , ƒ = ƒj+1, Δƒ, , Δ®, ® i Δƒ, , FIGURE 6, A typical spherical wedge in, the partition P of the solid T, , Δ¨, x, , r i  ® i sin ƒj, , y, ¨ = ¨k+1, ® i sin ƒj D ¨, ¨ = ¨k, , If you refer to Figure 6, you will see that Tijk is approximately a rectangular box, with dimensions ⌬r, ri ⌬f (the arc of a circle with radius ri that subtends an angle of

Page 183 :
1204, , Chapter 14 Multiple Integrals, , ⌬f) and ri sin fj ⌬u (the arc of a circle with radius ri sin fj and subtending an angle, of ⌬u). Thus, its volume ⌬V is, ⌬V  r2i sin fj ⌬r ⌬f ⌬u, Therefore, an approximation to a Riemann sum of f over T is, l, , m, , n, , 2, i sin f*, i sin f*, j cos u*, k , r*, i sin f*, j sin u*, k , r*, i cos f*, j )r*, j ⌬r ⌬f ⌬u, a a a f(r*, i1 j1 k1, , But this is a Riemann sum of the function, F(r, f, u)  f(r sin f cos u, r sin f sin u, r cos f)r2 sin f, and its limit is the triple integral, d, , b, , 冮冮冮, a, , c, , b, , F(r, f, u) r2 sin f dr df du, , a, , Therefore, we have the following formula for transforming a triple integral in rectangular coordinates into one involving spherical coordinates., , Triple Integral in Spherical Coordinates, , 冮冮冮, , d, , b, , f(x, y, z) dV , , T, , 冮冮冮, a, , c, , b, , f(r sin f cos u, r sin f sin u, r cos f)r2 sin f dr df du, , (4), , a, , where T is the spherical wedge, T  {(r, f, u) 冟 a  r  b, c  f  d, a  u  b}, , Equation (4) states that to transform a triple integral in rectangular coordinates to, one in spherical coordinates, make the substitutions, x  r sin f cos u,, , y  r sin f sin u,, , z  r cos f,, , and, , x2, , y2, , z 2  r2, , then make the appropriate change in the limits of integration, and replace dV by, r2 sin f dr df du. This element of volume can be recalled with the help of Figure 7., z, , d®, ® sin ƒ d¨, ®r, ƒ, , dƒ, , Δ®, ® dƒ, , FIGURE 7, The element of volume in spherical, coordinates is dV  r2 sin f dr df du., , ¨, x, , d¨, , y

Page 184 :
14.7, z, , Triple Integrals in Cylindrical and Spherical Coordinates, , 1205, , Equation (4) can be extended to include more general regions. For example, if T is, R-simple, that is, if the region T can be described by, , ®  h 2(ƒ, ¨), , T  {(r, f, u) 冟 h 1 (f, u)  r  h 2(f, u), c  f  d, a  u  b}, then, ®  h 1(ƒ, ¨), , 冮冮冮 f(x, y, z) dV, , 0, , T, , y, , d, , b, , x, , , , FIGURE 8, A r-simple region is bounded by, the surfaces r  h 1 (f, u) and, r  h 2(f, u), , 冮冮冮, c, , a, , h2(f, u), , f(r sin f cos u, r sin f sin u, r cos f)r2 sin f dr df du, , (5), , h1(f, u), , Observe that r-simple regions are precisely those regions that lie between two surfaces, r  h 1(f, u) and r  h 2(f, u) , as shown in Figure 8. To find the limits of integration, with respect to r, we draw a radial line emanating from the origin. The line first intersects the surface, r  h 1 (f, u) , giving the lower limit of integration, and then intersects the surface r  h 2(f, u) , giving the upper limit of integration., , z, , EXAMPLE 3 Evaluate 兰兰兰T x dV, where T is the part of the region in the first octant, , T, , lying inside the sphere x 2, , y2 z2  1, ® 1, , x2, or, , T  5 (r, f, u) 冟 0  r  1, 0  f  p2 , 0  u  p2 6, , x, , Furthermore, x  r sin f cos u. Therefore, using Equation (4), we obtain, p>2, , 0, , T, , 0, , 冮 冮 冮r, , 3, , 0, , 冮 冮, 0, , ®  cos ƒ, , , , 1, 8, , p>2, , 0, , p>2, , 冮 冮, , p, 16, , 0, , sin2 f cos u dr df du, , 0, , p>2, , , , 1, , p>2, , 0, , , , (r sin f cos u)r2 sin f dr df du, , 0, , p>2, , T, , 1, , p>2, , 冮冮冮 x dV  冮 冮 冮, , , z, , z 2  1., , Solution The solid T is shown in Figure 9. Since the boundary of T is part of a sphere,, let’s use spherical coordinates. In terms of spherical coordinates we can write, , y, , FIGURE 9, T is the part of the ball, x 2 y 2 z 2  1 lying in the first, octant., , y2, , p>2, , (1  cos 2f)cos u df du , , 0, , 冮, , p>2, , p>2, , cos u du , , 0, , p>2, , r1, 1, 1, c r4 sin2 f cos ud, df du , 4, 4, r0, , p, sin u `, 16, 0, , , , 1, 8, , 冮, , 冮 冮, 0, , p>2, , sin2 f cos u df du, , 0, , 0, , p>2, , cos ucf , , fp>2, 1, sin 2fd, du, 2, f0, , p, 16, , π, __, 4, , EXAMPLE 4 Find the center of mass of the solid T of uniform density bounded by, y, x, , FIGURE 10, The solid T is bounded below, by part of a cone and above, by part of a sphere., , the cone z  2x 2, , y 2 and the sphere x 2, , y2, , z 2  z. (See Figure 10.), , Solution We first express the given equations in terms of spherical coordinates. The, equation of the cone is, r cos f  2r2 sin2 f cos2 u, , r2 sin2 f sin2u  r sin f

Page 186 :
14.7, , 14.7, , 3, , 1., , 冮 冮冮, 0, , 2p, , 2., , 0, 2, , 冮 冮冮, 0, , r dz dr du, , 0, 2r, , r dz dr du, , 0, p>2, , 2, , 冮 冮 冮r, , 2, , 0, , 0, , 2 sec f, , p>4, , 冮 冮 冮, 0, , sin f dr df du, , 0, , 2p, , 4., , r2, , 1, , 2p, , 3., , 0, , 1207, , EXERCISES, , In Exercises 1–4, sketch the solid whose volume is given by the, integral, and evaluate the integral., p>2, , Triple Integrals in Cylindrical and Spherical Coordinates, , r2 sin f dr df du, , 0, , In Exercises 5–18, solve the problem using cylindrical coordinates., 5. Evaluate 兰兰兰T 2x 2 y 2 dV, where T is the solid bounded, by the cylinder x 2 y 2  1 and the planes z  1 and, z  3., 2, , 2, , 6. Evaluate 兰兰兰T ex y dV, where T is the solid bounded by the, cylinder x 2 y 2  4 and the planes z  0 and z  4., , 17. Find the moment of inertia about the z-axis of a homogeneous solid bounded by the cone z  2x 2 y 2 and the, paraboloid z  x 2 y 2., 18. Find the moment of inertia about the z-axis of a solid, bounded by the cylinder x 2 y 2  4 and the planes z  0, and z  3 if the mass density at any point on the solid is, directly proportional to its distance from the xy-plane., In Exercises 19–24, solve the problem by using spherical, coordinates., 19. Evaluate 兰兰兰B 2x 2, x 2 y 2 z 2  1., 2, , y2, 2, , z 2 dV, where B is the unit ball, , 2 3>2, , 20. Evaluate 兰兰兰B e(x y z ) dV, where B is the part of the unit, ball x 2 y 2 z 2  1 lying in the first octant., 21. Evaluate 兰兰兰T y dV, where T is the solid bounded by the, hemisphere z  21  x 2  y 2 and the xy-plane., 22. Evaluate 兰兰兰T x 2 dV, where T is the part of the unit ball, x 2 y 2 z 2  1 lying in the first octant., , 7. Evaluate 兰兰兰T y dV, where T is the part of the solid in the, first octant lying inside the paraboloid z  4  x 2  y 2., , 23. Evaluate 兰兰兰T xz dV, where T is the solid bounded above by, the sphere x 2 y 2 z 2  4 and below by the cone, , 8. Evaluate 兰兰兰T x dV, where T is the part of the solid in the, first octant bounded by the paraboloid z  x 2 y 2 and the, plane z  4., , 24. Evaluate 兰兰兰T z dV, where T is the solid bounded above by, the sphere x 2 y 2 z 2  4 and below by the cone, , 9. Evaluate 兰兰兰T (x 2 y 2) dV, where T is the solid bounded, by the cone z  4  2x 2 y 2 and the xy-plane., 2, , 10. Evaluate 兰兰兰T y dV, where T is the solid that lies within, the cylinder x 2 y 2  1 and between the xy-plane and the, paraboloid z  2x 2 2y 2., 11. Find the volume of the solid bounded above by the sphere, x 2 y 2 z 2  9 and below by the paraboloid, 8z  x 2 y 2., 12. Find the volume of the solid bounded by the paraboloids, z  x 2 y 2 and z  12  2x 2  2y 2., 13. A solid is bounded by the cylinder x 2 y 2  4 and the, planes z  0 and z  3. Find the center of mass of the solid, if the mass density at any point is directly proportional to its, distance from the xy-plane., , z  2x 2, , z  2x 2, , y 2., , y 2., , 25. Find the volume of the solid that is bounded above by the, plane z  1 and below by the cone z  2x 2 y 2., 26. Find the volume of the solid bounded by the cone, z  2x 2 y 2, the cylinder x 2 y 2  4, and the plane, z  0., 27. Find the volume of the solid lying outside the cone, z  2x 2 y 2 and inside the upper hemisphere, x 2 y 2 z 2  1., 28. Find the volume of the solid lying above the cone f  p>6, and below the sphere r  4 cos f., 29. Find the centroid of a homogeneous solid hemisphere of, radius a., , 14. A solid is bounded by the cone z  2x 2 y 2 and the, plane z  4. Find its center of mass if the mass density at, P(x, y, z) is directly proportional to the distance between P, and the z-axis., , 30. Find the centroid of the solid of Exercise 28., , 15. Find the center of mass of a homogeneous solid bounded by, the paraboloid z  4  x 2  y 2 and z  0., , 32. Find the center of mass of the solid of Exercise 31., , 16. Find the center of mass of a homogeneous solid bounded by, the paraboloids z  x 2 y 2 and z  36  3x 2  3y 2., , 31. Find the mass of a solid hemisphere of radius a if the mass, density at any point on the solid is directly proportional to, its distance from the base of the solid., 33. Find the mass of the solid bounded by the cone, z  2x 2 y 2 and the plane z  2 if the mass density at, any point on the solid is directly proportional to the square, of its distance from the origin., , V Videos for selected exercises are available online at www.academic.cengage.com/login.

Page 187 :
1208, , Chapter 14 Multiple Integrals, , 34. Find the center of mass of the solid of Exercise 33., , 1, , 21x2, , 1, , 21x2, , 冮 冮, , 冮, , 2x2 y2, , (x 2, , y 2) 3>2 dz dy dx, , 35. Find the moment of inertia about the z-axis of the solid of, Exercise 28, assuming that it has constant mass density., , 42., , 36. Find the moment of inertia with respect to the axis of symmetry for a solid hemisphere of radius a if the density at a, point is directly proportional to its distance from the center, of the base., , In Exercises 43 and 44, evaluate the integral by using spherical, coordinates., , 冮冮, , 43., , 0, , 37. Find the moment of inertia with respect to a diameter of the, base of a homogeneous solid hemisphere of radius a., 38. Show that the average distance from the center of a circle of, radius a to other points of the circle is 2a>3 and that the, average distance from the center of a sphere of radius a to, other points of the sphere is 3a>4., 39. Let T be a uniform solid of mass m bounded by the spheres, r  a and r  b, where 0  a  b. Show that the moment, of inertia of T about a diameter of T is, 2m b  a, a, b, 5 b 3  a3, 5, , I, , 5, , 40. a. Use the result of Exercise 39 to find the moment of inertia of a uniform solid ball of mass m and radius b about, a diameter of the ball., b. Use the result of Exercise 39 to find the moment of inertia of a hollow spherical shell of mass m and radius b, about a diameter of the shell., Hint: Find lim a→b I., , , In Exercises 41 and 42, evaluate the integral by using cylindrical, coordinates., 1, , 41., , 冮 冮, 1, , 21x2, , 0, , 14.8, , 冮, , 24x2 y2, , z dz dy dx, , 21x2, , 1, , y2, , 22x2 y2, , (x 2, , 2x2, , 0, , 3, , 29x2, , 3, , 29x2, , 冮 冮, , 44., , 冮, , 2x2, , 冮, , y2, , z 2)3>2 dz dy dx, , y2, 225x2 y2, , (x 2, , y2, , z 2)1>2 dz dy dx, , 4, , 45. The temperature (in degrees Fahrenheit) at a point (x, y, z), of a solid ball of radius 3 in. centered at the origin is given, by T(x, y, z)  20(x 2 y 2 z 2). What is the average temperature of the ball?, In Exercises 46–50, determine whether the statement is true or, false. If it is true, explain why. If it is false, explain why or give, an example that shows it is false., 46. The volume of the solid bounded above by the paraboloid, z  4  x 2  y 2 and below by the xy-plane in cylindrical, 2, coordinates is 兰02p 兰02 兰04r dz dr du., 47. 兰0p>2 兰02p 兰02 r2 sin f dr du df  16p, 3, 冟, 48. If T  5 (r, f, u) a  r  b, 0  f  p2 , 0  u  p2 6 , then, p, 3, 3, 兰兰兰T dV  6 (b  a )., 49. If T is a solid with constant density k, then its moment of, inertia about the z-axis is given by Iz  k 兰兰兰T r2 sin2 f dV., 50. If T  5 (r, f, u) 冟 0  r  a, 0  f  p2 , 0  u  2p 6 ,, then 兰兰兰T r cos u dV  0., , 0, , Change of Variables in Multiple Integrals, We often use a change of variable (a substitution) when we integrate a function of one, variable to transform the given integral into one that is easier to evaluate. For example, using the substitution x  sin u, we find, , 冮, , 1, , 21  x 2 dx , , 0, , 冮, , p>2, , cos2 u du , , 0, , , , 1, 2, , 冮, , p>2, , (1, , cos 2u) du, , 0, , p, 4, , Observe that the interval of integration is [0, 1] if we integrate with respect to x, and, it changes to C0, p2 D if we integrate with respect to u. More generally, the substitution, x  t(u) [so dx  t¿(u) du] enables us to write, , 冮, , b, , a, , where a  t(c) and b  t(d)., , d, , f(x) dx , , 冮 f(t(u))t¿(u) du, c, , (1)

Page 188 :
14.8, , Change of Variables in Multiple Integrals, , 1209, , As you have also seen on many occasions, a change of variables can be used to, help us to evaluate integrals involving a function of two or more variables. For example, in evaluating a double integral 兰兰R f(x, y) dA, where R is a circular region, it is, often helpful to use the substitution, x  r cos u, , y  r sin u, , to transform the original integral into one involving polar coordinates. In this instance, we have, , 冮冮 f(x, y) dA  冮冮 f(r cos u, r sin u) r dr du, R, , D, , where D is in the region in the ru-plane that corresponds to the region R in the xyplane., These examples raise the following questions:, 1. If an integral 兰兰 f(x, y) dA cannot be readily found when we are integrating, with respect to the variables x and y, can we find a substitution x  t(u, √),, y  h(u, √) that transforms this integral into one involving the variables u and √, that is more convenient to evaluate?, 2. What form does the latter integral take?, , Transformations, The substitutions that are used to change an integral involving the variables x and y, into one involving the variables u and √ are determined by a transformation or function T from the u√-plane to the xy-plane. This function associates with each point (u, √), in a region S in the u√-plane exactly one point (x, y) in the xy-plane. (See Figure 1.), The point (x, y), called the image of the point (u, √) under the transformation T, is written (x, y)  T(u, √) and is defined by the equations, x  t(u, √), , y  h(u, √), , (2), , where t and h are functions of two variables. The totality of all points in the xy-plane, that are images of all points in S is called the image of S and denoted by T(S). Figure 1 gives a geometric visualization of a transformation T that maps a region S in the, u√-plane onto a region R in the xy-plane., √, , y, T, S, , R, , (u, √), , FIGURE 1, T maps the region S in the u√-plane, onto the region R in the xy-plane., , (x, y), , u, , x, , A transformation T is one-to-one if no two distinct points in the u√-plane have the, same image. In this case it may be possible to solve Equation (2) for u and √ in terms, of x and y to obtain the equations, u  G(x, y), which defines the inverse transformation T, , √  H(x, y), 1, , from the xy-plane to the u√-plane.

Page 189 :
1210, , Chapter 14 Multiple Integrals, , EXAMPLE 1 Let T be a transformation defined by the equations, xu, , y√, , √, , Find the image of the rectangular region S  {(u, √) 冟 0  u  2, 0  √  1} under the, transformation T., Solution Let’s see how the sides of the rectangle S are transformed by T. Referring to, Figure 2a, observe 0  u  2 and √  0 on S1. Using the given equations describing, T, we see that x  u and y  0. This shows that S1 is mapped onto the line segment, 0  x  2 and y  0 (labeled T(S1) in Figure 2b). On S2, u  2 and 0  √  1, so, x  2 y, for 0  y  1. This gives the image of S2 under T as the line segment, T(S2). On S3, 0  u  2 and √  1, so x  u 1 and y  1, which means that S3 is, mapped onto the line segment T(S3) described by 1  x  3, y  1. Finally, on S4,, u  0 and 0  √  1, and this gives the image of S4 as the line segment x  y, for, 0  y  1. Observe that as the perimeter of S is traced in a counterclockwise direction, so too is the boundary of the image R  T(S) of S. The image of S under T is the, region inside and on the parallelogram R., √, , y, S3, , 1, , FIGURE 2, The region S in part (a) is transformed, onto the region R in part (b) by T., , S4, , S, , 0, , 1, , 1, S2, S1, , 2, , u, , (1, 1), T(S4), , 0, , (a), , T(S3), R, , (3, 1), T(S2), , 1 T(S1) 2, , 3, , x, , (b), , Change of Variables in Double Integrals, To see how a double integral is changed under the transformation T defined by Equation (2), let’s consider the effect that T has on the area of a small rectangular region, S in the u√-plane with vertices (u 0, √0), (u 0, ⌬u, √0) , (u 0, ⌬u, √0 ⌬√), and, (u 0, √0 ⌬√) as shown in Figure 3a. The image of S is the region R  T(S) in the, xy-plane shown in Figure 3b. The lower left-hand corner point of S, (u 0, √0), is mapped, onto the point (x 0, y0)  T(u 0, √0)  (t(u 0, √0), h(u 0, √0)) by T. On the side L 1 of S,, u 0  u  u 0 ⌬u and √  √0. Therefore, the image T(L 1) of L 1 under T is the curve, with equations, x  t(u, √0), , y  h(u, √0), , √, , y, (u0, √0, , Δ√), , (u0, , Δu, √0, , Δ√), T(L2), , L2, , S, , (u0, √0), , L1, , T, , R, T(L1), , 0, , FIGURE 3, The transformation T maps S onto R., , (a), , (u0, , Δu, √0), , (x0, y0), , u, , 0, (b), , x

Page 190 :
14.8, , Change of Variables in Multiple Integrals, , 1211, , or, in vector form,, , y, , r(u, √0)  t(u, √0)i, b, , with parameter interval [u 0, u 0, , R, , ⌬u]. As you can see from Figure 4, the vector, , a  r(u 0, , r(u0, √0), , r(u0, , h(u, √0)j, , ⌬u, √0)  r(u 0, √0), , provides us with an approximation of T(L 1). Similarly, we see that the vector, , a, Δu, √0), , b  r(u 0, √0, , 0, , x, , FIGURE 4, The vector, a  r(u 0, ⌬u, √0)  r(u 0, √0), , ⌬√)  r(u 0, √0), , provides us with an approximation of T(L 2)., But we can write, ac, , r(u 0, , ⌬u, √0)  r(u 0, √0), d ⌬u, ⌬u, , If ⌬u is small, as we have assumed, then the term inside the brackets is approximately, equal to ru(u 0, √0). So, a ⬇ ⌬u ru (u 0, √0), Similarly, we see that, b ⬇ ⌬√ r√ (u 0, √0), This suggests that we can approximate R by the parallelogram having ⌬u ru(u 0, √0), and ⌬√ r√ (u 0, √0) as adjacent sides. (See Figure 5.) The area of this parallelogram is, 冟 a  b 冟, or, 冟 (⌬u ru)  (⌬√ r√) 冟  冟 ru  r√ 冟 ⌬u ⌬√, , y, , Δ√r√(u0, √0), , R, , where the partial derivatives are evaluated at (u 0, √0). But, ru  tui, , Δuru(u0, √0), , FIGURE 5, The image region R is approximated, by the parallelogram with sides, ⌬u ru (u 0, √0) and ⌬√ r√(u 0, √0)., , x, i, u, , y, j, u, , where the partial derivatives are evaluated at (u 0, √0). Similarly,, , r(u0, √0), 0, , hu j , , x, , r√  t√i, , h√ j , , So, i, x, ru  r√ , u, x, √, , j, y, u, y, √, , x, i, √, , x, u, 0 ∞, x, √, 0, , k, , y, j, √, , y, x, u, u, ∞k  ∞, y, y, √, u, , x, √, ∞k, y, √, , Before proceeding, let’s define the following determinant, which is named after the, German mathematician Carl Jacobi (1804–1851)., , DEFINITION The Jacobian, The Jacobian of the transformation T defined by x  t(u, √) and y  h(u, √) is, x, (x, y), u, ∞, y, (u, √), u, , x, y x, √, x y, ∞, , y, u √, u √, √

Page 191 :
1212, , Chapter 14 Multiple Integrals, , In terms of the Jacobian we can write the approximation of the area ⌬A of R as, ⌬A ⬇ 冟 ru  r√ 冟 ⌬u ⌬√  `, , (x, y), ` ⌬u ⌬√, (u, √), , (3), , where the Jacobian is evaluated at (u 0, √0)., Now let R be the image (in the xy-plane) under T of the region S in the u√-plane;, that is, let R  T(S) as shown in Figure 6. Enclose S by a rectangle, and partition the, latter into mn rectangles Sij, where 1  i  m, 1  j  n. The images Sij are transformed onto images Rij in the xy-plane, as shown in Figure 6., R, , √, , S, , y, , Sij, T, , (xi, yj), , FIGURE 6, The images Sij in the u√-plane, are transformed onto the, images Rij in the xy-plane., , Rij, , (ui, √j), 0, , u, , 0, , x, , Suppose that f is continuous on R, and define F by, FR (x, y)  e, , f(x, y), 0, , if (x, y) 僆 R, if (x, y) 僆 R, , Using the approximation in Equation (3) on each subrectangle Rij, we can write the, double integral of f over R as, , 冮冮, , m, , f(x, y) dA  lim, , n, , a a FR(x i, yj) ⌬A, , m, n→⬁ i1 j1, , R, ,  lim, , m, n→⬁, , m, n, (x, y), a a FR(t(u i, √j), h(u i, √j)) ` (u, √) ` ⌬u ⌬√, i1 j1, , where the Jacobian is evaluated at (u i, √j). But the sum on the right is the Riemann sum, associated with the integral, (x, y), , 冮冮 f(t(u, √), h(u, √))` (u, √) ` du d√, S, , This discussion suggests the following result. Its proof can be found in books on, advanced calculus., , THEOREM 1 Change of Variables in Double Integrals, Let T be a one-to-one transformation defined by x  t(u, √), y  h(u, √) that maps, a region S in the u√-plane onto a region R in the xy-plane. Suppose that the boundaries of both R and S consist of finitely many piecewise smooth, simple, closed, curves. Furthermore, suppose that the first-order partial derivatives of t and h are, continuous functions. If f is continuous on R and the Jacobian of T is nonzero, then, (x, y), , 冮冮 f(x, y) dA  冮冮 f(t(u, √), h(u, √))` (u, √) ` du d√, R, , S, , (4)

Page 192 :
14.8, , Change of Variables in Multiple Integrals, , 1213, , Note Theorem 1 tells us that we can formally transform an integral 兰兰R f(x, y) dA, involving the variables x and y into an integral involving the variables u and √ by replacing x by t(u, √) and y by h(u, √) and the area element dA in x and y by the area element, dA  `, , (x, y), ` du d√, (u, √), , in u and √. If you compare Equation (4) with Equation (1), you will see that the absolute, value of the Jacobian of T plays the same role as the derivative t¿(u) of the “transformation” t defined by x  t(u) in the one-dimensional case., , EXAMPLE 2 Use the transformation T defined by the equations x  u, , evaluate 兰兰R (x, ple 1.), , √, y  √ to, y) dA, where R is the parallelogram shown in Figure 2b. (See Exam-, , Solution Recall that the transformation T maps the much simpler rectangular region, S  {(u, √) 冟 0  u  2, 0  √  1} onto R and that this is precisely the reason for, choosing this transformation. The Jacobian of T is, x, (x, y), u, ∞, y, (u, √), u, , x, √, 1, ∞`, y, 0, √, , 1, `1, 1, , Using Theorem 1, we obtain, , 冮冮 (x, , y) dA , , R, , 冮冮 [(u, , √](1) du d√, , √), , S, , 1, , , , 冮冮, 0, , 2, , (u, , 0, , 冮, , 1, , 0, , 1, , , , 2√) du d√ , , 冮 (2, , 4√) d√  C2√, , 0, , 1, c u2, 2, , u2, , 2u√d, , d√, u0, , 2√2 D 0  4, 1, , In Example 2 the transformation T was chosen so that the region S in the u√-plane, corresponding to the region R could be described more simply. This made it easier to, evaluate the transformed integral. In other instances the transformation is chosen so, that the corresponding integrand in u and √ is easier to integrate than the original integrand in the variables x and y, as the following example shows., , EXAMPLE 3 Evaluate, xy, b dA, y, , 冮冮 cosa x, R, , where R is the trapezoidal region with vertices (1, 0), (2, 0), (0, 2), and (0, 1)., Solution As it stands, this integral is difficult to evaluate. But observe that the form, of the integrand suggests that we make the substitution, uxy, , √x, , y

Page 193 :
1214, , Chapter 14 Multiple Integrals, , These equations define a transformation T 1 from the xy-plane to the u√-plane. If we, solve these equations for x and y in terms of u and √, we obtain the transformation T, from the u√-plane to the xy-plane defined by, x, , 1, (u, 2, , y, , √), , 1, (√  u), 2, , The given region R is shown in Figure 7., √, , y, T, T 1, , √2, S, , √  u, , 2, , √u, , 1, , √1, 2 1, , FIGURE 7, T maps S onto R, and, T 1 maps R onto S., , 0, , 1, , 2, , u, , 0, , (a), , R, 1, , x, , 2, , (b), , To find the region S in the u√-plane that is mapped onto R under the transformation T, observe that the sides of R lie on the lines, y  0,, , x  2,, , y, , x  0,, , and, , x1, , y, , Using the equations defining T 1, we see that the sides of S corresponding to these, sides of R are, √  u,, , √  2,, , √  u,, , √1, , and, , The region S is shown in Figure 7a., The Jacobian of T is, x, (x, y), u, ∞, y, (u, √), u, , x, 1, √, 2, ∞∞, y, 1, , √, 2, , 1, 2, 1, ∞, 1, 2, 2, , If we use Theorem 1 while viewing S as a u-simple region, we find, xy, b dA , y, , 冮冮 cosa x, R, , (x, y), , 冮冮 cosa √ b ` (u, √) ` du d√, u, , S, , 2, , , , 冮冮, 1, , √, , u, 1, 1, cosa b ⴢ a b du d√ , √, 2, 2, √, 2, ,  sin 1, , 冮, , 1, , 2, , u u√, c√ sina b d, d√, √ u√, , 冮 √ d√  2 sin 1, 3, , 1, , The next example shows how the formula for integration in polar coordinates can, be derived with the help of Theorem 1., , EXAMPLE 4 Suppose that f is continuous on a polar rectangle, R  {(r, u) 冟 a  r  b, a  u  b}

Page 194 :
14.8, , Change of Variables in Multiple Integrals, , 1215, , in the xy-plane. Show that, , 冮冮 f(x, y) dA  冮冮 f(r cos u, r sin u) r dr du, R, , S, , where S is the region in the ru-plane mapped onto R under the transformation T defined, by, x  t(r, u)  r cos u, y  h(r, u)  r sin u, Solution, , Observe that T maps the r-simple region, S  {(r, u) 冟 a  r  b, a  u  b}, , onto the polar rectangle R as shown in Figure 8. The Jacobian of T is, x, (x, y), r, ∞, y, (r, u), r, , x, u, cos u r sin u, `, ∞`, y, sin u, r cos u, u, ,  r cos2 u, , r sin2 u  r, , 0, , Using Theorem 1, we obtain, (x, y), , 冮冮 f(x, y) dA  冮冮 f(t(r, u), h(r, u))` (r, u) ` dr du, R, , S, , b, , , , 冮冮, a, , t 2(u), , f(r cos u, r sin u) r dr du, , t1(u), , as was to be shown., ¨, ∫, , ¨∫, ra, , FIGURE 8, T maps the region S onto, the polar rectangle R., , rb, , S, ¨å, a, , å, b, , r, , rb, R, , ra, , å, 0, , y, , ¨∫, , ¨å, , ∫, x, , 0, , Change of Variables in Triple Integrals, The results for a change of variables for double integrals can be extended to the case, involving triple integrals. Let T be a transformation from the u√w-space to the xyzspace defined by the equations, x  t(u, √, w),, , y  h(u, √, w),, , z  k(u, √, w), , and suppose that T maps a region S in uvw-space onto a region R in xyz-space. The, Jacobian of T is, x x x, u √ w, (x, y, z), y y y, , (u, √, w), u √ w, z z z, u √ w

Page 195 :
1216, , Chapter 14 Multiple Integrals, , The following is the analog of Equation (4) for triple integrals., , Change of Variables in Triple Integrals, (x, y, z), , 冮冮冮 f(x, y, z) dV  冮冮冮 f(t(u, √, w), h(u, √, w), k(u, √, w))` (u, √, w) ` du d√ dw, R, , (5), , S, , EXAMPLE 5 Use Equation (5) to derive the formula for changing a triple integral in, rectangular coordinates to one in spherical coordinates., Solution, , The required transformation is defined by the equations, x  r sin f cos u,, , y  r sin f sin u,, , z  r cos f, , where r, f, and u are spherical coordinates. The Jacobian of T is, sin f cos u, (x, y, z),  † sin f sin u, (r, f, u), cos f, , r cos f cos u, r cos f sin u, r sin f, , r sin f sin u, r sin f cos u †, 0, , Expanding the determinant by the third row, we find, (x, y, z), r cos f cos u,  cos f`, (r, f, u), r cos f sin u, , r sin f sin u, `, r sin f cos u, ,  cos f(r2 cos f sin f cos2 u,  r2 cos2 f sin f, , r sin f`, , sin f cos u r sin f sin u, `, sin f sin u r sin f cos u, , r2 cos f sin f sin2 u), , r sin f(r sin2 f cos2 u, , r sin2 f sin2 u), , r2 sin3 f  r2 sin f, , Since 0  f  p, we see that sin f  0, so, `, , (x, y, z), `  冟 r2 sin f 冟  r2 sin f, (r, f, u), , Using Equation (5), we obtain, , 冮冮冮 f(x, y, z) dV  冮冮冮 f(r sin f cos u, r sin f sin u, r cos f)r sin f dr df du, 2, , R, , S, , which is Equation (4) in Section 14.7, the formula for integrating a triple integral in, spherical coordinates., , 14.8, , CONCEPT QUESTIONS, , 1. a. Let T be a transformation defined by x  t(u, √) and, y  h(u, √). What is the Jacobian of T?, b. Write the Jacobian of the transformation T given by, x  t(u, √, w), y  h(u, √, w), and z  k(u, √, w)., , 2. a. Let T be the one-to-one transformation defined by, x  t(u, √) and y  h(u, √) that maps a region S in the, u√-plane onto a region R in the xy-plane. Write the formula for transforming the integral 兰兰R f(x, y) dA into an, integral involving u and √ over the region S., b. Repeat part (a) for the case of a triple integral.

Page 196 :
14.8, , 1. S  {(u, √) 冟 0  u  2, 0  √  1};, , x  u  √, y  √, , 2. S  {(u, √) 冟 0  u  1, 0  √  2};, yu√, , xu, , √,, , 4. S is the trapezoidal region with vertices (2, 0), (1, 0),, (0, 1), (0, 2); x  u √, y  u  √, 5. S  {(u, √) 冟 u 2, y  2u√, , √2  1, u  0, √  0};, , y  2u√, , 10. x  u ln √,, , y  √ ln u, , 11. x  u, , w,, , √, , 12. x  2u, , w,, , yu√, , y  u 2  √2,, , 2, , dA, where R is the region in the first quadrant, , 冮冮, , T is defined by x  u 2  √2 and y  2u√, where u, √  0., , 冮冮 y sin x dA, where R is the region bounded by the graphs, of x  y 2, x  0, and y  1;, y√, , 冮冮 (2x, , y) dA, where R is the parallelogram bounded by, , R, , w,, zu, , z  u  2√, , the lines x y  1, x, 2x  y  4, , 3w, , √2  2w 2, , y) dA, where R is the parallelogram bounded by the, , 22., , 冮冮 (x, , y) sin (2x  y) dA, where R is the parallelogram, , bounded by the lines y  x, y  x, y  2x  2, 23., , 冮冮 e, , (xy)>(x y), , by the lines x  0, y  0, and x, 24., , 冮冮 e, , (x y)>(xy), , R, , dA, where R is the trapezoidal region with, , vertices (2, 0), (1, 0), (0, 1), and (0, 2), 25., , 冮冮 2xy dA, where R is the region in the first quadrant, , 冮冮 xy dA, where R is the region in the first quadrant, R, , bounded by the ellipse, , R, , bounded by the ellipse 4x, x  3u and y  2√, 2, ,  xy, , 2, , 9y  36;, 2, , T is defined by, , 26., , 冮冮 ln(4x, , 2, , 25y 2, , y2, , a2, , b2, , 1, , 25y 2  1, , by the ellipse 4x 2, , y 2) dA, where R is the region bounded by, , x2, , 1) dA, where R is the region bounded, , R, , 27. Find the volume V of the solid E enclosed by the ellipsoid, , R, , the ellipse x 2  xy y 2  2; T is defined by, x  12u  12>3√ and y  12u, 12>3√, 17., , y1, , R, , the lines with equations y  2x, y  12 x 3, y  2x 3,, and y  12 x; T is defined by x  u  2√ and y  2u  √, , 冮冮 cos(x, , 1, y  2x, and, , dA, where R is the triangular region bounded, , R, , lines with equations y  2x, y  12 x  152, y  2x 10,, and y  12 x; T is defined by x  u 2√ and y  √  2u, , 16., , y  3, 2x  y  0, and, , R, , R, , 3y) dA, where R is the parallelogram bounded by, , T is defined by x  u 2 and, , In Exercises 21–26, evaluate the integral by making a suitable, change of variables., , In Exercises 13–20, evaluate the integral using the transformation T., , 冮冮 (2x, , 冮冮 xy, , bounded by the hyperbolas xy  1 and xy  2 and the lines, u, y  x and y  2x; T is defined by x  and y  √, √, 1, 2, dA, where R  {(x, y) 冟 x, y 2  1, y  0};, 19., 2, 2, 2x, y, R, , 21., , y  eu sin 2√, , 9. x  eu cos 2√,, , 18., , R, , y  u2  √, , √,, , 8. x  u 2  √2,, , 15., , y2,  1; T is defined by x  2u and, 9, , y  3√, , 20., , In Exercises 7–12, find the Jacobian of the transformation T, defined by the equations., 7. x  2u, , the ellipse, , x  u 2  √2;, , 6. S  5 (u, √) 冟 1  u  2, 0  √  p2 6 ;, x  u cos √, y  u sin √, , 14., , x2, 4, , 1217, , R, , 3. S is the triangular region with vertices (0, 0), (1, 1), (0, 1);, x  u 2√, y  2√., , 冮冮 (x, , Change of Variables in Multiple Integrals, , EXERCISES, , In Exercises 1–6, sketch the image R  T(S) of the set S under, the transformation T defined by the equations x  t(u, √),, y  h(u, √)., , 13., , 14.8, , 2, , 2, , 冮冮 B 1  4  9 dA, where R is the region bounded by, x, , y, , R, , V Videos for selected exercises are available online at www.academic.cengage.com/login., , x2, , y2, , z2, , 2, , 2, , c2, , a, , b, , 1, , Hint: V  兰兰兰E dV. Use the transformation x  au, y  b√, and, z  cw.

Page 197 :
1218, , Chapter 14 Multiple Integrals, , 28. Let E be the solid enclosed by the ellipsoid, x, , 2, , a2, , y, , 2, , b2, , z, , In Exercises 32 and 33, determine whether the statement is true, or false. If it is true, explain why. If it is false, explain why or, give an example that shows it is false., , 2, , c2, , 1, , 32. If T is defined by x  t(u, √) and y  h(u, √) and maps a, region S in the u√-plane onto a region R in the xy-plane,, then the area of R is the same as the area of S., , Find the mass of E if it has constant mass density d., Hint: Use the transformation of Exercise 27., , 33. If T is defined by x  t(u, √), y  h(u, √) and maps a region, S onto a region R, then, , 29. Find the moment of inertia, Ix, of the lamina that, has constant mass density r and occupies the disk, x 2 y 2  ax  0 about the x-axis., 30. Show that the moment of inertia of the solid of Exercise 28, about the z-axis is Iz  15 m(a 2 b 2) , where m  43 pdabc is, the mass of the solid., , 冮冮 (x, R, , 2, , y 2) dx dy , , 冮冮 (u, S, , 2, , √2)`, , (x, y), ` du d√, (u, √), , 31. Use Formula (5) to find the formula for changing a triple, integral in rectangular coordinates to one in cylindrical coordinates., , CHAPTER, , 14, , REVIEW, , CONCEPT QUESTIONS, In Exercises 1–12, fill in the blanks., 1. a. If f is a continuous function defined on a rectangle, R  [a, b]  [c, d], then the Riemann sum of f over R, with respect to a partition P  {Rij} is, , where, (x *, ij , y *, ij ) is a point in Rij., b. The double integral 兰兰R f(x, y) dA , if the, limit exists for all choices of, in Rij., c. If f(x, y)  0 on R, then 兰兰R f(x, y) dA gives the, of the solid lying directly above R and below, the surface, ., d. If D is a bounded region that is not rectangular, then, , where fD (x, y) , 兰兰D f(x, y) dA , if (x, y) is in D and fD (x, y) , if (x, y) is not, in D., 2. The following properties hold for double integrals:, a. 兰兰D cf(x, y) dA , b. 兰兰D [ f(x, y) t(x, y)] dA , c. If f(x, y)  0 on D, then 兰兰D f(x, y) dA, ., d. If f(x, y)  t(x, y) on D, then 兰兰D f(x, y) dA, e. If D  D1 傼 D2 and D1 傽 D2  , then, ., 兰兰D f(x, y) dA , , ., , 3. a. If R  [a, b]  [c, d], then the two iterated integrals of, and, ., f over R are, b. Fubini’s Theorem for a rectangular region, R  [a, b]  [c, d] states that 兰兰R f(x, y) dA is, equal to the, integrals in part (a)., 4. a. A y-simple region has the form R , and t2 are continuous functions on [a, b]., , , where t1, , b. An x-simple region has the form R , , where, h 1 and h 2 are continuous functions on [c, d]., c. Fubini’s Theorem for the y-simple region R of part (a),, states that 兰兰R f(x, y) dA , . If R is the x-simple, region R of part (b), then 兰兰R f(x, y) dA , ., 5. a. A polar rectangle is a set of the form R , b. If f is continuous on a polar rectangle R, then, ., 兰兰R f(x, y) dA , c. An r-simple region is a set of the form R , d. If f is continuous on an r-simple region R, then, ., 兰兰R f(x, y) dA , , ., , ., , 6. If a lamina occupies a region R in the plane and the mass, density of the lamina is r(x, y), then, a. The mass of the lamina is given by m , ., b. The moments of the lamina with respect to the x- and, and M y , . The, y-axes are M x , coordinates of the center of mass of the lamina are x , and y , ., c. The moments of inertia of the lamina with respect to the, ,, x-axis, the y-axis, and the origin are Ix , , and I0 , , respectively., Iy , d. If the moment of inertia of a lamina with respect to an, axis is I, then its radius of gyration with respect to the, axis is R , ., 7. a. If fx and fy are continuous on a region R in the xy-plane,, then the area of the surface z  f(x, y) over R is A , ., b. If t is defined in a region R in the xz-plane, then the area, of the surface y  t(x, z) is A , .

Page 198 :
Review Exercises, , and the moment of T about the xy-plane is M xy , ., c. The, of T is located at the, point (x, y, z), where x , ,y, , and, ., z, d. The moments of inertia of T about the x-, y-, and z-axes, are Ix , , Iy , , and Iz , ., , c. If h is defined in a region R in the yz-plane, then the area, of the surface x  h(y, z) is A , ., 8. a. If f is a continuous function defined on a rectangular, box B  [a, b]  [c, d]  [p, q], then the Riemann sum, of f over B with respect to a partition P  {Bijk} is, , where (x *, ijk, y *, ijk, z *, ijk) is a point in Bijk., b. The triple integral 兰兰兰B f(x, y, z) dV , if the, *, *, limit exists for all choices of (x *, in, ., ,, y, ,, z, ), B, ijk, ijk, ijk, ijk, c. If f is continuous on a bounded solid region T in space, B, is a rectangular box that contains T, Q  {B111, B112, p ,, Bijk, p , Blmn} is a partition of B, F is a function defined, by, F(x, y, z)  e, , f(x, y, z), 0, , if (x, y, z) is in T, if (x, y, z) is in B but not in T, , then a Riemann sum of f over T is, ., d. The triple integral of f over T is 兰兰兰T f(x, y, z) dV , provided that the limit exists for all choices of, (x *, ijk, y *, ijk, z *, ijk) in T., 9. a. If f is continuous on B  [a, b]  [c, d]  [p, q], then, 兰兰兰B f(x, y, z) dV is equal to any of six iterated integrals, depending on the, of integration. If we integrate with respect to x, y, and z, in that order, then, ., 兰兰兰B f(x, y, z) dV , b. If f is continuous on a z-simple region, T  {(x, y, z) 冟 (x, y) 僆 R, k 1(x, y)  z  k 2(x, y)},, where R is the projection of T onto the xy-plane,, then 兰兰兰T f(x, y, z) dV , ., 10. If r(x, y, z) gives the density at the point (x, y, z) of a solid, T, then, a. The mass of T is m , ., b. The moment of T about the yz-plane is M yz , ,, the moment of T about the xz-plane is M xz , ,, , 1219, , 11. a. If T is a z-simple region described by, T  {(x, y, z) 冟 (x, y) 僆 R, h 1(x, y)  z  h 2(x, y)}, where, R  {(r, u) 冟 a  u  b, t1 (u)  r  t2(u)}, then in, terms of cylindrical coordinates, 兰兰兰T f(x, y, z) dV , ., b. If T  {(r, f, u) 冟 a  r  b, c  f  d, a  u  b} is, a spherical wedge, then in terms of spherical coordinates,, ., 兰兰兰T f(x, y, z) dV , c. If T is r-simple, T  {(r, f, u) 冟 h 1 (f, u)  r  h 2(f, u),, c  f  d, a  u  b}, then 兰兰兰T f(x, y, z) dV , ., 12. a. If T is a transformation defined by x  t(u, √) and, (x, y), y  h(u, √), then the Jacobian of T is, , (u, √), ., b. If T maps S in the u√-plane onto a region R in the xyplane, then the formula for transforming the integral, 兰兰R f(x, y) dx dy into one involving u and √ is, ., 兰兰R f(x, y) dx dy , c. If T maps S in u√w-space onto R in xyz-space and is, defined by x  t(u, √, w), y  h(u, √, w), and, (x, y, z), z  k(u, √, w), then the Jacobian T is, , (u, √, w), , and the change of variable formula for triple, integrals is 兰兰兰R f(x, y, z) dx dy dz , ., , REVIEW EXERCISES, In Exercises 1–8 evaluate the iterated integral., 2, , 1., , 冮冮, , 2, , 冮冮, 0, , (2x, 2, , 冮冮4, , 4., , 1z, , x, , dx dy, , 0, y, , 0, , 3, , 冮冮冮, 1, , x, , 0, , 冮冮, , 0, , In Exercises 9–12, sketch the region of integration for the iterated integral., 2, , 9., , 冮冮, , 3, 1, x, , 1, , f(x, y) dy dx, , 10., , 1, , 2y dx dy, 11., , 冮冮, , 冮冮, 0, , ln x, p 1 cos u, , sin1 y, , f(x, y) dx dy, , 0, , f(r, u) r dr du, , 0, , 0, 12, , 1ln x dy dx, 12., , 2, , 冮 冮冮, 12, , 2x, , f(x, y, z) dz dx dy, , y2 0, , 2z) dy dx dz, , y, y, , 6., , 1, , 冮 冮 冮 (x, 2, , 冮冮, , 0, e 1>x, , y2, , y, , 0, , 8., , 3y) dy dx, , 21y2, , 0, , 2, , 7., , 0, , 1, , 1, , 0, , 冮 冮 x sin xy dy dx, 0, , x, , 2, , 5., , 2., , 1x, , 1, , 3., , 3xy 2) dx dy, , (2x, , 1, , 0, , 1, , p, , z, , dz dy dx, , In Exercises 13 and 14, reverse the order of integration, and, evaluate the resulting integral., 1, , 13., , 冮冮, 0, , y, , 1, , 1, , sin x 2 dx dy, , 14., , 冮冮, 0, , y, , 1y, , cos x, dx dy, x

Page 199 :
1220, , Chapter 14 Multiple Integrals, 28. The solid under the paraboloid z  4  x 2  y 2 and above, the triangular region in the xy-plane with vertices (0, 0) ,, (1, 1) , and (0, 1), , In Exercises 15–26, evaluate the multiple integral., 15., , 冮冮 (x, , 2, , 3y 2) dA, where, , 29. The solid bounded by the paraboloid z  x 2, der x 2 y 2  1, and the plane z  0, , R, , R  {(x, y) 冟 1  x  1, 0  y  2}, 16., , 冮冮(x, , 30. The solid under the paraboloid z  9  x 2  y 2 and above, the circular region x 2 y 2  4 in the xy-plane, , y) dA, where, , R, , R  {(x, y) 冟 0  x  1, 0  y  21  x 2}, 17., , 冮冮 y dA, where R is the region bounded by the parabola, x  y and the line x  2y  3, 2, , 冮冮 (x, , 2y) dA, where R is the region bounded by the, , R, , graphs of x  4  y 2, x  0, and y  0, 19., , 冮冮, , x dA, where R is the region in the first quadrant bounded, , by the ellipse 4x, 20., , 9y  36, 2, , 冮冮冮 xy dV, where, y}, , 冮冮冮 z dV, where R is the tetrahedron bounded by the planes, x, , z  6, x  0, y  0, and z  0, , 2y, , 冮冮冮 xyz dV, where T is the region bounded by the hemiT, , sphere z  21  x 2  y 2 and the plane z  0, , 冮冮冮, , z dV, where T is the region bounded by the cylinder, , T, , x2, 25., , 33. D is the region in the first quadrant bounded by the graphs, of y  x and y  x 3; r(x, y)  y, , 36. D is the region bounded by the semicircle y  24  x 2, and the x-axis; r(x, y)  x 2y, , T, , 24., , In Exercises 33–36, find the mass and the center of mass of, the lamina occupying the region D and having the given mass, density., , of y  1>x, y  x, and x  e, , T, , 23., , 冮冮冮, , z 2  1 and the planes y  x, y  2x, and z  0, x 2z dV, where T is the region bounded above by the, , In Exercises 37 and 38, find the moments of inertia Ix, Iy, and I0, of the lamina occupying the region D and having the given mass, density., 37. D is the region bounded by the triangle with vertices (0, 0),, (0, 1) , and (1, 1) ; r(x, y)  x 2 y 2, 38. D is the region bounded by the graphs of y  x and y  x 2;, r(x, y)  x, In Exercises 39 and 40, find the area of the surface S., 39. S is the part of the plane 2x, , paraboloid y  1  x  z , above the plane z  0, and to, the left by the plane y  0, , 冮冮冮 2x, T, , 1, 2, , y2, , z2, , 2, , dV, where T is the region bounded, , z  6 in the first octant., y 2 below the plane, , In Exercises 41 and 42, evaluate the integral by changing to, cylindrical or spherical coordinates., , 冮冮, , 24x2, , 3, , 29x2, , 2, , 2, , 3y, , 40. S is the part of the paraboloid z  x 2, z  4., , T, , 26., , y2, , 35. D is the region in the first quadrant bounded by the circle, x 2 y 2  1; r(x, y)  2x 2 y 2, , T  {(x, y, z) 冟 0  x  1, 0  y  x 2, 0  z  x, 22., , 32. The solid bounded above by the paraboloid, z  4  x 2  y 2 and below by the cone z  2x 2, , 冮冮 ln x dy dx, where R is the region bounded by the graphs, R, , 21., , 2, , 34. D is the region bounded by the parabola y  x 2 and the line, y  4; r(x, y)  x 2y, , R, , 2, , 31. The solid under the surface z  e(x y ), within the cylinder, x 2 y 2  1 and above the plane z  0, 2, , R, , 18., , y 2, the cylin-, , 41., , 0, , 42., , 0, , 2, , y 2) 3>2 dz dy dx, , 0, , 0, , 冮冮, , 1, , 冮 (x, , 0, , 冮, , 29x2 y2, , z2x 2, , y2, , z 2 dz dy dx, , 0, , In Exercises 27–32, find the volume of the solid., , 43. Express the triple integral 兰兰兰T f(x, y, z) dV as an iterated, integral in six different ways using different orders of integration, where T is the tetrahedron bounded by the planes, 2x 3y z  6, x  0, y  0, and z  0., , 27. The solid under the surface z  xy 2 and above the rectangular region R  {(x, y) 冟 0  x  1, 1  y  2}, , 44. Set up, but do not evaluate, the iterated integral giving the, mass of the solid bounded by the cone z  2x 2 y 2 and, , above by the hemisphere z  21  x 2  y 2 and below by, the plane z  0

Page 200 :
Challenge Problems, the sphere x 2 y 2 z 2  8 if the density of the solid at, any point P is r(x, y, z)  11 xz., 45. Find the Jacobian of the transformation T defined by the, equations x  u w 2, y  2u 2 √, and z  u 2  √2 2w., 46. Use the transformation x  u>√ and y  √ to evaluate, 兰兰R y cos xy dy dx, where R is the region bounded by the, hyperbolas xy  1 and xy  4 and the lines y  1 and, y  4., 47. Evaluate 兰兰R e(xy)>(x y) dA, where R is the triangular region, bounded by the lines y  x, x y  2, and y  0., , 1221, , In Exercises 48–53, state whether the statement is true or false., Give a reason for your answer., b, , 48., , 冮冮, , b, , 1, , 3, , a, , 49., , f(x)f(y) dx dy  c, , a, , 冮冮, , 冮, , b, , a, 3, , cos xy) dx dy , , (x, , 2, 1 y, , 0, , 50., , 2, , f(x) dxd, , 1, , 1, , 冮 冮 (x, 2, , cos xy) dy dx, , 0, , x, , 冮 冮 f(x, y) dx dy  冮 冮 f(x, y) dy dx, 0, , 0, , 0, , 0, , 51. If 兰兰D f(x, y) dA  0, then f(x, y)  0 for all (x, y) in D., 1, , 52., , 3, , 冮 冮x, , sin y 2 dy dx  0, , 3, , 1 0, 1 3, , 53., , 冮 冮 [1x, 0, , cos2(xy)] dx dy  6, , 1, , CHALLENGE PROBLEMS, 1. a. Use the definition of the double integral as a limit of a, Riemann sum to compute 兰兰R (3x 2 2y) dA, where, R  {(x, y) 冟 0  x  2, 0  y  1}., Hint: Take ⌬x  2>m, and ⌬y  1>n, so that x i  2i>m, where, 1  i  m, and yj  j>n, where 1  j  n., , b. Verify the result of part (a) by evaluating an appropriate, iterated integral., 2. The following figure shows a triangular lamina. Its mass, density at (x, y) is f(x, y)  cos(y 2). Find its mass., , 4. Using the result of Problem 3, show that the area of the, parallelogram determined by the vectors a  具a1, a2典 and, b  具b1, b2典 is 冟 a1b2  a2b1 冟., 5. Monte Carlo Integration This is a method that is used to find, the area of complicated bounded regions in the xy-plane., To describe the method, suppose that D is such a region, completely enclosed by a rectangle R  {(x, y) 冟 a  x  b,, c  y  d}, as shown in the figure. Using a random number, generator, we then pick points in R. If A(D) denotes the area, of D, then, , y, , N(D), A(D), ⬇, n, A(R), , (2, 1), , 1, , where N(D) denotes the number of points landing in D,, A(R)  (b  a)(d  c), and n is the number of points, picked. Then, 0, , 1, , 2, , x, , A(D) ⬇, , 3. Show that the area of the parallelogram shown in the figure, is (b  a)(d  c), where a  b and c  d., , (b  a)(d  c), N(D), n, , Use Monte Carlo integration with n  5000 to estimate the, area of the disk of radius 5., y, , y, , d, D, , d, , R, , c, c, 0, , a, , b, , x, , 0, , a, , b, , x

Page 201 :
1222, , Chapter 14 Multiple Integrals, , n, 2, 6. The expression 兺m, y 3j ) ⌬x ⌬y, where x i  i>m,, i1 兺j1 (x i, i  1, 2, p , m, and yj  1 ( j>n), j  1, 2, p , n, is the, Riemann sum of a function f(x, y) over a region associated, with a regular pattern., a. Write a double integral corresponding to this Riemann, sum., b. Write an iterated integral corresponding to this Riemann, sum., , 7. a. Suppose that f(x, y) is continuous in the triangular region, R  {(x, y) 冟 x  b, y  a, y  x}. Show that, b, , x, , b, , b, , 冮 c 冮 f(x, y) dyd dx  冮 c 冮 f(x, y) dxd dy, a, , a, , a, , y, , b. Use the result of part (a) to evaluate, 1, , 冮 c冮, 0, , where A(R) is the area of the region R in the xy-plane., z, d_, c, , d_, a, , R, , d_, b, y, , x, , 11. A thin rectangular metal plate has dimensions a ft by b ft, and a constant density of k slugs/ft2. The plate is placed, in the xy-plane as shown in the figure and is allowed to, rotate about the z-axis at a constant angular velocity of, v radians/sec., , 1, , z, , sin x 2 dxd dy, , y, , 8. Let f be a continuous function of one variable. Show that, x, , y, , 冮冮冮, a, , a, , z, , f(t) dt dz dy , , a, , 1, 2, , x, , 冮 (x  t), , 2, , f(t) dt, , a, , y, , x, , Hint: Use the result of Exercise 7., , a. Show that the kinetic energy of the plate is given by, , 9. a. Let R be a region in the xy-plane that is symmetric, with respect to the y-axis, and let f be a function that, satisfies the condition f(x, y)  f(x, y). Show that, 兰兰R f(x, y) dA  0., b. Use the result of part (a) to show that if a lamina with, uniform density r that occupies a plane region that is, symmetric with respect to a straight line L, then the centroid of the lamina lies on L., 10. In Exercise 6 in the Challenge Problems for Chapter 11, you, were asked to show that the area of the portion of the plane, ax by cz  d, where a, b, and c are positive constants,, in the first octant is given by, d 2 2a 2 b 2, 2abc, , c2, , b2, , E, , kv2, 2, , b, , a, , 冮 冮 (x, 0, , y 2) dx dy , , 2, , 0, , c2, , 1 2, (a, 3, , b 2)mv2, , where m  kab., Hint: The kinetic energy of a particle of mass m slugs and velocity √ ft/sec is 12 m√2 ft-lb., , b. Show that E  12Iv2, where I  23 (a 2, , b 2)m., , 12. The Schwartz inequality for functions of one variable holds, for multiple integrals. (See Exercise 9 in the Challenge, Problems for Chapter 4.) Thus,, `, , 冮冮 f(x, y)t(x, y) dA `  B 冮冮 [ f(x, y)] dA冮冮[t(x, y)], 2, , D, , Derive this formula again, this time using integration. Show, that the result can also be written as, A(R), 2a 2, c, , b, , a, , D, , 2, , dA, , D, , a. Use Schwartz’s inequality to prove that, `, , 冮冮24x, , 2, ,  y 2 dA ` , , 213, 3, , D, , where D is the triangle with vertices A(0, 0), B(1, 2),, and C(1, 0)., b. Find the exact value of the integral in part (a). How, accurate is the estimate?

Page 202 :
MODULE 4, Textbook : Calculus: Soo T Tan Brooks/Cole, Cengage Learning (2010) ISBN 0-534-46579-X), Sections 15.1-15.9

Page 203 :
1224, , Chapter 15 Vector Analysis, , 15.1, , Vector Fields, Figure 1 shows the airflow around an airfoil in a wind tunnel. The smooth curves, traced, by the individual air particles and made visible by kerosene smoke, are called streamlines., , FIGURE 1, A vector field associated with the, airflow around an airfoil, , FIGURE 2, A vector field associated with the flow, of blood in an artery, , To facilitate the analysis of this flow, we can associate a tangent vector with each, point on a streamline. The direction of the vector indicates the direction of flow of the, air particle, and the length of the vector gives the speed of the particle. If we assign a, tangent vector to each point on every streamline, we obtain what is called a vector field, associated with this flow., Another example of a vector field arises in the study of the flow of blood through, an artery. Here, the vectors give the direction of flow and the speed of the blood cells, (see Figure 2)., , DEFINITION Vector Field in Two-Dimensional Space, Let R be a region in the plane. A vector field in R is a vector-valued function, F that associates with each point (x, y) in R a two-dimensional vector, F(x, y)  P(x, y)i  Q(x, y)j, where P and Q are functions of two variables defined on R., y, , EXAMPLE 1 A vector field F in R2 (two-dimensional space) is defined by F(x, y) , , xi  yj. Describe F, and sketch a few vectors representing the vector field., x, , Solution The vector-valued function F associates with each point (x, y) in R2 its position vector r  xi  yj. This vector points directly away from the origin and has length, 冟 F(x, y) 冟  冟 r 冟  2x 2  y 2  r, , FIGURE 3, Some vectors representing the, vector field F(x, y)  xi  yj, , which is equal to the distance of (x, y) from the origin. As an aid to sketching some, vectors representing F, observe that each point on a circle of radius r centered at the, origin is associated with a vector of length r. Figure 3 shows a few vectors representing this vector field., , EXAMPLE 2 A vector field F in R2 is defined by F(x, y)  yi  xj. Describe F,, and sketch a few vectors representing the vector field.

Page 204 :
15.1, , Solution, , y, , Vector Fields, , 1225, , Let r  xi  yj be the position vector of the point (x, y). Then, F ⴢ r  (yi  xj) ⴢ (xi  yj),  yx  xy  0, , x, , and this shows that F is orthogonal to the vector r. This means that F(x, y) is tangent, to the circle of radius r  冟 r 冟 with center at the origin. Furthermore,, 冟 F(x, y) 冟  2(y)2  x 2  2x 2  y 2  r, , FIGURE 4, Some vectors representing the, vector field F(x, y)  yi  xj, , gives the length of the position vector. Therefore, F associates with each point (x, y) a, vector of length equal to the distance between the origin and (x, y) and direction that, is perpendicular to the position vector of (x, y). A few vectors representing this vector, field are sketched in Figure 4. As in Example 1, this task is facilitated by first sketching a few concentric circles centered at the origin., The “spin” vector field of Example 2 is used to describe phenomena as diverse as, whirlpools and the motion of a ferris wheel. It is called a velocity field., The definition of vector fields in three-dimensional space is similar to that in twodimensional vector fields., , DEFINITION Vector Field in Three-Dimensional Space, Let T be a region in space. A vector field in T is a vector-valued function F that, associates with each point (x, y, z) in T a three-dimensional vector, F(x, y, z)  P(x, y, z)i  Q(x, y, z)j  R(x, y, z)k, where P, Q, and R are functions of three variables defined on T., , Important applications of vector fields in three-dimensional space occur in the form, of gravitational and electric fields, as described in the following examples., , EXAMPLE 3 Gravitational Field Suppose that an object O of mass M is located at, the origin of a three-dimensional coordinate system. We can think of this object as, inducing a force field F in space. The effect of this gravitational field is to attract any, object placed in the vicinity of O toward it with a force that is governed by Newton’s, Law of Gravitation. To find an expression for F, suppose that an object of mass m is, located at a point (x, y, z) with position vector r  xi  yj  zk. Then, according to, Newton’s Law of Gravitation, the force of attraction of the object O of mass M on the, object of mass m has magnitude, GmM, 冟 r 冟2, and direction given by the unit vector r> 冟 r 冟, where G is the gravitational constant., Therefore, we can write, F(x, y, z)  , , , GM, r, 冟 r 冟3, GMx, (x  y  z ), 2, , 2, , 2 3>2, , i, , GMy, (x  y  z ), 2, , 2, , 2 3>2, , j, , GMz, (x  y 2  z 2)3>2, 2, , k

Page 205 :
1226, , Chapter 15 Vector Analysis, z, , The force exerted by the gravitational field F on a particle of mass m with position, vector r is mF. The vector field F is sketched in Figure 5., Observe that all the arrows point toward the origin and that the lengths of the arrows, decrease as one moves farther away from the origin. Physically, F(x, y, z) is the force, per unit mass that would be exerted on a test mass placed at the point P(x, y, z)., , m, , M, y, , x, , FIGURE 5, A gravitational force field, , EXAMPLE 4 Electric Field Suppose that a charge of Q coulombs is located at the, origin of a three-dimensional coordinate system. Then, according to Coulomb’s Law,, the electric force exerted by this charge on a charge of q coulombs located at a point, (x, y, z) with position vector r  xi  yj  zk has magnitude, k 冟 q 冟冟 Q 冟, (where k, the electrical constant, depends on the units used) and direction given by the, unit vector r> 冟 r 冟 for like charges Q and q (repulsion). Therefore, we can write the electric field E that is induced by Q as, E(x, y, z) , , , kQ, r, 冟 r 冟3, kQx, (x 2  y 2  z 2) 3>2, , i, , kQy, (x 2  y 2  z 2)3>2, , j, , kQz, (x 2  y 2  z 2) 3>2, , k, , The force exerted by the electric field E on a charge of q coulombs, located at (x, y, z),, is qE. Physically, E(x, y, z) is the force per unit charge that would be exerted on a test, charge placed at the point P(x, y, z)., , Conservative Vector Fields, Recall from our work in Section 13.6 that if f is a scalar function of three variables,, then the gradient of f, written §f or grad f, is defined by, §f(x, y, z)  fx (x, y, z)i  fy (x, y, z)j  fz (x, y, z)k, If f is a function of two variables, then, §f(x, y)  fx (x, y)i  fy (x, y)j, Since §f assigns to each point (x, y, z) the vector §f(x, y, z), we see that §f is a vector field that associates with each point in its domain a vector giving the direction of, greatest increase of f. (See Section 13.6.) The vector field §f is called the gradient, vector field of f., , EXAMPLE 5 Find the gradient vector field of f(x, y, z)  x 2  xy  y 2z 3., Solution, , The required gradient vector field is given by, , §f(x, y, z) , , , f, f, f, i, j, k, x, y, z,  2,  2,  2, (x  xy  y 2z 3)i , (x  xy  y 2z 3)j , (x  xy  y 2z 3)k, x, y, z, ,  (2x  y)i  (x  2yz 3)j  3y 2z 2k

Page 206 :
15.1, , Vector Fields, , 1227, , Before we proceed further, it should be pointed out that vector fields in both twoand three-dimensional space can be plotted with the help of most computer algebra, systems. The computer often scales the lengths of the vectors but still gives a good, visual representation of the vector field. The vector fields of Examples 1 and 2 and two, examples of vector fields in 3-space are shown in Figures 6a–6d., y, , y, , 6, , 6, , 4, , 4, , 2, , 2, , 0, , 0, , x, , x, , 2, , 2, , 4, , 4, , 6, , 6, 6 4 2, , 0, , 2, , 4, , 6 4 2, , 6, , 0, , 2, , 4, , 6, , (b) F(x, y)  yi  xj, , (a) F(x, y)  xi  yj, , 4, , 2, , 24, 0, , 4, 2, , 2, 0, , 1, , 4, , 0, 1, , FIGURE 6, Some computer-generated, graphs of vector fields, , 2, , 1, , 0, 0, , 4 2, , 1, 1, , 0, , 2, , 4, , xi  yj  zk, (d) F(x, y, z)  _____________, (x 2 y 2  z 2 )3/2, , z, (c) F(x, y, z)  yi  xj  _ k, 2, , Not all vector fields are gradients of scalar functions, but those that are play an, important role in the physical sciences., , DEFINITION Conservative Vector Field, A vector field F in a region R is conservative if there exists a scalar function f, defined in R such that, F  §f, The function f is called a potential function for F., , The reason for using the words conservative and potential in this definition will be, apparent when we discuss the law of conservation of energy in Section 15.4.

Page 207 :
1228, , Chapter 15 Vector Analysis, , Vector fields of the form, F(x, y, z) , , k, r, 冟 r 冟3, , are called inverse square fields. The gravitational and electric fields in Examples 3, and 4 are inverse square fields. The next example shows that these fields are conservative., , EXAMPLE 6 Find the gradient vector field of the function, f(x, y, z)  , , k, 2x  y 2  z 2, 2, , and hence deduce that the inverse square field F is conservative., Solution, , The gradient vector field of f is given by, , §f(x, y, z)  fx (x, y, z)i  fy (x, y, z)j  fz (x, y, z)k, , , , kx, (x  y  z ), 2, , 2, , 2 3>2, , i, , ky, (x  y 2  z 2)3>2, 2, , j, , kz, (x 2  y 2  z 2) 3>2, , k, , k, r, 冟 r 冟3, , where r  xi  yj  zk. This shows that the inverse square field, F(x, y, z) , , k, r, 冟 r 冟3, , is the gradient of the potential function f and is therefore conservative., Note In Example 6 we were able to show that an inverse square field F is conservative because we were given a potential function f such that F  §f. In Section 15.4 we, will learn how to find the potential function f for a conservative vector field. We will, also learn how to determine whether a vector field is conservative without knowing its, potential function., , 15.1, , CONCEPT QUESTIONS, , 1. a. What is a vector field in the plane? In space? Give examples of each., b. Give three examples of vector fields with a physical, interpretation., , 2. a. What is a conservative vector field? Give an example., b. What is a potential function? Give an example.

Page 208 :
15.1, , 15.1, , Vector Fields, , 1229, , EXERCISES, , In Exercises 1–6, match the vector field with one of the plots labeled (a)–(f)., (a), , (b), , y, , y, , 3, 4, 2, , 3, , 1, , 2, 1, , 3 2, , 1, , 1, , 2, , 3, , x, , 1, , 6 5 4 3 2 1, 1, , 3, , 4, , 5, , 6, , x, , 2, 3, , 2, , 4, , 3, (c), , 1 2, , (d), , y, , y, 4, , 2, , 3, 2, , 1, , 1, 2, , 1, , 1, , x, , 2, , 1, , 5 4 3 2 1, 1, , 1, , 2, , 3, , 4, , x, , 5, , 2, , 2, , 3, 4, , (e), , (f), , y, , y, 2, , 2, , 1, 1, 4 3 2 1, 2, , 1, , 1, , x, , 2, , 1, 1, , 2, , 3, , x, , 4, , 2, , 1, 2, , 2. F(x, y) , , 1. F(x, y)  yi, 3. F(x, y)  , 4. F(x, y)  , , y, x y, 2, , 2, , i, , y, 2x  y, 2, , 2, , x, x  y2, 2, , i, , 5. F(x, y)  , , 2x  y 2, , x, 2x  y, 2, , 2, , i, , y, 2x  y 2, 2, , j, , 1, 6. F(x, y)   xi  yj, 2, , j, , x, 2, , x, i, 冟x冟, , j, , In Exercises 7–18, sketch several vectors associated with the, vector field F., 7. F(x, y)  2i, 9. F(x, y)  xi  yj, , V Videos for selected exercises are available online at www.academic.cengage.com/login., , 8. F(x, y)  i  j, 10. F(x, y)  yi  xj

Page 209 :
1230, , Chapter 15 Vector Analysis, , 11. F(x, y)  xi  2yj, 13. F(x, y) , 14. F(x, y) , , 12. F(x, y)  xi  3yj, , x, 2x 2  y 2, y, 2x 2  y 2, , 15. F(x, y, z)  cj,, , i, i, , y, 2x 2  y 2, x, 2x 2  y 2, , j, j, , c a constant, , 16. F(x, y, z)  zk, , 17. F(x, y, z)  i  j  k, , 18. F(x, y, z)  xi  yj  zk, In Exercises 19–22, match the vector field with one of the plots, labeled (a)–(d)., 19. F(x, y, z)  i  j  2k, , 20. F(x, y, z)  xi  yj  2k, , 21. F(x, y, z)  xi  yj  zk, 22. F(x, y, z) , , x, , y, , i, , 2x  y  z, 2x  y 2  z 2, z, , k, 2x 2  y 2  z 2, 2, , 2, , 2, , 2, , j, , (b), , (a), , 1, , 0, , 1, 1, , 1, , 1, , 0, , 0, , 1, , 1, , 1, , 0, , 0, , 1, , 0 1, , 1, , 1, , 1, , 0, , 0, , 1, , 1, 1, , 0, , 1, , 0, , 1, , 0, , 23. F(x, y) , , 1, 1, (x  y)i , (x  y)j, 10, 10, , 24. F(x, y)  2xyi  2x 2yj, 1, 25. F(x, y, z)  (yi  xj  zk), 5, 26. F(x, y, z)  , , xi  yj  zk, 2x 2  y 2  z 2, , 29. f(x, y, z)  xyz, , 30. f(x, y, z)  xy 2  yz 3, , 31. f(x, y, z)  y ln(x  z), , 32. f(x, y, z)  tan1 (xyz), , 33. Velocity of a Particle A particle is moving in a velocity field, V(x, y, z)  2xi  (x  3y)j  z 2k, At time t  2 the particle is located at the point (1, 3, 2)., a. What is the velocity of the particle at t  2?, b. What is the approximate location of the particle at, t  2.01?, 34. Velocity of Flow The following figure shows a lateral section of, a tube through which a liquid is flowing. The velocity of flow, may vary from point to point, but it is independent of time., a. Assuming that the flow is from right to left, sketch vectors emanating from the indicated points representing the, speed and direction of fluid flow. Give a reason for your, answer. (The answer is not unique.), , 35. Show that the vector field F(x, y)  yi is not a gradient, vector field of a scalar function f., Hint: If F is a gradient vector field of f, then f> x  y and, , 1, , f> y  0. Show that f cannot exist., , 36. Is F(x, y)  yi  xj a gradient vector field of a scalar, function f ? Explain your answer., In Exercises 37–40, determine whether the statement is true or, false. If it is true, explain why. If it is false, explain why or give, an example that shows it is false., 1, , 0, , 1, , cas In Exercises 23–26, use a computer algebra system to plot the, , vector field., , 28. f(x, y)  e2x sin 3y, , 1, , (d), , (c), , 27. f(x, y)  x 2y  y 3, , b. Explain why it is a bad idea to seek shelter in a tunnel, when a tornado is approaching., , 1, , 1, , In Exercises 27–32, find the gradient vector field of the scalar, function f. (That is, find the conservative vector field F for the, potential function f of F.), , 37. If F is a vector field in the plane, then G  cF defined by, G(x, y)  cF(x, y), where c is a constant, is also a vector, field., 38. If F is a velocity field in space, then 冟 F(x, y, z) 冟 gives the, speed of a particle at the point (x, y, z), and, F(x, y, z)> 冟 F(x, y, z) 冟, where 冟 F(x, y, z) 冟  0, is a unit vector, giving its direction., 39. A constant vector field F(x, y, z)  ai  bj  ck is a gradient vector field., 40. All the vectors of the vector field F(x, y)  x 2i  y 2j point, outward in a radial direction from the origin.

Page 210 :
15.2, , 15.2, , Divergence and Curl, , 1231, , Divergence and Curl, In this section we will look at two ways of measuring the rate of change of a vector, field F: the divergence of F at a point P and the curl of F at P. The divergence and, curl of a vector field play a very important role in describing fluid flow, heat conduction, and electromagnetism., , Divergence, Suppose that F is a vector field in 2- or 3-space and P is a point in its domain. For the, purpose of this discussion, let’s also suppose that the vector field F describes the flow, of a fluid in 2- or 3-space. Then the divergence of F at P, written div F(P), measures, the rate per unit area (or volume) at which the fluid departs or accumulates at P. Let’s, consider several examples., , EXAMPLE 1, a. Figure 1a shows the vector field F(x, y)  xi  yj described in Example 1, of Section 15.1. Let P be a point in the plane, and let N be a neighborhood, of P with center P. Referring to Figure 1b, observe that an arrow entering N, along a streamline is matched by one that emerges from N and has a greater, length (because it is located farther away from the origin). This shows that, more fluid leaves than enters a neighborhood of P. We will show in Example 2a, that the vector field F is “divergent” at P; that is, the divergence of F at P is, positive., y, P, N, P, x, N, , FIGURE 1, , (a) The vector field F(x, y)  x i  yi, , (b) Flow through a neighborhood of P, (enlarged and not to scale), , b. Figure 2a shows the vector field F(x, y)  yi for x  0 and y  0. Observe that, the streamlines are parallel to the x-axis and that the lengths of the arrows on, each horizontal line are constant. We can think of F as describing the flow of a, river near one side of a riverbank. The velocity of flow is near zero close to the, bank (the x-axis) and increases as we move away from it. You can see from Figure 2b that the amount of fluid flowing into the neighborhood N of P is matched, by the same amount that exits N. Consequently, we expect the “divergence” at P, to be zero. We will show that this is the case in Example 2b.

Page 211 :
1232, , Chapter 15 Vector Analysis, y, P, , 0, , FIGURE 2, , 1, , 2, , P, , 3, , 4, , N, , x, , (a) The vector field F(x, y)  yi, , (b) Flow through a neighborhood of P, (enlarged and not to scale), , c. Figure 3a shows the vector field, F(x, y) , , 1, i, x1, , for x  0 and y  0. Observe that the streamlines are parallel to the x-axis and, that the lengths of the arrows on each horizontal line get smaller as x increases., From Figure 3b you can see that the “flow” into a neighborhood N of P is greater, than the flow that emerges from N. In this case, more fluid enters the neighborhood than leaves it, and the “divergence” is negative. We will show that our intuition is correct in Example 2., y, , N, , 1, , P, , P, N, 0, , FIGURE 3, , 1, , x, , 2, , 1, (a) The vector field F(x, y)  x  1 i, , (b) Flow through a neighborhood of P, (enlarged and not to scale), , Up to now, we have looked at the notion of “divergence” intuitively. The divergence of a vector field can be defined as follows., , DEFINITION Divergence of a Vector Field, Let F(x, y, z)  Pi  Qj  Rk be a vector field in space, where P, Q, and R, have first-order partial derivatives in some region T. The divergence of F is the, scalar function defined by, div F , , Q, P, R, , , x, y, z, , (1), , (We will justify this definition of divergence in Section 15.8.) In two-dimensional space,, F(x, y)  Pi  Qj, , and, , div F , , Q, P, , x, y

Page 212 :
15.2, , Divergence and Curl, , 1233, , As an aid to remembering Equation (1), let’s introduce the vector differential operator § (read “del”) defined by, §, , , , , i, j, k, x, y, z, , If we let § operate on a scalar function f(x, y, z), we obtain, §f(x, y, z)  a, , , , , i, j, kb f(x, y, z), x, y, z, , , , , , , f(x, y, z)i , f(x, y, z)j , f(x, y, z)k, x, y, z, , , , f, f, f, (x, y, z)i , (x, y, z)j , (x, y, z)k, x, y, z, , which is the gradient of f. If we take the “dot product” of § with the vector field, F(x, y, z) Pi  Qj  Rk, we obtain, §ⴢF a, , , , , , i, j, kb ⴢ (Pi  Qj  Rk), x, y, z, , Q, , , , P, R, P, Q, R, , , x, y, z, x, y, z, , which is the divergence of the vector field F. Thus, we can write the divergence of F, symbolically as, div F  § ⴢ F, , (2), , Let’s apply the definition of divergence to the vector fields that we discussed in, Example 1., , EXAMPLE 2 Find the divergence of (a) F(x, y)  xi  yj, (b) F(x, y)  yi, and, 1, i. Reconcile your results with the intuitive observations that, x1, were made in Example 1., (c) F(x, y) , , Solution, , , (x) , (y)  1  1  2. Here, div F 0, as expected., x, y, , , b. Here, F  yi  0j, so div F , (y) , (0)  0. In this case, div F  0, as, x, y, was observed in Example 1b., c. With F  (x  1)1i  0j we find, a. div F , , div F , , , , 1, (x  1)1 , (0)  (x  1)2  , x, y, (x  1)2, , and div F  0, as we concluded intuitively in Example 1c., We turn now to an example involving a vector field whose streamlines are not so, easily visualized.

Page 213 :
1234, , Chapter 15 Vector Analysis, , EXAMPLE 3 Find the divergence of F(x, y, z)  xyzi  x 2y 2zj  xy 2k at the point, , (1, 1, 2)., Solution, , div F , , ,  2 2, , (xyz) , (x y z) , (xy 2), x, y, z, ,  yz  2x 2yz, In particular, at the point (1, 1, 2) we find, div F(1, 1, 2)  (1)(2)  2(1)2 (1)(2)  6, The divergence of the vector field F(x, y)  yi of Examples 1b and 2b is zero. In, general, if div F  0, then F is called incompressible. In electromagnetic theory a vector field F that satisfies § ⴢ F  0 is called solenoidal. For example, the electric field, E in Example 4 is solenoidal. We will study the divergence of a vector field in greater, detail in Section 15.8., , EXAMPLE 4 Show that the divergence of the electric field E(x, y, z) , , r  xi  yj  zk, is zero., Solution, , kQ, r, where, 冟 r 冟3, , We first write, , E(x, y, z) , , kQx, (x  y  z ), 2, , 2, , 2 3>2, , i, , kQy, (x  y  z ), 2, , 2, , 2 3>2, , j, , kQz, (x  y 2  z 2) 3>2, 2, , k, , Then, div E  kQe, , y, , x, , , z, c 2, d, c 2, d, c 2, df, 2, 2, 3>2, 2, 2, 3>2, 2, x (x  y  z ), y (x  y  z ), z (x  y  z 2)3>2, , But, , x, , c 2, d, [x(x 2  y 2  z 2)3>2], 2, 2, 3>2, x (x  y  z ), x, 3,  (x 2  y 2  z 2) 3>2  x ⴢ a b (x 2  y 2  z 2)5>2 (2x), 2,  (x 2  y 2  z 2) 5>2[(x 2  y 2  z 2)  3x 2], , , 2x 2  y 2  z 2, (x 2  y 2  z 2) 5>2, , Similarly, we find, y, x 2  2y 2  z 2, , c 2, d 2, 2, 2, 3>2, y (x  y  z ), (x  y 2  z 2) 5>2, and, , x 2  y 2  2z 2, , z, c 2, d, , z (x  y 2  z 2) 3>2, (x 2  y 2  z 2) 5>2, , Therefore,, div E  kQc, , 2x 2  y 2  z 2, (x 2  y 2  z 2) 5>2, , , , x 2  2y 2  z 2, (x 2  y 2  z 2)5>2, , , , x 2  y 2  2z 2, (x 2  y 2  z 2) 5>2, , d0

Page 214 :
15.2, Axis, , Divergence and Curl, , 1235, , Curl, We now turn our attention to the other measure of the rate of change of a vector field, F. Let F be a vector field in 3-space, and let P be a point in its domain. Once again,, let’s think of the vector field as one that describes the flow of fluid. Suppose that a, small paddle wheel, like the one shown in Figure 4, is immersed in the fluid at P. Then, the curl of F, written curl F, is a measure of the tendency of the fluid to rotate the, device about its vertical axis at P. Later, we will show that the paddle wheel will rotate, most rapidly if its axis coincides with the direction of curl F at P and that its maximum rate of rotation at P is given by the length of curl F at P., , FIGURE 4, A paddle wheel, , EXAMPLE 5, a. Consider the vector field F(x, y, z)  yi for x  0 similar to that of Example 1b., This field is shown in Figure 5a. Notice that the positive z-axis points vertically, out of the page. Suppose that a paddle wheel is planted at a point P. Referring, to Figure 5b, you can see that the arrows in the upper half of the circle with, center at P are longer than those in the lower half. This shows that the net, clockwise flow of the fluid is greater than the net counterclockwise flow. This, will cause the paddle to rotate in a clockwise direction, as we will show in, Example 6., y, , 5, 4, 3, 2, 1, 0, , P, P, , 1 2 3 4 5 6 7 8 9 10 11 12 13 14, , (a) The vector field F(x, y, z)  yi, , FIGURE 5, , x, , (b) Flow through a neighborhood of P, at which a paddle wheel is located, (enlarged and not to scale), , b. Consider the vector field F(x, y, z)  yi  xj shown in Figure 6a. Observe, that it is similar to the spin vector field of Example 2 in Section 15.1. Again,, the positive z-axis points vertically out of the page. If a paddle wheel is placed, at the origin, it is easy to see that it will rotate in a counterclockwise direction., Next, suppose that the paddle wheel is planted at a point P other than the origin. If you refer to Figure 6b, you can see that the circle with center at P is, divided into two arcs by the points of tangency of the two half-lines starting, from the origin. Notice that the arc farther from the origin is longer than the, one closer to the origin and that the flow on the larger arc is counterclockwise,, whereas the flow on the shorter arc is clockwise. Furthermore, the arrows emanating from the longer arc are longer than those emanating from the shorter, arc. This shows that the amount of fluid flowing in the counterclockwise direction is greater than that flowing in the clockwise direction. Therefore, the paddle wheel will rotate in a counterclockwise direction, as we will show in, Example 6.

Page 215 :
1236, , Chapter 15 Vector Analysis, y, , y, , P, , P, x, , 0, , x, , 0, (a) The vector field F(x, y, z)  yi  xj, , FIGURE 6, , (b) Flow through a neighborhood of P at, which a paddle wheel is located (enlarged, and not to scale), , c. Consider the vector field F(x, y, z)  xi  yj shown in Figure 7a. Note that it is, similar to the vector field in Example 1 in Section 15.1. Suppose that a paddle, wheel is placed at a point P. Then referring to Figure 7(b) and using an argument, involving symmetry, you can convince yourself that the paddle wheel will not, rotate. Again, we will show in Example 6 that this is true., y, , y, P, N, , P, x, N, 0, (a) The vector field F(x, y, z)  xi  yj, , FIGURE 7, , x, , (b) Flow through a neighborhood of P at, which a paddle wheel is located (enlarged, and not to scale), , The following definition provides us with an exact way to measure the curl of a, vector field., , DEFINITION Curl of a Vector Field, Let F(x, y, z)  Pi  Qj  Rk be a vector field in space, where P, Q, and R, have first-order partial derivatives in some region T. The curl of F is the vector, field defined by, curl F  §, , Fa, , Q, Q, R, P, R, P, , bi  a, , bj  a, , bk, y, z, z, x, x, y, , (We will justify this definition in Section 15.9.)

Page 216 :
15.2, , Divergence and Curl, , 1237, , As in the case of the cross product of two vectors, we can remember the expression for the curl of a vector field by writing it (formally) in determinant form:, , curl F  §, , a, , i, , F∞, x, P, , j, , y, Q, , k, , ∞, z, R, , Q, Q, R, P, R, P, , bi  a, , bj  a, , bk, y, z, z, x, x, y, , Let’s apply this definition to the vector fields that we discussed in Example 5., , EXAMPLE 6 Find the curl of (a) F(x, y, z)  yi for x  0, (b) F(x, y, z)  yi  xj,, and (c) F(x, y, z)  xi  yj. Reconcile your results with the intuitive observations that, were made in Example 5., Solution, a., curl F  §, , c, , i, , F∞, x, y, , j, , y, 0, , k, , ∞, z, 0, , , , , , , , (0) , (0)di  c (0) , (y)d j  c (0) , (y)dk  k, y, z, x, z, x, y, , This shows that curl F is a (unit) vector that points vertically into the page. Applying the right-hand rule, we see that this result tells us that at any point in the vector, field, the paddle wheel will rotate in a clockwise direction, as was observed earlier., b., curl F  §, , c, , i, , F∞, x, y, , j, , y, x, , k, , ∞, z, 0, , , , , , , , (0) , (x)di  c (0) , (y)d j  c (x) , (y)dk, y, z, x, z, x, y, ,  2k, The result tells us that curl F points vertically out of the page, so the paddle, wheel will rotate in a counterclockwise direction when placed at any point in the, vector field F., c., curl F  §, , c, , i, , F∞, x, x, , j, , y, y, , k, , ∞, z, 0, , , , , , , , (0) , (y)di  c (0) , (x)d j  c (y) , (x)dk  0, y, z, x, z, x, y, , This shows that a paddle wheel placed at any point in F will not rotate, as, observed earlier.

Page 217 :
1238, , Chapter 15 Vector Analysis, , The vector field F in Example 6c has the property that curl F  0 at any point P., In general, if curl F  0 at a point P, then F is said to be irrotational at P. This means, that there are no vortices or whirlpools there., , EXAMPLE 7, a. Find curl F if F(x, y, z)  xyi  xzj  xyz 2k., b. What is curl F(1, 2, 1)?, Solution, a. By definition,, curl F  §, , c, , i, , F ∞, x, xy, , j, , y, xz, , k, , ∞, z, xyz 2, , , , , , , , (xyz 2) , (xz)di  c (xyz 2) , (xy)d j  c (xz) , (xy)dk, y, z, x, z, x, y, ,  (xz 2  x)i  yz 2j  (z  x)k,  x(z 2  1)i  yz 2j  (z  x)k, b. curl F(1, 2, 1)  (1)(12  1)i  (2)(12)j  [1  (1)]k  2j  2k, The div and curl of vector fields enjoy some algebraic properties as illustrated in, the following examples. Other properties can be found in the exercises at the end of, this section., , EXAMPLE 8 Let f be a scalar function, and let F be a vector field. If f and the components of F have first-order partial derivatives, show that, div( f F)  f div F  F ⴢ §f, Solution Let’s write F  Pi  Qj  Rk, where P, Q, and R are functions of x, y,, and z. Then, f F  f(Pi  Qj  Rk)  f Pi  fQj  fRk, so the left-hand side of the given equation reads, div( f F)  § ⴢ ( f F)  a, , , , , , i, j, kb ⴢ ( fPi  fQj  fRk), x, y, z, , , , , ( fP) , ( fQ) , ( fR), x, y, z, , f, , f, Q, f, f, P, R, , Pf, , Qf, , R, x, x, y, y, z, z, ,  fa, , Q, f, f, f, P, R, , , ba P, Q, Rb, x, y, z, x, y, z, ,  f( § ⴢ F)  ( §f ) ⴢ F,  f div F  F ⴢ §f, which is equal to the right-hand side.

Page 218 :
15.2, , Divergence and Curl, , 1239, , EXAMPLE 9 Let F  Pi  Qj  Rk be a vector field in space, and suppose that P,, Q, and R have continuous second-order partial derivatives. Show that, div curl F  0, Solution, , Direct computation shows that, , div curl F  § ⴢ ( §, a, , F), , Q, Q, , , , R, P, R, P, i, j, kb ⴢ c a, , bi  a, , bj  a, , bkd, x, y, z, y, z, z, x, x, y, , , , Q,  R,  P, R,  Q, P, a, , b, a, , b, a, , b, x y, z, y z, x, z x, y, , , ,  2Q,  2Q,  2R,  2P,  2R,  2P, , , , , , x y, x z, y z, y x, z x, z y, , 0, Here we have used the fact that the mixed derivatives are equal because, by assumption, they are continuous., , 15.2, , CONCEPT QUESTIONS, , 1. a. Define the divergence of a vector field F and give a formula for finding it., b. Define the curl of a vector field F, and give a formula for, finding it., c. Suppose that F is the velocity vector field associated with, the airflow around an airfoil. Give an interpretation of, § ⴢ F and § F., , 15.2, , 2. a. What is meant by a vector field F that is incompressible?, Give a physical example of an (almost) incompressible, field., b. Repeat part (a) for an irrotational vector field., , EXERCISES, , In Exercises 1–4, you are given the vector field F and a plot, of the vector field in the xy-plane. (The z-component of F is 0.), (a) By studying the plot of F, determine whether div F is positive, negative, or zero. Justify your answer. (b) Find div F, and, reconcile your result with your answer in part (a). (c) By studying the plot of F, determine whether a paddle wheel planted at, a point in the field will rotate clockwise, rotate counterclockwise,, or not rotate at all. Justify your answer. (d) Find curl F, and, reconcile your result with your answer in part (c)., x, 1. F(x, y, z) , i,, 冟x冟, , x0, , 5, 4, 3, 2, 1, 432 1, 2, 3, 4, 5, , y, 2, 1, 3 2 1, 1, , 2. F(x, y, z)  xj, , 1, , 2, , 3 x, , 2, , V Videos for selected exercises are available online at www.academic.cengage.com/login., , 1 2 3 4x

Page 219 :
1240, , Chapter 15 Vector Analysis, , 3. F(x, y, z) , , x, 2x  y, 2, , 2, , i, , y, 2x  y, 2, , 2, , 18. Show that the vector field F(x, y, z)  f(x)i  t(y)j  h(z)k,, where f, t and h are differentiable, is irrotational., , j, , y, 3, , In Exercises 19–26, prove the property for vector fields F and G, and scalar fields f and t. Assume that the appropriate partial, derivatives exist and are continuous., , 2, 1, 3 2 1, 1, , 19. div(F  G)  div F  div G, 1, , 20. curl(F  G)  curl F  curl G, , 3 x, , 2, , 21. curl( f F)  f curl F  ( §f ), , 2, 3, , 23. div(F, 4. F(x, y, z)  , , y, 2x 2  y 2, , i, , x, 2x 2  y 2, , 24. div( §f, , j, , 2, 1, 2 1, 1, , G)  G ⴢ curl F  F ⴢ curl G, §t)  0, , F)  § ( § ⴢ F)  §2F, where, 2, , 2, 2, §2F  a 2  2  2 bF, x, y, z, , y, 3, , 3, , F, , 22. curl( §f )  0, , 25. §, , (§, , 26. §, , [ §f  ( §, , F)]  §, , (§, , F), , Hint: Use the results of Exercises 20 and 22., 1, , 2, , 27. Show that there is no vector field F in space such that, curl F  xyi  yzj  xyk., , 3 x, , Hint: See Example 9., , 2, , 28. Find the value of the constant c such that the vector field, , 3, , G(x, y, z)  (2x  3y  z 2)i  (cy  z)j  (x  y  2z)k, In Exercises 5–12, find (a) the divergence and (b) the curl of the, vector field F., 5. F(x, y, z)  yzi  xzj  xyk, , 30. Let f be a differentiable function, r  xi  yj  zk, and, r  冟 r 冟., a. Find curl[ f(r)r] by interpreting it geometrically., b. Verify your answer to part (a) analytically., , 7. F(x, y, z)  x 2y 3i  xz 2k, 8. F(x, y, z)  yz 2i  x 2zj, 9. F(x, y, z)  sin xi  x cos yj  sin zk, 10. F(x, y, z)  x cos yi  y tan xj  sec zk, , In Exercises 31–34, let r  xi  yj  zk and r  冟 r 冟., , 11. F(x, y, z)  ex cos yi  ex sin yj  ln zk, 12. F(x, y, z)  exyz i  cos(x  y)j  ln(x  z)k, In Exercises 13–15, let F be a vector field, and let f be a scalar, field. Determine whether each expression is meaningful. If so, state, whether the expression represents a scalar field or a vector field., f, ( §f ), , 29. Show that F  (cos x)yi  (sin y)xj is not a gradient vector, field., Hint: See Exercise 22., , 6. F(x, y, z)  x 2yi  xy 2j  xyzk, , 13. a. §, c. §, , is the curl of some vector field F., , b. § ⴢ f, d. grad F, , 14. a. div( §f ), c. § (grad f ), , b. grad( §f ), d. curl(curl F), , 15. a. § ( § F), c. § ⴢ ( § ⴢ F), , b. § ⴢ ( §f ), d. § [ §, , ( §f )], , 16. Find div F if F  grad f, where f(x, y, z)  2xy 2z 3., 17. Show that the vector field F(x, y, z)  f(y, z)i  t(x, z)j , h(x, y)k, where f, t, and h are differentiable, is incompressible., , 31. Show that §r  r>r., 32. Show that §(1>r)  r>r 3., 33. Show that §(ln r)  r>r 2., 34. Show that §r n  nr n2r., In Exercises 35–38, the differential operator §2 (called the, 2, 2, 2, Laplacian) is defined by §2  § ⴢ §  2  2  2 ., x, y, z, 2, 2, f, f, , ,  2f, It acts on f to produce the function §2f  2  2  2 ., x, y, z, Assume that f and t have second-order partial derivatives., 35. Show that § ⴢ ( §f )  §2f., 36. Show that §2( ft)  f §2t  t§2f  2§f ⴢ §t., 37. Show that §2r 3  12r, where r  冟 r 冟 and r  xi  yj  zk.

Page 220 :
15.3, 1, 38. Show that §2 a b  0, where r  冟 r 冟 and r  xi  yj  zk., r, 39. Angular Velocity of a Particle A particle located at the point P is, rotating about the z-axis on a circle of radius R that lies in, the plane z  h, as shown in the figure. Suppose that the, angular speed of the particle is a constant v. Then this rotational motion can be described by the vector w  vk,, which gives the angular velocity of P., z, , Line Integrals, , 1241, , Show that, a. §2E , b. §2H , , 1  2E, c2 t 2, 1  2H, c2 t 2, , Hint: Use Exercise 25., , In Exercises 41–48, determine whether the statement is true or, false. If it is true, explain why. If it is false, explain why or give, an example that shows it is false., 41. If F is a nonconstant vector field, then div F  0., , (0, 0, h), w, , R, ¨, , 42. If F(x, y)  0 and div F  0 for all x and y, then the, streamlines of F must be closed curves., , v(t), P, , 43. If the streamlines of a vector field F are straight lines, then, div F  0., , r(t), y, , 44. If the streamlines of a vector field F are concentric circles,, then curl F  0., , x, , a. Show that the velocity v(t) of P is given by v  w, , r., , Hint: The position of P is r(t)  R cos vti  R sin vtj  hk., , b. Show that v  vyi  vxj., c. Show that curl v  2w. This shows that the angular, velocity of P is one half the curl of its tangential, velocity., , 40. Maxwell’s Equations Maxwell’s equations relating the electric, field E and the magnetic field H, where c is the speed of, light, are given by, § ⴢ E  0,, §, , 15.3, , E, , 45. If the streamlines of a vector field F are straight lines, then, curl F  0., 46. The curl of a “spin” field is never equal to 0., 47. There is no nonzero vector field F such that div F  0 and, curl F  0, simultaneously., 48. If curl F  0, then F must be a constant vector field., , § ⴢ H  0,, , 1 H, ,, c t, , §, , H, , 1 E, c t, , Line Integrals, Line Integrals, Once again recall that the mass of a thin, straight wire of length (b  a) and linear, mass density f(x) is given by, , y, , m, , 冮, , b, , f(x) dx, , a, , y  f(x), b, , m  y f(x) dx, a, , 0, , a, , b, , FIGURE 1, The mass of a wire of length (b  a), and linear mass density f(x) is, 兰ab f(x) dx., , x, , which has the same numerical value as the area under the graph of f on [a, b]. (See, Figure 1.), Instead of being straight, suppose that the wire takes the shape of a plane curve C, described by the parametric equations x  x(t) and y  y(t), where a t b, or, equivalently, by the vector equation r(t)  x(t)i  y(t)j with parameter interval [a, b]. (See, Figure 2a.) Furthermore, suppose that the linear mass density of the wire is given by, a continuous function f(x, y). Then one might conjecture that the mass of the curved, wire should be numerically equal to the area of the region under the graph of z  f(x, y), with (x, y) lying on C. (See Figure 2b.)

Page 221 :
1242, , Chapter 15 Vector Analysis, y, , z, C, , (x, y, f (x, y)), , y, x, , 0, , (x, y), , x, , FIGURE 2, , (a) The curve C gives the shape of a, wire with linear density f (x,y)., , C, , (b) The region under the graph of f along C, , But how do we define this area, and how do we compute it? As we will now see,, this area can be defined in terms of an integral called a line integral, even though the, term “curve integral” would seem more appropriate., Let C be a smooth plane curve defined by the parametric equations, x  x(t),, , y  y(t),, , a, , t, , b, , or, equivalently, by the vector equation r(t)  x(t)i  y(t)j, and let P be a regular partition of the parameter interval [a, b] with partition points, a  t0  t1  t2 p  tn  b, If x k  x(t k) and yk  y(t k) , then the points Pk (x k, yk) divide C into n subarcs +, P0P1 ,, +, P1P2 , p , +, Pn1Pn of lengths ⌬s1, ⌬s2, p , ⌬sn, respectively. (See Figure 3.) Next, we, pick any evaluation point t *, k in the subinterval [t k1, t k]. This point is mapped onto, the point P *, Pk1Pk . If f is any function of two variables, k (x *, k , y*, k ) lying in the subarc +, with domain that contains the curve C, then we can evaluate f at the point (x *, k , y*, k ),, *, *, *, *, obtaining f(x k , y k ) . If f is positive, we can think of the product f(x k , y k ) ⌬sk as representing the area of a curved panel with a curved base of length ⌬sk and constant, height f(x *, k , y*, k ) . (See Figure 4.) This panel is an approximation of the area under the, Pk1Pk . Therefore, the sum, curve z  f(x, y) on the subarc +, n, k , y*, k ) ⌬sk, a f(x *, , (1), , k1, , t, , y, , b  tn, , Pn, , tk, tk*, , Pk, Pk*, , f (x k*, yk*), , Pk1, , tk1, , C, Pk, , t2, t1, , P0, , a  t0, , (a) A parameter interval, , Pk–1, , P1, , 0, , FIGURE 3, , P k*, , P2, , (b) The point Pk (xk , yk ) corresponds, to the point tk ., , x, , FIGURE 4, The product f(x *, k , y*, k ) ⌬sk gives the area, of a curved panel with a curved base of, length ⌬sk and with constant height.

Page 222 :
15.3, , Line Integrals, , 1243, , gives an approximation of the area under the curve z  f(x, y) and along the curve C., If we let n → ⬁ , then it seems reasonable to expect that this sum will approach the, area under the curve z  f(x, y) along the curve C. This observation suggests the following definition., , DEFINITION Line Integral, If f is defined in a region containing a smooth curve C with parametric representation r(t), where a t b, then the line integral of f along C is, , 冮, , C, , n, , f(x, y) ds  lim a f(x *, k , y*, k ) ⌬sk, n→⬁, , (2), , k1, , provided that the limit exists., , Note Observe that over a small piece of a curved wire represented by the segment, +, Pk1Pk , the linear density of the wire does not vary by much. Therefore, we may, assume that the linear mass density of the wire in the segment +, Pk1Pk is approximately, f(x k*, y k*) , so the mass of this segment is approximately f(x k*, y k*) ⌬sk. (This is also the, area of a typical panel.) Adding the masses of all the segments of the wire leads to the, sum (1). Taking the limit as n → ⬁ in (1) then gives the mass of the wire., In general, it can be shown that if f is continuous, then the limit in Equation (2), always exists, and the line integral can be evaluated as an ordinary definite integral, with respect to a single variable by using the following formula., , 冮, , f(x, y) ds , , 冮, , b, , f(x(t), y(t))2[x¿(t)]2  [y¿(t)]2 dt, , (3), , a, , C, , Notes, 1. Equation (3) is easier to remember by observing that the element of arc length is, given by ds  冟 r¿(t) 冟 dt  2[x¿(t)]2  [y¿(t)]2 dt., 2. If C is given by the interval [a, b], then C is just the line segment joining (a, 0) to, (b, 0). So C can be described by the parametric equations x  t and y  0, where, a t b. In this case, Equation (3) becomes 兰C f(x, y) ds  兰ab f(t, 0) dt , 兰ab t(x) dx, where t(x)  f(x, 0). So the line integral reduces to an integral of a, function defined on an interval [a, b], as expected., , EXAMPLE 1 Evaluate 兰C (1  xy) ds, where C is the quarter-circle described by, r(t)  cos ti  sin tj, 0 t p2 , as shown in Figure 5., y, , 1, C, , FIGURE 5, The curve C is described, by r(t)  cos ti  sin tj,, where 0 t p2 ., , 0, , 1, , x

Page 223 :
1244, , Chapter 15 Vector Analysis, , Solution Here, x(t)  cos t and y(t)  sin t, so x¿(t)  sin t and y¿(t)  cos t., Therefore, using Equation (3), we obtain, , 冮, , (1  xy) ds , , 冮, , p>2, , 冮, , p>2, , 冮, , p>2, , (1  cos t sin t)2[x¿(t)]2  [y¿(t)]2 dt, , 0, , C, , , , (1  cos t sin t)2(sin t)2  (cos t)2 dt, , 0, , y, , , , C4, , 0, , C3, , , C2, C1, x, , 0, , FIGURE 6, A piecewise-smooth curve composed, of four smooth curves (n  4), , y, , (1  cos t sin t) dt  ct , , 1 2 p>2, sin td, 2, 0, , p, 1, 1,   (p  1), 2, 2, 2, , A curve is piecewise-smooth if it is made up of a finite number of smooth curves, C1, C2, p , Cn connected at consecutive endpoints as shown in Figure 6. If f is continuous in a region containing C, then it can be shown that, , 冮 f(x, y) ds  冮, C, , f(x, y) ds , , C1, , 冮, , f(x, y) ds  p , , C2, , 冮, , f(x, y) ds, , Cn, , EXAMPLE 2 Evaluate 兰C 2x ds, where C consists of the arc C1 of the parabola y  x 2, , 1, C2, , from (0, 0) to (1, 1) followed by the line segment C2 from (1, 1) to (0, 0) ., , (1, 1), , yx, , Solution The curve C is shown in Figure 7. C1 can be parametrized by taking x  t,, where t is a parameter. Thus,, , y  x2, , C1:, , C1, (0, 0), , 1, , x, , x(t)  t,, , y(t)  t 2,, , 0, , t, , 1, , Therefore,, , 冮, , FIGURE 7, C is composed of two smooth curves, C1 and C2., , 2x ds , , 冮, , 1, , 2x2[x¿(t)]2  [y¿(t)]2 dt, , 0, , C1, , 2, , 冮, , 1, , t21  4t 2 dt, , 0, , 1, 1 2, 515  1,  c2a b a b (1  4t 2)3>2 d , 8 3, 6, 0, , C2 can be parametrized by taking x  1  t. Thus,, C2:, , x(t)  1  t,, , y(t)  1  t,, , 0, , Therefore,, , 冮, , C2, , 2x ds , , 冮, , 1, , 2x 2[x¿(t)]2  [y¿(t)]2 dt, , 0, , 2, , 冮, , 1, , (1  t) 11  1 dt, , 0, ,  c212 at , , 1 2 1, t b d  12, 2, 0, , t, , 1

Page 224 :
15.3, , Line Integrals, , 1245, , Putting these results together, we have, , 冮 2x ds  冮, C, , 冮, , 2x ds , , C1, , 2x ds , , C2, , 515  1,  12, 6, , As we saw earlier, the mass of a thin wire represented by C that has linear mass, density r(x, y) is given by, m, , 冮 r(x, y) ds, C, , The center of mass of the wire is located at the point (x, y), where, x, , 1, m, , 冮 xr(x, y) ds, , y, , C, , 1, m, , 冮 yr(x, y) ds, , (4), , C, , EXAMPLE 3 The Mass and Center of Mass of a Wire A thin wire has the shape of a, semicircle of radius a. The linear mass density of the wire is proportional to the distance from the diameter that joins the two endpoints of the wire. Find the mass of the, wire and the location of its center of mass., y, , Solution If the wire is placed on a coordinate system as shown in Figure 8, then it, coincides with the curve C described by the parametric equations x  a cos t and, y  a sin t, where 0 t p. Its linear mass density is given by r(x, y)  ky, where, k is a positive constant. Since x¿(t)  a sin t and y¿(t)  a cos t, we see that the mass, of the wire is, , C, Center of, mass, , a, , 0, , a, , x, , m, , 冮, , r(x, y) ds , , C, , FIGURE 8, The curve C has parametric equations, x  a cos t and y  a sin t, where, 0 t p., , , , 冮, , 冮, , ky ds , , 冮, , p, , ky2[x¿(t)]2  [y¿(t)]2 dt, , 0, , C, , p, , ka sin t2(a sin t)2  (a cos t)2 dt, , 0, ,  ka 2, , 冮, , 0, , p, , sin t dt  Cka 2 cos tD 0  2ka 2, p, , Next, we note that by symmetry, x  0. Using Equation (4), we obtain, y, , , , 1, m, a, 2, , 冮, , yr(x, y) ds , , C, , 冮, , 0, , p, , sin2 t dt , , a, 4, , 1, 2ka 2, , 冮, , 冮, , p, , ky 2 ds , , 0, , 1, 2a 2, , 冮, , p, , a(a sin t)2 dt, , 0, , p, , (1  cos 2t) dt, , 0, , p, a, 1, 1, ct  sin 2td  pa, 4, 2, 4, 0, , Therefore, the center of mass of the curve is located at 1 0, pa, 4 2 . (See Figure 8.), , Line Integrals with Respect to Coordinate Variables, The line integrals that we have dealt with up to now are taken with respect to arc, length. Two other line integrals are obtained by replacing ⌬sk in Equation (2) by, ⌬x k  x(t k)  x(t k1) and ⌬yk  y(t k)  y(t k1) . In the first instance we have the line

Page 225 :
1246, , Chapter 15 Vector Analysis, , integral of f along C with respect to x,, n, , 冮, , C, , f(x, y) dx  lim a f(x *, k , y*, k ) ⌬x k, n→⬁, k1, , and in the second instance we have the line integral of f along C with respect to y,, , 冮, , C, , n, , f(x, y) dy  lim a f(x *, k , y*, k ) ⌬yk, n→⬁, k1, , Line integrals with respect to both coordinate variables can also be evaluated as, ordinary definite integrals with respect to a single variable. In fact, since x  x(t) and, y  y(t), we see that dx  x¿(t) dt and dy  y¿(t) dt. This leads to the following formulas:, , 冮, , f(x, y) dx , , 冮, , f(x, y) dy , , 冮, , b, , 冮, , b, , f(x(t), y(t))x¿(t) dt, , (5a), , f(x(t), y(t))y¿(t) dt, , (5b), , a, , C, , a, , C, , Thus, if P and Q are continuous functions of x and y, then, , 冮 P(x, y) dx  Q(x, y) dy  冮 P(x, y) dx  冮 Q(x, y) dy, C, , C, , C, , can be evaluated as an ordinary integral of a single variable using the formula, , 冮, , P(x, y) dx  Q(x, y) dy , , (6), , EXAMPLE 4 Evaluate 兰C y dx  x 2 dy, where (a) C is the line segment C1 from, , y, 2, , (1, 1) to (4, 2) , (b) C is the arc C2 of the parabola x  y 2 from (1, 1) to (4, 2), and, (c) C is the arc C3 of the parabola x  y 2 from (4, 2) to (1, 1) . (See Figure 9.), , (4, 2), , C2, C3, , 1, , Solution, a. C1 can be described by the parametric equations, , C1, , 1, , [P(x(t), y(t))x¿(t)  Q(x(t), y(t))y¿(t)] dt, , a, , C, , 0, , 冮, , b, , 1, , 2, , 3, , (1, 1), , FIGURE 9, The curves C1, C2, and C3, , 4, , x  1  3t,, , x, , y  1  3t,, , 0, , t, , 1, , (See Section 10.5.) We have dx  3 dt and dy  3 dt, so Equation (6) gives, , 冮, , y dx  x 2 dy , , 冮, , 1, , (1  3t)(3 dt)  (1  3t)2(3 dt), , 0, , C1, ,  27, , 冮, , 1, , 0, , 1, 1 1 45, (t 2  t) dt  27c t 3  t 2 d , 3, 2 0, 2, , b. A parametric representation of C2 is obtained by letting y  t. Thus,, C2 :, , x  t 2,, , y  t,, , 1, , t, , 2, , Then dx  2t dt and dy  dt, so Equation (6) gives, , 冮, , y dx  x 2 dy , , C2, , , , 冮, , 2, , 1, 2, , t(2t dt)  (t 2)2 dt, , 2, 1 2, 63, (2t 2  t 4) dt  c t 3  t 5 d , 3, 5, 5, 1, 1, , 冮

Page 226 :
15.3, , Line Integrals, , 1247, , c. C3 can be parametrized by taking y  t. Thus,, x  t 2,, , C3:, , y  t,, , 2, , t, , 1, , Then dx  2t dt and dy  dt, so Equation (6) gives, , 冮, , y dx  x 2 dy , , C3, , 冮, , 1, , 2, ,  1, y, B, A, C, x, , 0, y, B, A, , 冮, , C, x, , FIGURE 10, C is the curve consisting of the, points of C but traversed in the, opposite direction., , 1, , 2, 1 1, 63, (2t 2  t 4) dt  c t 3  t 5 d  , 3, 5, 5, 2, 2, , 冮, , Example 4 sheds some light on the nature of line integrals. First of all, the results, of parts (a) and (b) suggest that the value of a line integral depends not only on the, endpoints, but also on the curve joining these points. Second, the results of parts (b), and (c) seem to suggest that reversing the direction in which a curve is traced changes, the sign of the value of the line integral., This latter observation turns out to be true in the general case. For example, suppose that the orientation of the curve C (the direction in which it is traced as t, increases) is reversed. Let C denote precisely the curve C with its orientation reversed, (so that the curve is traced from B to A instead of from A to B as shown in Figure 10)., Then, , C, , 0, , (t)(2t dt)  (t 2) 2 (dt), , P dx  Q dy  , , 冮 P dx  Q dy, C, , In contrast, note that the value of a line integral taken with respect to arc length does, not change sign when C is reversed. These results follow because the terms x¿(t) and, y¿(t) change sign but ds does not when the orientation of C is reversed., , Line Integrals in Space, The line integrals in two-dimensional space that we have just considered can be, extended to line integrals in three-dimensional space. Suppose that C is a smooth space, curve described by the parametric equations, x  x(t),, , y  y(t),, , z  z(t),, , a, , t, , b, , or, equivalently, by the vector equation r(t)  x(t)i  y(t)j  z(t)k, and let f be a function of three variables that is defined and continuous on some region containing C. We, define the line integral of f along C (with respect to arc length) by, n, , 冮 f(x, y, z) ds  lim a f(x *, y*, z *) ⌬s, k, , n→⬁ k1, , C, , k, , k, , k, , This integral can be evaluated as an ordinary integral by using the following formula,, which is the analog of Equation (3) for the three-dimensional case., , 冮, , C, , f(x, y, z) ds , , 冮, , b, , f(x(t), y(t), z(t)), , a, , dy 2, dx 2, dz 2, b  a b  a b dt, B dt, dt, dt, a, , (7), , If we make use of vector notation, Equation (7) can be written in the equivalent form, , 冮, , C, , f(x, y, z) ds , , 冮, , a, , b, , f(r(t))冟 r¿(t) 冟 dt

Page 227 :
1248, , Chapter 15 Vector Analysis, , EXAMPLE 5 Evaluate 兰C kz ds, where k is a constant and C is the circular helix with, parametric equations x  cos t, y  sin t, and z  t, where 0 t 2p., Solution, , With x¿(t)  sin t, y¿(t)  cos t, and z¿(t)  1, Equation (7) gives, , 冮, , kz ds , , 冮, , 2p, , 冮, , 2p, , kt 2[x¿(t)]2  [y¿(t)]2  [z¿(t)]2 dt, , 0, , C, , , , kt 2sin2 t  cos2 t  1 dt, , 0, ,  12k, , 冮, , 0, , 2p, , 1 2p, t dt  12kc t 2 d  212kp2, 2 0, , Note In Example 5, suppose that C represents a thin wire whose linear mass density, is directly proportional to its height. Then our calculations tell us that its mass is, 212kp2 units., Line integrals along a curve C in space with respect to x, y, and z are defined in, much the same way as line integrals along a curve in two-dimensional space. For example, the line integral of f along C with respect to x is given by, n, , 冮, , f(x, y, z) dx  lim a f(x *, k , y*, k , z*, k ) ⌬x k, n→⬁, , 冮, , f(x, y, z) dx , , C, , k1, , so, , 冮, , b, , f(x(t), y(t), z(t))x¿(t) dt, , (8), , a, , C, , If the line integrals with respect to x, y, and z occur together, we have, , 冮 P(x, y, z) dx  Q(x, y, z) dy  R(x, y, z) dz, C, , , , 冮, , a, , b, , cP(x(t), y(t), z(t)), , dy, dx, dz,  Q(x(t), y(t), z(t)),  R(x(t), y(t), z(t)) d dt, dt, dt, dt, , (9), , EXAMPLE 6 Evaluate 兰C y dx  z dy  x dz, where C consists of part of the twisted, cubic C1 with parametric equations x  t, y  t 2, and z  t 3, where 0 t 1, followed by the line segment C2 from (1, 1, 1) to (0, 1, 0) ., Solution The curve C is shown in Figure 11. Integrating along C1, we have dx  dt,, dy  2t dt, and dz  3t 2 dt. Therefore,, , z, , 冮, , y dx  z dy  x dz , , C1, , 冮, , 1, , t 2 dt  t 3(2t dt)  t(3t 2) dt, , 0, , C1, , (1, 1, 1), 0, , 冮, , 1, , , , C2, , (t 2  3t 3  2t 4) dt, , 0, , (0, 1, 0), , 1, 3, 2 1 89,  c t 3  t 4  t 5d , 3, 4, 5 0 60, , y, , x, , FIGURE 11, The curve C is composed of C1 and, C2 traversed in the directions shown., , Next, we write the parametric equations of the line segment from (1, 1, 1) to (0, 1, 0)., On C2:, , x  1  t,, , y  1,, , z  1  t,, , 0, , t, , 1

Page 228 :
15.3, , Line Integrals, , 1249, , Then dx  dt, dy  0, and dz  dt. Therefore,, , 冮, , y dx  z dy  x dz , , 冮, , 1, , 冮, , 1, , 1(dt)  (1  t)(0)  (1  t)(dt), , 0, , C2, , , , 0, , 1, 1, 3, (t  2) dt  c t 2  2td  , 2, 2, 0, , Finally, putting these results together, we have, , 冮 y dx  z dy  x dz  60  2  60, 89, , 3, , 1, , C, , Line Integrals of Vector Fields, Up to now, we have considered line integrals involving a scalar function f. We now, turn our attention to the study of line integrals of vector fields. Suppose that we, want to find the work done by a continuous force field F in moving a particle from, a point A to a point B along a smooth curve C in space. Let C be represented parametrically by, x  x(t),, , y  y(t),, , z  z(t),, , a, , t, , b, , or, equivalently, by the vector equation r(t)  x(t)i  y(t)j  z(t)k with parameter, interval [a, b]. Take a regular partition P of the parameter interval [a, b] with partition, points, a  t0  t1  t2  p  tn  b, If x k  x(t k), yk  y(t k), and z k  z(t k), then the points Pk (x k, yk, z k) divide C into n, subarcs +, P0P1 , +, P1P2 , p , +, Pn1Pn of lengths ⌬s1, ⌬s2, p , ⌬sn, respectively. (See Figure 12.) Furthermore, because r is smooth, the unit tangent vector T(t) at any point on, the subarc +, Pk1Pk will not exhibit an appreciable change in direction and may be, approximated by T(t k*). Also, because F is continuous, the force F(x(t), y(t), z(t)) for, t k1 t t k is approximated by F(x k*, y k*, z k*) . Therefore, we can approximate the, work done by F in moving the particle along the curve from Pk1 to Pk by the work, done by the component of the constant force F(x k*, y k*, z k*) in the direction of the line, segment (approximated by T(t *, k ) ) from Pk1 to Pk, that is, by, ⌬Wk  F(x *, k , y*, k , z*, k ) ⴢ T(t *, k ) ⌬sk, z, , t, b  tn, , T(tk*), , tk, tk*, , Pk, Pk*, Pk1, , tk1, t2, t1, a  t0, , (a) A parameter interval, , Pn, , A  P0, P2, x, , FIGURE 12, , Constant force in the direction, of T(x k*) times displacement, , P1, , (b) The partition P of [a, b] breaks the, curve C into n subarcs., , y

Page 229 :
1250, , Chapter 15 Vector Analysis, , Here, we have used the fact that the length of the line segment from Pk1 to Pk is approximately ⌬sk. So the total work done by F in moving the particle from A to B is, n, , W ⬇ a F(x *, k , y*, k , z*, k ) ⴢ T(t *, k ) ⌬sk, k1, , This approximation suggests that we define the work W done by the force field, F as, n, , W  lim a F(x *, k , y*, k , z*, k ) ⴢ T(t *, k ) ⌬sk , n→⬁, k1, , 冮 F ⴢ T ds, , (10), , C, , Since T(t)  r¿(t)> 冟 r¿(t) 冟, Equation (10) can also be written in the form, W, , 冮, , b, , 冮, , b, , a, , , , cF(r(t)) ⴢ, , r¿(t), d 冟 r¿(t) 冟 dt, 冟 r¿(t) 冟, , F(r(t)) ⴢ r¿(t) dt, , a, , The last integral is usually written in the form 兰C F ⴢ dr. In words, it says that the work, done by a force is given by the line integral of the tangential component of the force, with respect to arc length. Although this integral was defined in the context of work, done by a force, integrals of this type occur frequently in many other areas of physics, and engineering., , DEFINITION Line Integral of Vector Fields, Let F be a continuous vector field defined in a region that contains a smooth, curve C described by a vector function r(t), a t b. Then the line integral, of F along C is, , 冮, , C, , F ⴢ dr , , 冮, , C, , F ⴢ T ds , , 冮, , b, , F(r(t)) ⴢ r¿(t) dt, , (11), , a, , z, , Note We remind you that dr is an abbreviation for r¿(t) dt and that F(r(t)) is an abbreviation for F(x(t), y(t), z(t)) ., , (0, 1, ), p, _, 2, , EXAMPLE 7 Find the work done by the force field F(x, y, z)  yi  xj  zk in, moving a particle along the helix C described by the parametric equations x  cos t,, y  sin t, and z  t from (1, 0, 0) to 1 0, 1, p2 2 . (See Figure 13.), , C, –1, 1, , 1, , y, , Since x(t)  cos t, y(t)  sin t, and z(t)  t, we see that, F(r(t))  F(x(t), y(t), z(t))  yi  xj  zk,  sin ti  cos tj  tk, , x, , FIGURE 13, The curve C is described by, r(t)  cos ti  sin tj  tk, 0, , Solution, , Furthermore, observe that the vector equation of C is, t, , p, 2., , r(t)  x(t)i  y(t)j  z(t)k  cos ti  sin tj  tk, , 0, , t, , p, 2

Page 230 :
15.3, , Line Integrals, , 1251, , from which we have, r¿(t)  sin ti  cos tj  k, Therefore, the work done by the force is, W, , 冮, , F ⴢ dr , , F(r(t)) ⴢ r¿(t) dt, , 0, , C, , , , 冮, , p>2, , 冮, , p>2, , 冮, , p>2, , (sin ti  cos tj  tk) ⴢ (sin ti  cos tj  k) dt, , 0, , , , (sin2 t  cos2 t  t) dt , , 0, , 冮, , 0, , p>2, , (1  t) dt  ct , , 1 2 p>2 p, p, t d,  a1  b, 2 0, 2, 4, , We close this section by pointing out the relationship between line integrals of vector, fields and line integrals of scalar fields with respect to the coordinate variables. Suppose, that a vector field F in space is defined by F  P(x, y, z)i  Q(x, y, z)j  R(x, y, z)k., Then by Equation (11) we have, , 冮, , F ⴢ dr , , 冮, , b, , 冮, , b, , 冮, , F(r(t)) ⴢ r¿(t) dt , , a, , C, , , , b, , (Pi  Qj  Rk) ⴢ (x¿(t)i  y¿(t)j  z¿(t)k) dt, , a, , [P(x(t), y(t), z(t))x¿(t)  Q(x(t), y(t), z(t))y¿(t)  R(x(t), y(t), z(t))z¿(t)] dt, , a, , But the integral on the right is just the line integral of Equation (9). Therefore, we have, shown that, , 冮 F ⴢ dr  冮 P dx  Q dy  R dz, C, , where F  Pi  Qj  Rk, , (12), , C, , You are urged to rework Example 7 with the aid of Equation (12)., As a consequence of Equation (12), we have the result, , 冮, , C, , F ⴢ dr  , , 冮 F ⴢ dr, C, , (see page 1247). This result also follows from the equation, , 冮, , C, , F ⴢ dr  , , 冮 F ⴢ T ds, C, , and we observe that even though line integrals with respect to arc length do not change, sign when the direction traversed is reversed, the unit vector T does change sign when, C is replaced by C., , EXAMPLE 8 Let F(x, y)  18 (x  y)i  18 (x  y)j be the force field shown in Figure 14. Find the work done on a particle that moves along the quarter-circle of radius 1, centered at the origin (a) in a counterclockwise direction from (1, 0) to (0, 1) and, (b) in a clockwise direction from (0, 1) to (1, 0).

Page 231 :
1252, , Chapter 15 Vector Analysis, y, , y, , 1, , 1, , 0, , FIGURE 14, The force field, F(x, y)  18 (x  y)i  18 (x  y)j, , x, , 1, , 1 x, , 0, , (a) The direction of the path, goes against the direction of F., , (b) The direction of the path, has the same direction as the, direction of F., , Solution, a. The path of the particle may be represented by r(t)  cos ti  sin tj for, 0 t p2 . Since x  cos t and y  sin t, we find, 1, 1, F(r(t))   (cos t  sin t)i  (cos t  sin t)j, 8, 8, and, r¿(t)  sin ti  cos tj, Therefore, the work done by the force on the particle is, , 冮, , p>2, , 冮, , F ⴢ dr , , F(r(t)) ⴢ r¿(t) dt , , 0, , C, , , , 1, 8, , 冮, , p>2, , dt  , , 0, , 冮, , 1, 8, , p>2, , (cos t sin t  sin2 t  cos2 t  sin t cos t) dt, , 0, , p, 16, , b. Here, we can represent the path by r(t)  sin ti  cos tj for 0, x  sin t and y  cos t. So, , 冮 F ⴢ dr  冮, , p>2, , F(r(t)) ⴢ r¿(t) dt , , 0, , C, , , , 1, 8, , 冮, , 0, , p>2, , dt , , 1, 8, , 冮, , t, , p, 2., , Then, , p>2, , (sin t cos t  cos2 t  sin2 t  sin t cos t) dt, , 0, , p, 16, , In Example 8, observe that the work done by F on the particle in part (a) is negative because the force field opposes the motion of the particle., , 15.3, , CONCEPT QUESTIONS, , 1. a. Define the line integral of a function f(x, y, z) along a, smooth curve C with parametric representation r(t),, where a t b., b. Write a formula for evaluating the line integral of part (a)., 2. a. Define the line integral of a function f(x, y, z) along a, smooth curve with respect to x, with respect to y, and, with respect to z., , b. Write formulas for evaluating the line integrals for the, integrals of part (a)., c. Write a formula for evaluating 兰C P dx  Q dy  R dz., 3. a. Define the line integral of a vector field F along a, smooth curve C., b. If F is a force field, what does the line integral in part (a), represent?

Page 232 :
15.3, , 15.3, , EXERCISES, , In Exercises 1–22, evaluate the line integral over the given, curve C., 1., , 冮 (x  y) ds;, , C: r(t)  3ti  4tj,, , 0, , t, , 14., , 冮 (x, , 2, , C, , 1, ,  2y) ds; C: r(t)  ti  (t  1)j, 0, , t, , 2, , 15., , C, , 3., , 冮 y ds;, 冮, , C: r(t)  2ti  t 3 j, 0, , 0, t, , 1, 16., , (x  y ) ds;, 3, , C: r(t)  t i  tj,, 3, , 0, , t, , 1, , 冮 (xy, , 2, ,  yx 2) ds, where C is the upper semicircle, , 17., , 冮 xyz, , 1, 2, , ds, where C is the line segment joining (1, 1, 0) to, , 冮 xy, , 2, , ds; C: r(t)  cos 2ti  sin 2tj  3tk, 0, , p, 2, , t, , C, , y  24  x 2, , 冮, , t, , (2, 3, 1), , C, , 6., , C: r(t)  (1  t)i  2tj  (1  t)k,, , C, , C, , 5., , 冮 xyz ds;, C, , C, , 4., , 冮 (2x  y) dx  2y dy, where C consists of the elliptical, path 9x 2  16y 2  144 from (4, 0) to (0, 3) and the circular, path x 2  y 2  9 from (0, 3) to (3, 0), , C, , 2., , 1253, , Line Integrals, , 18., (x  y ) ds, where C is the right half of the circle, 2, , C, , x 2  y2  9, , 冮 (8x  27z) ds;, , C: r(t)  ti  2t 2 j  3t 3 k,, , 0, , t, , 1, , C, , 2, , 19., , 冮 (x  y) dx  xy dy  y dz;, , C: r(t)  et i  et j  2e2t k,, , C, , 7., , 冮 2xy ds, where C is the line segment joining (2, 1) to, , 0, , C, , 20., , (1, 3), , t, , 1, , 冮 x dx  y, , 2, , dy  yz dz; C: r(t)  ti  cos tj  sin tk,, , C, , 8., , 冮 (x, , 2, , 0, ,  2y) ds, where C is the line segment joining (1, 1), , C, , 21., , to (0, 3), 9., , 2, , 10., , 1, , 22., , 冮 (x  3y ) dy;, 2, , 11., , t, , 1, , 冮 xy dx  (x  y) dy, where C consists of the line segment, C, , from (1, 2) to (3, 4) and the line segment from (3, 4) to, (4, 0), 12., , 冮 (y  x) dx  y, , 2, , dy, where C consists of the line segment, , C, , from (0, 0) to (1, 0), and the line segment from (1, 0) to, (2, 4), 13., , 2, , dz, where C consists of the line, , 冮 (x  y  z) dx  (x  y) dy  xz dz, where C consists, C, , C: r(t)  (1  2t)i  (1  3t)j,, , of the line segment from (0, 0, 0) to (1, 1, 1) and the line, segment from (1, 1, 1) to (1, 2, 3), , C, , 0, , 冮 xy dx  yz dy  x, , segment from (0, 0, 0) to (1, 1, 0) and the line segment from, (1, 1, 0) to (2, 3, 5), , C: r(t)  (1  2t)i  (1  3t)j,, , C, , t, , p, 4, , C, , 冮 (x  3y ) dx;, 0, , t, , 冮 y dx  x dy, where C consists of the arc of the parabola, C, , y  4  x 2 from (2, 0) to (0, 4) and the line segment, from (0, 4) to (2, 0), , 23. A thin wire has the shape of a semicircle of radius a. Find, the mass and the location of the center of mass of the wire if, it has a constant linear mass density k., 24. A thin wire in the shape of a quarter-circle r(t) , a cos ti  a sin tj, 0 t p2 , has linear mass density, p(x, y)  k(x  y), where k is a positive constant. Find the, mass and the location of the center of mass of the wire., 25. A thin wire has the shape of a semicircle x 2  y 2  a 2,, y  0. Find the center of mass of the wire if the linear mass, density of the wire at any point is proportional to its distance, from the line y  a., 26. A thin wire of constant linear mass density k takes the shape, of an arch of the cycloid x  a(t  sin t), y  a(1  cos t),, 0 t 2p. Determine the mass of the wire, and find the, location of its center of mass., , V Videos for selected exercises are available online at www.academic.cengage.com/login.

Page 233 :
1254, , Chapter 15 Vector Analysis, C: r(t)  ti  t 2j,, , 27. A thin wire of constant linear mass density k has the shape, of the astroid x  cos3 t, y  sin3 t, 0 t p2 . Determine, the location of its center of mass., , 32. F(x, y)  ln xi  y 2j;, , 28. A thin wire has the shape of the helix x  a cos t,, y  a sin t, z  bt, 0 t 3p. Find the mass and the, center of mass of the wire if it has constant linear mass, density k., , 34. F(x, y)  xi  (y  1)j, where C is an arch of the cycloid, x  t  sin t, y  1  cos t, 0 t 2p, , 冮 xr(x, y, z) ds, y  m 冮 yr(x, y, z) ds,, 1, z  冮 zr(x, y, z) ds, where m  冮 r(x, y, z) ds, m, , Hint: x , , 1, m, , 1, , C, , C, , C, , C, , 29. The vector field F(x, y)  (x  y)i  (x  y)j is shown in, the figure. A particle is moved from the point (2, 0) to the, point (2, 0) along the upper semicircle of radius 2 with center at the origin., , 1, , t, , 2, , 33. F(x, y)  xe i  yj, where C is the part of the parabola, y  x 2 from (1, 1) to (2, 4), y, , 35. F(x, y, z)  x 2i  y 2j  z 2k;, 0 t 1, , C: r(t)  ti  t 2j  t 3k,, , 36. F(x, y, z)  (x  2y)i  2zj  (x  y)k, where C is the, line segment from (1, 3, 2) to (1, 2, 4)., 37. Walking up a Spiral Staircase A spiral staircase is described by, the parametric equations, x  5 cos t,, , y  5 sin t,, , z, , 16, t,, p, , 0, , t, , p, 2, , 2, , where the distance is measured in feet. If a 90-lb girl walks, up the staircase, what is the work done by her against gravity in walking to the top of the staircase?, , 1, , Note: You can also obtain the answer using elementary physics., , y, , 0, , x, , 1, 2, 2 1, , 0, , 1, , 2, , a. By inspection, determine whether the work done by F on, the particle is positive, zero, or negative., b. Find the work done by F on the particle., y, , x, i, j, 2x  y 2, 2x 2  y 2, is shown in the figure. A particle is moved once around the, circle of radius 2 with center at the origin in the counterclockwise direction., , 30. The vector field F(x, y)  , , 38. A particle is moved along a path from (0, 0) to (1, 2) by the, force F  2xy 2i  3yx 2j. Which of the following polygonal, paths results in the least work?, a. The path from (0, 0) to (1, 0) to (1, 2), b. The path from (0, 0) to (0, 2) to (1, 2), c. The path from (0, 0) to (1, 2), 39. Newton’s Second Law of Motion Suppose that the position of a, particle of varying mass m(t) in 3-space at time t is r(t)., According to Newton’s Second Law of Motion, the force, acting on the particle at r(t) is, , 2, , y, 2, , F(r(t)) , , a. Show that F(r(t)) ⴢ r¿(t)  m¿(t)√2 (t)  m(t)√(t)√¿(t),, where √  冟 r¿ 冟 is the speed of the particle., b. Show that if m is constant, then the work done by the, force in moving the particle along its path from t  a to, t  b is, , 1, 0, , W, , x, , 1, , d, [m(t)v(t)], dt, , m 2, [√ (b)  √2(a)], 2, , Note: The function W(t)  12 m√2(t) is the kinetic energy of the, , 2, , particle., 2 1, , 0, , 1, , 2, , a. By inspection, determine whether the work done by F on, the particle is positive, zero, or negative., b. Find the work done by F on the particle., In Exercises 31–36, find the work done by the force field F on a, particle that moves along the curve C., 31. F(x, y)  (x 2  y 2)i  xyj;, 0 t 1, , C: r(t)  t 2i  t 3j,, , 40. Work Done by an Electric Field Suppose that a charge of Q, coulombs is located at the origin of a three-dimensional, coordinate system. This charge induces an electric field, E(x, y, z) , , cQ, r, 冟 r 冟3, , where r  xi  yj  zk and c is a constant (see Example 4, in Section 15.1). Find the work done by the electric field on, a particle of charge q coulombs as it is moved along the, path C: r(t)  ti  2tj  (1  4t)k, where 0 t 1.

Page 234 :
15.3, 41. Work Done by an Electric Field The electric field E at any point, (x, y, z) induced by a point charge Q located at the origin is, given by, E, , Qr, 4pe0 冟 r 冟3, , Line Integrals, , 1255, , In words, the line integral of the tangential component of the, magnetic field around a closed loop C is proportional to the, current I passing through any surface bounded by the loop., The constant m0 is called the permeability of free space. By, taking the loop to be a circle of radius r centered on the, wire, show that the magnitude B  冟 B 冟 of the magnetic field, at a distance r from the center of the wire is, , where r  具x, y, z典 and e0 is a positive constant called the, permittivity of free space., m0I, B, a. Find the work done by the field when a particle of, 2pr, charge q coulombs is moved from A(2, 1, 0) to D(0, 5, 5), along the indicated paths., cas In Exercises 43 and 44, plot the graph of the vector field F and, the curve C on the same set of axes. Guess at whether the line, z, integral of F over C is positive, negative, or zero. Verify your, D(0, 5, 5), answer by evaluating the line integral., 4, , 1, 1, (x  y)i  (x  y)j; C is the curve, 2, 2, r(t)  2 sin ti  2 cos tj, 0 t p, , 43. F(x, y) , , C(0, 5, 0), 4, x, , 4, , A(2, 1, 0), , y, , B(2, 5, 0), , (i) The straight line segment from A to D., (ii) The polygonal path from A(2, 1, 0) to B(2, 5, 0) to, C(0, 5, 0) and then to D(0, 5, 5)., b. Is there any difference in the work done in part (a) and, part (b)?, 42. Magnitude of a Magnetic Field The following figure shows a, long straight wire that is carrying a steady current I. This, current induces a magnetic field B whose direction is circumferential; that is, it circles around the wire. Ampere’s, Law states that, , 冮 B ⴢ dr  m I, 0, , C, , B, , I, , 44. F(x, y) , 1, , t, , 1, 1, xi  yj; C is the curve r(t)  ti  (1  t 2)j,, 4, 2, 1, , In Exercises 45–48, determine whether the statement is true or, false. If it is true, explain why. If it is false, explain why or give, an example that shows it is false., 45. If F(x, y)  xi  yj, then 兰C F ⴢ dr  0, where C is any, circular path centered at the origin., 46. If f(x, y) is continuous and C is a smooth curve, then, 兰C f(x, y) ds   兰C f(x, y) ds., 47. If C is a smooth curve defined by r(t)  x(t)i  y(t)j with, a, , t, , b, then 兰C xy dy  12 xy 2 兩 ., ta, tb, , 48. If f(x, y) is continuous and C is a smooth curve defined by, r(t)  x(t)i  y(t)j with a t b, then, C 兰C f(x, y) dsD 2  C 兰C f(x, y) dxD 2  C 兰C f(x, y) dyD 2.

Page 235 :
1256, , Chapter 15 Vector Analysis, , 15.4, , Independence of Path and Conservative Vector Fields, The gravitational field possesses an important property that we will demonstrate in the, following example., z, , EXAMPLE 1 Work Done on a Particle by a Gravitational Field Consider the gravitational field F induced by an object of mass M located at the origin (see Example 3,, Section 15.1):, , B, C, , F(x, y, z)  , , A, , , , y, x, , GM, r, 冟 r 冟3, GMx, , i, , (x  y  z ), 2, , 2, , 2 3>2, , GMy, (x  y  z ), 2, , 2, , 2 3>2, , j, , GMz, (x  y 2  z 2)3>2, 2, , k, , Suppose that a particle with mass m moves in the gravitational field F from the point, A(x(a), y(a), z(a)) to the point B(x(b), y(b), z(b)) along a smooth curve C defined by, , FIGURE 1, The particle moves from A to B along, the path C in the gravitational field., , r(t)  x(t)i  y(t)j  z(t)k, with parameter interval [a, b]. (See Figure 1.) What is the work W done by F on the, particle?, Solution To find W, we note that the particle moving in the gravitational field F is, subjected to a force of mF, so the work done by the force on the particle is, , W, , 冮, , mF ⴢ dr , , 冮, , b, , mF(r(t)) ⴢ r¿(t) dt, , a, , C, ,  GMm, , 冮, , b, , a, , ⴢc, , c, , x, (x  y  z ), 2, , 2, , 2 3>2, , i, , y, (x  y  z ), 2, , 2, , 2 3>2, , j, , z, (x  y  z 2)3>2, 2, , 2, , kd, , dy, dx, dz, i, j, kd dt, dt, dt, dt, ,  GMm, , 冮, , a, , b, , c, , x, (x  y  z ), 2, , 2, , 2 3>2, , y, dy, dx, z, dz,  2,  2, d dt, 2, 2 3>2 dt, 2, 2 3>2 dt, dt, (x  y  z ), (x  y  z ), , But the expression inside the brackets can be written as, f dx, f dy, f dz, d, f(x, y, z) , , , dt, x dt, y dt, z dt, where, sf(x, y, z) , , 1, 2x  y 2  z 2, 2, , as you can verify. (Also, see Example 6 in Section 15.1.) Using this result, we can, write, W  GMm, , 冮, , a, , b, , tb, d, 1, GMm, a, b, dt, , , `, dt 2x 2  y 2  z 2, 2x 2  y 2  z 2 ta, ,  GMm f(x, y, z) `, , tb, ,  GMm[f(x(b), y(b), z(b))  f(x(a), y(a), z(a))], ta

Page 236 :
15.4, , Note, , Independence of Path and Conservative Vector Fields, , 1257, , Don’t worry about finding the potential function, f(x, y, z) , , 1, 2x  y 2  z 2, 2, , for the gravitational field F. We will develop a systematic method for finding potential, functions f of gradient fields §f later in this section., Example 1 shows that the work done on a particle by a gravitational field F depends, only on the initial point A and the endpoint B of a curve C and not on the curve itself., We say that the value of the line integral along the path C is independent of the path., (A path is a piecewise-smooth curve.), More generally, we say that the line integral 兰C F ⴢ dr is independent of path if, , 冮, , C1, , F ⴢ dr , , 冮, , F ⴢ dr, , C2, , for any two paths C1 and C2 that have the same initial and terminal points., Observe that the gravitational field F happens to be a conservative vector field with, potential function f ; that is, F  §f. Also, Example 1 seems to suggest that if F  §f, is a gradient vector field with potential function f, then, , 冮 F ⴢ dr  冮, C, , §f ⴢ dr  f(x(b), y(b), z(b))  f(x(a), y(a), z(a)), , (1), , C, , This expression reminds us of Part 2 of the Fundamental Theorem of Calculus which, states that, , 冮, , b, , F¿(x) dx  F(b)  F(a), , a, , where F is continuous on [a, b]. The Fundamental Theorem of Calculus, Part 2, tells us, that if the derivative of F in the interior of the interval [a, b] is known, then the integral, of F¿ over [a, b] is given by the difference of the values of F (an antiderivative of F¿), at the endpoints of [a, b]. If we think of §f as some kind of derivative of f, then Equation (1) says that if we know the “derivative” of f, then the line integral of §f is given, by the difference of the values of the potential function f (“antiderivative” of §f ) at the, endpoints of the curve C., We now show that Equation (1) is indeed true for all conservative vector fields., We state and prove the result for a function f of two variables and a curve C in the, plane., , THEOREM 1 Fundamental Theorem for Line Integrals, Let F(x, y)  §f(x, y) be a conservative vector field in an open region R, where, f is a differentiable potential function for F. If C is any piecewise-smooth curve, lying in R given by, r(t)  x(t)i  y(t)j, , a, , t, , b, , then, , 冮 F ⴢ dr  冮, C, , C, , §f ⴢ dr  f(x(b), y(b))  f(x(a), y(a))

Page 237 :
1258, , Chapter 15 Vector Analysis, , PROOF We will give the proof for a smooth curve C. Since F(x, y)  §f  fx(x, y)i , fy (x, y)j, we see that, , 冮, , F ⴢ dr , , C, , 冮, , §f ⴢ dr , , , , 冮, 冮, , b, , a, , , , a, , c, , A(1, 2), , d, [ f(x(t), y(t))] dt, dt, tb, ,  f(x(b), y(b))  f(x(a), y(a)), ta, , a. Prove that F is conservative by showing that it is the gradient of the potential, function f(x, y)  x 2y., b. Use the Fundamental Theorem for Line Integrals to evaluate 兰C F ⴢ dr, where C, is any piecewise-smooth curve joining the point A(1, 2) to the point B(3, 1), (See Figure 2.), , C, , 2, , 0, , Use the Chain Rule., , EXAMPLE 2 Let F(x, y)  2xyi  x 2j be a force field., , B(3, 1), 1, , 1, , dr, dt, dt, , f dx, f dy, , d dt, x dt, y dt, ,  f(x(t), y(t)) `, , y, , §f ⴢ, , a, , C, b, , 冮, , b, , 1, , 2, , x, , 3, , Solution, , FIGURE 2, C is a piecewise smooth curve joining, A to B., ,  2,  2, (x y)i , (x y)j  2xyi  x 2j  F(x, y), we conclude that, x, y, F is indeed conservative., b. Thanks to the Fundamental Theorem for Line Integrals, we do not need to know, the rule defining the curve C; the integral depends only on the coordinates of the, endpoints A and B of the curve. We have, a. Since §f(x, y) , , y, , 冮 F ⴢ dr  f(3, 1)  f(1, 2)  x y `, , (3, 1), , 2, , C, , C, r(t), , Line Integrals Along Closed Paths, , r(a)  r(b), 0, , FIGURE 3, On the closed curve C, the tip of r(t), starts at r(a), traverses C, and ends up, back at r(b)  r(a)., , (1, 2), ,  (3)2(1)  (1)2 (2)  7, , x, , A path is closed if its terminal point coincides with its initial point. If a curve C has parametric representation r(t) with parameter interval [a, b], then C is closed if r(a)  r(b)., (See Figure 3.), The following theorem gives an alternative method for determining whether a line, integral is independent of path., , THEOREM 2, Suppose that F is a continuous vector field in a region R. Then 兰C F ⴢ dr is independent of path if and only if 兰C F ⴢ dr  0 for every closed path C in R.

Page 238 :
15.4, , Independence of Path and Conservative Vector Fields, , 1259, , PROOF Suppose that 兰C F ⴢ dr is independent of path in R, and let C be any closed, path in R. We can pick any two points A and B on C and regard C as being made up, of the path from A to B and the path from B to A. (See Figure 4a.) Then, , 冮 F ⴢ dr  冮, C, , F ⴢ dr , , C1, , 冮, , F ⴢ dr , , C2, , 冮, , F ⴢ dr , , C1, , 冮, , C2, , F ⴢ dr, , where C2 is the path C2 traversed in the opposite direction. But both C1 and C2, have the same initial point A and the same terminal point B. Since the line integral is, assumed to be independent of path, we have, , 冮, , F ⴢ dr , , C1, , 冮, , C2, , F ⴢ dr, , and this implies that 兰C F ⴢ dr  0., y, , y, , C2, B, , C2, B, A, , C1, , A, C1, x, , 0, , FIGURE 4, C is a closed path in an open region R., , (b) C is made up of C1 and C2., , (a) C is made up of C1 and C2., , Conversely, suppose that 兰C F ⴢ dr  0 for every closed path C in R. Let A and B, be any two points in R and let C1 and C2 be any two paths in R connecting A to B,, respectively. (See Figure 4b.) Let C be the closed path composed of C1 followed by, C2. Then, , y, R1, , x, , 0, , B, , 0, , 冮 F ⴢ dr  冮, C, , A, , C1, , F ⴢ dr , , 冮, , C2, , F ⴢ dr , , 冮, , C1, , F ⴢ dr , , 冮, , F ⴢ dr, , C2, , so 兰C1 F ⴢ dr  兰C2 F ⴢ dr, which shows that the line integral is independent of path., x, , 0, , (a) The plane region R1 is connected., y, , As a consequence of Theorem 1, we see that if a body moves along a closed path, that ends where it began, then the work done by a conservative force field on the body, is zero., , B, , Independence of Path and Conservative Vector Fields, , R2, A, , 0, , x, , (b) The region R2 is not connected,, since it is impossible to find a path, from A to B lying strictly within R2., , FIGURE 5, , The Fundamental Theorem for Line Integrals tells us that the line integral of a conservative vector field is independent of path. A question that arises naturally is: Is a, vector field whose integral is independent of path necessarily a conservative vector, field? To answer this question, we need to consider regions that are both open and, connected. A region is open if it doesn’t contain any of its boundary points. It is connected if any two points in the region can be joined by a path that lies in the region., (See Figure 5.) The following theorem provides an answer to part of the first question, that we raised.

Page 239 :
1260, , Chapter 15 Vector Analysis, , THEOREM 3 Independence of Path and Conservative Vector Fields, Let F be a continuous vector field in an open, connected region R. The line integral 兰C F ⴢ dr is independent of path if and only if F is conservative, that is, if, and only if F  §f for some scalar function f., , PROOF If F is conservative, then the Fundamental Theorem for Line Integrals implies, , y, , that the line integral is independent of path. We will prove the converse for the case in, which R is a plane region; the proof for the three-dimensional case is similar. Suppose, that the integral is independent of path in R. Let (x 0, y0) be a fixed point in R, and let, (x, y) be any point in R. If C is any path from (x 0, y0) to (x, y), we define the function, f by, , (x1, y), (x, y), R, , f(x, y) , , C2, , 冮, , F ⴢ dr , , (x0, y0), x, , 0, , FIGURE 6, The path C consists of an arbitrary, path C1 from (x 0, y0) to (x 1, y), followed by the horizontal line, segment from (x 1, y) to (x, y)., , F ⴢ dr, , (x0, y0), , C, , C1, , 冮, , (x, y), , Since R is open, there exists a disk contained in R with center (x, y). Pick any point, (x 1, y) in the disk with x 1  x. Now, by assumption, the line integral is independent, of path, so we can choose C to be the path consisting of any path C1 from (x 0, y0) to, (x 1, y) followed by the horizontal line segment C2 from (x 1, y) to (x, y), as shown in, Figure 6. Then, f(x, y) , , 冮, , F ⴢ dr , , C1, , 冮, , F ⴢ dr , , 冮, , (x1, y), , F ⴢ dr , , (x0, y0), , C2, , 冮, , F ⴢ dr, , C2, , Since the first of the two integrals on the right does not depend on x, we have, , , f(x, y)  0 , x, x, , 冮, , F ⴢ dr, , C2, , If we write F(x, y)  P(x, y)i  Q(x, y)j, then, , 冮, , F ⴢ dr , , C2, , 冮, , P(x, y) dx  Q(x, y) dy, , C2, , Now C2 can be represented parametrically by x(t)  t, y(t)  y, where x 1 t x and, y is a constant. This gives dx  x¿(t) dt  dt and dy  0 since y is constant on C2., Therefore,, , , f(x, y) , x, x, y, , , , (x, y), C1, , (x, y1), , FIGURE 7, The path C consists of an arbitrary, path from (x 0, y0) to (x, y1), followed by the vertical line, segment from (x, y1) to (x, y)., , P(x, y) dx  Q(x, y) dy, , C2, , 冮, , x, , P(t, y) dt  P(x, y), , x1, , upon using the Fundamental Theorem of Calculus, Part 1. Similarly, by choosing C to, be the path with a vertical line segment as shown in Figure 7, we can show that, , f(x, y)  Q(x, y), y, , (x0, y0), 0, , , x, , 冮, , x, , Therefore,, F  Pi  Qj , that is, F is conservative., , f, f, i, j  §f, x, y

Page 240 :
15.4, , Independence of Path and Conservative Vector Fields, , 1261, , Determining Whether a Vector Field Is Conservative, Although Theorem 3 provides us with a good characterization of conservative vector, fields, it does not help us to determine whether a vector field is conservative, since it, is not practical to evaluate the line integral of F over all possible paths. Before stating, a criterion for determining whether a vector field is conservative, we look at a condition that must be satisfied by a conservative vector field., , THEOREM 4, If F(x, y)  P(x, y)i  Q(x, y)j is a conservative vector field in an open region, R and both P and Q have continuous first-order partial derivatives in R, then, Q, P, , x, y, at each point (x, y) in R., , PROOF Because F  Pi  Qj is conservative in R, there exists a function f such that, F  §f, that is,, Pi  Qj  fxi  fy j, This equation is equivalent to the two equations, P  fx, , and, , Q  fy, , Since Py and Q x are continuous by assumption, it follows from Clairaut’s Theorem in, Section 13.3 that, Q, P,  fxy  fyx , y, x, The converse of Theorem 4 holds only for a certain type of region. To describe this, region, we need the notion of a simple curve. A plane curve described by r  r(t) is, a simple curve if it does not intersect itself anywhere except possibly at its endpoints;, that is, r(t 1)  r(t 2) if a  t 1  t 2  b. (See Figure 8.), r(b), , r(a), r(b), , C1, , r(a)  r(b), , C3, , FIGURE 8, C1 is simple, C2 is not simple,, C3 is simple and closed, and, C4 is closed but not simple., , C4, , C2, r(a)  r(b), , r(a), (a), , (b), , (c), , (d), , A connected region R in the plane is a simply-connected region if every simple, closed curve C in R encloses only points that are in R. As is illustrated in Figure 9, a, simply-connected region not only is connected, but also does not have any hole(s).

Page 241 :
1262, , Chapter 15 Vector Analysis, , FIGURE 9, R1 is simply-connected; R2 is not, simply-connected because the simple, closed curve shown encloses points, outside R2; R3 is not simply-connected, because it is not connected., , R3, R1, , R2, , (a), , (b), , (c), , The following theorem, which is a partial converse of Theorem 4, gives us a test, to determine whether a vector field on a simply-connected region in the plane is conservative., , THEOREM 5 Test for a Conservative Vector Field in the Plane, Let F  Pi  Qj be a vector field in an open simply-connected region R in the, plane. If P and Q have continuous first-order partial derivatives on R and, Q, P, , x, y, , (2), , for all (x, y) in R, then F is conservative in R., , The proof of this theorem can be found in advanced calculus books., , EXAMPLE 3 Determine whether the vector field F(x, y)  (x 2  2xy  1)i , (y 2  x 2)j is conservative., Solution, , Here, P(x, y)  x 2  2xy  1 and Q(x, y)  y 2  x 2. Since, Q, P,  2x , y, x, , for all (x, y) in the plane, which is open and simply-connected, we conclude by Theorem 5 that F is conservative., , EXAMPLE 4 Determine whether the vector field F(x, y)  2xy 2i  x 2yj is conservative., Solution, , Here, P(x, y)  2xy 2 and Q(x, y)  x 2y. So, P,  4xy, y, , and, , Q,  2xy, x, , Since P>y  Q>x except along the x- or y-axis, we see that Equation (2) of Theorem 5 is not satisfied for all points (x, y) in any open simply-connected region in the, plane. Therefore, F is not conservative., , Finding a Potential Function, Once we have ascertained that a vector field F is conservative, how do we go about, finding a potential function f for F? One such technique is utilized in the following, example.

Page 242 :
15.4, , Independence of Path and Conservative Vector Fields, , 1263, , EXAMPLE 5 Let F(x, y)  2xyi  (1  x 2  y 2)j., a. Show that F is conservative, and find a potential function f such that F  §f., b. If F is a force field, find the work done by F in moving a particle along any path, from (1, 0) to (2, 3)., Solution, a. Here, P(x, y)  2xy and Q(x, y)  1  x 2  y 2. Since, Q, P,  2x , y, x, for all points in the plane, we see that F is conservative. Therefore, there exists a, function f such that F  §f. In this case the equation reads, 2xyi  (1  x 2  y 2)j , , f, f, i, j, x, y, , This vector equation is equivalent to the system of scalar equations, f,  2xy, x, , (3), , f,  1  x 2  y2, y, , (4), , Integrating Equation (3) with respect to x, (so that y is treated as a constant), we, have, f(x, y)  x 2y  t(y), , (5), , where t(y) is the constant of integration. (Remember that y is treated as a constant, so the most general expression of a constant here involves a function of y.), To determine t(y), we differentiate Equation (5) with respect to y, obtaining, f,  x 2  t¿(y), y, , (6), , Comparing Equation (6) with Equation (4) leads to, x 2  t¿(y)  1  x 2  y 2, or, t¿(y)  1  y 2, Integrating Equation (7) with respect to y gives, t(y)  y , , 1 3, y C, 3, , where C is a constant. Finally, substituting t(y) into Equation (5) gives, f(x, y)  x 2y  y , , 1 3, y C, 3, , the desired potential function., b. Since F is conservative, we know that the work done by F in moving a particle, from (1, 0) to (2, 3) is independent of the path connecting these two points., , (7)

Page 243 :
1264, , Chapter 15 Vector Analysis, , Using Equation (1), we see that the work done by F is, W, , 冮 F ⴢ dr  冮, C, , §f ⴢ dr  f(2, 3)  f(1, 0), , C, ,  c(22)(3)  3 , , 1 3, 1, (3 )d  c(12)(0)  0  (0)d  6, 3, 3, , Note In Example 5a you may also integrate Equation (4) first with respect to y and, proceed in a similar manner., The following theorem provides us with a test to determine whether a vector field, in space is conservative. Theorem 6 is an extension of Theorem 5, and its proof will, be omitted., , THEOREM 6 Test for a Conservative Vector Field in Space, Let F  Pi  Qj  Rk be a vector field in an open, simply connected region, D in space. If P, Q, and R have continuous first-order partial derivatives in space,, then F is conservative if curl F  0 for all points in D. Equivalently, F is conservative if, Q, R, , ,, y, z, , R, P, , ,, x, z, , and, , Q, P, , x, y, , The following example illustrates how to find a potential function for a conservative vector field in space., , EXAMPLE 6 Let F(x, y, z)  2xyz 2i  x 2z 2j  2x 2yzk., a. Show that F is conservative, and find a function f such that F  §f., b. If F is a force field, find the work done by F in moving a particle along any path, from (0, 1, 0) to (1, 2, 1)., Solution, a. We compute, i, , curl F  ∞, x, 2xyz 2, , j, , y, x 2z 2, , k, , ∞, z, 2x 2yz, ,  (2x 2z  2x 2z)i  (4xyz  4xyz)j  (2xz 2  2xz 2)k, 0, Since curl F  0 for all points in R3, we see that F is a conservative vector field, by Theorem 6. Therefore, there exists a function f such that F  §f. In this case, the equation reads, 2xyz 2i  x 2z 2j  2x 2yzk , , f, f, f, i, j, k, x, y, z

Page 244 :
15.4, , Independence of Path and Conservative Vector Fields, , 1265, , This vector equation is equivalent to the system of three scalar equations, f,  2xyz 2, x, , (8), , f,  x 2z 2, y, , (9), , f,  2x 2yz, z, , (10), , Integrating Equation (8) with respect to x (so that y and z, are treated as constants), we have, f(x, y, z)  x 2yz 2  t(y, z), , (11), , where t(y, z) is the constant of integration. To determine t(y, z), we differentiate, Equation (11) with respect to y, obtaining, t, f,  x 2z 2 , y, y, , (12), , Comparing Equation (12) with Equation (9) leads to, x 2z 2 , , t, y, ,  x 2z 2, , or, t, y, , 0, , (13), , Integrating Equation (13) with respect to y (so that z, is treated as a constant), we, obtain t(y, z)  h(z), so, f(x, y, z)  x 2yz 2  h(z), , (14), , Differentiating Equation (14) with respect to z, and comparing the result with, Equation (10), we have, f,  2x 2yz  h¿(z)  2x 2yz, z, Therefore, h¿(z)  0 and h(z)  C, where C is a constant. Finally, substituting the, value of h(z) into Equation (14) gives, f(x, y, z)  x 2yz 2  C, as the desired potential function., b. Since F is conservative, we know that the work done by F in moving a particle, from (0, 1, 0) to (1, 2, 1) is independent of the path connecting these two, points. Therefore, the work done by F is, W, , 冮 F ⴢ dr  冮, C, , §f ⴢ dr  f(1, 2, 1)  f(0, 1, 0), , C, ,  (1)2(2)(1)2  0  2

Page 245 :
1266, , Chapter 15 Vector Analysis, , Conservation of Energy, z, F, C, A, r(t), 0, , B, , y, , The Fundamental Theorem for Line Integrals can be used to derive one of the most, important laws of physics: the Law of Conservation of Energy. Suppose that a body of, mass m is moved from A to B along a piecewise-smooth curve C such that its position, at any time t is given by r(t), a t b, and suppose that the body is subjected to the, action of a continuous conservative force field F. (See Figure 10.) To find the work, done by the force on the body, we use Newton’s Second Law of Motion to write, F  ma  mv¿(t)  mr⬙(t), where v(t)  r¿(t) and a(t)  r⬙(t) are the velocity and, acceleration of the body at any time t, respectively. The work done by the force F on, the body as it is moved from A to B along C is, , x, , FIGURE 10, The path C of a body from A  r(a), to B  r(b), , W, , 冮, , F ⴢ dr , , 冮, , F(r(t)) ⴢ r¿(t) dt, , a, , C, , , , 冮, , b, , b, , mr⬙(t) ⴢ r¿(t) dt, , a, , , , , m, 2, , 冮, , b, , m, 2, , 冮, , b, , a, , a, , d, [r¿(t) ⴢ r¿(t)] dt, dt, , Use Theorem 2 in Section 12.2., , d, 冟 r¿(t) 冟2 dt, dt, , , , m, b, C 冟 r¿(t) 冟2 D a, 2, , , , m, 1 冟 r¿(b) 冟2  冟 r¿(a) 冟2 2, 2, , , , 1, 1, m 冟 v(b) 冟2  m 冟 v(a) 冟2, 2, 2, , Use the Fundamental Theorem for Line Integrals., , Since v(t)  r¿(t), , Since the kinetic energy K of a particle of mass m and speed √ is 12 m√2, we can write, W  K(B)  K(A), , (15), , which says that the work done by the force field on the body as it moves from A to B, along C is equal to the change in kinetic energy of the body at A and B., Since F is conservative, there is a scalar function f such that F  §f. The potential energy P of a body at the point (x, y, z) in a conservative force field is defined to, be P(x, y, z)  f(x, y, z), so we have F   §P. Consequently, the work done by F, on the body as it is moved from A to B along C is given by, W, , 冮 F ⴢ dr  冮, C, , §P ⴢ dr, , C, ,  CP(r(t)) D a  [P(r(b))  P(r(a))], b, ,  P(A)  P(B), , Comparing this equation with Equation (15), we see that, P(A)  K(A)  P(B)  K(B), which states that as the body moves from one point to another in a conservative force, field, then the sum of its potential energy and kinetic energy remains constant. This is, the Law of Conservation of Energy and is the reason why certain vector fields are, called conservative.

Page 246 :
15.4, , 15.4, , Independence of Path and Conservative Vector Fields, , 1267, , CONCEPT QUESTIONS, , 1. State the Fundamental Theorem for Line Integrals., 2. a. Explain what it means for the line integral 兰C F ⴢ dr to, be independent of path?, b. If 兰C F ⴢ dr is independent of path for all paths C in an, open, connected region R, what can you say about F?, , 15.4, , 3. a. How do you determine whether a vector field, F  P(x, y)i  Q(x, y)j is conservative?, b. How do you determine whether a vector field, F  P(x, y, z)i  Q(x, y, z)j  R(x, y, z)k is conservative?, , EXERCISES, , In Exercises 1–10, determine whether F is conservative. If so,, find a function f such that F  §f., 1. F(x, y)  (4x  3y)i  (3x  2y)j, 2. F(x, y)  (2x  4y)i  (2x  3y )j, 2, , 2, , In Exercises 21 and 22, find the work done by the force field F, on a particle moving along a path from P to Q., 21. F(x, y)  2 1yi , , x, j; A(1, 1) , B(2, 9), 1y, , 3. F(x, y)  (2x  y 2)i  (x 2  y)j, , 22. F(x, y)  ex cos yi  ex sin yj; A(0, 0) , B(1, p), , 4. F(x, y)  (x 2  y 2)i  2xyj, , 23. Show that the line integral 兰C yz dx  xz dy  xyz dz is not, independent of path., , 5. F(x, y)  y cos xi  (2y sin x  3)j, 2, , 6. F(x, y)  (x cos y  sin y)i  (cos y  x sin y)j, 7. F(x, y)  (ex  2y cos 2x)i  (sin 2x  yex)j, 8. F(x, y)  (tan y  2xy)i  (x sec2 y  x 2)j, y, 9. F(x, y)  ax 2  bi  (y 2  ln x)j, x, 10. F(x, y)  (ex cos y  y sec2 x)i  (tan x  ex cos y)j, In Exercises 11–18, (a) show that F is conservative and find, a function f such that F  §f, and (b) use the result of part (a), to evaluate 兰C F ⴢ dr, where C is any path from A(x 0, y0) to, B(x 1, y1)., , 24. Show that the following line integral is not independent of, path: 兰C ey sin z dx  xey sin z dy  xey cos z dz, In Exercises 25–32, determine whether F is conservative. If so,, find a function f such that F  §f., 25. F(x, y, z)  yzi  xz j  xyk, 26. F(x, y, z)  2xy 2zi  2x 2yz j  x 2y 2k, 27. F(x, y, z)  2xyi  (x 2  z 2)j  xyk, 28. F(x, y, z)  sin yi  (x cos y  cos z)j  sin zk, 29. F(x, y, z)  ex cos zi  z sinh yj  (cosh y  ex sin z)k, , 11. F(x, y)  (2y  1)i  (2x  3)j;, , A(0, 0) and B(1, 1), , y, 30. F(x, y, z)  zexz i  ln z j  axexz  bk, z, , 12. F(x, y)  (x  2y)i  (y  2x)j;, , A(0, 0) and B(1, 1), , 31. F(x, y, z)  z cos(x  y)i  z sin(x  y)j  cos(x  y)k, , 13. F(x, y)  (2xy 2  2y)i  (2x 2y  2x)j; A(1, 1) and, B(1, 2), 14. F(x, y)  2xy 3i  (3x 2y 2  1)j;, 15. F(x, y)  xe i  x e j;, 2y, , 2 2y, , A(1, 1) and B(2, 0), , A(0, 0) and B(1, 1), , A(0, 0) and B 1 1, p2 2, , 16. F(x, y)  2x sin yi  x 2 cos yj;, , 17. F(x, y)  e sin yi  (e cos y  y)j; A(0, 0) and B(0, p), x, , x, , 18. F(x, y)  (x  tan, , 1, , y)i , , xy, 1  y2, , j; A(0, 0) and B(1, 1), , In Exercises 19 and 20, evaluate 兰C F ⴢ dr for the vector field F, and the path C. (Hint: Show that F is conservative, and pick a, simpler path.), 19. F(x, y)  (2xy 2  cos y)i  (2x 2y  x sin y)j, C: r(t)  (1  cos t)i  sin tj, 0 t p, 20. F(x, y)  (e  y sin x)i  (xe  2y cos x)j, C: r(t)  4 cos ti  3 sin tj, 0 t p, y, , 2, , y, , 32. F(x, y, z) , , x, x, 1, i 2 j 2k, yz, y z, yz, , In Exercises 33–36, (a) show that F is conservative, and find a, function f such that F  §f, and (b) use the result of part (a) to, evaluate 兰C F ⴢ dr, where C is any curve from A(x 0, y0, z 0) to, B(x 1, y1, z 1) ., 33. F(x, y, z)  yz 2i  xz 2j  2xyzk; A(0, 0, 1) and B(1, 3, 2), 34. F(x, y, z)  2xy 2z 3i  2x 2yz 3j  3x 2y 2z 2k; A(0, 0, 0) and, B(1, 1, 1), 35. F(x, y, z)  cos yi  (z 2  x sin y)j  2yzk; A(1, 0, 0), and B(2, 2p, 1), y, 36. F(x, y, z)  ey i  (xey  ln z)j  a bk; A(0, 1, 1) and, z, B(1, 0, 2), 37. Evaluate 兰C (2xy 2  3) dx  (2x 2y  1) dy, where C is the, curve x 4  6xy 3  4y 2  0 from (0, 0) to (2, 1)., , V Videos for selected exercises are available online at www.academic.cengage.com/login.

Page 247 :
1268, , Chapter 15 Vector Analysis, , 38. Evaluate 兰C (3x 2y  e y) dx  (x 3  xe y  2y) dy, where C, is the curve of Exercise 37., , 42. Let, F(x, y, z) , , 39. Let, E(x, y, z) , , kQ, r, 冟 r 冟3, , x  y2, , i, , 44. If F is a nonconservative vector field, then 兰C F ⴢ dr  0, whenever C is a closed path., 45. If F has continuous first-order partial derivatives in space, and C is any smooth curve, then 兰C §f ⴢ dr depends only, on the endpoints of C., , x, x 2  y2, , j, , 46. If F  Pi  Q j is in an open connected region R and, Q, P, for all (x, y) in R, then 兰C F ⴢ dr  0 for any, , x, y, smooth curve C in R., , Q, P, a. Show that, ., , x, y, b. Show that 兰C F ⴢ dr is not independent of path by computing 兰C1 F ⴢ dr and 兰C2 F ⴢ dr, where C1 and C2 are, the upper and lower semicircles of radius 1, centered at, the origin, from (1, 0) to (1, 0)., c. Do your results contradict Theorem 5? Explain., , 15.5, , 47. If F(x, y) is continuous and C is a smooth curve, then, 兰C F ⴢ dr   兰C F ⴢ dr., 48. If F has first-order partial derivatives in a simply-connected, region R, then 兰C F ⴢ dr  0 for every closed path in R., , Green’s Theorem, , y, , Green’s Theorem for Simple Regions, R, , C, , x, , 0, , FIGURE 1, A plane region R bounded by a simple, closed plane curve C, y, , Green’s Theorem, named after the English mathematical physicist George Green (1793–, 1841), relates a line integral around a simple closed plane curve C to a double integral, over the plane region R bounded by C. (See Figure 1.), Before stating Green’s Theorem, however, we need to explain what is meant by the, orientation of a simple closed curve. Suppose that C is defined by the vector function, r(t), where a t b. Then C is traversed in the positive or counterclockwise direction if the region R is always on the left as the terminal point of r(t) traces the boundary curve C. (See Figure 2.), , THEOREM 1 Green’s Theorem, C, , Let C be a piecewise-smooth, simple closed curve that bounds a region R in the, plane. If P and Q have continuous partial derivatives on an open set that contains R, then, , R, , Q, , P, , 冯 P dx  Q dy  冮冮 c x  y d dA, , r(t), , C, , 0, , k, , 43. The region R  {(x, y) 冟 0  x 2  y 2  1} is simplyconnected., , 41. Let, y, , z, (y 2  z 2)2, , In Exercises 43–48, determine whether the statement is true or, false. If it is true, explain why. If it is false, explain why or give, an example that shows it is false., , 40. Find the work that is done by the force field, F(x, y, z)  y 2zi  2xyz j  xy 2k on a particle moving along, a path from P(1, 1, 1) to Q(2, 1, 3) ., , 2, , j, , a. Show that curl F  0., b. Is F conservative? Explain., , where k is a constant, and let r  xi  yj  zk be the electric field induced by a charge Q located at the origin. (See, Example 4 in Section 15.1.) Find the work done by E in, moving a charge of q coulombs from the point A(1, 3, 2), along any path to the point B(2, 4, 1)., , F(x, y) , , y, (y 2  z 2)2, , x, , FIGURE 2, The curve C traversed in the positive or, counterclockwise direction, , (1), , R, , where the line integral over C is taken in the positive (counterclockwise) direction.

Page 248 :
15.5, , Note, , Historical Biography, , 冯 P dx  Q dy, , (1793–1841), , 冯, , or, , P dx  Q dy, , C, , C, , is sometimes used to indicate that the line integral over a simple closed curved C is, taken in the positive, or counterclockwise, direction., , Born a miller’s son in Nottingham, George, Green worked in his father’s grain mill for, most of the first forty years of his life. He, did receive some formal schooling when he, was 8 to 9 years old, but Nottingham had, limited educational resources, and Green, quickly surpassed the education that was, available there. He studied on his own,, though it is not quite clear how he got, access to the current mathematical works., However, in 1828 Green published “An Essay, on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism.” This work included the theorem, that is now known as Green’s Theorem. The, essay was sold to only 51 people, many of, whom are believed to have been friends of, Green’s, who probably did not understand, the importance of the work. Eventually,, Green’s talents were recognized by, acquaintances who were more connected, to academia, and he enrolled as an undergraduate at Cambridge in 1833 at the age, of 40. Green graduated in 1837 with the, fourth highest scores in his class. He, stayed on at Caius College, Cambridge and, was elected a fellow in 1839. During his, time at Cambridge he made significant, contributions to areas such as optics,, acoustics, and hydrodynamics. Green’s, health was poor, and he died in Nottingham in1841. Because of the limited contact, he had with his scientific contemporaries,, most of Green’s work was not appreciated, during his lifetime., , Since it is not easy to prove Green’s Theorem for general regions, we will prove it, only for the special case in which the region R is both a y-simple and an x-simple, region. (See Section 14.2.) Such regions are called simple or elementary regions., , PROOF OF GREEN’S THEOREM FOR SIMPLE REGIONS Let R be a simple region with, boundary C as shown in Figure 3. Since, , 冯 P dx  Q dy  冯 P dx  冯 Q dy, C, , C, , C, , we can consider each integral on the right separately. Since R is a y-simple region, it, can be described as, R  {(x, y) 冟 a, , x, , b, f1(x), , y, , f2 (x)}, , where f1 and f2 are continuous on [a, b]. Observe that the boundary C of R consists of, the curves C1 and C2 that are the graphs of the functions f1 and f2 as shown in the figure. Therefore,, , 冯 P dx  冮, C, , P dx , , C1, , 冮, , P dx, , C2, , where C1 and C2 are oriented as shown in Figure 3., Observe that the point (x, f1 (x)) traces C1 as x increases from a to b, whereas the, point (x, f2 (x)) traces C2 as x decreases from b to a. Therefore,, , 冯 P dx  冯, C, , P dx , , C1, , , , 冮, , b, , 冮, , b, , 冮, , b, , 冯, , P dx, , C2, , P(x, f1 (x)) dx , , a, , , , , 冮, , a, , 冮, , b, , P(x, f2(x)) dx, , b, , P(x, f1 (x)) dx , , a, , C2, , y  f 2(x), , 1269, , The notation, , GEORGE GREEN, , y, , Green’s Theorem, , P(x, f2(x)) dx, , a, , [P(x, f1 (x))  P(x, f2(x))] dx, , (2), , a, , R, , Next, we find, C1, 0, , a, , b, , P, , 冮冮 y dA  冮 冮, , y  f 1(x), b, , FIGURE 3, The simple region R viewed as a, y-simple region, , x, , a, , R, , , , 冮, , f2(x), , f1(x), , P, (x, y) dy dx, y, , b, , [P(x, f2(x))  P(x, f1 (x))] dx, , (3), , a, , where the last equality is obtained with the aid of the Fundamental Theorem of Calculus. Comparing Equation (3) with Equation (2), we see that, P, , 冯 P dx  冮冮 y dA, C, , R, , (4)

Page 249 :
1270, , Chapter 15 Vector Analysis, , By viewing R as an x-simple region (Figure 4),, , y, d, , R  {(x, y) 冟 c, , x  g2(y), , x  g1(y), R, , C1, , y, , d, t1 (y), , x, , t2 (y)}, , you can show in a similar manner that, , C2, , Q, , 冯 Q dy  冮冮 x dA, , c, , C, , x, , 0, , FIGURE 4, The simple region R viewed as an, x-simple region, , (5), , R, , (See Exercise 48.) Adding Equation (4) and Equation (5), we obtain Equation (1), the, conclusion of Green’s Theorem for the case of a simple region., , EXAMPLE 1 Evaluate 养C x 2 dx  (xy  y 2) dy, where C is the boundary of the, , region R bounded by the graphs of y  x and y  x 2 and is oriented in a positive direction., y, , Solution The region R is shown in Figure 5. Observe that R is simple. Using Green’s, Theorem with P(x, y)  x 2 and Q(x, y)  xy  y 2, we have, , (1, 1), , 1, , 冯, , C, , yx, , C, , yx, , 2, , R, , x 2 dx  (xy  y 2) dy , , R, , , 0, , 冮, , 1, , 0, , x, , 1, , 冮冮, , c, , Q, P, , d dA , x, y, , yx, 1, 1, c y2d, dx , 2, 2, yx2, , 1, , x, , 冮 冮 (y  0) dy dx, 0, , 冮, , x2, , 1, , (x 2  x 4) dx, , 0, , 1, , , , FIGURE 5, The curve C is the boundary of, the region R., , 1 1 3 1 5, 1, a x  x b` , 2 3, 5, 15, 0, , EXAMPLE 2 Evaluate 养C (y 2  tan x) dx  (x 3  2xy  1y) dy, where C is the, , circle x 2  y 2  4 and is oriented in a positive direction., , Solution The simple region R bounded by C is the disk R  {(x, y) 冟 x 2  y 2 4}, shown in Figure 6. Using Green’s Theorem with P(x, y)  y 2  tan x and Q(x, y) , x 3  2xy  1y, we find, , y, x2  y2  4, R, 2, , 0, , 2, , x, , C, , FIGURE 6, The region R is the disk bounded, by the circle x 2  y 2 4., , Q,  3, , (x  2xy  1y)  3x 2  2y, x, x, , P,  2, , (y  tan x)  2y, y, y, , and, , and so, , 冯 (y, C, , 2, ,  tan x) dx  (x 3  2xy  1y) dy , , Q, , P, , 冮冮 c x  y d dA  冮冮 3x, R, , 冮 冮, , 2, , 2p, , 2, , 0, , 3, , 冮 冮, 冮, , (r cos u)2r dr du, , 0, , 0, , 3, , dA, , R, , 2p, , 3, , 2, , r 3 cos2 u dr du, , 0, , 0, , 2p, , r2, 1, c r 4 cos2 ud, du, 4, r0, , Use polar, coordinates.

Page 250 :
15.5, ,  12, , 冮, , 1271, , Green’s Theorem, , 2p, , cos2 u du, , 0, , 6, , 冮, , 2p, , (1  cos 2u) du, , 0, ,  6cu , , 2p, 1, sin 2ud  12p, 2, 0, , The results obtained in Examples 1 and 2 can be verified by evaluating the given, line integrals directly without the benefit of Green’s Theorem, but this entails much, more work than evaluating the corresponding double integrals. In certain situations,, however, the opposite is true; that is, it is easier to evaluate a line integral than it is to, evaluate the corresponding double integral. This fact is exploited in the following formulas based on Green’s Theorem for finding the area of a plane region., , THEOREM 2 Finding Area Using Line Integrals, Let R be a plane region bounded by a piecewise-smooth simple closed curve C., Then the area of R is given by, A, , 冯 x dy   冯 y dx  2 冯 x dy  y dx, 1, , C, , C, , (6), , C, , PROOF Taking P(x, y)  0 and Q(x, y)  x, Green’s Theorem gives, Q, , P, , 冯 x dy  冮冮 c x  y d dA  冮冮 1 dA  A, C, , R, , R, , Similarly, by taking P(x, y)  y and Q(x, y)  0, we have, Q, , P, , 冯 y dx  冮冮 c x  y d dA  冮冮 1 dA  A, C, , R, , R, , Finally, with P(x, y)  12 y and Q(x, y)  12 x, we have, , 冯, , C, , , , 1, 1, y dx  x dy , 2, 2, , Q, , P, , 冯 P dx  Q dy  冮冮 c x  y d dA  冮冮 a 2  2 b dA  A, C, , 1, , R, , EXAMPLE 3 Find the area enclosed by the ellipse, , 1, , R, , x2, a2, , , , y2, b2, ,  1., , Solution The ellipse C can be represented by the parametric equations x  a cos t, and y  b sin t, where 0 t 2p. Also observe that the ellipse is traced in the counterclockwise direction as t increases from 0 to 2p. Using Equation (6), we have, A, , , 1, 2, , 冯, , ab, 2, , x dy  y dx , , C, , 冮, , 0, , 2p, , dt  pab, , 1, 2, , 冮, , 0, , 2p, , (a cos t)(b cos t) dt  (b sin t)(a sin t) dt

Page 251 :
1272, , Chapter 15 Vector Analysis, , Green’s Theorem for More General Regions, y, , So far, we have proved Green’s Theorem for the case in which R is a simple region,, but the theorem can be extended to the case in which the region R is a finite union of, simple regions. For example, the region R shown in Figure 7 is not simple, but it can, be written as R  R1 傼 R2, where R1 and R2 are both simple. The boundary of R1 is, C1 傼 C3, and the boundary of R2 is C2 傼 C4, where C3 and C4 are paths along the, crosscut traversed in the indicated directions., Applying Green’s Theorem to each of the regions R1 and R2 gives, , C1, , C3, , R1, , C4, R2, C2, x, , 0, , FIGURE 7, The region R is the union of two simple, regions R1 and R2., , 冯, , P dx  Q dy , , 冯, , P dx  Q dy , , C1 傼C3, , 冯, , x, , FIGURE 8, The region R is a union of three simple, regions R1, R2, and R3., , x2  y2  9, , R, , 1, , 0, , P, , R2, , 冯, , 冯, , P dx  Q dy , , C2 傼C4, , P dx  Q dy , , C1 傼C2, , Q, , P, , 冮冮 c x  y d dA, R, , which is Green’s Theorem for the region R  R1 傼 R2 with boundary C  C1 傼 C2., A similar argument enables us to establish Green’s Theorem for the general case, in which R is the union of any finite number of nonoverlapping, except perhaps for the, common boundaries, simple regions (see Figure 8)., , EXAMPLE 4 Evaluate 养C (ex  y 2) dx  (x 2  3xy) dy, where C is the positively oriented closed curve lying on the boundary of the semiannular region R bounded by the, upper semicircles x 2  y 2  1 and x 2  y 2  9 and the x-axis as shown in Figure 9., , y, , 3, , P dx  Q dy , , C1 傼C3, , R2, , 0, , Q, , 冮冮 c x  y d dA, , Adding these two equations and observing that the line integrals along C3 and C4 cancel each other, we obtain, , R3, , R1, , P, , R1, , and, C2 傼C4, , y, , Q, , 冮冮 c x  y d dA, , 1, , x, 3, x2  y2  1, , FIGURE 9, The region R is divided into two, simple regions by the crosscut that, lies on the y-axis., , Solution The region R is not simple, but it can be divided into two simple regions by, means of the crosscut that is the intersection of R and the y-axis. Also notice that in, polar coordinates,, R  {(r, u) 冟 1 r 3, 0 u p}, Using Green’s Theorem with P(x, y)  ex  y 2 and Q(x, y)  x 2  3xy, we have, Q,  2, , (x  3xy)  2x  3y, x, x, , and, , P,  x, , (e  y 2)  2y, y, y, , and so, , 冯 (e, , x, ,  y 2) dx  (x 2  3xy) dy , , C, , Q, , R, , R, , 3, , p, , , , 冮 冮 (2r cos u  r sin u)r dr du, 0, , , , 冮, , 1, , p, , 0, , , , P, , 冮冮 c x  y d dA  冮冮 (2x  y) dA, 3, 1, (2 cos u  sin u) c r 3 d du, 3, 1, , 26, 52, p, C2 sin u  cos uD 0 , 3, 3, , Use polar, coordinates.

Page 252 :
15.5, y, R1, R, C2, R2, C1, x, , 0, , FIGURE 10, The annular region R can be divided, into two simple regions using two, crosscuts., , Green’s Theorem, , 1273, , Green’s Theorem can be extended to even more general regions. Recall that a region, R is simply-connected if for every simple closed curve C that lies in R, the region, bounded by C is also in R. Thus, as was noted earlier, a simply-connected region “has, no holes.” For example, a rectangle is simply-connected, but an annulus (a ring bounded, by two concentric circles) is not. Also, multiply-connected regions may have one or, more holes in them and also may have boundaries that consist of two or more simple, closed curves. For example, the annular region R shown in Figure 10 has a boundary, C consisting of two simple closed curves C1 and C2. Observe that C is traversed in the, positive direction provided that C1 is traversed in the counterclockwise direction and, C2 is traversed in the clockwise direction (so that the region R always lies to the left, as the curve is traced)., The region R can be divided into two simple regions, R1 and R2, by means of two, crosscuts, as shown in Figure 10. Applying Green’s Theorem to each of these subregions of R, we obtain, Q, , Q, , P, , Q, , P, , P, , 冮冮 c x  y d dA  冮冮 c x  y d dA  冮冮 c x  y d dA, R, , R1, , , , R2, , 冮冮 P dx  Q dy  冮冮 P dx  Q dy, R1, , R2, , where R1 and R2 denote the boundaries of R1 and R2, respectively. Since the line, integrals along the crosscuts are traversed in opposite directions, they cancel out, and, we have, Q, , P, , 冮冮 c x  y d dA  冯, , P dx  Q dy , , C1, , R, , 冯, , P dx  Q dy , , C2, , 冯 P dx  Q dy, C, , which is Green’s Theorem for the region R. Observe that the second line integral above, is traversed in the clockwise direction., , EXAMPLE 5 Let C be a smooth, simple, closed curve that does not pass through the, origin. Show that, y, x,  2, dx  2, dy, 2, x, , y, x, , y2, C, , 冯, , is equal to zero if C does not enclose the origin but is equal to 2p if C encloses the, origin., y, , Solution Suppose that C does not enclose the origin. (See Figure 11.) Using Green’s, Theorem with P(x, y)  y>(x 2  y 2) and Q(x, y)  x>(x 2  y 2) so that, Q, (x 2  y 2)(1)  x(2x), y2  x 2, , , x, (x 2  y 2) 2, (x 2  y 2) 2, , R, C, , 0, , FIGURE 11, C does not enclose the origin., , and, , x, , (x 2  y 2)(1)  (y)(2y), y2  x 2, Q, P, , , , 2, 2, 2, 2, 2, 2, y, x, (x  y ), (x  y ), , we obtain, , 冯 x, C, , y, 2, , y, , 2, , dx , , x, x y, 2, , 2, , dy , , Q, , P, , 冮冮 c x  y d dA  冮冮 0 dA  0, , Here, R denotes the region enclosed by C., , R, , R

Page 253 :
1274, , Chapter 15 Vector Analysis, y, , a, 0, , x, , R, C, , C, , Next, suppose that C encloses the origin. Since P and Q are not continuous in the, region enclosed by C, Green’s Theorem is not directly applicable. Let C¿ be a counterclockwise-oriented circle with center at the origin and radius a chosen small enough, so that C¿ lies inside C. (See Figure 12.) Then both P and Q have continuous partial, derivatives in the annular region bounded by C and C¿. Applying Green’s Theorem to, the multiply-connected region R with its positively oriented boundary C 傼 (C¿), we, obtain, , 冯 P dx  Q dy  冯, , FIGURE 12, C encloses the origin., , C¿, , P dx  Q dy , , C, , Q, , P, , 冮冮 c x  y d dA  冮冮 0 dA  0, R, , R, , or, upon reversing the direction of traversal of the second line integral,, , 冯 P dx  Q dy  冯, C, , P dx  Q dy  0, , C¿, , Therefore,, , 冯 P dx  Q dy  冯, C, , P dx  Q dy, , C¿, , Up to this point, we have shown that the required line integral is equal to the line, integral taken over the circle C¿ in the counterclockwise direction. To evaluate this integral, we represent the circle by the parametric equations x  a cos t and y  a sin t,, where 0 t 2p. We obtain, , 冯, , C¿, , , , y, x y, 2, , dx , 2, , x, x y, 2, , 2, , dy , , 冮, , 2p, , 冮, , 2p, , , , 0, , , , (a sin t)(a sin t), (a cos t)  (a sin t), 2, , 2, , dt , , (a cos t)(a cos t), (a cos t)2  (a sin t)2, , dt, , 1 dt  2p, , 0, , Therefore,, , 冯 x, C, , y, 2, , y, , 2, , dx , , x, x  y2, 2, , dy  2p, , Vector Form of Green’s Theorem, The vector form of Green’s Theorem has two useful versions: one involving the curl, of a vector field and another involving the divergence of a vector field., Suppose that the curve C, the plane region R, and the functions P and Q satisfy the, hypothesis of Green’s Theorem. Let F  Pi  Qj be a vector field. Then, , 冯 F ⴢ T ds  冯 P dx  Q dy, C, , C, , Recalling that P and Q are functions of x and y, we have, , curl F  §, , i, , F∞, x, P, , j, , y, Q, , k, Q, , P, ∞a, , bk, z, x, y, 0, , Remember that P and Q are, functions of x and y.

Page 254 :
15.5, , Green’s Theorem, , 1275, , so, (curl F) ⴢ k  a, , Q, Q, P, P, , bk ⴢ k , , x, y, x, y, , Therefore, Green’s Theorem can be written in the vector form, , 冯 F ⴢ T ds  冮冮curl F ⴢ k dA, C, , (7), , R, , Equation (7) states that the line integral of the tangential component of F around, a closed curve C is equal to the double integral of the normal component to R of, curl F over the region R enclosed by C., Next, let the curve C be represented by the vector equation r(t)  x(t)i  y(t)j,, a t b. Then the outer unit normal vector to C is, n(t) , , x¿(t), y¿(t), i, j, 冟 r¿(t) 冟, 冟 r¿(t) 冟, , which you can verify by showing that n(t) ⴢ T(t)  0, where, T(t) , , y¿(t), x¿(t), i, j, 冟 r¿(t) 冟, 冟 r¿(t) 冟, , is the unit tangent vector to C. (See Figure 13.) We have, , y, T(t) n(t), , 冯, , F ⴢ n ds , , 冮, , b, , 冮, , b, , R, , , , 冮, , b, , a, , r(t), , C, , 0, , (F ⴢ n)(t) 冟 r¿(t) 冟 dt, , a, , C, , , , x, , c, , P(x(t), y(t))y¿(t), Q(x(t), y(t))x¿(t), , d 冟 r¿(t) 冟 dt, 冟 r¿(t) 冟, 冟 r¿(t) 冟, , P(x(t), y(t))y¿(t) dt , , a, , FIGURE 13, n(t) is the outer normal vector to C., , , , 冮, , b, , Q(x(t), y(t))x¿(t) dt, , a, , 冯 P dy  Q dx, C, , But by Green’s Theorem,, , , , , 冯 P dy  Q dx  冮冮 c x (P)  y (Q)d dA, C, , R, , , , P, , Q, , 冮冮 a x  y b dA, R, , Observing that the integrand of the last integral is just the divergence of F, we obtain, the second vector form of Green’s Theorem:, , 冯 F ⴢ n ds  冮冮 div F dA, C, , (8), , R, , Equation (8) states that the line integral of the normal component of F around a closed, curve C is equal to the double integral of the divergence of F over R.

Page 255 :
1276, , Chapter 15 Vector Analysis, , 15.5, , CONCEPT QUESTIONS, , 1. State Green’s Theorem., , 15.5, , 2. Write three line integrals that give the area of a region, bounded by a piecewise smooth curve C., , EXERCISES, , In Exercises 1–4, evaluate the line integral (a) directly and, (b) by using Green’s Theorem, where C is positively oriented., 1., , 冯, , 11., , x, , 2xy dx  3xy 2 dy, where C is the square with vertices, , C, , 12., , (0, 0), (1 , 0), (1, 1), and (0, 1), , 冯, , (x 2  y) dx  21  y 2 dy, where C is the astroid, , C, 2>3, ,  y 2>3  a 2>3, , 冯 6xy dx  (3x, , 2, ,  ln(1  y)) dy, where C is the cardioid, , C, , 2., , 冯x, , 2, , r  1  cos u, , dx  xy dy, where C is the triangle with vertices, , C, , 13., , (0, 0), (1, 0), and (0, 1), 3., , 冯y, , 2, , (x  ex sin y) dx  (x  ex cos y) dy, where C is the, y2, x2, ellipse, , 1, 9, 4, C, , dx  (x 2  2xy) dy, where C is the boundary of the, , C, , region bounded by the graphs of y  x and y  x 3 lying in, the first quadrant, 4., , 冯, , 冯 2x dx  3y dy, where C is the circle x, , 2, ,  y2  a2, , C, , 14., , 冯, , y, , dx  (x  tan1 x) dy, where C is the right-hand, 1  x2, loop of the lemniscate r 2  cos 2u, C, , 15., , 冯, , (y dx  x dy), where C is the boundary of the annular, , C, , region formed by circles x 2  y 2  1 and x 2  y 2  4, , In Exercises 5–16, use Green’s Theorem to evaluate the line, integral along the positively oriented closed curve C., 5., , 冯x, , 16., 3, , dx  xy dy, where C is the triangle with vertices, , (0, 0), (1 , 1), and (0, 1), , 冯, , (x 2  y 2) dx  2xy dy, where C is the square with, , C, , vertices ( 1,, 7., , 冯, , 1), , (x 2y  x 3) dx  2xy dy, where C is the boundary of the, , C, , region bounded by the graphs of y  x and y  x 2, 8., , 冯, , 2, , (y 3  cos x) dx  e y dy, where C is the boundary of, , C, , 9., , 3, , 17. Use Green’s Theorem to find the work done by the force, F(x, y)  (x 2  y 2)i  2xyj in moving a particle in the positive direction once around the triangle with vertices (0, 0),, (1, 0), and (0, 1)., 18. Use Green’s Theorem to find the work done by the force, F(x, y)  3yi  2xj in moving a particle once around the, y2, x2, ellipse, ,  1 in the clockwise direction., 4, 9, , the region bounded by the parabolas y  x 2 and x  y 2, , In Exercises 19–22, use one of the formulas on page 1271 to find, the area of the indicated region., , 冯, , 19. The region enclosed by the astroid x 2>3  y 2>3  a 2>3, , (y 2  cos x) dx  (x  tan1 y) dy, where C is the, , C, , boundary of the region bounded by the graphs of, y  4  x 2 and y  0, 10., ,  x) dy, where C is the boundary of, y2, x2, ,  1 and the, the region lying between the ellipse, 4, 9, circle x 2  y 2  1, 2, , C, , C, , 6., , 冯 3x y dx  (x, , 冯, , x 2y dx  y 3 dy, where C consists of the line segment, , C, , from (1, 0) to (1, 0) and the upper half of the circle, x 2  y2  1, , 20. The region bounded by an arc of the cycloid, x  a(t  sin t), y  a(1  cos t), and the x-axis, 21. The region enclosed by the curve x  a sin t and, y  b sin 2t, 22. The region enclosed by the curve x  cos t and y  4 sin3 t,, where 0 t 2p, , V Videos for selected exercises are available online at www.academic.cengage.com/login.

Page 256 :
15.5, 23. a. Plot the curve C defined by x  t(1  t 2) and, y  t 2 (1  t 3) , where 0 t 1., b. Find the area of the region enclosed by the curve C., , 2, , Hint: Use Green’s Theorem, noting that C1 傼 C2, where C2 is the, straight path from (3, 0) to (3, 0), is a closed path., y, , 25. Swallowtail Catastrophe, a. Plot the swallowtail catastrophe defined by x  2t(1  t 2), and y  12 t 2(3t 2  2), where 1 t 1., b. Find the area of the region enclosed by the swallowtail, catastrophe., , 冮, , F ⴢ dr  3p,, , C2, , where F(x, y)  P(x, y)i  Q(x, y)j, and that, Q, P, , b  6 for all (x, y) in the region R bounded by, a, x, y, the circles C1 and C2, and oriented in a counterclockwise, direction. Use Green’s Theorem to find, , 冮, , 2, , C1, , B(3, 0), , A(3, 0), 0, , 3, , C2, , Hint: See the hint in Exercise 28., y, 3, , F ⴢ dr., 2, , y, 1, , C2, , C1, , D, , 0, , C1, 1, , x, , 2, , R, , 27. Refer to the figure below. Suppose that, , 冯, , 冯, , B, , F ⴢ dr  2p and, , A, , F ⴢ dr  3p, where F(x, y)  P(x, y)i  Q(x, y)j,, , C3, , Q, P, , b  6 for all (x, y) in the region R, x, y, lying inside the curve C1 and outside the curves C2 and C3., Use Green’s Theorem to find, , 冯, , 31., , C1, R, , 1, , C3, , 4 3 2 1, 1, , 1, , C2, , 3, , 4, , x, , 5, , 1, [(x 1y2  x 2y1)  (x 2y3  x 3y2)  p, 2,  (x n1yn  x nyn1)  (x ny1  x 1yn)], , 32., , y, , C1, , 2, , 2, , In Exercises 31 and 32, use the result of Exercise 30 to find the, area of the shaded region., , F ⴢ dr., , y, 3, , 1, , A, , 30. a. Let C be the line segment joining the points (x 1, y1) and, (x 2, y2). Show that 兰C y dx  x dy  x 1y2  x 2y1., b. Use the result of part (a) to show that the area of a polygon with vertices (x 1, y1), (x 2, y2), p , (x n, yn) (appearing, in the counterclockwise order) is, , C2, , and that a, , C, , E, F, , 0, , x, , 3, , 29. Evaluate 兰C1 (x 2  2y) dx  (3x  sinh y) dy, where C1 is, the path ABCDEF shown in the figure., , C1, , 2, , 1277, , 28. Evaluate 兰C (x 2  2y) dx  1 4x  e y 2 dy, where C1 is the, 1, semi-elliptical path from A to B shown in the figure., , 24. a. Plot the deltoid defined by x  14 (2 cos t  cos 2t) and, y  14 (2 sin t  sin 2t) , where 0 t 2p., b. Find the area of the region enclosed by the deltoid., , 26. Refer to the following figure. Suppose that, , Green’s Theorem, , 5, , 5, 4, 3, 2, 1, 0, 2, , 1 2 3 4 5, , x, , y, 5, 4, 3, 2, 1, _4, , 0, , 1 2 3 4 5, , x, , _4, , 2, 3, , 2, , 3, , 4 x, , In Exercises 33 and 34, use the result of Exercise 30 to find the, area of the polygon., 33. Pentagon with vertices (0, 0) , (2, 0) , (3, 1), (1, 3) , and, (1, 1)., 34. Hexagon with vertices (0, 0) , (3, 0) , (4, 1) , (2, 4) , (0, 3), and, (2, 1).

Page 257 :
1278, , Chapter 15 Vector Analysis, , 35. Let R be a plane region of area A bounded by a piecewisesmooth simple closed curve C. Use Green’s Theorem to, show that the centroid of R is (x, y), where, x, , 冯x, , 1, 2A, , 2, , y, , dy, , C, , 冯y, , 1, 2A, , 2, , 36. The triangle with vertices (0, 0), (1, 0), and (1, 1)., 37. The region bounded by the graphs of y  0 and y  9  x 2., 38. A plane lamina with constant density r has the shape of, a region bounded by a piecewise-smooth simple closed, curve C. Show that its moments of inertia about the axes are, , 冯y, , 3, , r, Iy , 3, , dx, , C, , 冯x, , 3, , dy, , C, , 39. Use the result of Exercise 38 to find the moment of inertia, of a circular lamina of radius a and constant density r about, a diameter., 40. Show that if f and t have continuous derivatives, then, , 冯 f(x) dx  t(y) dy  0, , 冯, , for every piecewise-smooth simple closed curve C., 41. Let C be a piecewise-smooth simple closed curve that, encloses a region R of area A. Show that, (ay  b) dx  (cx  d) dy  (c  a)A, , (cos x  x 3y) dx  (x 4  ey) dy  0, , C, , where C is the boundary of the square with vertices, (1, 1), (1, 1), (1, 1), and (1, 1)., b. Note that, ,  4, (x  e y) , (cos x  x 3y), x, y, Does this contradict Theorem 4 of Section 15.4?, Explain., c. Evaluate the line integral of part (a), taking C to be the, boundary of the square with vertices (0, 0), (1, 0), (1, 1),, and (0, 1)., 46. Can Green’s Theorem be applied to evaluate, , 冯, , x, , C, , C, , 冯, , 冯, , 45. a. Use Green’s Theorem to show that, , dx, , C, , In Exercises 36 and 37, use the result of Exercise 35 to find the, centroid of the region., , r, Ix  , 3, , Q, P, in R but, , P dx  Q dy  0., y, x, C, b. Does this contradict Green’s Theorem? Explain., a. Show that, , 2(x  2)  y, 2, , 2, , dx , , y, 2(x  2)2  y 2, , dy, , where C is the circle of radius 1 centered at the origin?, Explain., 47. Show that if P( y) and Q(x) have continuous derivatives,, then, , 冯 P(y) dx  Q(x) dy  2CQ(t)  P(t) D, C, , t1, t1, , C, , 42. Let C be a piecewise-smooth simple closed curve that does, not pass through the origin. Evaluate, , 冯, , x, , C, , x y, 2, , 2, , dx , , y, x  y2, 2, , 48. Refer to the proof of Green’s Theorem. Show that by viewing R as an x-simple region, we have, , dy, , (a) where C does not enclose the origin and (b) where C, encloses the origin., 43. Let P(x, y)  , a. Show that, , 冯, , y, x y, 2, , 2, , and Q(x, y) , , x, x y, 2, , 2, , (P dx  Q dy)  0, where C is the circle, , of radius 1 centered at the origin., Q, P, , b. Verify that, ., y, x, c. Do parts (a) and (b) contradict each other? Explain., 44. Let R be the region bounded by the circles of radius 1 and 3, centered at the origin, and let C be the circle of radius 2, centered at the origin described by r(t)  2 cos ti  2 sin tj,, where 0 t 2p. Let, y, x 2  y2, , and, , Q(x, y) , , Q, , 冯 Q dy  冮冮 x dA, C, , R, , ., , C, , P(x, y)  , , where C is the rectangular path that is traced in a counterclockwise direction with vertices (1, 1), (1, 1), (1, 1),, and (1, 1)., , x, x 2  y2, , In Exercises 49–51, determine whether the statement is true or, false. If it is true, explain why. If it is false, explain why or give, an example that shows it is false., 49. If a and b are constants, then 养C a dx  b dy  0, where C, is a simple closed curve., 50. If C is a piecewise-smooth simple closed curve that bounds, a region R in the plane, then 养C xy 2 dx  (x 2y  x) dy is, equal to the area of R., 51. The work done by the force field F(x, y)  12 yi  12 xj on a, particle that moves once around a piecewise-smooth simple, closed curve in a counterclockwise direction is numerically, equal to the area of the region bounded by the curve.

Page 258 :
15.6, , 15.6, , Parametric Surfaces, , 1279, , Parametric Surfaces, Why We Use Parametric Surfaces, z, 4π, , 1, , 1, (x, y), , y, , x, , FIGURE 1, The helicoid shown here, is not the graph of a, function z  f(x, y)., , In Chapter 13 we studied surfaces that are graphs of functions of two variables. However, not every surface is the graph of a function z  f(x, y). Consider, for example,, the helicoid shown in Figure 1. Observe that the point (x, y) in the xy-plane is associated with more than one point on the helicoid, so this surface cannot be the graph of, a function z  f(x, y)., Just as we found it useful to describe a curve in the plane (and in space) as the, image of a line under a vector-valued function r rather than as the graph of a function,, we will now see that a similar situation exists for surfaces. Instead of a single parameter, however, we will use two parameters and view a surface in space as the image of, a plane region. More specifically, we have the following., , DEFINITION Parametric Surface, Let, r(u, √)  x(u, √)i  y(u, √)j  z(u, √)k, be a vector-valued function defined for all points (u, √) in a region D in the, u√-plane. The set of all points (x, y, z) in R3 satisfying the parametric equations, x  x(u, √),, , y  y(u, √),, , z  z(u, √), , as (u, √) ranges over D is called a parametric surface S represented by r. The, region D is called the parameter domain., , Thus, as (u, √) ranges over D, the tip of the vector r(u, √) traces out the surface S, (see Figure 2). Put another way, we can think of r as mapping each point (u, √) in D, onto a point (x(u, √), y(u, √), z(u, √)) on S in such a way that the plane region D is bent,, twisted, stretched, and/or shrunk to yield the surface S., z, , y, , S, , r, D, , r(u, √), , (u, √), , FIGURE 2, The function r maps D, onto the surface S., , 0, , 0, y, , x, , x, , EXAMPLE 1 Identify and sketch the surface represented by, r(u, √)  2 cos ui  2 sin uj  √k, with parameter domain D  {(u, √) 冟 0, Solution, , (x, y, z), , u, , 2p, 0, , √, , 3}., , The parametric equations for the surface are, x  2 cos u,, , y  2 sin u,, , z√

Page 259 :
1280, , Chapter 15 Vector Analysis, , Eliminating the parameters u and √ in the first two equations, we obtain, x 2  y 2  4 cos2 u  4 sin2 u  4, Observe that the variable z is missing in this equation, so it represents a cylinder with, the z-axis as its axis. (See Section 11.6.) Furthermore, the trace in the xy-plane is a circle of radius 2, and we conclude that the cylinder is a circular cylinder. Finally, because, 0 √ 3, the third equation z  √ tells us that 0 z 3. Thus, the required surface, is the truncated cylinder shown in Figure 3., z, , √, , (x, y, z), , 3, , r, r(u, √), , 3, D, , FIGURE 3, The function r “bends” the rectangular, region D into a cylinder., , 2, 0, , u, , 2π, , 2, , y, , x, , There is another way of visualizing the way r maps the domain D onto a surface, S. If we fix u by setting u  u 0, where u 0 is a constant, and allow √ to vary so that the, points (u 0, √) lie in D, then we obtain a vertical line segment L 1 lying in D. When, restricted to L 1, the function r becomes a function involving one parameter √ whose, domain is the parameter interval L 1. Therefore, r(u 0, √) maps L 1 onto a curve C1 lying, on S (see Figure 4)., √, , z, C2, , L1, D, √0, , C1, , r, , u  u0, , √  √0 (u0, √0), 0, , L2, , FIGURE 4, r maps L 1 onto C1 and L 2 onto C2., , 0, , u0, , u, , r(u0, √), , y, , x, , Similarly, by holding √ fixed, say, √  √0, where √0 is a constant, the tip of the resulting vector r(u, √0) traces the curve C2 as u is allowed to assume values in the parameter interval L 2. The curves C1 and C2 are called grid curves., By way of illustration, if we set u  u 0 in Example 1, then both x  2 cos u 0 and, y  2 sin u 0 are constant. So the vertical line u  u 0 is mapped onto the vertical line, segment (2 cos u 0, 2 sin u 0, √) , 0 √ 3. Similarly, you can verify that a horizontal, line segment √  √0 in D is mapped onto a circle on the cylinder at a height of √0 units, from the xy-plane., , EXAMPLE 2 Use a computer algebra system (CAS) to generate the surface represented by, r(u, √)  sin u cos √i  sin u sin √j  cos uk

Page 260 :
15.6, , Parametric Surfaces, , 1281, , with parameter domain D  {(u, √) 冟 0 u p, 0 √ 2p}. Identify the curves on, the surface that correspond to the curves with u held constant and those with √ held, constant., Solution The required surface is the unit sphere centered at the origin. (See Figure 5a.), You can verify that this is the case by eliminating u and √ in the parametric equations, x  sin u cos √,, , y  sin u sin √,, , z  cos u, , to obtain the rectangular equation x  y  z  1 for the sphere. Fixing u  u 0,, where u 0 is a constant, leads to the equations, 2, , x  sin u 0 cos √,, , 2, , 2, , y  sin u 0 sin √,, , z  cos u 0, , We have, x 2  y 2  sin2 u 0 cos2 √  sin2 u 0 sin2 √,  sin2 u 0 (cos2 √  sin2 √)  sin2 u 0, The system of equations, x 2  y 2  sin2 u 0, z  cos u 0, , v, , for a fixed u 0 lying in [0, p] or, equivalently, the vector-valued function, r(u 0, √)  sin u 0 cos √i  sin u 0 sin √j  cos u 0k, represents a circle of radius sin u 0 on the sphere that is parallel to the xy-plane. Thus,, if we think of the sphere as a globe then the horizontal line segments in the domain of, r are mapped onto the latitudinal lines, or parallels. (See Figure 5b.) Similarly, we can, show that the vertical line segments in the domain of r with √  √0, where √0 is a constant, are mapped by, r(u, √0)  sin u cos √0i  sin u sin √0 j  cos uk, onto the meridians of longitude—great circles on the surface of the globe passing, through the poles., , 1.0, , 0.5, , z, , 1.0, 0.5, 0.0, , 1.0, Parallels (u  u0), 0.5, 0.0, y, 0.5, 1.0, 1.0, , FIGURE 5, , (a), , x, , 0.5, , 0.0, , 0.5, , 1.0, , Meridians (√  √0), , (b), , Finding Parametric Representations of Surfaces, We now turn our attention to finding vector-valued function representations of surfaces., We begin by showing that if a surface is the graph of a function f(x, y), then it has a, simple parametric representation.

Page 261 :
1282, , Chapter 15 Vector Analysis, , EXAMPLE 3, a. Find a parametric representation for the graph of a function f(x, y)., b. Use the result of part (a) to find a parametric representation for the elliptic paraboloid z  4x 2  y 2., z, z  f(x, y), , (x, y, f(x, y)), , x  x(u, √)  u,, , S, r(u, √), , f (u, √)k, , 0, y(√), , ui  √j, , Solution, a. Suppose that S is the graph of z  f(x, y) defined on a domain D in the xy-plane., (See Figure 6.) We simply pick x and y to be the parameters; in other words, we, write the desired parametric equations as, , D, , x(u), , z  z(u, √)  f(u, √), , and take the domain of f to be the parameter domain. Equivalently, we obtain the, vector-valued representation by writing, r(u, √)  ui  √j  f(u, √)k, b. The surface is the graph of the function f(x, y)  4x 2  y 2. So we can let x and y, be the parameters. Thus, the required parametric equations are, , (x, y)(u, √), , FIGURE 6, The vector r(u, √)  ui  √j  f(u, √)k, by the rule for vector addition., , y  y(u, √)  √,, , x  u,, , y  √,, , z  4u 2  √2, , and the corresponding vector-valued function is, r(u, √)  ui  √j  (4u 2  √2)k, The parameter domain is D  {(u, √) 冟 ⬁  u  ⬁, ⬁  √  ⬁}., , EXAMPLE 4 Find a parametric representation for the cone x  2y 2  z 2., Solution The surface is the graph of the function f(y, z)  2y 2  z 2. So we can let, y and z be the parameters. Thus, the required parametric equations are, x  2u 2  √2,, , y  u,, , z√, , and the corresponding vector-valued function is, r(u, √)  2u 2  √2 i  uj  √k, The parameter domain is D  {(u, √) 冟 ⬁  u  ⬁, ⬁  √  ⬁}., , EXAMPLE 5, a. Find a parametric representation of the plane that passes through the point P0 with, position vector r0 and contains two nonparallel vectors a and b., b. Using the result of part (a), find a parametric representation of the plane passing, through the point P0 (3, 1, 1) and containing the vectors a  2i  5j  k and, b  3i  2j  3k. (This is the plane in Example 6 in Section 11.5.), , z, , P0, , b, a, r0, , √b, , P0 P, P, , ua, r, , O, y, x, , FIGURE 7, r  r0  P0P៝  r0  ua  √b, , Solution, a. Let P be a point lying on the plane, and let r  OP៝. Since P0P៝ lies in the plane, determined by a and b, there exist real numbers u and √ such that P0P៝  ua  √b., (See Figure 7.) Furthermore, we see that r  r0  P0P៝  r0  ua  √b. Finally,, since any point on the plane is located at the tip of r for an appropriate choice of, u and √, we see that the required representation is, r(u, √)  r0  ua  √b

Page 262 :
15.6, , Parametric Surfaces, , 1283, , The parameter domain is D  {(u, √) 冟 ⬁  u  ⬁, ⬁  √  ⬁}., b. The required representation is, r(u, √)  (3i  j  k)  u(2i  5j  k)  √(3i  2j  3k),  (2u  3√  3)i  (5u  2√  1)j  (u  3√  1)k, with domain D  {(u, √) 冟 ⬁  u  ⬁, ⬁  √  ⬁}., Note The representation in Example 5b is by no means unique. For example, an, equation of the plane in question is 13x  3y  11z  47. (See Example 6 in Section, 11.5.) Solving this equation for z in terms of x and y, we obtain z  f(x, y) , 1, 11 (47  13x  3y) . Thus, the plane is the graph of the function f, and this observation, leads us to the representation, r(u, √)  ui  √j  a, , 47  13u  3√, bk, 11, , The next two examples involve surfaces that are not graphs of functions., , EXAMPLE 6 Find a parametric representation for the cone x 2  y 2  z 2., Solution The cone has a simple representation r 2  z 2 in cylindrical coordinates. This, suggests that we choose r and u as parameters. Writing u for r and √ for u, we have, x  u cos √,, , y  u sin √,, , zu, , as the required parametric equations. In vector form we have, r(u, √)  u cos √i  u sin √j  uk, , EXAMPLE 7 Find a parametric representation for the helicoid shown in Figure 1., Solution, , Recall that the parametric equations for a helix are, x  a cos u,, , y  a sin u,, , zu, , where u and z are in cylindrical coordinates. This suggests that we let u denote r and, √ denote u. Then the parametric equations for the helicoid are, x  u cos √,, , z, , y  u sin √,, , z√, , with parameter domain D  {(u, √) 冟 1 u 1, 0 √ 4p}. In vector form we, have, r(u, √)  u cos √i  u sin √j  √k, a, f(u), √, b, , y  f(x), , x, , FIGURE 8, S is obtained by revolving the graph, of f between x  a and x  b about, the x-axis., , y, , We now turn our attention to finding the parametric representation for surfaces of, revolution. Suppose that a surface S is obtained by revolving the graph of the function, y  f(x) for a x b about the x-axis, where f(x)  0. (See Figure 8.) Letting u, denote x and √ denote the angle shown in the figure, we see that if (x, y, z) is any point, on S, then, x  u,, y  f(u) cos √,, z  f(u) sin √, (1), or, equivalently,, r(u, √)  ui  f(u) cos √j  f(u) sin √k, The parameter domain is D  {(u, √) 冟 a, , u, , b, 0, , √, , 2p}.

Page 263 :
1284, , Chapter 15 Vector Analysis, , EXAMPLE 8 Gabriel’s Horn Find a parametric representation for the surface obtained by revolving the graph of f(x)  1>x, where 1 x  ⬁ , about the x-axis., , z, 1, , Solution, 1, , 1, , Using Equation (1), we obtain the parametric equations, x  u,, , y, , y, , 1, cos √,, u, , z, , with parametric domain D  {(u, √) 冟 1 u  ⬁, 0 √, is a portion of Gabriel’s Horn as shown in Figure 9., , 1, sin √, u, 2p}. The resulting surface, , x, , FIGURE 9, Gabriel’s Horn, , Tangent Planes to Parametric Surfaces, Suppose that S is a parametric surface represented by the vector function, r(u, √)  x(u, √)i  y(u, √)j  z(u, √)k, and P0 is a point on the surface S represented by the vector r(u 0, √0), where (u 0, √0) is, a point in the parameter domain D of r. If we fix u by putting u  u 0 and allow √ to, vary, then the tip of r(u 0, √) traces the curve C1 lying on S. (See Figure 10.) The tangent vector to C1 at P0 is given by, r√(u 0, √0) , , y, x, z, (u 0, √0)i , (u 0, √0)j , (u 0, √0)k, √, √, √, , Similarly, by holding √ fast, √  √0, and allowing u to vary, the tip of r(u, √0) traces, the curve C2 lying on S, with tangent vector at P0 given by, ru (u 0, √0) , , y, x, z, (u 0, √0)i , (u 0, √0)j , (u 0, √0)k, u, u, u, , If ru(u, √) r√ (u, √)  0 for each (u, √) in the parameter domain of r, then the surface S is said to be smooth. For a smooth surface the tangent plane to S at P0 is the, plane that contains the tangent vectors ru (u 0, √0) and r√ (u 0, √0) and thus has a normal, vector given by, n  ru(u 0, √0), , r√(u 0, √0), , √, , n  ru, , z, , r√, , r√(u0, √0 ), D, √0, , √  √0, , ru(u0, √0 ), , r, , u  u0, , C2, , (u0, √0 ), 0, , FIGURE 10, , 0, , u0, , u, , C1, , P0, , r(u0, √0 ), y, , x, , EXAMPLE 9 Find an equation of the tangent plane to the helicoid, r(u, √)  u cos √i  u sin √j  √k, of Example 7 at the point where u  12 and √  p4 .

Page 264 :
15.6, , Solution, , Parametric Surfaces, , 1285, , We start by finding the partial derivatives of r. Thus,, ru(u, √)  cos √i  sin √j, r√ (u, √)  u sin √i  u cos √j  k, , So, 1 p, 12, 12, ru a , b , i, j, 2 4, 2, 2, 1 p, 1 12, 1 12, 12, 12, r√ a , b   ⴢ, i ⴢ, jk, i, jk, 2 4, 2, 2, 2, 2, 4, 4, A normal vector to the tangent plane is, 1 p, n  ru a , b, 2 4, , i, 1 p, 12, r√ a , b  5, 2 4, 2, 12, , 4, , j, 12, 2, 12, 4, , k, 05 , , 12, 12, 1, i, j k, 2, 2, 2, , 1, , Since any normal vector will do, let’s take n  12i  12j  k., Next note that the point 1 12 , p4 2 in the parameter domain is mapped onto the point, with coordinates, x0 , , 1, p, 1 12, 12, cos  ⴢ, , ,, 2, 4, 2, 2, 4, , y0 , , 1, p, 12, sin , ,, 2, 4, 4, , Therefore, an equation of the required tangent plane at, 12 ax , , 1 124, 124, p4 2 is, , z0 , , p, 4, , 12, 12, p, b  12 ay , b  1az  b  0, 4, 4, 4, 12x  12y  z , , p, 0, 4, , or, 412x  412y  4z  p  0, , Area of a Parametric Surface, In Section 14.5, we learned how to find the area of a surface that is the graph of a function z  f(x, y). We now take on the task of finding the areas of parametric surfaces,, which are more general than the surfaces (graphs) defined by functions., For simplicity, let’s assume that the parametric surface S defined by, r(u, √)  x(u, √)i  y(u, √)j  z(u, √)k, has parameter domain R that is a rectangle. (See Figure 11.) Let P be a regular partition of R into n  mn subrectangles R11, R12, p , Rmn. The subrectangle Rij is mapped, by r onto the patch Sij with area denoted by ⌬Sij. Since the subrectangles Rij are nonoverlapping, except for their common boundaries, so are the patches Sij, and so the area of, S is given by, m, , n, , S  a a ⌬Sij, i1 j1

Page 265 :
1286, , Chapter 15 Vector Analysis, , y, , S, , z, , Pij, , Sij, , R, Rij, r, , Δ√, , 0, , FIGURE 11, The subrectangle Rij is, mapped onto the patch Sij., , (ui, √j), , Δu, , y, , x, , 0, , x, , Next, let’s find an approximation of ⌬Sij. Let (u i, √j) be the corner of Rij closest to, the origin with image the point Pij represented by r(u i, √j) as shown in Figure 12. For, the sake of clarity, both Rij and Sij are shown enlarged. The sides of Sij with corner represented by r(u i, √j) are approximated by a and b, where a  r(u i  ⌬u, √j)  r(u i, √j), and b  r(u i, √j  ⌬√)  r(u i, √j). So ⌬Sij may be approximated by the area of the parallelogram with a and b as adjacent sides, that is,, ⌬Sij ⬇ 冟 a, , b冟, z, , (ui, √j  Δ√), , Sij, , (ui  Δu, √j  Δ√), Pij, Rij, , r(ui, √j), , (ui, √j), , b, a, , r(ui  Δu, √j), , (ui  Δu, √j), y, , FIGURE 12, , x, , But we can write, ac, , r(u i  ⌬u, √j)  r(u i, √j), ⌬u, , d ⌬u, , If ⌬u is small, as we assume, then the term inside the brackets is approximately equal, to ru(u i, √j). So, a ⬇ ⌬u ru(u i, √j), Similarly, we see that, b ⬇ ⌬√ r√(u i, √j), Therefore,, ⌬Sij ⬇ 冟 [⌬u ru (u i, √j)] [⌬√ r√ (u i, √j)] 冟,  冟 ru (u i, √j), , r√(u i, √j) 冟 ⌬u ⌬√, , and the area of S may be approximated by, m, , n, , a a 冟 ru(u i, √j), i1 j1, , r√(u i, √j) 冟 ⌬u ⌬√

Page 266 :
15.6, , Parametric Surfaces, , 1287, , Intuitively, the approximation gets better and better as m and n get larger and larger., But the double sum is the Riemann sum of 冟 ru r√ 冟, and we are led to define the area, of S as, m, , lim, , n, , a a 冟 ru(u i, √j), , r√ (u i, √j) 冟 ⌬u ⌬√, , m, n→⬁ i1 j1, , Alternatively, we have the following definition., , DEFINITION Surface Area (Parametric Form), Let S be a smooth surface represented by the equation, r(u, √)  x(u, √)i  y(u, √)j  z(u, √)k, with parameter domain D. If S is covered just once as (u, √) varies throughout, D, then the surface area of S is, A(S) , , 冮冮 冟 r, , r√ 冟 dA, , u, , (2), , D, , EXAMPLE 10 Find the surface area of a sphere of radius a., Solution, tion, , The sphere centered at the origin with radius a is represented by the equar(u, √)  a sin u cos √i  a sin u sin √j  a cos uk, , with parameter domain D  {(u, √) 冟 0, , u, , p, 0, , √, , 2p}. We find, , ru(u, √)  a cos u cos √i  a cos u sin √j  a sin uk, r√ (u, √)  a sin u sin √i  a sin u cos √j, and so, i, j, k, r√  † a cos u cos √ a cos u sin √ a sin u †, a sin u sin √ a sin u cos √, 0, , ru, ,  a 2 sin2 u cos √i  a 2 sin2 u sin √j  a 2 sin u cos uk, Therefore, 冟 ru, , r√ 冟  2a 4 sin4 u cos2 √  a 4 sin4 u sin2 √  a 4 sin2 u cos2 u,  2a 4 sin4 u  a 4 sin2 u cos2 u  2a 4 sin2 u,  a 2 sin u, , since sin u  0 for 0, A, , 冮冮, , 冟 ru, , D, , , , 冮, , 0, , 2p, , u, , p. Using Equation (2), the area of the sphere is, r√ 冟 dA , , 2p, , 冮 冮, 0, , p, , 0, , Ca 2 cos uD u0 d√ , up, , a 2 sin u du d√, , 冮, , 0, , 2p, , 2a 2 d√  2a 2(2p)  4pa 2

Page 267 :
1288, , Chapter 15 Vector Analysis, , EXAMPLE 11 Find the area of one complete turn of the helicoid of width one represented by the equation r(u, √)  u cos √i  u sin √j  √k with parameter domain, D  {(u, √) 冟 0 u 1, 0 √ 2p}. (Refer to Figure 1.), Solution, , We first find, ru  cos √i  sin √j, r√  u sin √i  u cos √j  k, , and so, i, r√  † cos √, u sin √, , ru, , j, k, sin √ 0 †, u cos √ 1, ,  sin √i  cos √j  (u cos2 √  u sin2 √)k,  sin √i  cos √j  uk, Therefore,, 冟 ru, , r√ 冟  2sin2 √  cos2 √  u 2  21  u 2, , So, the required area is, A, , 冮冮, , 冟 ru, , 冮, , 2p, , 冮, , 2p, , 0, , , , 0, , 2p, , 1, , 冮 冮 21  u, 0, , D, , , , r√ 冟 dA , , 2, , du d√, , 0, , u1, u, 1, c 21  u 2  ln(u  21  u 2)d, d√, 2, 2, u0, , c, , 1, 1, 12  ln(1  12)d d√, 2, 2, , Use Formula 37 from the, Table of Integrals., ,  p[ 12  ln(1  12)] ⬇ 7.212, , 15.6, 1. a., b., 2. a., b., , CONCEPT QUESTIONS, , Define a parametric surface., What are the grid curves of a parametric surface?, What is a smooth surface?, Explain how you would find an equation of the tangent, plane to a smooth surface with representation r(u, √) at, the point represented by r(u 0, √0)., , 3. Write a double integral giving the area of a surface S, defined by a vector function r(u, √), where (u, √) lies in the, parameter domain D of r.

Page 268 :
15.6, , 15.6, , 1289, , Parametric Surfaces, , EXERCISES, u, 14. r(u, √)  c2 cos u  √ cosa b d i, 2, u, u,  c2 sin u  √ cosa b d j  √ sina bk;, 2, 2, 0 u 2p, 12, , In Exercises 1–4, match the equation with one of the graphs, labeled (a)–(d). Give a reason for your choice., 1. r(u, √)  2 cos ui  2 sin uj  √k, 2. r(u, √)  u cos √i  u sin √j  uk, 3. r(u, √)  u cos √i  u sin √j  u 2k, , 1, 2, , √, , Note: This is a representation for the Möbius strip., , 4. r(u, √)  u cos √i  u sin √j  √k, (a), , In Exercises 15–22, find a vector representation for the surface., , (b), , 15. The plane that passes through the point (2, 1, 3) and contains the vectors 2i  j  k and i  2j  k, , z, , z, , 16. The plane 2x  3y  z  6, 17. The lower half of the sphere x 2  y 2  z 2  1, 18. The upper half of the ellipsoid 9x 2  4y 2  36z 2  36, y, , y, , 19. The part of the cylinder x 2  y 2  4 between z  1 and, z3, , x, , x, (c), , 20. The part of the cylinder 9y 2  4z 2  36 between x  0 and, x3, , (d), z, , z, , 21. The part of the paraboloid z  9  x 2  y 2 inside the cylinder x 2  y 2  4, 22. The part of the plane z  x  2 that lies inside the cylinder, x 2  y2  1, y, , y, , x, , In Exercises 23–26, find a vector equation for the surface, obtained by revolving the graph of the function about the indicated axis. Graph the surface., , x, , 23. y  1x,, x, , 24. y  e ,, , In Exercises 5–8, find an equation in rectangular coordinates,, and then identify and sketch the surface., 6. r(u, √)  (u 2  √2)i  uj  √k, 8. r(u, √)  2 cos √ cos ui  2 cos √ sin uj  3 sin √k, , 0, , u, , x, , 1;, , x-axis, , 0, , y, , 3;, , p, , z, , p;, , y-axis, z-axis, , 28. r(u, √)  ui  (u  √ )j  √k;, 2, , graph the surface represented by the vector function., , 10. r(u, √)  ui  (√  1)j  (√3  √)k;, 2 √ 1, , 0, , 27. r(u, √)  (u  √)i  (u  √)j  √2k;, , cas In Exercises 9–14, use a computer algebra system (CAS) to, , 9. r(u, √)  (u  √)i  (u  √)j  (u 2  √2)k;, 1 √ 1, , x-axis, , In Exercises 27–32, find an equation of the tangent plane to the, parametric surface represented by r at the specified point., , 2, , √, , 4;, , 25. x  9  y ,, 26. y  cos z,, , 0, , x, , 2, , 5. r(u, √)  (u  √)i  3√j  (u  √)k, 7. r(u, √)  3 sin ui  2 cos uj  √k,, , 0, , 2, , (2, 5, 1), , 29. r(u, √)  u cos √i  2u sin √j  u k;, 2, , 1, 1,, , 11. r(u, √)  cos u sin √i  sin u sin √j  (1  cos √)k;, 0 u 2p, 0 √ 2p, , u, , (2, 0, 1), u  1,, , √p, , 1,, 30. r(u, √)  cos u sin √i  sin u sin √j  cos √k;, p, √, 4, , u, , 31. r(u, √)  ue√ i  √euj  u√k;, , u  0,, , √  ln 2, , 32. r(u, √)  u√i  u ln √j  √k;, , u  1,, , √1, , 12. r(u, √)  √ cos u sin √i  √ sin u sin √j  u cos √k;, 0 u 2p, 0 √ p, , In Exercises 33–40, find the area of the surface., , 13. r(u, √)  √ cos ui  √ sin uj  √ k;, 0 √ 1, , 33. The part of the plane r(u, √)  (u  2√  1)i , (2u  3√  1)j  (u  √  2)k; 0 u 1, 0, , 2, , 0, , u, , 2p,, , V Videos for selected exercises are available online at www.academic.cengage.com/login., , p, ,, 2, , √, , 2

Page 269 :
1290, , Chapter 15 Vector Analysis, , 34. The part of the plane 2x  3y  z  1 that lies above the, rectangular region [1, 2] [1, 3], 35. The part of the plane z  8  2x  3y that lies inside the, cylinder x 2  y 2  4, 36. The part of the paraboloid r(u, √)  u cos √i , u sin √j  u 2k; 0 u 3, 0 √ 2p, , 45. In Section 5.4 we defined the area of the surface of revolution obtained by revolving the graph of a nonnegative, smooth function f on [a, b] about the x-axis as, S  2p, , 38. The part of the sphere r(u, √)  a sin u cos √i , a sin u sin √j  a cos uk that lies in the first octant, , Use Equation (1) to derive this formula., 46. If the circle with center at (a, 0, 0) and radius b, where, 0  b  a, in the xz-plane is revolved about the z-axis,, we obtain a torus represented parametrically by, x  (a  b cos √)cos u, , 39. The surface r(u, √)  sin u cos √i  sin u sin √j  uk;, 0 u p, 0 √ 2p, 40. The part of the surface r(u, √)  u i  u√j  √ k;, 0 u 1, 0 √ 2, 1, 2, , x2, a2, , , , y2, b2, , , , z2, c2, , y  (a  b cos √)sin u, , 2, , z  b sin √, , cas 41. a. Show that the vector equation r(u, √)  a sin u cos √i , , b sin u sin √j  c cos uk, where 0 u, 0 √ 2p, represents the ellipsoid, , f(x)21  [ f ¿(x)]2 dx, , a, , 37. The part of the cone r(u, √)  u cos √i  u sin √j  uk;, 1 u 2, 0 √ p2, , 2, , 冮, , b, , p and, , with parametric domain D  {(u, √) 冟 0 u 2p,, 0 √ 2p}. (See the figure below.) Find the surface, area of the torus., z, , 1, , b. Use a CAS to graph the ellipsoid with a  3, b  4, and, c  5., c. Use a CAS to find the approximate surface area of the, ellipsoid of part (b)., , u, , cas 42. a. Show that the vector equation r(u, √)  a sin u cos √i , 3, , a sin3 u sin3 √j  a cos3 uk, where 0 u p and, 0 √ 2p, represents the astroidal sphere, x 2>3  y 2>3  z 2>3  a 2>3., b. Use a CAS to graph the astroidal sphere with a  1., c. Find the area of the astroidal sphere with a  1., , 43. Find the area of the part of the cone z  2x 2  y 2 that is, cut off by the cylinder x 2  (y  1) 2  1., 44. In Section 13.7 we showed that the tangent plane to the, graph S of a function f(x, y) at the point (a, b, f(a, b)) is, given by the equation, z  f(a, b)  fx (a, b)(x  a)  fy (a, b)(y  b), (See Equation (4) in Section 13.7.) Show that parametrizing, S by r(x, y)  xi  yj  f(x, y)k yields the same tangent, plane., , 15.7, , y, , 3, , (x, y, z), x (a, 0, 0), , In Exercises 47 and 48, determine whether the statement is true, or false. If it is true, explain why. If it is false, explain why or, give an example that shows it is false., 47. The surface described by r(u, √)  u cos √i  u sin √j  uk,, where 2 u 2 and 0 √ 2p, is smooth., 48. If r(u, √)  2 sin u cos √i  2 sin u sin √j  2 cos uk,, where 0 u p2 and 0 √ p2 , then, p>2, , 冮 冮, 0, , p>2, , 冟 ru, , r√ 冟 du d√  2p, , 0, , Surface Integrals, Surface Integrals of Scalar Fields, As we saw in Section 14.1, the mass of a thin plate lying in a plane region can be, found by evaluating the double integral 兰兰R s(x, y) dA, where s(x, y) is the mass density of the plate at any point (x, y) in R. Now, instead of a flat plate, let’s suppose that

Page 270 :
15.7, , Surface Integrals, , 1291, , we have a plate that takes the form of a curved surface. How do we determine the mass, of this plate?, For simplicity, let’s suppose that the thin plate has the shape of the surface S that, is the graph of a continuous function t of two variables defined by z  t(x, y). To further simplify our discussion, suppose that the domain of t is a rectangular region, R  {(x, y) 冟 a x b, c y d}. A typical surface is shown in Figure 1., , z, Sij, , z = t(x, y), S, , 0, a, , c, d, , FIGURE 1, A thin plate that takes the shape of, a surface S defined by z  t(x, y), , y, , b, , Rij, , x, , Let the mass density of the plate at any point on S be s(x, y, z), where s is a nonnegative continuous function defined on an open region containing S, and let P  {Rij}, be a partition of R into N  mn subrectangles. Corresponding to each subrectangle, Rij, there is a part of S, Sij, that lies directly above Rij with area ⌬Sij. Then, ⌬Sij  2[tx(x i, yj)]2  [ty(x i, yj)]2  1 ⌬A, , (1), , where (x i, yj) is the corner of Rij closest to the origin and ⌬A is the area of Rij. If m, and n are large so that the dimensions of Rij are small, then the continuity of t and s, implies that s(x, y, z) does not differ appreciably from s(x i, yj, t(x i, yj)) . Therefore,, the mass of the part of the plate that lies on S and directly above Rij is, ⌬m ij ⬇ s(x i, yj, t(x i, yj)) ⌬Sij, , Constant mass density ⴢ surface area, , Using Equation (1), we see that the mass of the plate is approximately, m, , n, , 2, 2, a a s(x i, yj, t(x i, yj))2[tx (x i, yj)]  [ty (x i, yj)]  1 ⌬A, i1 j1, , The approximation should improve as m and n approach infinity. This suggests that we, define the mass of the plate to be, m, , n, , lim a a s(x i, yj, t(x i, yj))2[tx(x i, yj)]2  [ty(x i, yj)]2  1 ⌬A, n, m→⬁, i1 j1, , Using the definition of the double integral, we see that the required mass, m, is, m, , 冮冮 s(x, y, z) dS  冮冮 s(x, y, t(x, y))2[t (x, y)], , 2, , x, , S, ,  [ty (x, y)]2  1 dA (2), , R, , if we assume that both tx and ty are continuous on R., The integral that appears in Equation (2) is a surface integral. More generally,, we can define the surface integral of a function f over nonrectangular regions as, follows.

Page 271 :
1292, , Chapter 15 Vector Analysis, , DEFINITION Surface Integral of a Scalar Function, Let f be a function of three variables defined in a region in space containing a, surface S. The surface integral of f over S is, , 冮冮, , m, , n, , f(x, y, z) dS  lim a a f(x i, yj, t(x i, yj)) ⌬Sij, n, m→⬁, i1 j1, , S, , We also have the following formulas for evaluating a surface integral depending on, the way S is defined., , THEOREM 1 Evaluation of Surface Integrals, (for Surfaces That Are Graphs), 1. If S is defined by z  t(x, y) and the projection of S onto the xy-plane is R, (Figure 2a), then, , 冮冮 f(x, y, z) dS  冮冮 f(x, y, t(x, y))2[t (x, y)], , 2, , x, , S, ,  [ty (x, y)]2  1 dA, , (3), , R, , 2. If S is defined by y  t(x, z) and the projection of S onto the xz-plane is R, (Figure 2b), then, , 冮冮 f(x, y, z) dS  冮冮 f(x, t(x, z), z)2[t (x, z)], , 2, , x, , S, ,  [tz(x, z)]2  1 dA, , (4), , R, , 3. If S is defined by x  t(y, z) and the projection of S onto the yz-plane is R, (Figure 2c), then, , 冮冮 f(x, y, z) dS  冮冮 f(t(y, z), y, z)2[t (y, z)], , 2, , y, , S, ,  [tz(y, z)]2  1 dA, , (5), , R, , z, , z, , z, y = t(x, z), , R, , z = t(x, y), , S, , S, , R, , S, y, , y, , R, , y, , x, , x, , x, , (a), , (b), , (c), , x = t(y, z), , FIGURE 2, The surfaces S and their projections onto the coordinate planes, , Note, , If we take f(x, y, z)  1, then each of the formulas gives the area of S., , EXAMPLE 1 Evaluate 兰兰S x dS, where S is the part of the plane 2x  3y  z  6, that lies in the first octant.

Page 272 :
15.7, , Surface Integrals, , 1293, , Solution The plane S is shown in Figure 3a, and its projection onto the xy-plane, is shown in Figure 3b. Using Equation (3) with f(x, y, z)  x and z  t(x, y) , 6  2x  3y, we have, , 冮冮 f(x, y, z) dS  冮冮 x dS  冮冮 x2[t (x, y)], , 2, , x, , S, , S, , R, , 冮冮 x2(2), , , ,  [ty(x, y)]2  1 dA, , 2, ,  (3)2  1 dA  114, , R, , 3, ,  114, , 冮冮, 0, , 冮 CxyD, 0, ,  114, , 冮, , R, , 2(2>3)x, , x dy dx, , View R as y-simple., , 0, , 3, ,  114, , 冮冮 x dA, , 0, , 3, , y2(2>3)x, y0, , a2x , , dx, , 3, 2 2, 2, x b dx  114 cx 2  x 3 d  3 114, 3, 9, 0, , y, , z, 6, , 2, , 2x  3y  z  6, , 2x  3y  6, 1, , R, , 2, , R, , y, , 3, , 1, , 2, , 3, , x, , x, (a) The surface S, , (b) The projection R of S onto the xy-plane, viewed as y-simple, , FIGURE 3, , EXAMPLE 2 Find the mass of the surface S composed of the part of the paraboloid, y  x 2  z 2 between the planes y  1 and y  4 if the density at a point P on S is, inversely proportional to the distance between P and the axis of symmetry of S., Solution The surface S is shown in Figure 4a, and its projection onto the xz-plane is, shown in Figure 4b. Using Equation (4) with f(x, y, z)  s(x, y, z)  k(x 2  z 2)1>2,, where k is the constant of proportionality and y  t(x, z)  x 2  z 2, we have, m, , 冮冮 s(x, y, z) dS  冮冮 k(x, S, , 2, ,  z 2)1>2 dS, , S, , 冮冮 (x, , k, , 2, ,  z 2)1>2 2[tx (x, z)]2  [tz(x, z)]2  1 dA, , 2, ,  z 2)1>2 2(2x)2  (2z)2  1 dA, , 2, ,  z 2)1>2 24x 2  4z 2  1 dA, , R, , 冮冮 (x, , k, , R, , 冮冮 (x, , k, , R

Page 273 :
1294, , Chapter 15 Vector Analysis, z, , z, yx z, 2, , 2, , R, , R, y4, , 2, , 4, , x, , FIGURE 4, , 1, , 1, , 2, , x, , y, , y1, , (b) The projection R of S onto the xz-plane, , (a) The surface S, , Changing to polar coordinates, x  r cos u and z  r sin u, we obtain, 2p, , mk, , 2, , 冮 冮 a r b24r, 0, ,  2k, , 1, , 2p, , 2, ,  1 r dr du  2k, , 1, , 冮, , 0, , 2p, , 2, , 冮 冮 2r, 0, , 2, , , , 1, , r2, r, 1, c 2r 2  14  ln ` r  2r 2  14 ` d, du, 2, 8, r1, ,  kc 117 , , 1, 1, 4  117, 15  lna, bd, 2, 4, 2  15, ,  kpc2117  15 , , 1 12 2 2 dr du, Use Formula 37 from, the Table of Integrals., , 2p, , 冮, , du, , 0, , 1, 4  117, lna, bd, 2, 2  15, , Parametric Surfaces, If a surface S is represented by a vector equation, r(u, √)  x(u, √)i  y(u, √)j  z(u, √)k, with parameter domain D, then an element of surface area is given by, 冟 ru(u, √), , r√ (u, √) 冟 dA, , as we saw in Section 15.6. This leads to the following formula for evaluating a surface, integral in which the surface is defined parametrically., , THEOREM 2 Evaluation of Surface Integrals (for Parametric Surfaces), If f is a continuous function in a region that contains a smooth surface S with, parametric representation, r(u, √)  x(u, √)i  y(u, √)j  z(u, √)k, , (u, √) 僆 D, , then the surface integral of f over S is, , 冮冮 f(x, y, z) dS  冮冮 f(r(u, √)) 冟 r, , u, , S, , D, , where f(r(u, √))  f(x(u, √), y(u, √), z(u, √))., , r√ 冟 dA, , (6)

Page 274 :
15.7, , Surface Integrals, , 1295, , Note You can show that if S is the graph of a function z  t(x, y), then Equation (3), follows from Equation (6) by putting r(u, √)  ui  √j  t(u, √)k. (See Exercise 46.), , EXAMPLE 3 Evaluate, , xy, , 冮冮 12z  1 dS,, , where S is the surface represented by, , S, , r(u, √)  (u  √)i  (u  √)j  (u 2  √2)k, where 0, Solution, , u, , 1 and 0, , √, , 2., , We first find, ru(u, √)  i  j  2uk, r√(u, √)  i  j  2√k, i, r√  † 1, 1, , ru, , j k, 1 2u †, 1 2√, ,  2[(u  √)i  (u  √)j  k], so, , 冟 ru, , r√ 冟  22(u  √)2  (u  √)2  1  222u 2  2√2  1, , Therefore,, , 冮冮, S, , xy, dS , 12z  1, , 2, , 冮冮, 0, , (u  √)  (u  √), , 1, , 22(u 2  √2)  1, , 0, , 2, , 4, , 冮冮, 0, , √ du d√, , 冮 Cu√D, 0, , 4, , 1, , 0, , 2, , 4, , 冮, , ⴢ 222u 2  2√2  1 du d√, , 0, , 2, , u1, u0, , d√, 2, , √ d√  2√2 `  8, 0, , Oriented Surfaces, , S, , n, –n, , FIGURE 5, Unit inner and outer normals to, an (orientable) closed surface S, , One of the most important applications of surface integrals involves the computation, of the flux of a vector field across an oriented surface. Before explaining the notion of, flux, however, we need to elaborate on the meaning of orientation., A surface S is orientable or two-sided if it has a unit normal vector n that varies, continuously over S, that is, if the components of n are continuous at each point (x, y, z), on S. Closed surfaces (surfaces that are boundaries of solids) such as spheres are examples of orientable surfaces. There are two possible choices of n for orientable surfaces:, the unit inner normal that points inward from S and the unit outer normal that points, outward from S (see Figure 5). By convention, however, the positive orientation for, a closed surface S is the one for which the unit normal vector points outward from S., An example of a nonorientable surface is the Möbius strip, which can be constructed, by taking a long, rectangular strip of paper, giving it a half-twist, and then taping the short, edges together to produce the surface shown in Figure 6. If you take a unit normal n starting at P (see Figure 6), then you can move it along the surface in such a way that upon, returning to the starting point (and without crossing any edges), it will point in a direction precisely opposite to its initial direction. This shows that n does not vary continuously on a Möbius strip, and accordingly, the strip is not orientable.

Page 275 :
1296, , Chapter 15 Vector Analysis, , n, , Start, P, , d, , FIGURE 6, The Möbius strip can be constructed, by using a rectangular strip of paper., , c, d b, , a, , b, , a c, , Surface Integrals of Vector Fields, Suppose that F is a continuous vector field defined in a region R in space. We can think, of F(x, y, z) as giving the velocity of a fluid at a point (x, y, z) in R, and S as a smooth,, oriented surface lying in R. If S is flat and F is a constant field, then the flux, or rate, of flow (volume of fluid crossing S per unit time), is equal to, , F, F•n, , F ⴢ n A(S), , n, S, , FIGURE 7, If S is flat and F is constant, then, the flux is equal to the volume, of the prism., z, , z  t (x, y), S, c, , a, b, , d, y, , R, , (the normal component of F with respect to S times the area of S). Geometrically, the, flux is given by the volume of fluid in the prism in Figure 7., More generally, suppose that S is the graph of a function of two variables defined, by z  t(x, y), where, for simplicity, we assume that the domain of t is a rectangular, region R  {(x, y) 冟 a x b, c y d}. (See Figure 8.) Let P  {Rij} be a partition of R into N  mn subrectangles. Corresponding to each subrectangle Rij there is, the part of S that lies directly above Rij with area ⌬Sij. As in Section 14.5, let (x i, yj) be, the corner of Rij closest to the origin, and let (x i, yj, t(x i, yj)) be the point directly above, it, as shown in Figure 9. Let n ij denote the unit normal vector to S at (x i, yj, t(x i, yj))., If m and n are large so that the dimensions of Rij are small, then the continuity of F, implies that F(x, y, z) does not differ appreciably from F(x i, yj, t(x i, yj)) on Rij., Furthermore, the continuity of t implies that Sij may be approximated by Tij, the, parallelogram that is part of the tangent plane to S at the point (x i, yj, t(x i, yj)) lying, directly above Rij. But the flux of F across (the flat) Tij is approximately, , x, , F(x i, yj, t(x i, yj)) ⴢ n ij (⌬Tij), , FIGURE 8, A smooth surface S defined by, z  t(x, y) for (x, y) in R, , where ⌬Tij is the area of Tij. Since ⌬Tij ⬇ ⌬Sij, we see that the flux of F across S may, be approximated by the sum, m, , nij, (xi, yj, t(xi, yj)), , n, , a a F(x i, yj, t(x i, yj)) ⴢ n ij ⌬Sij, , F, , i1 j1, , m, , Sij, , n, ,  a a F(x i, yj, t(x i, yj)) ⴢ n ij2[tx (x i, yj)]2  [ty(x i, yj)]2  1 ⌬A, , Tij, , i1 j1, , This last equality follows upon using Equation (1). We can expect that the approximation will get better as the partition P becomes finer. This observation leads to the following definition., , DEFINITION Surface Integral of a Vector Field, (xi, yj), Rij, , FIGURE 9, , Let F be a continuous vector field defined in a region containing an oriented surface S with unit normal vector n. The surface integral of F across S in the, direction of n is, , 冮冮 F ⴢ dS  冮冮 F ⴢ n dS , S, , S, , m, , lim, , n, , a a F(x i, yj, f(x i, yj)) ⴢ n ij ⌬Sij, , m, n→⬁ i1 j1

Page 276 :
15.7, , Surface Integrals, , 1297, , Thus, the surface integral (also called flux integral) of a vector field F across an, oriented surface S is the integral of the normal component of F over S. If the fluid has, density r(x, y, z) at (x, y, z), then the flux integral, , 冮冮 rF ⴢ n dS, S, , gives the mass of the fluid flowing across S per unit time., To obtain a formula for finding the flux integral in terms of t(x, y), recall from, Section 13.6 that the normal to the surface z  t(x, y) is given by §G, where, G(x, y, z)  z  t(x, y). Therefore, the unit normal to S is, n, , tx(x, y)i  ty (x, y)j  k, §G(x, y, z), , 冟§G(x, y, z) 冟, 2[tx (x, y)]2  [ty (x, y)]2  1, , Furthermore, in Section 14.5 we showed that the “element of area” dS is given by, dS  2[tx (x, y)]2  [ty (x, y)]2  1 dA, so, F ⴢ [tx (x, y)i  ty (x, y)j  k], , 冮冮 F ⴢ n dS  冮冮 2[t (x, y)], , 2, , S, , D, , , , x, ,  [ty(x, y)]  1, 2, , ⴢ 2[tx(x, y)]2  [ty(x, y)]2  1 dA, , 冮冮 F ⴢ [t (x, y)i  t (x, y)j  k] dA, x, , y, , D, , where D is the projection of S onto the xy-plane., If F(x, y, z)  P(x, y, z)i  Q(x, y, z)j  R(x, y, z)k, then we can write, , 冮冮 F ⴢ n dS  冮冮 (Pt, , x, , S, ,  Qty  R) dA, , D, , THEOREM 3 Evaluation of Surface Integrals (for Graphs), If F  Pi  Qj  Rk is a continuous vector field in a region that contains a, smooth oriented surface S given by z  t(x, y) and D is its projection onto the, xy-plane, then, , 冮冮 F ⴢ dS  冮冮 (Pt, , x, , S, ,  Qty  R) dA, , (7), , D, , Before looking at the next example, we note the following property of surface integrals: If S  S1 傼 S2 傼 p 傼 Sn, where each of the surfaces is smooth and intersect, only along their boundaries, then, , 冮冮 F ⴢ dS  冮冮 F ⴢ dS  p  冮冮 F ⴢ dS, S, , S1, , Sn, , EXAMPLE 4 Evaluate 兰兰S F ⴢ dS, where F(x, y, z)  xi  yj  zk and S is the surface that is composed of the part of the paraboloid z  1  x 2  y 2 lying above the, xy-plane and the disk D  {(x, y) 冟 0 x 2  y 2 1}.

Page 277 :
1298, , Chapter 15 Vector Analysis, z, , S1, z = 1 – x 2 – y2, S2, , Solution The (closed) surface S together with a few vectors from the vector field F, is shown in Figure 10. Writing the equation of the surface S1 in the form t(x, y) , 1  x 2  y 2, we find that tx  2x and ty  2y. Observe that the projection of S onto, the xy-plane is D  {(x, y) 冟 0 x 2  y 2 1}. Also, P(x, y, z)  x, Q(x, y, z)  y, and, R(x, y, z)  z. So using Equation (7), we obtain, , 冮冮 F ⴢ dS  冮冮 (Pt, , x, , y, , S1, ,  Qty  R) dA, , D, , x, , , , FIGURE 10, The part of the paraboloid, z  1  x 2  y 2 that lies above, the xy-plane and is oriented so that, the unit normal vector n points, upward. The unit normal for the, disk D  S2 points downward., , 冮冮 [x(2x)  y(2y)  z] dA, D, , , , 冮冮 (2x, , 2, ,  2y 2  z) dA, , 2, ,  2y 2  (1  x 2  y 2)] dA, , D, , , , 冮冮 [2x, , z  1  x 2  y2, , D, , , , 冮冮 (1  x, , 2, ,  y 2) dA, , D, , 2p, , , , 1, , 冮 冮 (1  r )r dr du, 2, , 0, , , , 冮, , Use polar coordinates., , 0, , 2p, , 0, , r1, 1, 1, c r 2  r 4d, du , 2, 4, r0, , 冮, , 0, , 2p, , 3, 3, du  p, 4, 2, , Next, observe that the normal for the surface S2 is n  k. (Remember that the normal for a closed surface, by convention, points outward.) So we have, , 冮冮 F ⴢ dS  冮冮 F ⴢ (k) dS  冮冮 (z) dA  冮冮 0 dA  0, S2, , S2, , D, , D, , because z  0 on S2. Therefore,, , 冮冮 F ⴢ dS  冮冮 F ⴢ dS  冮冮 F ⴢ dS  2 p  0 , 3, , S, , S1, , 3p, 2, , S2, , Notes, 1. If the vector field F of Example 4 describes the velocity of a fluid flowing, through the paraboloidal surface S, then the integral 兰兰S F ⴢ n dS that we have, just evaluated tells us that the fluid is flowing out through S at the rate of 3p>2, cubic units per unit time., 2. In Example 4, if we had wanted the paraboloid to be oriented so that the normal, pointed downward, then we would simply have picked the normal to be n. In, this case the fluid flows into S at the rate of 3p>2 cubic units per unit time., , Parametric Surfaces, If an oriented surface S is a smooth surface represented by a vector equation, r(u, √)  x(u, √)i  y(u, √)j  z(u, √)k

Page 278 :
15.7, , Surface Integrals, , 1299, , with parameter domain D, then the normal to S is, n, , ru, 冟 ru, , r√, r√ 冟, , Therefore,, , 冮冮 F ⴢ dS  冮冮 F ⴢ n dS  冮冮 F ⴢ 冟 r, , ru, , S, , S, , , , u, , S, , 冮冮 cF(r(u, √)) ⴢ 冟 r, , ru, u, , D, , , , r√, dS, r√ 冟, , 冮冮 F(r(u, √)) ⴢ (r, , u, , r√, d 冟 ru, r√ 冟, , r√ 冟 dA, , r√) dA, , D, , THEOREM 4 Evaluation of Surface Integrals of a Vector Field, (for Parametric Surfaces), If F is a continuous vector field in a region that contains a smooth, oriented surface S with parametric representation, r(u, √)  x(u, √)i  y(u, √)j  z(u, √)k, , (u, √) 僆 D, , then the surface integral of f over S is, , 冮冮 F ⴢ dS  冮冮 F(r(u, √)) ⴢ (r, , u, , S, , r√) dA, , (8), , D, , where, F(r(u, √))  F(x(u, √), y(u, √), z(u, √)), , EXAMPLE 5 Find the flux of the vector field F(x, y, z)  yi  xj  2zk across the, unit sphere x 2  y 2  z 2  1., Solution, , The unit sphere has parametric representation, r(u, √)  sin u cos √i  sin u sin √j  cos uk, , with parameter domain D  {(u, √) 冟 0 u p, 0 √, Example 10 in Section 15.6 and taking a  1, we find, ru, , 2p}. Proceeding as in, , r√  sin2 u cos √i  sin2 u sin √j  sin u cos uk, , Therefore,, F(r(u, √)) ⴢ (ru, , r√)  (sin u sin √i  sin u cos √j  2 cos uk), ⴢ (sin2 u cos √i  sin2 u sin √j  sin u cos uk),  sin3 u sin √ cos √  sin3 u cos √ sin √  2 cos2 u sin u,  2(sin3 u sin √ cos √  cos2 u sin u)

Page 279 :
1300, , Chapter 15 Vector Analysis, , Using Equation (8), the flux across the sphere is, , 冮冮 F ⴢ dS  冮冮 F(r(u, √)) ⴢ (r, , r√) dA, , u, , S, , D, , 2p, , 2, , 冮 冮, 0, , 2, , 冮, , p, , (sin3 u sin √ cos √  cos2 u sin u) du d√, , 0, , p, , sin3 u du, , 0, , 冮, , 2p, , sin √ cos √ d√  2, , 0, , 冮, , p, , cos2 u sin u du, , 0, , 冮, , 2p, , d√, , 0, , The first term on the right is equal to zero because, , Historical Biography, , 冮, , 2p, , sin √ cos √ d√ , , 0, , 1 2 2p, sin √ `  0, 2, 0, , so, , 冮冮 F ⴢ dS  2冮, , 2, , cos u sin u du, , 0, , S, , Bettmann/Corbis, , p, , 冮, , 2p, , d√, , 0, , p, 1,  2a cos3 ub ` ⴢ 2p, 3, 0, , CHARLES-AUGUSTIN DE COULOMB, , , , (1736–1806), Born to a wealthy family, Charles-Augustin, de Coulomb spent his early years in, Angoulême in southwestern France. His, family later moved to Paris, where he, attended good schools and received a solid, education. In 1760, Coulomb entered the, two-year military engineering program at, the Ecole de Genie at Mézières. At the conclusion of those studies, he was commissioned as a second lieutenant in the, infantry, where he served in the engineering corps. For the next twenty years, Coulomb served in a variety of posts and, was involved in a wide range of military, engineering projects. From 1764 to 1772, Coulomb was put in charge of constructing, the new Fort Bourbon in Martinique in the, West Indies. Upon his return to France, he, started writing works on applied mechanics, and he began publishing important, works in 1773. This period culminated with, his work on friction, Theorie des machines, simples, which won him a prize in 1781. This, recognition changed the direction of his, life. He was elected to the mechanics section of the Academie des Sciences, and he, focused on his work as a physicist instead, of engineering. Between 1785 and 1791 he, wrote seven important treatises on electricity and magnetism, in which he developed the theory of attraction and repulsion between electrical charges that is now, known as Coulomb’s Law., , 8p, 3, , We have used an example involving fluid flow to illustrate the concept of the surface integral of a vector field. But these integrals have wider applications in the physical sciences. For example, if E is the electric field induced by an electric charge of q, coulombs located at the origin of a three-dimensional coordinate system, then by, Coulomb’s Law (Example 4 in Section 15.1),, E, , q, r, ⴢ, 4pe0 冟 r 冟3, , where e0 is a constant called the permittivity of free space. If S is a sphere of radius r, centered at the origin, then the surface integral, , 冮冮 E ⴢ n dS, S, , is the flux of E passing through S., Yet another application of surface integrals can be found in the study of heat flow., Suppose that the temperature at a point (x, y, z) in a homogeneous body is T(x, y, z)., Empirical results suggest that heat will flow from points at higher temperatures to those, at lower temperatures. Since the temperature gradient §T points in the direction of, maximum increase of the temperature, we see that the flow of heat can be described, by the vector field, q  k §T, where k is a constant of proportionality known as the thermal conductivity of the body., The rate at which heat flows across a surface S in the body is given by the surface, integral, , 冮冮 q ⴢ n dS  k冮冮 §T ⴢ n dS, S, , S

Page 280 :
15.7, , Surface Integrals, , 1301, , EXAMPLE 6 Rate of Flow of Heat Across a Sphere The temperature at a point, P(x, y, z) in a medium with thermal conductivity k is inversely proportional to the distance between P and the origin. Find the rate of flow of heat across a sphere S of radius, a, centered at the origin., Solution, , We have, T(x, y, z) , , c, 2x  y 2  z 2, 2, , where c is the constant of proportionality. Then the flow of heat is, q  k§T  kc, , , cx, (x  y  z ), 2, , ck, (x  y 2  z 2)3>2, 2, , 2, , 2 3>2, , cy, , i, , (x  y  z ), 2, , 2, , 2 3>2, , j, , cz, (x  y 2  z 2)3>2, 2, , kd, , (xi  yj  zk), , The outward unit normal to the sphere x 2  y 2  z 2  a 2 at the point (x, y, z) is, n, , 1, (xi  yj  zk), a, , So the rate at which heat flows across S is, , 冮冮 q ⴢ n dS  冮冮 (x, S, , S, , , , ck, a, , ck, 2, , y z ), 2, , 冮冮 2x, S, , , , , 15.7, , ck, a2, ck, a2, , 1, (xi  yj  zk) ⴢ c (xi  yj  zk)d dS, a, , 1, 2, ,  y2  z2, , 冮冮 dS, , dS, , Since x 2  y 2  z 2  a 2 on S, , S, , A(S) , , ck, a2, , (4pa 2)  4pck, , CONCEPT QUESTIONS, , 1. a. Define the surface integral of a scalar function f over a, surface that is the graph of a function z  f(x, y)., b. How do you evaluate the integral of part (a)?, c. How do you evaluate the surface integral if the surface is, represented by a vector function r(u, √)?, 2. What is an orientable surface? Give an example of a surface, that is not orientable., , 15.7, , 2 3>2, , 3. a. Define the surface (flux) integral of a vector field F over, an oriented surface S with a unit normal n., b. How do you evaluate the surface integral if the surface is, the graph of a function z  t(x, y)?, c. How do you evaluate the surface integral if the surface is, represented by the vector function r(u, √)?, , EXERCISES, , In Exercises 1–14, evaluate 兰兰S f(x, y, z) dS., 1. f(x, y, z)  x  y; S is the part of the plane, 3x  2y  z  6 in the first octant, V Videos for selected exercises are available online at www.academic.cengage.com/login., , 2. f(x, y, z)  xy; S is the part of the plane, 2x  3y  z  6 in the first octant

Page 281 :
1302, , Chapter 15 Vector Analysis, , is the part of the surface, above the rectangular region, R  {(x, y) 冟 0 x 2, 0 y, , In Exercises 19–28, evaluate 兰兰S F ⴢ dS, that is, find the flux of, F across S. If S is closed, use the positive (outward) orientation., , 3. ;, , 1}, , 19. F(x, y, z)  2xi  2yj  zk; S is the part of the paraboloid, z  4  x 2  y 2 above the xy-plane; n points upward, , 4. f(x, y, z)  xz ; S is the part of the surface, y  x 2  2z to the right of the square region, R  {(x, z) 冟 0 x 1, 0 z 1}, 2, , 20. F(x, y, z)  3xi  3yj  2k; S is the part of the paraboloid z  x 2  y 2 between the planes z  0 and z  4;, n points downward, , 5. f(x, y, z)  x  2y  z; S is the part of the plane, y  z  4 inside the cylinder x 2  y 2  1, , 21. F(x, y, z)  xi  yj  zk; S is the part of the plane, 3x  2y  z  6 in the first octant; n points upward, , 6. f(x, y, z)  xz; S is the part of the plane y  z  4, inside the cylinder x 2  y 2  4, 7. f(x, y, z)  x 2z; S is the part of the cone z  2x 2  y 2, inside the cylinder x 2  y 2  1, 8. f(x, y, z)  x  2y  3z; S is the part of the cone, x  2y 2  z 2 between the planes x  1 and x  4, 9. f(x, y, z)  xyz; S is the part of the cylinder x 2  y 2  4, in the first octant between z  0 and z  4, 10. f(x, y, z)  z 2; S is the hemisphere z  29  x 2  y 2, y, ; S is the surface with, 14z  5, vector representation r(u, √)  ui  √j  (√2  1)k,, 0 u 1, 1 √ 1, , 11. f(x, y, z)  x , , 22. F(x, y, z)  6zi  2xj  yk; S is the part of the plane, 2x  3y  6z  12 in the first octant; n points upward, 23. F(x, y, z)  yi  xj  2zk; S is the hemisphere, z  24  x 2  y 2; n points upward, 24. F(x, y, z)  xi  yj  zk; S is the hemisphere, z  29  x 2  y 2; n points upward, Hint: First evaluate 兰兰S– F ⴢ n dS, where S is the part of the hemisphere z  29  x 2  y 2 inside the cylinder x 2  y 2  a 2, where, 0  a  3. Then take the limit as a → 3., , 25. F(x, y, z)  2i  3j  k; S is the part of the cone, z  2x 2  y 2 inside the cylinder x 2  y 2  1; n points, upward, , 12. f(x, y, z)  x  z; S is the surface with vector representation, r(u, √)  u sin √i  u cos √j  u 2k, 0 u 1, 0 √ p2, , 26. F(x, y, z)  xi  2yj  zk; S is the part of the cone, y  2x 2  z 2 inside the cylinder x 2  z 2  1; n points, to the right, , 13. f(x, y, z)  z21  x 2  y 2; S is the helicoid with vector, representation r(u, √)  u cos √i  u sin √j  √k,, 0 u 1, 0 √ 2p, , 27. F(x, y, z)  y 3i  x 2j  zk; S is the boundary of the, cylindrical solid bounded by x 2  y 2  9, z  0, and z  3, , 14. f(x, y, z)  z; S is the part of the torus with vector, representation r(u, √)  (a  b cos √)cos ui , (a  b cos √)sin uj  b sin √k, 0 u 2p, 0 √, , 28. F(x, y, z)  x 2i  xyj  xzk; S is the surface of the tetrahedron with vertices (0, 0, 0), (1, 0, 0), (0, 2, 0), and, (0, 0, 3), , p, 2, , In Exercises 15–18, find the mass of the surface S having the, given mass density., 15. S is the part of the plane x  2y  3z  6 in the first, octant; the density at a point P on S is directly proportional, to the square of the distance between P and the yz-plane., 16. S is the part of the paraboloid z  x  y between the, planes z  1 and z  4; the density at a point P on S is, directly proportional to the distance between P and the axis, of symmetry of S., 2, , 2, , 17. S is the hemisphere x 2  y 2  z 2  4, z  0; the density at, a point P on S is directly proportional to the distance, between P and the xy-plane., 18. S is the part of the sphere x 2  y 2  z 2  1 that lies above, the cone z  2x 2  y 2; the density at a point P on S is, directly proportional to the distance between P and the, xy-plane., , In Exercises 29 and 30, a thin sheet has the shape of the surface, S. If its density (mass per unit area) at the point (x, y, z) is, r(x, y, z), then its center of mass is (x, y, z), where, x, , 1, m, , 冮冮 xr(x, y, z) dS,, S, , 1, z, m, , y, , 1, m, , 冮冮 yr(x, y, z) dS,, S, , 冮冮 zr(x, y, z) dS, S, , and m is the mass of the sheet. Find the center of mass of the, sheet., 29. S is the upper hemisphere x 2  y 2  z 2  a 2, z  0,, r(x, y, z)  k, where k is a constant., 30. S is the part of the paraboloid z  4  12 x 2  12 y 2, z  0,, r(x, y, z)  k, where k is a constant.

Page 282 :
15.7, In Exercises 31 and 32, a thin sheet has the shape of a surface, S. If its density (mass per unit area) at the point (x, y, z) is, r(x, y, z), then its moment of inertia about the z-axis is, Iz  兰兰S (x 2  y 2)r(x, y, z) dS., 31. Show that the moment of inertia of a spherical shell of uniform density about its diameter is 23 ma 2, where m is its mass, and a is its radius., 32. Find the moment of inertia of the conical shell z 2  x 2  y 2,, where 0 z 2, if it has constant density k., In Exercises 33 and 34 the electric charge density at a point, (x, y, z) on S is s(x, y, z), and the total charge on S is given, by Q  兰兰S s(x, y, z) dS., 33. Electric Charge Find the total charge on the part of the plane, 2x  3y  z  6 in the first octant if the charge density at a, point P on the surface is directly proportional to the square, of the distance between P and the xy-plane., 34. Electric Charge Find the total charge on the part of the hemisphere z  225  x 2  y 2 that lies directly above the, plane region R  {(x, y) 冟 x 2  y 2 9} if the charge density, at a point P on the surface is directly proportional to the, square of the distance between P and the xy-plane., 35. Flow of a Fluid The flow of a fluid is described by the vector, field F(x, y, z)  2xi  2yj  3zk. Find the rate of flow of, the fluid upward through the surface S that is the part of the, plane 2x  3y  z  6 in the first octant., 36. Flow of a Liquid The flow of a liquid is described by the vector field F(x, y, z)  xi  yj  3zk. If the mass density of, the fluid is 1000 (in appropriate units), find the rate of flow, (mass per unit time) upward of the liquid through the surface S that is part of the paraboloid z  9  x 2  y 2 above, the xy-plane., Hint: The flux is 兰兰S rF ⴢ n dS, where r is the mass density function., , 37. Rate of Flow of Heat The temperature at a point (x, y, z) in a, homogeneous body with thermal conductivity k  5 is, T(x, y, z)  x 2  y 2. Find the rate of flow of heat across, the cylindrical surface x 2  y 2  1 between the planes, z  0 and z  1., , 1303, , xz-plane. Write a double integral similar to Equation (7), that gives 兰兰S F ⴢ dS., b. Use the result of part (a) to evaluate 兰兰S F ⴢ dS, where, F(x, y, z)  yi  z j  3yz 2k and S is the part of the, cylinder x 2  y 2  4 that lies in the first octant between, z  0 and z  3 and oriented away from the origin., 40. Flux of an Electric Field Find the flux of the electric field, q, r, across the sphere x 2  y 2  z 2  a 2. Is the, E, 4pe0 冟 r 冟3, flux independent of the radius of the sphere? Give a physical, interpretation., 41. Suppose that the density at each point of a thin spherical, shell of radius R is proportional to the (linear) distance from, the point to a fixed point on the sphere. Find the total mass, of the shell., 42. Mass of a Ramp Suppose that the density at each point, of a spiral ramp represented by the vector equation, r(u, √)  u cos √i  u sin √j  √k, where 0 u 3 and, 0 √ 6p, is proportional to the distance of the point, from the central axis of the ramp. What is the total mass of, the ramp?, 43. Suppose that f is a nonnegative real-valued function defined, on the interval [a, b] and f has a continuous derivative in, (a, b) . Show that the area of the surface of revolution S, obtained by revolving the graph of f about the x-axis is, given by, 2p, , 冮, , b, , f(x)21  [ f ¿(x)]2 dx, , a, , Hint: First find a parametric representation of S (see Section 15.6)., , 44. Find the flux of F(x, y, z)  2xzi  yzj  z 2k out of a unit, cube T  {(x, y, z) 冟 0 x 1, 0 y 1, 0 z 1}., Hint: The flux out of the cube is the sum of the fluxes across the, sides of the cube., z, , 38. Rate of Flow of Heat The temperature at a point P(x, y, z) in a, medium with thermal conductivity k is proportional to the, square of the distance between P and the origin. Find the, rate of flow of heat across a sphere S of radius a, centered, at the origin., 39. a. Suppose that F  Pi  Qj  Rk is a continuous vector, field in a region that contains a smooth oriented surface S, given by y  t(x, z) and D is its projection onto the, , Surface Integrals, , y, x, , Four of the six unit normal vectors are shown.

Page 283 :
1304, , Chapter 15 Vector Analysis, 47. Let F and G be continuous vector fields defined on a, smooth, oriented surface S. If a and b are constants,, show that, , 45. a. Let f be a function of three variables defined on a region, in space containing a surface S. Suppose that S is the, graph of the function z  t(x, y) that is represented, implicitly by the equation F(x, y, z)  0, where F is, differentiable. Show that, , 冮冮 f(x, y, z) dS  冮冮, S, , D, , f 2F 2x  F 2y  F 2z, 冟 Fz 冟, , 冮冮 (aF  bG) ⴢ dS  a冮冮 F ⴢ dS  b冮冮 F ⴢ dS, S, , dA, , S, , In Exercises 48 and 49, determine whether the statement is true, or false. If it is true, explain why. If it is false, explain why or, give an example that shows it is false., , where D is the projection of S onto the xy-plane., b. Re-solve Example 1 using the result of part (a)., , 48. If f(x, y, z)  0, then 兰兰S f dS  A(S) , where A(S) is the, area of S., , 46. Show that if the surface S is the graph of a function, z  t(x, y), then Equation (3) follows from Equation (6), by putting r(u, √)  ui  √j  t(u, √)k., , 15.8, , S, , 49. If F is a constant vector field and S is a sphere, then, 兰兰S F ⴢ dS  0., , The Divergence Theorem, Recall that Green’s Theorem can be written in the form, , 冯 F ⴢ n ds  冮冮 div F dA, C, , Equation (8) in Section 15.5, , R, , where F is a vector field in the plane, C is an oriented, piecewise-smooth, simple closed, curve that bounds a region R, and n is the outer normal vector to C. The theorem states, that the line integral of the normal component of a vector field in two-dimensional, space around a simple closed curve is equal to the double integral of the divergence of, the vector field over the plane region bounded by the curve., , The Divergence Theorem, The Divergence Theorem generalizes this result to the case involving vector fields in, three-dimensional space. This theorem, also called Gauss’s Theorem in honor of the, German mathematician Karl Friedrich Gauss (1777–1855), relates the surface integral, of the normal component of a vector field F in three-dimensional space over a closed, surface S to the triple integral of the divergence of F over the solid region T bounded, by S., Although the Divergence Theorem is true for very general surfaces, we will restrict, our attention to the case in which the solid regions T are simultaneously x-, y-, and, z-simple. These regions are called simple solid regions. Examples are regions bounded, by spheres, ellipsoids, cubes, and tetrahedrons., , THEOREM 1 The Divergence Theorem, Let T be a simple solid region bounded by a closed piecewise-smooth surface, S, and let n be the unit outer normal to S. If F  Pi  Qj  Rk is a vector, field, where P, Q, and R have continuous partial derivatives on an open region, containing T, then, , 冮冮 F ⴢ dS  冮冮 F ⴢ n dS  冮冮冮 div F dV, S, , S, , T, , (1)

Page 285 :
1306, , Chapter 15 Vector Analysis, , Therefore,, , 冮冮 Rk ⴢ n dS  冮冮 Rk ⴢ n dS  冮冮 Rk ⴢ n dS, S, , S1, , (6), , S2, , To evaluate the integrals on the right-hand side of Equation (6), observe that the outer, normal n points downward on S1. Writing t1(x, y, z)  z  k 1 (x, y), we find, , n, , §t1(x, y, z), , 冟 §t1(x, y, z) 冟, , k 1, k 1, i, jk, x, y, k 1 2, k 1 2, b a, b 1, B x, y, a, , Therefore, using Equation (6), we obtain, , 冮冮 Rk ⴢ n dS  冮冮 R(x, y, k (x, y)) ⴢ, 1, , S1, , D, , , , 1, k 1 2, k 1 2, b a, b 1, B x, y, a, , k 1 2, k 1 2, b a, b  1 dA, B x, y, a, , 冮冮 R(x, y, k (x, y)) dA, 1, , D, , On S2 the outer normal n points upward. Writing t2 (x, y, z)  z  k 2(x, y), we see that, , n, , §t2 (x, y, z), , 冟 §t2 (x, y, z) 冟, , , , k 2, k 2, i, jk, x, y, , k 2 2, k 2 2, b a, b 1, B x, y, a, , so, , 冮冮 Rk ⴢ n dS  冮冮 R(x, y, k (x, y)) ⴢ, 2, , S2, , D, , , , 1, k 2 2, k 2 2, b a, b 1, B x, y, a, , k 2 2, k 2 2, b a, b  1 dA, B x, y, a, , 冮冮 R(x, y, k (x, y)) dA, 2, , D, , Therefore, Equation (6) becomes, , 冮冮 Rk ⴢ n dS  冮冮 [R(x, y, k (x, y))  R(x, y, k (x, y))] dA, 2, , S, , 1, , D, , Comparing this with Equation (5), we have, R, , 冮冮 Rk ⴢ n dS  冮冮冮 z dV, S, , T, , so Equation (4) is established. Equations (2) and (3) are proved in a similar manner by, viewing R as x-simple and y-simple, respectively., , EXAMPLE 1 Compute 兰兰S F ⴢ n dS given that, F(x, y, z)  (x  sin z)i  (2y  cos x)j  (3z  tan y)k, and S is the unit sphere x 2  y 2  z 2  1.

Page 286 :
15.8, , The Divergence Theorem, , 1307, , Solution To evaluate the integral directly would be a difficult task. Applying the Divergence Theorem, we have, , 冮冮 F ⴢ n dS  冮冮冮 div F dV, S, , T, , But, §ⴢF, , , , , (x  sin z) , (2y  cos x) , (3z  tan y)  1  2  3  6, x, y, z, , and the solid T is the unit ball B bounded by the unit sphere x 2  y 2  z 2  1. Therefore,, , 冮冮 F ⴢ n dS  冮冮冮 § ⴢ F dV  冮冮冮 6 dV  6V(B)  6 c 3 p(1) d  8p, 4, , S, , B, , 3, , B, , EXAMPLE 2 Let T be the solid bounded by the cylinder x 2  y 2  4 and the planes, , z  0 and z  3, and let S be the surface of T. Calculate the outward flux of the vector field F(x, y, z)  xy 2i  yz 2j  zx 2k over S., z, z3, , Solution The surface S is shown in Figure 2. The flux of F over S is given by, 兰兰S F ⴢ n dS, which by the Divergence Theorem can also be found by evaluating, 兰兰兰T § ⴢ F dV. Now, , n, 3, , div F , , x2  y2  4, , n, , , , , (xy 2) , (yz 2) , (zx 2)  y 2  z 2  x 2, x, y, z, , Therefore,, , 冮冮 F ⴢ n dS  冮冮冮 div F dV  冮冮冮 (x, , –2, 2, , 2, , S, , y, , n, , T, , 2, ,  y 2  z 2) dV, , T, , Using cylindrical coordinates to evaluate the triple integral, we have, , x, , FIGURE 2, The surface S and some of the outer, normals to S, , 冮冮, , 2p, , F ⴢ n dS , , 冮 冮冮, 0, , S, , 2, , 0, , 冮 冮, , 2, , 2p, , 2, , 0, , z, , , , 0, , 冮 冮, 0, , n, , , , z  1  y2, , x2z, , 冮, , (r 2  z 2)r dz dr du, , 0, , 2p, , , , 3, , cr 3z , , 1 3 z3, rz d, dr du, 3, z0, , (3r 3  9r) dr du , , 0, , 冮, , 0, , 2p, , r2, 3, 9, c r 4  r 2d, du, 4, 2, r0, , 2p, , 30 du  60p, , 0, , EXAMPLE 3 Let T be the region bounded by the parabolic cylinder z  1  y 2 and, , n, 1, 2, , n, , x, , FIGURE 3, The surface S and some of the outer, normals to S, , y, , the planes z  0, x  0, and x  z  2, and let S be the surface of T. If F(x, y, z) , xy 2i  1 13 y 3  cos xz 2 j  xey k, find 兰兰S F ⴢ n dS., Solution, , The surface S is shown in Figure 3. We first compute, , §ⴢF, , ,  1 3, , (xy 2) , a y  cos xzb , (xey)  y 2  y 2  2y 2, x, y 3, z

Page 288 :
15.8, , The Divergence Theorem, , 1309, , of the Divergence Theorem for the solid T that lies between S1 and S. Using Equation (7), we obtain, , 冮冮冮 div E dV  冮冮 E ⴢ dS  冮冮 E ⴢ dS, T, , S1, , S, , But we showed that div E  0 in Example 4 in Section 15.2. So we have, , 冮冮 E ⴢ dS  冮冮 E ⴢ dS  冮冮 E ⴢ n dS, S, , S1, , S1, , To evaluate the integral on the right, note that the unit normal to the sphere S1 is, n  r> 冟 r 冟. Therefore,, Eⴢn, , , , q, q rⴢr, r, r, ⴢa b, 3, 冟, 冟, 4pe0 冟 r 冟, r, 4pe0 冟 r 冟4, q, 4pe0 冟 r 冟2, , r ⴢ r  冟 r 冟2, , q, 4pe0a 2, , because 冟 r 冟  a on the sphere S1. Therefore, we have, , 冮冮 E ⴢ dS  冮冮 E ⴢ dS  4pe a 冮冮 dS  4pe a, q, , q, , 2, , S, , 0, , S1, , S1, , 2, , A(S1), , 0, , q, , q, , (4pa 2) , 2, e0, 4pe0a, The result in Example 4 shows that the flux across any closed surface that contains, the charge q is q>e0. This is intuitively clear, since any closed surface enclosing the, charge q would trap the same number of field lines., Furthermore, by using the principle of superposition (the field induced by several, electric charges is the vector sum of the fields due to the individual charges), it can be, shown that for any closed surface S,, , 冮冮 E ⴢ n dS  e, , Q, , n, Br, , S, , P0 (x0 , y0 , z0 ), , 0, , where Q is the total charge enclosed by S. This is one of the most important laws in, electrostatics and is known as Gauss’s Law., , Interpretation of Divergence, Sr, , FIGURE 6, Br is a ball of radius r centered at, P0(x 0, y0, z 0)., , For a physical interpretation of the divergence of a vector field F, we can think of F, as representing the velocity field associated with the flow of a fluid. Let P0 (x 0, y0, z 0), be a point in the fluid, and let Br be a ball with radius r, centered at P0, and having the, sphere Sr for its boundary, as shown in Figure 6. If r is small, then the continuity of

Page 289 :
1310, , Chapter 15 Vector Analysis, , div F guarantees that (div F)(P) ⬇ (div F)(P0) for all points P in Br. Therefore, using, the Divergence Theorem, we have, , 冮冮 F ⴢ n dS  冮冮冮 div ⴢ F dV ⬇ 冮冮冮 (div F)(P ) dV, 0, , Sr, , Br, , Br, ,  (div F)(P0), , 冮冮冮 dV  (div F)(P )V(B ), 0, , r, , Br, , where V(Br)  43 pr 3. This approximation improves as r → 0, and we have, (div F)(P0)  lim, r→0, , 1, V(Br), , 冮冮 F ⴢ n dS, , (8), , Br, , Equation (8) tells us that we can interpret div F(P0) as the rate of flow outward of the, fluid per unit volume at P0—hence the term divergence. In general, if div F(P) 0,, the net flow is outward near P, and P is called a source. If div F(P)  0, the net flow, is inward near P, and P is called a sink. Finally, if the fluid is incompressible and there, are no sources or sinks present, then no fluid exits or enters Br and, accordingly,, div F(P)  0 at every point P., , 15.8, , CONCEPT QUESTIONS, , 1. State the Divergence Theorem., 2. Suppose that the vector field F is associated with the flow of, a fluid and P is a point in the domain of F. Give a physical, , 15.8, , interpretation for div F(P). What happens to the flow at P if, div F(P) 0? If div F(P)  0? If div F(P)  0?, , EXERCISES, 7. F(x, y, z)  (x 3  cos y)i  (y 3  sin xz)j  (z 3  2ex)k;, S is the surface of the region bounded by the cylinder, y 2  z 2  1 and the planes x  0 and x  3, , In Exercises 1–4, verify the Divergence Theorem for the given, vector field F and region T., 1. F(x, y, z)  xi  yj  zk; T is the cube bounded by the, planes x  0, x  1, y  0, y  1, z  0, and z  1, , 8. F(x, y, z)  sin yi  (x 2y  ez)j  (2x 2z  ex)k; S is the, surface of the region bounded by the cylinder z  4  x 2, and the planes z  0, y  0 and y  z  5, , 2. F(x, y, z)  2xyi  y 2j  3yzk; T is the cube bounded by, the planes x  0, x  2, y  0, y  2, z  0, and z  2, 3. F(x, y, z)  yi  zj  3yz 2k; T is the region bounded by, the cylinder x 2  y 2  4 in the first octant between z  0, and z  3, 4. F(x, y, z)  xi  yj  2z 2k; T is the region bounded by, the paraboloid z  x 2  y 2 and the plane z  1, In Exercises 5–18, use the Divergence Theorem to find the flux of, F across S; that is, calculate 兰兰S F ⴢ n dS., 5. F(x, y, z)  xy 2i  2yzj  3x 2y 3k; S is the surface of the, cube bounded by the planes x  1, y  1, and z  1, 6. F(x, y, z)  2xzi  y 2j  yzk; S is the surface of the rectangular box bounded by the planes x  0, x  2, y  0,, y  3, z  1, and z  1, , 9. F(x, y, z)  2xyi  y 2j  (x 2  yz)k; S is the surface of, the tetrahedron bounded by the planes x  y  z  1, x  0,, y  0, and z  0, 10. F(x, y, z)  x 2i  xz 2j  (2xz  sin xy)k; S is the, surface of the tetrahedron bounded by the planes, x  2y  3z  6, x  0, y  0, and z  0, 11. F(x, y, z)  xi  2yj  3zk; S is the sphere, x 2  y2  z2  9, 12. F(x, y, z)  (x  y 2)i  (2x 2  z 2  y)j  cos xy k;, S is the surface of the region bounded by the cylinder, x 2  z 2  1 and the planes y  0 and y  1, 13. F(x, y, z)  xzi  yzj  xyk; S is the ellipsoid, 9x 2  4y 2  36z 2  36, , V Videos for selected exercises are available online at www.academic.cengage.com/login.

Page 290 :
15.9, , Stokes’ Theorem, , 1311, , 14. F(x, y, z)  (x  yez)i  (y  tan xz)j  (x  cosh y)k;, S is the surface of the region bounded by the cone, y  2x 2  z 2 and the plane y  4, , 23. Show that 兰兰兰T §f dV  兰兰S f n dS., , 15. F(x, y, z)  (x 3  1)i  (yz 2  cos xz)j  (2y 2z  etan x)k;, S is the sphere x 2  y 2  z 2  1, , 24. Show that if f and t have continuous second-order partial, derivatives, then, , Hint: Apply the Divergence Theorem to F  f c, where c is a constant vector., , 16. F(x, y, z)  x 3i  3yz 2j  2z 3k; S is the surface of the, region bounded by the paraboloid y  9  x 2  z 2 and the, xz-plane, 17. F(x, y, z)  xzi  x yj  (y z  1)k; S is the surface of, the region that lies between the cylinders x 2  y 2  1 and, x 2  y 2  4 and between the planes z  1 and z  3, 2, , 2, , 18. F(x, y, z)  yz 2i  (y 3  xz)j  (y 2z  10x)k; S is the, surface of the region between the spheres x 2  y 2  z 2  1, and x 2  y 2  z 2  4, , 冮冮冮 ( f § t  §f ⴢ §t) dV  冮冮 ( f §t) ⴢ n dS, 2, , T, , 25. Show that if f and t have continuous second-order partial, derivatives, then, , 冮冮冮 ( f § t  t§ f ) dV  冮冮 ( f §t  t §f ) ⴢ n dS, 2, , 冮冮 curl F ⴢ n dS  0, , F(x, y, z) , , 22. Show that if f has continuous second-order partial derivatives, then, 2, , T, , where §2f , ,  2f, , n, , xi  yj  zk, (x 2  y 2  z 2)3>2, , across the ellipsoid (x 2>9)  (y 2>16)  (z 2>4)  1., In Exercises 27–29, determine whether the statement is true or, false. If it is true, explain why. If it is false, explain why or give, an example that shows it is false., 27. If div F  0, then 兰兰S F ⴢ dS  0 for every closed surface S., , S, , 冮冮冮 § f dV  冮冮 D, , S, , 26. Find the flux of the vector field, , 19. Show that the volume of T is given by V(T)  13 兰兰S r ⴢ n dS,, where r  xi  yj  zk., 21. Show that if F has continuous second-order derivatives, then, , 2, , T, , In Exercises 19–25, assume that S and T satisfy the conditions of, the Divergence Theorem., , 20. Show that 兰兰S a ⴢ n dS  0, where a is a constant vector., , S, , f dS, , 28. If F is a constant vector field and S is a cube, then, 兰兰S F ⴢ dS  0., 29. If 冟 F(x, y, z) 冟 1 for all points (x, y, z) in a solid region T, bounded by a closed surface S, then 兰兰兰T div F dV A(S),, where A(S) is the area of S., , S, , , ,  2f, , , ,  2f, , and Dn f is the directional, x 2, y 2, z 2, derivative of f in the direction of an outer normal n of S., , 15.9, , Stokes’ Theorem, Stokes’ Theorem, In this section we consider another generalization of Green’s Theorem to higher dimensions. We start with the following version of Green’s Theorem:, , 冯 F ⴢ T ds  冮冮curl F ⴢ k dA, C, , Equation (7) in Section 15.5, , R, , where the plane curve C is an oriented piecewise-smooth simply closed curve that, bounds a region R. The theorem states that the line integral of the tangential component of a vector field in two-dimensional space around a closed curve is equal to the, double integral of the normal component to R of the curl of the vector field over the, plane region bounded by the curve., Stokes’ Theorem generalizes this version of Green’s Theorem to three-dimensional, space. Named after the English mathematical physicist George G. Stokes (1819–1903),

Page 291 :
1312, , Chapter 15 Vector Analysis, z, , Stokes’ Theorem relates the line integral of the tangential component of a vector field, in three-dimensional space around a simple closed curve in space to the surface integral of the normal component of the curl of the vector field over any surface that has, the closed curve as its boundary. (See Figure 1.), The orientation of the surface induces an orientation on C that is determined by, using the right-hand rule: Imagine grasping the normal vector n to S with your right, hand in such a way that your thumb points in the direction of n. Then your fingers will, point toward the positive direction of C. (See Figure 2.), , n, , S, , C, , 0, , y, x, , THEOREM 1 Stokes’ Theorem, , FIGURE 1, The curve C is a boundary of the, surface S., , Let S be an oriented piecewise-smooth surface that has a unit normal vector n and, is bounded by a simple, closed, positively oriented curve C. If F  Pi  Qj  Rk, is a vector field, where P, Q, and R have continuous partial derivatives in an open, region containing S, then, , n, , 冯 F ⴢ dr  冯 F ⴢ T ds  冮冮 curl F ⴢ dS  冮冮 curl F ⴢ n dS, , S, , C, , C, , S, , (1), , S, , In words, the line integral of the tangential component of F around a simple closed, curve C is equal to the surface integral of the normal component of the curl of F over, any surface S with C as its boundary., Stokes’ Theorem provides us with the following physical interpretation: If F is a, force field, then the work done by F along C is equal to the flux of curl F across S., The proof of Stokes’ Theorem can be found in more advanced textbooks., , C, , FIGURE 2, The orientation of S induces an, orientation on C., , EXAMPLE 1 Verify Stokes’ Theorem for the case in which F(x, y, z)  3zi , 2xj  y 2k, S is the part of the paraboloid z  4  x 2  y 2 with z  0, and C is the, trace of S on the xy-plane., Solution, culating, , i, , curl F  ∞, x, 3z, , z, , n, S, , z  4  x 2  y2, , –2, 2, , C, , 2, , The surface S and the curve C are sketched in Figure 3. We begin by cal-, , y, , j, , y, 2x, , k, , ∞  2yi  3j  2k, z, y2, , Next, writing t(x, y)  4  x 2  y 2, we find that tx  2x and ty  2y. Also,, observe that the projection of S onto the xy-plane is R  {(x, y) 冟 x 2  y 2 4}. So, using Equation (7) of Section 15.7 with P(x, y)  2y, Q(x, y)  3, and R(x, y)  2,, we obtain, , 冮冮 curl F ⴢ dS  冮冮 (Pt, , x, , x, , FIGURE 3, The outer normal n to S induces the, positive direction for C as shown., , S, ,  Qty  R) dA, , R, , , , 冮冮 (4xy  6y  2) dA, R

Page 293 :
1314, , Chapter 15 Vector Analysis, , Historical Biography, , Then, writing t(x, y)  x  2, we find tx  1 and ty  0. Next, observe that the, projection of S onto the xy-plane is R  {(x, y) 冟 x 2  y 2 1}. So using Equation (7), of Section 15.7 with P(x, y)  2, Q(x, y)  sin z, and R(x, y)  2x, we obtain, , Hulton Archive/Getty Images, , 冮冮 curl F ⴢ dS  冮冮 (Pt, , x, , S, ,  Qty  R) dA, , R, , 冮冮 (2  2x) dx, , , , R, , 2, , GEORGE GABRIEL STOKES, (1819–1903), George Stokes’s father was an Irish Protestant minister, as was his maternal grandfather. He was the youngest of six children,, and his three older brothers became, priests. Consequently, Stokes had a very, religious upbringing in which rigorous, study was encouraged. He attended school, in Dublin until the age of 16, after which he, moved to England and entered Bristol College to prepare for study at Cambridge., While at Bristol, Stokes was taught mathematics by Francis Newman, brother of John, Henry Newman (who later became Cardinal, Newman). Stokes excelled in his studies, and entered Pembroke College, Cambridge,, in 1837. Stokes graduated first in his class, in 1841 and was awarded a fellowship at, Pembroke College. During his fellowship, period, Stokes began his study of fluid, dynamics, having been influenced by the, recent work of George Green (page 1269)., Stokes published very highly regarded, work during that time and was chosen for, the position of Lucasian Professor of Mathematics at Cambridge in 1849. However,, this position paid very poorly, and he was, forced to take a second position as a, physics professor in London. Stokes continued to publish papers on fluid dynamics, but also published important work on the, wave theory of light. In 1854 he published, a prize examination question at Cambridge, that included the theorem that now bears, his name. Stokes was also made secretary, and later president of the Royal Society, and served in many capacities at Cambridge. As a mathematician and physicist, he continued to contribute important, works until his death in 1903., , 冮冮 (1  x) dA, R, , Changing to polar coordinates, we obtain, , 冮冮, , 2p, , curl F ⴢ dS  2, , 冮 冮, 0, , S, , 2, , 冮, , 1, , 2p, , (1  r cos u)r dr du  2, , 0, , 2p, , 0, , 冮 冮, 0, , r1, 1, 1, c r 2  r 3 cos ud, du  2, 2, 3, r0, , 1, , (r  r 2 cos u) dr du, , 0, , 冮, , 2p, , 0, , 1, 1, a  cos ub du, 2, 3, , 2p, , 1, 1,  2c u  sin ud  2p, 2, 3, 0, so by Stokes’ Theorem we have, , 冯 F ⴢ dr  冮冮 curl F ⴢ dS  2p, C, , S, , EXAMPLE 3 Evaluate 兰兰S curl F ⴢ dS, where F(x, y, z)  yzi  xzj  z 3k and S, , is the part of the sphere x 2  y 2  z 2  8 lying inside the cone z  2x 2  y 2. (See, Figure 5.), z, , S, , 2 √2, , z  √x 2  y 2, , n, x 2  y2  z2  8, , C, , –2 √ 2, 2 √2, , 2 √2, , y, , x, , FIGURE 5, The surface S is enclosed by C; the outer normal n, induces the positive direction of C shown., , Solution The boundary of S is the curve whose equation is obtained by solving the, equations x 2  y 2  z 2  8 and z  2x 2  y 2 simultaneously. Squaring the second, equation and substituting this result into the first equation give 2z 2  8, or z  2 (since, z 0). Therefore, the boundary of S is the circle C with equations x 2  y 2  4 and

Page 294 :
15.9, , Stokes’ Theorem, , 1315, , z  2. Since C is easily parametrized, we can use Stokes’ Theorem to help us find the, value of the given surface integral by evaluating the line integral 养C F ⴢ dr. A vector, equation of C is, r(t)  2 cos ti  2 sin tj  2k, , 0, , t, , 2p, , so, r¿(t)  2 sin ti  2 cos tj, Furthermore, we have, F(r(t))  (2 sin t)(2)i  (2 cos t)(2)j  (23)k  4 sin ti  4 cos tj  8k, Therefore,, , 冮冮 curl F ⴢ dS  冯 F ⴢ dr, C, , S, , , , 冯 F(r(t)) ⴢ r¿(t) dt, C, , , , 冮, , 2p, , 冮, , 2p, , (4 sin ti  4 cos tj  8k) ⴢ (2 sin ti  2 cos tj) dt, , 0, , , , (8 sin2 t  8 cos2 t) dt, , 0, ,  8, , 冮, , 2p, , dt  16p, , 0, , Interpretation of Curl, Sr, Cr, , n, , P0, , FIGURE 6, Cr is a circular disk of, radius r at P0(x 0, y0, z 0)., , Just as the divergence theorem can be used to give a physical interpretation of the divergence of a vector field, Stokes’ Theorem can be used to give a physical interpretation, of the curl vector. Once again, we think of F as representing the velocity field in fluid, flow. Let P0(x 0, y0, z 0) be a point in the fluid, and let S be a circular disk with radius r, centered at P0 and boundary Cr, as shown in Figure 6. Let n be a unit vector normal, to Sr at P0. Applying Stokes’ Theorem to the vector field F on the surface Sr, we obtain, , 冯 F ⴢ dr  冯 F ⴢ T ds  冮冮curl F ⴢ dS, C, , C, , (2), , Sr, , Since F ⴢ T is the component of F tangent to Cr, we see that 养Cr F ⴢ dr is a measure, of the tendency of the fluid to move around Cr, and accordingly, this line integral is, called the circulation of F around Cr . By taking r small, we see that the circulation, of F around Cr is a measure of the tendency of the field to rotate around the axis determined by n., Next, for small r the continuity of curl F implies that (curl F)(P) ⬇ (curl F)(P0) for, all points P in Sr. Therefore, if r is small, we can write, , 冮冮 curl F ⴢ dS  冮冮 (curl F)(P ) ⴢ n dS, 0, , Sr, , Sr, , ⬇ (curl F)(P0) ⴢ n, , 冮冮 dS  (curl F)(P ) ⴢ n(pr ), 2, , 0, , Sr

Page 295 :
1316, , Chapter 15 Vector Analysis, , So for small r, Equation (2) gives, , 冯 F ⴢ dr ⬇ (curl F)(P ) ⴢ n(pr ), 2, , 0, , C, , The approximation improves as r → 0, and we have, (curl F)(P0) ⴢ n  lim, r→0, , 1, pr 2, , 冯 F ⴢ dr, , (3), , C, , Equation (3) gives the relationship between the curl and the circulation. It tells us that, we can think of 冟 curl F(P0) 冟 as a measure of the magnitude of the tendency of the fluid, to rotate about the axis determined by n. It also tells us that we can think of curl F(P0), as determining the axis about which the circulation of F is greatest near P0., We now summarize the types of line and surface integrals and the major theorems, associated with these integrals., , Summary of Line and Surface Integrals, Line Integrals, a. Element of arc length: ds  冟 r¿(t) 冟 dt,  2[x¿(t)]2  [y¿(t)]2  [z¿(t)]2 dt, , 冮, , b. Line integral of a scalar function:, , f(x, y, z) ds , , f(x(t), y(t), z(t)) 0r¿(t) 0 dt, , a, , C, , c. Line integral of a vector field:, , 冮, , b, , 冮 F ⴢ dr  冮 F ⴢ T ds  冮, C, , b, , F(r(t)) ⴢ r¿(t) dt, , a, , C, , Surface Integrals, a. Element of surface area:, (i) If S is the graph of z  t(x, y), then dS  2t2x  t2y  1 dA., (ii) If S is represented parametrically by r(u, √), then dS  冟 ru rv 冟 dA., b. Surface integral of a scalar function:, (i) If S is the graph of z  t(x, y), then, , 冮冮 f(x, y, z) dS  冮冮 f(x, y, t(x, y))2t, ,  t2y  1 dA, , 2, x, , S, , R, , (ii) If S is represented parametrically by r(u, √), then, , 冮冮 f(r(u, √)) dS  冮冮 f(x(u, √), y(u, √), z(u, √)) 冟 r, , u, , S, , r√ 冟 dA, , R, , c. Surface integral of a vector field:, (i) If S is the graph of z  t(x, y), then, , 冮冮 F ⴢ dS  冮冮 F ⴢ n dS  冮冮 F ⴢ (t i  t j  k) dA, x, , S, , S, , y, , R, , (ii) If S is represented parametrically by r(u, √), then, , 冮冮 F ⴢ dS  冮冮 F ⴢ n dS  冮冮 F ⴢ (r, , u, , S, , S, , D, , r√) dA

Page 296 :
15.9, , 1317, , Stokes’ Theorem, , Summary of Major Theorems Involving Line, Integrals and Surface Integrals, z, , 1. Fundamental Theorem for Line Integrals, , 冮, , §f ⴢ dr  f(r(b))  f(r(a)), , r(b), , C, , C, r(a), , 0, y, , x, y, , 2. Green’s Theorem, Q, , P, , 冯 P dx  Q dy  冮冮 a x  y b dA, C, , R, , C, , R, , x, , 0, z, , 3. Divergence Theorem, , 冮冮 F ⴢ dS  冮冮冮 div F dV, S, , n, , S, T, , T, , n, 0, y, x, z, , 4. Stokes’ Theorem, , 冯 F ⴢ dr  冮冮curl F ⴢ dS, C, , n, S, , S, , 0, , C, y, , x, , 15.9, , CONCEPT QUESTIONS, , 1. State Stokes’ Theorem., 2. Suppose that F  Pi  Qj  Rk is a vector field in threedimensional space such that P, Q, and R have continuous, partial derivatives. Let S1 be the hemisphere, , z  21  x 2  y 2, and let S2 be the paraboloid, z  1  x 2  y 2. Explain why, , 冮冮 curl F ⴢ dS  冮冮 curl F ⴢ dS, S1, , S2

Page 297 :
1318, , Chapter 15 Vector Analysis, , 15.9, , EXERCISES, , In Exercises 1–4, verify Stokes’ Theorem for the given vector, field F and the surface S, oriented with the normal pointing, upward., 1. F(x, y, z)  2zi  3xj  2yk; S is the part of the paraboloid z  9  x 2  y 2 with z  0, 2. F(x, y, z)  2yi  2xj  zk; S is the part of the plane, z  1 lying within the cone z  2x 2  y 2, 3. F(x, y, z)  yi  zj  xk; S is the part of the plane, 2x  2y  z  6 lying in the first octant, 4. F(x, y, z)  (x  2y)i  yz 2j  y 2zk; S is the hemisphere, z  21  x 2  y 2, In Exercises 5–10, use Stokes’ Theorem to evaluate, 兰兰S curl F ⴢ dS., 5. F(x, y, z)  2yi  xz 2j  x 2yez k; S is the hemisphere, z  24  x 2  y 2 oriented with normal pointing upward, 6. F(x, y, z)  5yzi  2xzj  3xz 2k; S is the part of the, paraboloid z  x 2  y 2 lying below the plane z  4 and, oriented with normal pointing downward, 1, , 7. F(x, y, z)  xyzi  2xj  tan y k; S is the part of the, hemisphere z  24  x 2  y 2 lying inside the cylinder, x 2  y 2  1 and oriented with normal pointing upward, 2, , 8. F(x, y, z)  xyi  yzj  xzk; S is the part of the cylinder, z  21  x 2 lying in the first octant between y  0 and, y  1 and oriented with normal pointing in the positive, x-direction, , 14. F(x, y, z)  2zi  xyj  4yk; C is the ellipse obtained, by intersecting the plane y  z  4 with the cylinder, x 2  y 2  4, oriented in a counterclockwise direction, when viewed from above, 15. F(x, y, z)  xeyi  yexj  (xyz)k; C is the path consisting, of straight line segments joining the points (0, 0, 0), (0, 1, 0),, (2, 1, 0) , (2, 0, 0) , (2, 0, 1) , (0, 0, 1) , and (0, 0, 0) in that, order, y, bi  tan1 xj  xyk; C is the curve, 1  x2, obtained by intersecting the cylinder x 2  y 2  1 with the, hyperbolic paraboloid z  y 2  x 2, oriented in a counterclockwise direction when viewed from above, , 16. F(x, y, z)  a, , 17. Find the work done by the force field F(x, y, z) , (ex  z)i  (x 2  cosh y)j  (y 2  z 3)k on a particle, when it is moved along the triangular path that is obtained, by intersecting the plane 2x  2y  z  2 with the coordinate planes and oriented in a counterclockwise direction, when viewed from above., 18. Find the work done by the force field F(x, y, z) , xy 2 i  (x>z)j  (2x  y)k on a particle when it is, moved along the rectangular path with vertices A(0, 0, 3) ,, B(2, 0, 3) , C(2, 4, 3) , D(0, 4, 3) , and A(0, 0, 3) in that order., 19. Ampere’s Law A steady current in a long wire produces a, magnetic field that is tangent to any circle that lies in the, plane perpendicular to the wire and whose center lies on the, wire. (See the figure below.), , 9. F(x, y, z)  z sin xi  2xj  ex cos zk; S is the part of the, ellipsoid 9x 2  9y 2  4z 2  36 lying above the xy-plane, and oriented with normal pointing upward, , J, , 10. F(x, y, z)  yz 2i  xzj  z 3k; S is the part of the cone, z  2x 2  y 2 between the planes z  1 and z  3 and, oriented with normal pointing upward, , B, B, , B, C, , In Exercises 11–16, use Stokes’ Theorem to evaluate 养C F ⴢ dr., 11. F(x, y, z)  (y  z)i  (z  x)j  (x  y)k; C is the, boundary of the part of the plane 2x  3y  z  6 in the, first octant, oriented in a counterclockwise direction when, viewed from above, , Let J denote the vector that points in the direction of the, current and has magnitude measured in amperes per square, meter. This vector is called the electric current density. One, of Maxwell’s equations states that curl B  m0J, where B, denotes the magnetic field intensity and m0 is a constant, called the permeability of free space. Using Stokes’ Theorem, show that 养C B ⴢ dr  m0I, where C is any closed, curve enclosing the curve and I is the net current that passes, through any surface bounded by C. This is Ampere’s Law., (See Exercise 42 in Section 15.3.), , 12. F(x, y, z)  yi  zj  xk; C is the boundary of the triangle with vertices (0, 0, 0), (1, 0, 0), and (0, 1, 1) oriented in, a counterclockwise direction when viewed from above, 13. F(x, y, z)  3xzi  exzj  2xyk; C is the circle obtained, by intersecting the cylinder x 2  z 2  1 with the plane, y  3 oriented in a counterclockwise direction when viewed, from the right, , 20. Let S be a sphere, and suppose that F satisfies the conditions, of Stokes’ Theorem. Show that 兰兰S curl F ⴢ dS  0., , V Videos for selected exercises are available online at www.academic.cengage.com/login.

Page 298 :
15.9, 21. Let f and t have continuous partial derivatives, and let C and, S satisfy the conditions of Stokes’ Theorem. Show that:, a., , 冯, , ( f §t) ⴢ dr , , 冯, , ( f §f ) ⴢ dr  0, , C, , b., , 冮冮 ( §f, , §t) ⴢ dS, , S, , c., , C, , 冯, , ( f §t  t§f ) ⴢ dr  0, , C, , 22. Evaluate 养C (2xy  z 2) dx  (x 2  1) dy  2xz dz, where C, is the curve, r(t)  (1  cos t)i  (1  sin t)j  (1  sin t  cos t)k, 0 t 2p, 23. Let F(x, y, z)  f(r)r, where r  xi  yj  zk, f is a differentiable function, and r  冟 r 冟. Evaluate 养C F ⴢ dr, where C, is the boundary of the triangle with vertices (4, 2, 0) ,, (1, 5, 2) , and (1, 1, 5) , traced in a counterclockwise direction when viewed from above the plane., 24. Let F(x, y, z)  xyi  (4x  yz)j  (xy  1z)k, and let C, be a circle of radius r lying in the plane x  y  z  5., (See the following figure.) If 养C F ⴢ dr  13p, where C is, oriented in the counterclockwise direction when viewed, from above the plane, what is the value of r?, z, 5, , C, 5, , y, , 5, x, , 25. Use Stokes’ Theorem to evaluate 养C ex cos z dx  2xy 2 dy , cot 1 y dz, where C is the circle x 2  y 2  4 and z  0., Hint: Find a surface S with C as its boundary and such that C, is oriented counterclockwise when viewed from above., , 26. Find 兰兰S curl F ⴢ dS if F(x, y, z)  (x  y  z  2)i , (y cos z  4)j  xzk, where S is the surface with the, normal pointing outward and having the boundary, x 2  y 2  1, z  0, shown in the figure., z, n, , x, , 1, , 1319, , 27. Let S be the oriented piecewise-smooth surface that is the, boundary of a simple solid region T. If F  Pi  Qj  Rk, is a vector field, where P, Q, and R have continuous secondorder partial derivatives in an open region containing T,, show that, , 冮冮 curl F ⴢ dS  0, S, , 28. Suppose that f has continuous second-order partial derivatives in a simply connected set D. Use Stokes’ Theorem to, show that, , 冯, , §f ⴢ dr  0, , C, , for any simple piecewise-smooth closed curve C lying in D., 29. Refer to Example 2. Evaluate 养C F ⴢ dr directly (that is,, without using Stokes’ Theorem), where F(x, y, z) , cos zi  x 2j  2yk and C is the curve of intersection, of the plane x  z  2 and the cylinder x 2  y 2  1., 30. Refer to Exercise 16. Evaluate 养C F ⴢ dr directly, (that is, without using Stokes’ Theorem), where, y, bi  tan1 xj  xyk and C is, F(x, y, z)  a, 1  x2, the curve obtained by intersecting the cylinder x 2  y 2  1, with the hyperbolic paraboloid z  y 2  x 2, oriented in a, counterclockwise direction when viewed from above., In Exercises 31 and 32, determine whether the statement is true, or false. If it is true, explain why. If it is false, explain why or, give an example that shows it is false., , r, , 1, , Stokes’ Theorem, , y, , 31. If F  Pi  Qj  Rk, where P, Q, and R have continuous, partial derivatives in three-dimensional space and S1 and S2, are the upper and lower hemisphere z  24  x 2  y 2, and z  24  x 2  y 2, respectively, then, 兰兰S1 curl F ⴢ dS  兰兰S2 curl F ⴢ dS., 32. If F has continuous partial derivatives on an open region, containing an oriented surface S bounded by a piecewisesmooth simple closed curve C and curl F is tangent to S,, then 养C F ⴢ dr  0.

Page 299 :
1320, , Chapter 15 Vector Analysis, , CHAPTER, , 15, , REVIEW, , CONCEPT REVIEW, In Exercises 1–18, fill in the blanks., 1. A vector field on a region R is a function F that associates, with each point (x, y) in two-dimensional space a twodimensional, ; if R is a subset of three-dimensional, space, then F associates each point (x, y, z) with a threedimensional, ., 2. A vector field F is conservative if there exists a scalar function f such that F , . The function f is called a, function for F., 3. a. The divergence of F (describing the flow of fluid) at a, point P measures the rate of flow per unit area (or volume) at which the fluid, or, at P., b. The divergence of F  Pi  Qj  Rk in threedimensional space is defined by div F , ;, if div F(P)  0, more fluid, a neighborhood, of P than, from it; if div F(P)  0, the amount, of fluid entering a neighborhood of P, the, amount departing from it; if div F(P) 0, more fluid, a neighborhood of P than, it., 4. a. If F describes fluid flow, then the curl of F measures the, tendency of the fluid to, a paddle wheel., b. The curl of F  Pi  Qj  Rk is defined by curl F , ., 5. a. If C is a smooth curve, then the line integral of f along C, is 兰C f(x, y) ds , ., b. The formula for evaluating the line integral of part (a) is, ., 兰C f(x, y) ds , c. The line integral of f along C with respect to x is, defined to be 兰C f(x, y) dx , with formula, , where r(t)  x(t)i  y(t)j,, 兰C f(x, y) dx , a t b., d. The line integral of f along C with respect to y is, defined to be 兰C f(x, y) dy , with formula, ., 兰C f(x, y) dy , 6. The formula for evaluating the line integral of f along a, curve C (with respect to arc length) parametrized by, r(t)  x(t)i  y(t)j  z(t)k, a t b, is 兰C f(x, y, z) ds , ., 7. a. If F is a continuous vector field and C is a smooth, curve described by r(t), a t b, then the formula for, evaluating the line integral of F along C is 兰C F ⴢ dr , ., b. If F is a force field, then 兰C F ⴢ dr gives the, done by F on a particle as it moves along C from t  a, to t  b., , 8. a. If 兰C1 F ⴢ dr  兰C2 F ⴢ dr for any two paths having the, same initial and terminal points, then the line integral, ., 兰C F ⴢ dr is, b. The Fundamental Theorem for Line Integrals states that, if F  §f, where f is a, function for F and C, is any piecewise-smooth curve described by r(t),, ., a t b, then 兰C F ⴢ dr  兰C §f ⴢ dr , 9. a. The line integral 兰C F ⴢ dr is independent of path if and, only if 兰C F ⴢ dr  0 for every, path C., b. If F is continuous on an open,, region R, then, the line integral 兰C F ⴢ dr is independent of path if and, only if F is, ., 10. a. If F(x, y)  P(x, y)i  Q(x, y)j is a conservative vector, field in an open region R and both P and Q have continuous first-order partial derivatives in R, then, at, each point in R., b. If F  Pi  Qj is defined on an open,, region, R in the plane and P and Q have continuous first-order, derivatives on R and, , for all (x, y) in R, then F, is conservative in R., 11. If F  Pi  Qj  Rk, where P, Q, and R have continuous, first-order partial derivatives in space, then F is conservative, if and only if,  0, or, in terms of the partial deR, R, rivatives of P, Q, and R,, ,, ,, , , y, x, Q, and, ., , x, 12. Green’s Theorem states that if C is a, , simple, curve that bounds a region R in the plane and P, and Q have continuous partial derivatives on an open set, containing R, then 兰C P dx  Q dy , ., 13. If R is a plane region bounded by a piecewise-smooth, simple closed curve C, then the area of R is given by A , ., , , 14. If F  Pi  Qj, then Green’s Theorem has the vector, forms 养C F ⴢ T ds  养C P dx  Q dy , and, ., 养C F ⴢ n ds , 15. a. A parametric surface S can be represented by the vector, equation r(u, √) , , where (u, √) lies in its, domain; this vector equation is equivalent to, the three, x  x(u, √), y  y(u, √),, and z  z(u, √)., b. If S is the graph of the function z  f(x, y), then a vector, representation of S is r(u, √) , .

Page 300 :
Review Exercises, c. If S is a surface obtained by revolving the graph of a, nonnegative function f(x), where a x b, about the, x-axis, then a vector representation of S is r(u, √) , with parameter domain D , ., 16. If a parametric surface S is represented by, r(u, √)  x(u, √)i  y(u, √)j  z(u, √)k with parameter, domain D, then its surface area is A(S) , ., , 1321, , 17. a. If S is defined by z  f(x, y) and the projection of S onto, the xy-plane is R, then 兰兰S F(x, y, z) dS , ., b. If S is defined by r(u, √)  x(u, √)i  y(u, √)j  z(u, √)k, with parameter domain D, then 兰兰S F(x, y, z) dS , ., 18. a. The positive orientation of a surface S is the one for, which the unit normal vector points, from S., b. The flux of a vector field F across an oriented surface S, in the direction of the unit normal n is, ., , REVIEW EXERCISES, In Exercises 1–4, find (a) the divergence and (b) the curl of the, vector field F., , 13., , 3. F(x, y, z)  ex sin yi  ex cos yj  ezk, , 14., , In Exercises 5–14, evaluate the line integral., y ds, where C is the arch of the parabola y  1x from, , ment joining (0, 0, 0) to (1, 0, 0) and the line segment from, (1, 0, 0) to (2, 1, 3)., , 16. F(x, y, z)  yzi  zj  xk, where C is part of the helix, given by x  2t, y  2 sin t, z  2 cos t, 0 t p2, , 2, , 冮, , xy 2 ds, where C is the curve given by, , C, , r(t)  sin ti  cos tj  tk, 0, , p, 2, , t, , 冮 xyz ds, where C is the line segment from (1, 1, 0) to, C, , (2, 3, 4), , 冮, , 3, x 2y dx  x 3y dy, where C is the graph of y  1, x from, , C, , (1, 1) to (8, 2), 10., , In Exercises 15 and 16, find the work done by the force field F, in moving a particle along the curve C., , 冮 (1  x ) ds, where C is the upper semicircle centered at, the origin and joining (0, 1) to (0, 1), , 9., , dz, where C consists of the line seg-, , 15. F(x, y, z)  xyi  (y  z)j  z 2k, where C is the line segment from (1, 1, 1) to (2, 3, 5), , C, , 8., , 2, , (1, 1) to (4, 2), , C, , 7., , 冮 z dx  x dy  x, C, , 4. F(x, y, z)  ln(x 2  y 2)i  x cos yj  z 2k, , 6., , dy  zex dz, where C is the line segment, , joining (0, 0, 0) to (1, 1, 2), , 2. F(x, y, z)  xy cos yi  y sin xj  xzk, , 冮, , y, , C, , 1. F(x, y, z)  xy 2i  yz 2j  zx 2k, , 5., , 冮 xy dx  e, , 冮 xy dx  xy dy, where C is the quarter-circle from (0, 1), 2, , In Exercises 17 and 18, show that F is a conservative vector, field and find a function f such that F  §f., 17. F(x, y)  (4xy  3y 2)i  (2x 2  6xy)j, 18. F(x, y, z)  (y 2  2xz)i  (2xy  z 2)j  (x 2  2yz)k, In Exercises 19 and 20, show that F is conservative and use this, result to evaluate 兰C F ⴢ T ds for the given curve C., 19. F(x, y)  (2xy  y 3)i  (x 2  3xy 2)j; C is the elliptical, path 9x 2  25y 2  225 from (5, 0) to (0, 3) traversed in a, counterclockwise direction, 20. F(x, y, z)  (2xy  yz 2)i  (x 2  xz 2)j  2xyzk; C is the, twisted cubic x  t, y  t 2, z  t 3 from (0, 0, 0) to (1, 1, 1), , C, , to (1, 0), centered at the origin, 11., , 冮 yz dx  y cos x dy  y dz, where C is the curve x  t,, C, , y  cos t, z  sin t, 0, 12., , 冮 xe, , y, , p, 2, , t, , 21., , 冯, , (y 2  sec x) dx  (x 2  y 5) dy, where C is the boundary, , C, , of the region enclosed by the graphs of y  4  x 2 and, y  x  2, , dx  cos y dy  z dz,, 2, , C, , C: r(t)  ti  t 2j  t 3k, 0, , In Exercises 21–24, use Green’s Theorem to evaluate the line, integral along the positively oriented closed curve C., , t, , 1

Page 301 :
1322, 22., , Chapter 15 Vector Analysis, , 冯 xy dx  (x, , 2, ,  2y) dy, where C is the boundary of the, , C, , region enclosed by the graphs of y  1x, y  0, and x  4, 23., , 冯, , C, 2, , (x 2y  ex) dx  (ey  xy 2) dy, where C is the circle, , x y 1, 24., , 冯, , 2, , (2y  cosh x) dx  (x  sinh y) dy, where C is the, , C, , ellipse 9x 2  4y 2  36, In Exercises 25–28, evaluate the surface integral., 25., , 冮冮 (y  xz) dS, where S is the part of the plane, S, , 2x  2y  3z  6 in the first octant, 26., , 冮冮 z dS, where S is the part of the paraboloid, , 34. F(x, y, z)  y 2zi  2xzj  cos zk; S is the part of the, sphere x 2  y 2  z 2  16 below the plane z  2 with an, outward normal, In Exercises 35 and 36, use Stokes’ Theorem to evaluate, 养C F ⴢ T ds., 35. F(x, y, z)  (2x  y)i  (3x  z)j  (y  z)k; C is the, curve obtained by intersecting the plane 2x  y  z  6, with the coordinate planes, oriented clockwise when viewed, from the top, 36. F(x, y, z)  yi  xzj  xyzk; C is the curve obtained by, intersecting the surface z  x 2y with the planes x  0,, x  1, y  0, and y  1, oriented counterclockwise when, viewed from above, 37. Find the work done by the force field F(x, y) , 2xy 3i  3x 2y 2j when a particle is moved from (0, 0) to, (2, 4) along the path shown in the figure., y, , S, , z  4  x 2  y 2 inside the cylinder x 2  y 2  1, 27., , 冮冮 F ⴢ n dS, where F(x, y, z)  xi  yj  zk and S is the, , 2, , S, , part of the paraboloid y  1  x 2  z 2 lying to the right of, the xz-plane; n points to the right, 28., , 冮冮 F ⴢ n dS, where F(x, y, z)  yi  xj  zk and S is the, S, , part of the paraboloid z  5  x 2  y 2 lying above the, plane z  1; n points upward, , (2, 4), , 4, , (0, 0), , 2, , 38. Find the work done by the force field F(x, y, z) , 2xyi  (x 2  2yz 2)j  2y 2zk when a particle is moved from, (2, 0, 0) to (0, 3, 0) along the path shown in the figure., , In Exercises 29 and 30, find the mass of the surface S having the, given mass density., , z, (0, 0, 3), , 29. S is the part of the plane x  y  z  1 in the first octant;, the density at a point P on S is directly proportional to the, square of the distance between P and the yz-plane., , (0, 2, 3), , 30. S is the part of the paraboloid z  x 2  y 2 lying inside, the cylinder x 2  y 2  2; the density at a point P on S is, directly proportional to the distance between P and the, xy-plane., In Exercises 31 and 32, use the Divergence Theorem to find the, flux of F across S; that is, calculate 兰兰S F ⴢ n dS., 31. F(x, y, z)  xi  yj  zk; S is the surface of the cylinder, x 2  y 2  4 bounded by the planes z  0 and z  3;, n points outward, 32. F(x, y, z)  y 2i  yz 2j  z 3k; S is the unit sphere centered, at the origin; n points outward, In Exercises 33 and 34, use Stokes’ Theorem to evaluate, 兰兰S curl F ⴢ n dS., 33. F(x, y, z)  (x  y 2  2)i  2xyj  (x 2  yz 2)k; S is the, part of the paraboloid z  4  x 2  y 2 above the xy-plane, with an outward normal, , x, , 4, , (0, 3, 0) y, , (2, 0, 0), x, , 39. Let, F(x, y) , , xy, x y, 2, , 2, , i, , xy, x 2  y2, , j, , Evaluate 养C F ⴢ dr, where C is the path shown in the figure., y, , x

Page 302 :
Challenge Problems, 40. Let F(x, y, z)  1 4y  12 z 2 2 i  2xzj  x 2k, and let C be, a simple closed curve lying in the plane 2x  2y  z  6, (see the following figure) and oriented in the counterclockwise direction when viewed from above the plane. If, 养C F ⴢ dr  24, what is the area of the region enclosed, by C?, , 1323, , In Exercises 41–48, state whether the statement is true or false., (Assume that all differentiability conditions are met.) Give a, reason for your answer., 41. If F is a vector field, then curl (curl F) is a vector field., 42. If F and G are vector fields, then div (F, field., , z, , 43. If f is a scalar field, then § ⴢ [ §, 6, , G) is a scalar, , ( §f )] is undefined., , 44. If F is a vector field and f is a scalar field, then, div[ §f (curl F)] is undefined., , n, , 45. If f has continuous partial derivatives at all points and, §f  0, then f is a constant function., 46. If §, , C, 3, , 3, , F  0, then F is a constant vector field., , 47. If div F  0, then 兰兰S F ⴢ n dS  0 for every closed, surface S., , y, , 48. If 养C F ⴢ dr  0 for every closed path C, then curl F  0., , x, , CHALLENGE PROBLEMS, a. Show that a parametric representation of this curve is, , 1. Find and sketch the domain of the vector-valued function, F(x, y) , , 1, 24  x 2  4y 2, , i, , 1, 24x 2  4y 2  1, , x, , j, , 3at, , y, , 1  t3, , 3at 2, 1  t3, , Hint: Use the parameter t  y>x., , 2. Find the domain of, , b. Find the area of the region enclosed by the loop of the, curve., , F(x, y, z)  2冟 x 冟  y  1 i,  ln(1  冟 x 冟  y)j , , ln ln(z  1), 1z  3, , 3. The curve with equation x  y  3axy, where a is a, nonzero constant, is called the folium of Descartes., 3, , 3, , y, , k, , 4. Let, P(x, y) , , xy, x y, 2, , 2, , and, , Q(x, y) , , xy, x 2  y2, , Evaluate 养C P dx  Q dy, where C is the curve shown in the, figure., y, C, , a, , 0, a, , x, x

Page 303 :
1324, , Chapter 15 Vector Analysis, , 5. Let F(x, y, z)  y cos xi  (x  sin x)j  cos zk, and, let C be the curve represented by r(t)  (1  cos t)i , (1  sin t)j  (1  sin t  cos t)k for 0 t 2p. Evaluate 养C F ⴢ dr., 6. A differential equation of the form, P(x, y) dx  Q(x, y) dy  0, Q, P, , is called exact if, for all (x, y). Show that the, x, y, equation, (2y 2  6xy  2) dx  (4xy  3x 2  3) dy  0, is exact and solve it., , 7. Let T be a one-to-one transformation defined by x  t(u, √),, y  h(u, √) that maps a region S in the u√-plane onto a, region R in the xy-plane. Use Green’s Theorem to prove the, change of variables formula, (x, y), , 冮冮 dA  冮冮 ` (u, √) ` du d√, R, , S, , for the case in which f(x, y)  1. (Compare with, Formula (4) in Section 14.8.)