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UNIT, Chapter, Multiple Integrals, INTRODUCTION, When a function f(x) is integrated with respect to x between the limits a and b, we, get the difinite integral f(x)dx., a, If the integrand is a funtion f(x,y) and if it is integrated with respect to x and y, repeatedly between the limits xo and x, (for x) and between the limits, Yo, and, У, (for, y), we get a double integral that is denoted by the symbol|| f(x, y)dx dy., Yo Xo, Extending the concept of double integral one step further, we get the triple, integral, || f(x, y, z) dx dy dz., Zo Yo Xo, EVALUATION OF DOUBLE AND TRIPLE INTEGRALS, To evaluate, || f(x, y) dx dy, we first integrate f(x, y) with respect to x partially, i.e., Yo Xo, treating y as a constant temporarily, between x, and x1. The resulting function got, after the inner integration and substitution of limits will be a function of y. Then we, integrate this function of y with respect to y between the limits y, and y, as usual., The order in which the integrations are performed in the double integral is, illustrated in Fig. 5.1., Scanned by CamScanner, 5

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5.2, Unit Ill: Multiple Integrals and Vector Calculus, f(x,, y)dx, dy, Yo, Fig. 5.1, Note a Since the resulting function got after evaluating the inner integral is to, þe a function of y, the limits xo and x, may be either constants or functions of y., The order in which the integrations are performed in a triple integral is illustrated, in Fig. 5.2., f(x, y, z)dx, dy, dz, Yo, Fig. 5.2, When we first perform the innermost integration with respect to x, we treat y and, z as constants temporarily. The limits x, and x, may be constants or functions of, and z, so that the resulting function got after the innermost integration may be a, function of y and z. Then we perform the middle integration with respect to y,, treating z as a constant temporarily. The limits yo and yı may be constants or, functions of z, so that the resulting function got after the middle integration may be, a function of z only: Finally we perform the outermost integration with respect to z, between the constant limits, y, and, Z1., Note a, Sometimes, ||f(x, y)dr dy is also denoted as, dy f (x, y)dr and, Yo Xo, Yo, Z1, X,, f (x, y, z)dr dy dz is also denoted as, dz | dy | f (x, y, z) dr. If these, Zo Yo Xo, notations are used to denote the double and triple integrals, the integrations are, performed from right to left in order., Zo, Yo, REGION OF INTÉGRATION, d $2(y), Consider the double integral f (x, y) dx dy. As stated above x varies from, c (y), $10) to 0,(y) and y varies from c to d., i.e. 6,(y) Sx< ¢2v) and csysd., Scanned by CamScanner

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Chapter 5: Multiple Integrals, 5.3, These inequalities determine a region in the xy-plane, whose boundaries are the, curves x= ,(y), x ,(y) and the lines y c, y = d and which is shown in Fig. 5.3., This region ABCD is known as the region of integration of the above double integral., y= d, C, X= ,(Y), X = 2(), A, y = c, Fig. 5.3, b $2(x), Similarly, for the double integral S(x, y)dy dx, the region of integration, a ,(x), ABCD, whose boundaries are the curves y =, x = b, is shown in Fig. 5.4., $1(x), y = 02(x) and the lines x = a,, y, y = 4,(x), X = a, X = b, y = 0,(x), Fig. 5.4, 22 V2(2) 2(y. z), For the triple integral S, y, 2) dr dy dz, the inequalities 4V. 2) S x, 21 v, (z) (y, z), <$2(Y, z); Y¡(z) <ys 2(2); z, S z S z, hold good. These inequalities determine a, domain in space whose boundaries are the surfaces x = ¢,y, z) x = 0,y, z),, y = y,(z), y = V2(2), z = z1. and z = Z2. This domain is called the domain of integration, of the above triple integral., WORKED EXAMPLE 5(a), 2 1, Example 5.1 Verify that |(x² + y²) dx dy =, |(x² + y²) dy dx., %3D, 1 0, 0 1, L.S., + y²)d.x \dy, %3D, Scanned by CamScanner

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5.4, Unit I1: Multiple Integrals and Vector Calculus, dy, Note y is treated a constant during inner integration with respect to x., 2, 8, + y?, dy =, 3, 3, /1, R.S. =, |(x? + y?)dy dx, (), 2-עך, dx, 3, Jy=D1, Note a x is treated a constant during inner integration with respect to y., 8, 7, + -x, 3, x² +, dx =, %3D, %3D, 3, 3, or, Thus the two double integrals are equal., Note a From the above problem we note the following fact: If the limits of integra-, tion in a double integral are constants, then the order of integration is immaterial,, provided the relevant limits are taken for the concerned variable and the integrand is, continuous in the region of integration. This result holds good for a triple integral, also., 2n n a, Example 5.2 Evaluate, |r* sin o dr do d0., 0 0 0, a, The given integral = | de | dø r* sin o dr, %3D, 2n, de, sin o do, %3!, 0,, as, de sin o dø, 5, 2n, (-cos ø)% de, %3D, 2n, =- a, ,5, de, Scanned by CamScanner

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Chapter 5: Multiple Integrals, 5.5, Vity?, Example 5.3 Evaluate|, dr dy, 1+x? + y?, ton (x), The given integral, a?+ x2, d dy, (1+ y)+ x², -1, tan,, dy, +, 1, dy, + y?, | log y+ V1+ y² )|, 4, =4 log (1+ /7), + C, %3D, Example 5.4 Evaluate ||, xy (x + y) dr dy., 0 x, Since the limits for the inner integral are functions of x, the variable of inner integra-, tion should be y. Effecting this change, the given integral I becomes, I =, | | xy (x + y) dy dr, 38, 21 .6, 28,7,2, 28x6, dr, + x, 3, %3D, y =x, 1, +, 2, .5/2, dr, 3, %3D, 2, = 19, 188, 8., 21, 6, 0,, 168, 2, +, 8 21, 3, 1, %3D, 6., 56, %3D, 3, %3D, |, Scanned by CamScanner, /-