Ex.%2051-53%20in%2016.2

Ex.%2051-53%20in%2016.2 - 886 C H A P T E R 16 M U LTI P L...

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886 CHAPTER 16 MULTIPLE INTEGRATION (ET CHAPTER 15) The double integral of f over D 2 is ZZ D 2 x y 2 dA = Z 5 3 Z x + 8 x + 1 2 x y 2 dydx = Z 5 3 x y ¯ ¯ ¯ ¯ x + 8 y = x + 1 2 dx = Z 5 3 Ã x x + 8 + x 1 2 x + 1 2 ± = Z 5 3 x x + 8 + 2 Z 5 3 x x + 1 = Z 5 3 ( x + 8 ) 8 x + 8 + 2 Z 5 3 ( x + 1 ) 1 x + 1 = Z 5 3 µ 1 8 x + 8 ² + 2 Z 5 3 µ 1 1 x + 1 ² = Z 5 3 8 Z 5 3 8 x + 2 Z 5 3 2 Z 5 3 x + 1 = 2 + 8ln ( 8 x ) ¯ ¯ ¯ ¯ 5 3 + 4 2ln ( x + 1 ) ¯ ¯ ¯ ¯ 5 3 = 6 + 8 ( ln 3 ln 5 ) 2 ( ln 6 ln 4 ) = 6 + 8ln3 8ln5 2ln6 + 2ln4 = 6 + 6ln3 + 2ln2 (3) Finally we substitute (2) and (3) in (1) to obtain the following solution: D x y 2 = 3 1 4 ln 5 + 6 + + = 9 8 . 25 ln 5 + 2 . 314 51. Calculate the double integral f ( x , y ) = sin y y over the region D in Figure 23. D yy = x 2 1 x x 2 y = FIGURE 23 SOLUTION To describe D as a horizontally simple region, we rewrite the equations of the lines with x as a function of y ,thatis, x = y and x = 2 y . The inequalities for D are 1 y 2 , y x 2 y x y D x = y x = 2 y 1 2 We now compute the double integral of f ( x , y ) = sin y y over D by the following iterated integral: D sin y y = Z 2 1 Z 2 y y sin y y dx dy = Z 2 1 sin y y x ¯ ¯ ¯ ¯ 2 y x = y dy = Z 2 1 sin y y ( 2 y y ) = Z 2 1 sin y y · ydy = Z 2 1 sin =− cos y ¯ ¯ ¯ ¯ 2 1 = cos 1 cos 2 0 . 956 52. Evaluate D xdA for D in Figure 24.
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This note was uploaded on 03/11/2008 for the course MATH 32B taught by Professor Rogawski during the Spring '08 term at UCLA.

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Ex.%2051-53%20in%2016.2 - 886 C H A P T E R 16 M U LTI P L...

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