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### Ex.%2026%20in%2016.2

Course: MATH 32B, Spring 2008
School: UCLA
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Word Count: 259

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H 870 C A P T E R 16 M U LTI P L E I N T E G R AT I O N (ET CHAPTER 15) 25. f (x, y) = cos(2x + y), - x , 1 y 2x 2 2 SOLUTION The vertically simple region D defined by the given inequalities is shown in the figure: y y = 2x 1 y 2x 1 0 1 2 2 x We compute the double integral of f (x, y) = cos(2x + y) over D as an iterated integral, as stated in Theorem 2. This gives cos(2x + y) d A = = /2 1/2 1...

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H 870 C A P T E R 16 M U LTI P L E I N T E G R AT I O N (ET CHAPTER 15) 25. f (x, y) = cos(2x + y), - x , 1 y 2x 2 2 SOLUTION The vertically simple region D defined by the given inequalities is shown in the figure: y y = 2x 1 y 2x 1 0 1 2 2 x We compute the double integral of f (x, y) = cos(2x + y) over D as an iterated integral, as stated in Theorem 2. This gives cos(2x + y) d A = = /2 1/2 1 2x D cos(2x + y) d y d x = /2 1/2 sin(2x + y) /2 1/2 2x dx y=1 /2 1/2 (sin(2x + 2x) - sin(2x + 1)) d x = (sin(4x) - sin(2x + 1)) d x =- cos 22 + 1 cos 42 cos(2x + 1) /2 cos 2 cos 2 cos 4x + + - - + =- 4 2 4 2 4 2 x=1/2 1 cos( + 1) cos 2 = - -0.416 =- + 4 2 4 y 26. f (x, y) = x , SOLUTION y x = ey 1 1 x ey x 1 e 0 y 1, 1 x ey y The double integral of f (x, y) = x over the horizontally simple region D is computed using Theorem 2. The limits of integration of the iterated integral are determined by the inequalities 0 y 1, We obtain the following integral: 1 x ey ey 1 ey y 1 1 1 y y3 1 1 dA = dx dy = y ln e y - y ln 1 d y = y ln x dy = y2 d y = = 3 0 3 0 1 x 0 0 0 D x x=1 27. f (x, y) = 2x y, SOLUTION 0 y 1, y2 x y The intersection points of the graphs x = y and x = y 2 are (0, 0) (1, 1). The horizontally simple region D is shown in the figure: y x=y x = y2 1 y2 x y x 1
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