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# Ex.%2026%20in%2016.2 - 870 C H A P T E R 16 M U LTI P L E I...

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870 CHAPTER 16 MULTIPLE INTEGRATION (ET CHAPTER 15) 25. f ( x , y ) = cos ( 2 x + y ) , π 2 x 2 ,1 y 2 x SOLUTION The vertically simple region D defned by the given inequalities is shown in the fgure: x y π 2 1 0 y = 2 x 1 y 2 x 1 2 We compute the double integral oF f ( x , y ) = cos ( 2 x + y ) over D as an iterated integral, as stated in Theorem 2. This gives ZZ D cos ( 2 x + y ) dA = Z / 2 1 / 2 Z 2 x 1 cos ( 2 x + y ) dydx = Z / 2 1 / 2 sin ( 2 x + y ) ¯ ¯ ¯ ¯ 2 x y = 1 dx = Z / 2 1 / 2 ( sin ( 2 x + 2 x ) sin ( 2 x + 1 )) = Z / 2 1 / 2 ( sin ( 4 x ) sin ( 2 x + 1 )) =− cos 4 x 4 + cos ( 2 x + 1 ) 2 ¯ ¯ ¯ ¯ / 2 x = 1 / 2 cos 4 2 4 + cos ³ 2 2 + 1 ´ 2 µ cos 2 4 + cos 2 2 ± 1 4 + cos ( + 1 ) 2 cos 2 4 0 . 416 26. f ( x , y ) = y x ,0 y 1, 1 x e y 1 x e y x y x = e y e 1 1 The double integral oF f ( x , y ) = 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
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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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