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Ex 44-45 in 1 - S E C T I O N 16.2 Double Integrals over More General Regions(ET Section 15.2 881 Trying to evaluate the iterated integral in the

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SECTION 16.2 Double Integrals over More General Regions (ET Section 15.2) 881 Trying to evaluate the iterated integral in the original order of integration gives: Z 9 0 Z y 0 x 3 dx dy ( 3 x 2 + y ) 1 / 2 = Z 9 0 Ã Z y 0 x 3 ( 3 x 2 + y ) 1 / 2 dx ± dy The inner integral is much more difFcult to compute than the integrals involved in the previous computation. 44. Z 1 1 Z 1 x 2 1 x 2 q 1 y 2 dydx SOLUTION The limits of integration determine the vertically simple region D described by the following inequalities: 1 x 1 , p 1 x 2 y p 1 x 2 x 1 x 2 y 1 x 2 y 1 1 The domain D (shown in the Fgure) is the unit circle. Since f ( x , y ) = f ( x , y ) and the domain is symmetric with respect to the x -axis, the double integral over the circle equals twice the integral over the upper half of the circle. Similarly, since f ( x , y ) = f ( x , y ) ( f does not depend on x )and D is symmetric with respect to the y -axis, the double integral over D is four times the integral over the part of the circle in the Frst quadrant. That is,
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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 44-45 in 1 - S E C T I O N 16.2 Double Integrals over More General Regions(ET Section 15.2 881 Trying to evaluate the iterated integral in the

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