Ex.%2044-45%20in%2016.2

# Ex.%2044-45%20in%2016.2 - S E C T I O N 16.2 Double...

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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 original order of integration gives: 9 0 y 0 x 3 dx dy ( 3 x 2 + y ) 1 / 2 = 9 0 y 0 x 3 ( 3 x 2 + y ) 1 / 2 dx dy The inner integral is much more difficult to compute than the integrals involved in the previous computation. 44. 1 1 1 x 2 1 x 2 1 y 2 dy dx SOLUTION The limits of integration determine the vertically simple region D described by the following inequalities: 1 x 1 , 1 x 2 y 1 x 2 x 1 x 2 y 1 x 2 y 1 1 The domain D (shown in the figure) 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 first quadrant. That is,

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• Spring '08
• Rogawski
• Integrals, Multiple integral, y2, horizontally simple region

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