Ch16Rev Ex 16-20

Ch16Rev Ex 16-20 - 1032 C H A P T E R 16 M U LTI P L E I N...

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1032 CHAPTER 16 MULTIPLE INTEGRATION (ET CHAPTER 15) 15. Express Z 3 3 Z 9 x 2 0 f ( x , y ) dydx as an iterated integral in the order dx dy . SOLUTION The limits of integration correspond to the inequalities describing the domain D : 3 x 3 , 0 y 9 x 2 . A quick sketch veriFes that this is the region under the upper part of the parabola y = 9 x 2 ,tha tis ,thepar ttha above the x -axis. Therefore, the double integral can be rewritten as the following sum: Z 3 3 Z 9 x 2 0 f ( x , y ) = Z 9 0 Z 9 y 9 y f ( x , y ) 16. Let D be the domain between y = x and y = x . Calculate ZZ D xydA as an iterated integral in the order and . In the order : The inequalities describing D as a horizontally simple region are obtained by Frst rewriting the equations of the curves with x as a function of y , x = y and x = y 2 , respectively. The points of intersection are found solving the equation y = y 2 y ( 1 y ) = 0 y = 0 , y = 1 We obtain the following inequalities for D (see Fgure): D : 0 y 1 , y 2 x y x y 1 1 0 y 2 x y We now compute the double integral as the following iterated integral: D = Z 1 0 Z y y 2 xydx dy = Z 1 0 x 2 y 2 ¯ ¯ ¯ ¯ x = y x = y 2 dy = Z 1 0 ± y · y 2 2 y 4 · y 2 ! = Z 1 0 ± y 3 2 y 5 2 ! = y 4 8 y 6 12 ¯ ¯ ¯ ¯ 1 0 = 1 8 1 12 = 1 24 In the order : D is described as a vertically simple region by the following inequalities (see Fgure): D : 0 x 1 , x y x x y 1 1 0 x y x The corresponding iterated integral is D = Z 1 0 Z x x xydydx = Z 1 0 xy 2
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This note was uploaded on 04/13/2008 for the course MATH 32B taught by Professor Rogawski during the Winter '08 term at UCLA.

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Ch16Rev Ex 16-20 - 1032 C H A P T E R 16 M U LTI P L E I N...

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