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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 C H A P T E R 16 MULTIPLE INTEGRATION (ET CHAPTER 15) 15. Express 3 3 9 x 2 0 f ( x , y ) dy dx 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 verifies that this is the region under the upper part of the parabola y = 9 x 2 , that is, the part that is above the x -axis. Therefore, the double integral can be rewritten as the following sum: 3 3 9 x 2 0 f ( x , y ) dy dx = 9 0 9 y 9 y f ( x , y ) dx dy 16. Let D be the domain between y = x and y = x . Calculate D xy d A as an iterated integral in the order dx dy and dy dx . SOLUTION In the order dx dy : The inequalities describing D as a horizontally simple region are obtained by first rewriting the equations of the curves with x as a function of y , that is, 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 figure): 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 xy d A = 1 0 y y 2 xy dx dy = 1 0 x 2 y 2 x = y x = y 2 dy = 1 0 y · y 2 2 y 4 · y 2 dy = 1 0 y 3 2 y 5 2 dy = y 4 8 y 6 12 1 0 = 1 8 1 12 = 1 24 In the order dy dx : D is described as a vertically simple region by the following inequalities (see figure): D : 0 x

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