notes001 (85)

# notes001 (85) - 2 Solution The region of integration is...

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October 25, 2010 NAME______________________________ correct notation) in order to get full credit. 1) Find the volume of the region bounded above by the plane z = y= 2 and below by the rectangle R = f ( x; y ) j 0 x 4 , 0 y 2 g . Solution: The volume is V = ZZ R 1 2 y dA = Z 4 0 Z 2 0 1 2 y dy dx = Z 4 0 1 4 y 2 y =2 y =0 dx = Z 4 0 1 dx = x j x =4 x =0 = 4 : 2) Sketch the region of integration and evaluate Z 2 p ln(3) 0 Z p ln(3) y= 2 e x 2 dx dy by reversing the order of integration (Fubini±s Theorem). The answer is

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Unformatted text preview: 2 . Solution: The region of integration is shown below: 1 By Fubini&s Theorem, Z 2 p ln(3) Z p ln(3) y= 2 e x 2 dx dy = Z p ln(3) Z 2 x e x 2 dy dx . The inner integral is Z 2 x e x 2 dy = ye x 2 & & & y =2 x y =0 = 2 xe x 2 . We now have Z p ln(3) Z 2 x e x 2 dy dx = Z p ln(3) 2 xe x 2 dx = e x 2 & & & x = p ln(3) x =0 = e ln(3) & e = 2 . 2...
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notes001 (85) - 2 Solution The region of integration is...

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