notes001 (84)

notes001 (84) - 1 4 ( e 8 & 1) . Solution: The...

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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 elliptical paraboloid z = 16 x 2 y 2 and below by the square R = f ( x; y ) j 0 ± x ± 2 , 0 ± y ± 2 g . Solution: The volume is V = ZZ R 16 x 2 y 2 ± dA = Z 2 0 Z 2 0 16 x 2 y 2 ± dy dx = Z 2 0 ² 16 y x 2 y 1 3 y 3 ³´ ´ ´ ´ y =2 y =0 dx = Z 2 0 ² 32 2 x 2 8 3 ³ dx = ² 32 x 2 3 x 3 8 3 x ³´ ´ ´ ´ x =2 x =0 = 64 16 3 16 3 = 160 3 . 2) Sketch the region of integration and evaluate Z 2 0 Z 4 x 2 0 xe 2 y 4 y dy dx by reversing the order of integration (Fubini±s Theorem). The answer is
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Unformatted text preview: 1 4 ( e 8 & 1) . Solution: The region of integration is shown below: 1 By Fubini&s Theorem, Z 2 Z 4 & x 2 xe 2 y 4 & y dy dx = Z 4 Z p 4 & y xe 2 y 4 & y dx dy . The inner integral is Z p 4 & y xe 2 y 4 & y dx = 1 2 & e 2 y 4 & y x 2 x = p 4 & y x =0 = 1 2 e 2 y . We now have Z 4 Z p 4 & y xe 2 y 4 & y dx dy = 1 2 Z 4 e 2 y dy = 1 4 e 2 y y =4 y =0 = 1 4 e 8 & 1 . 2...
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notes001 (84) - 1 4 ( e 8 & 1) . Solution: The...

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