100hw6soln

# 100hw6soln - Math 100 Homework 6 fall 2010 1 The solid G...

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fall 2010 1. The solid G collects points ( x, y, z ) satisfying 0 z y and 0 y 4 - x 2 and - 2 x 2 . Therefore, R G ydV = R 2 x = - 2 R 4 - x 2 y =0 R y z =0 ydzdydx = R 2 x = - 2 R 4 - x 2 y =0 y 2 dydx = R 2 x = - 2 1 3 (4 - x 2 ) 3 dx = 4096 / 105 . 2. The solid G collects points ( x, y, z ) satisfying 3 x 2 + y 2 z 8 - x 2 - y 2 and - p 2 - y 2 / 2 x p 2 - y 2 / 2 and - 2 y 2 . Therefore, Z G f ( x, y, z ) dV = Z 2 - 2 Z 2 - y 2 / 2 - 2 - y 2 / 2 Z 8 - x 2 - y 2 3 x 2 + y 2 f ( x, y, z ) dzdxdy. 3. Since F × G ( x, y, z ) = ( - x 2 y 2 z, x 2 y 2 , xy 2 z 2 - x 2 yz ) , we have ∇· ( F × G )( x, y, z ) = - 2 xy 2 z + 2 x 2 y + 2 xy 2 z - x 2 y = x 2 y for all x, y, z. 4. Let C 1 , C 2 , C 3 be the line segments parametrized by r 1 ( t ) = ( t, t ) for 0 t 1; r 2 ( t ) = ( t + 1 , 1 - t ) for 0 t 1; r 3 ( t ) = (2 - 2 t, 0) for 0 t 1 respectively. Then, R C 1 ydx - xdy = R 1 0 ( t, - t ) · (1 , 1) dt = R 1 0 0 dt = 0;

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## This note was uploaded on 12/25/2011 for the course MATH 101 taught by Professor Ching during the Spring '11 term at HKU.

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100hw6soln - Math 100 Homework 6 fall 2010 1 The solid G...

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