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Question #eb959

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Apr 13, 2017

#9 Using Polar Coordinates,

#V=int_0^(2pi) int_2^4 (16-r^2)rdrd theta=72pi#

#10 By converting to Spherical Coordinates,

#int_0^(2pi)int_0^(pi/2) int_0^1 e^(r^3) r^2sin phi dr d phi d theta=2/3 (e-1)pi#

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