July 7th, 2017, 05:07 PM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 120 Thanks: 2  Tricky Spherical coordinates...
Consider the region S inside the cylinder x^2+y^2 = 1, outside the sphere x^2+y^2+z^2 = 1, and with z between 0 and 1. A view from above and a cross section (looking along the positive x axis) are shown below. Set up one or more triple integrals to integrate a function f(x; y; z) over S using spherical coordinates, but DO NOT EVALUATE. 
July 7th, 2017, 05:09 PM  #2 
Senior Member Joined: Jan 2017 From: Toronto Posts: 120 Thanks: 2 
Answer: $\displaystyle \int_{0}^{2 \pi} \int_{0}^{\pi / 4} \int_{1}^{sec \phi } \rho^2 sin \phi d \rho d \phi d \theta + \int_{0}^{2 \pi} \int_{\pi / 4}^{\pi / 2} \int_{1}^{csc \phi } \rho^2 sin \phi d \rho d \phi d \theta $ I guess I don't understand why the first time inner upper limit is $\displaystyle sec \phi $ while the second term inner upper limit is $\displaystyle csc \phi $ Last edited by zollen; July 7th, 2017 at 05:11 PM. 
July 8th, 2017, 03:07 PM  #3 
Senior Member Joined: Jan 2017 From: Toronto Posts: 120 Thanks: 2 
Both terms upper limits should have been $\displaystyle sec \phi $. I don't understand why the second term upper limit is $\displaystyle csc \phi $ 

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coordinates, sphereical, spherical, tricky 
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