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July 9th, 2017, 01:08 PM  #1 
Member Joined: Jan 2017 From: Toronto Posts: 84 Thanks: 1  Solving cylinder with spherical coordinate triple integration
Consider the region R within the cylinder x^2 + y^2 <= 4, bounded below by z = 0 and above by z = 2  y. Assume a mass density = z. Set up and evaluate the integral representing the mass of the solid. This is easy with cylinderical coordinates: $\displaystyle \int_{0}^{ 2\pi} \int_{0}^{2} \int_{0}^{2rsin(\theta)} z r dz dr d \theta $ However I would like to solve this problem by using sphereical coordinates: $\displaystyle \int_{0}^{2\pi} \int_{0}^{\pi/2} \int_{2}^{???} \rho cos(\phi) \rho^2 sin(\phi) $ To solve the upper limit of ???: $\displaystyle z = 2  r sin(\theta) ==> \rho cos(\phi) = 2  \rho sin(\phi) sin(\theta) , \rho = 2 / (cos(\phi) + sin(\phi) sin(\theta)) $ However $\displaystyle \rho$ would be less than 2, which does not make sense because I expected $\displaystyle \rho$ should be at least 2 or more Any tips would be much appreciated. Thanks. Last edited by zollen; July 9th, 2017 at 01:12 PM. 
July 13th, 2017, 01:49 PM  #2 
Member Joined: Dec 2016 From:  Posts: 54 Thanks: 10 
Why is $\rho$ less than 2? the denominator is a sum but not of two positive terms necessarily.

July 13th, 2017, 02:31 PM  #3  
Member Joined: Jan 2017 From: Toronto Posts: 84 Thanks: 1  Quote:
2/(cos(\phi) + sin(\phi)cos(\theta)) <= 2 $ I would not be able to integrate with this upper limit since it is possible for the upper limit to be less than the lower limit.  

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coordinate, cylinder, integration, solving, spherical, triple 
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