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July 9th, 2017, 02:08 PM   #1
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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:

$\displaystyle \int_{0}^{ 2\pi} \int_{0}^{2} \int_{0}^{2-rsin(\theta)} z r dz dr d \theta$

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.
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Last edited by zollen; July 9th, 2017 at 02:12 PM.

 July 13th, 2017, 02:49 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, 03:31 PM   #3
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Quote:
 Originally Posted by nietzsche Why is $\rho$ less than 2? the denominator is a sum but not of two positive terms necessarily.
$\displaystyle 2/(cos(\phi) + sin(\phi)cos(\theta)) <= 2$

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.

 Tags coordinate, cylinder, integration, solving, spherical, triple

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