# Moment of inertia of a spherical segment

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• Sep 1st 2010, 01:29 AM
bobred
Moment of inertia of a spherical segment
Am I going about this the right way?
There is a sphere of radius 3 and a region that lies between the planes $z=1$ and $z=2$ and has a density of $cz$. We are asked to work in cylindrical coordinates.

Let $\rho=\sqrt{9-z^2}$, is the following the right formula?

$I=\displaystyle{\int_R}cz\rho^2\,dV$ where $dV=\rho\,d\theta\,d\rho\,dz$

Would this be the integral?

$I=2\pi c z\displaystyle{\int^2_1}\displaystyle{\int^{\sqrt{ 9-z^2}}_0}\rho^3\,d\rho\,dz$

Thanks
• Sep 1st 2010, 03:24 AM
Ackbeet
A few questions:

From the way you're writing, I'm guessing the sphere is centered at the origin. Is that correct?

The moment of inertia must always be measured about some axis. That axis matters greatly, because you can get widely varying results depending on which axis you're finding the moment of inertia about. I'm guessing you're finding the moment of inertia about the z axis. Is that correct?
• Sep 1st 2010, 03:28 AM
bobred
Hi
Sorry, yes the sphere is centered at the origin and I need to find the moment of inertia about the z axis.
• Sep 1st 2010, 03:37 AM
Ackbeet
Ok, so I agree that your integral should be

$\displaystyle{I=\int_{1}^{2} \int_{0}^{2\pi} \int_{0}^{\sqrt{9-z^{2}}} cz\rho^{2}\,\rho\,d\rho\,d\theta\,dz.}$

However, your simplifications of this integral are a bit off. The $\theta$ integral is easy. We get

$\displaystyle{I=2c\pi\int_{1}^{2} z \int_{0}^{\sqrt{9-z^{2}}} \rho^{3}\,d\rho\,dz.}$ You can't pull the $z$ out of the $z$ integral! So what do you get from here?
• Sep 1st 2010, 04:04 AM
bobred
Here we go

$\displaystyle{z \int_{0}^{\sqrt{9-z^{2}}} \rho^{3}\,d\rho}=z\displaystyle{(\frac{1}{4}(\sqrt {9-z^2})^4)}=\frac{1}{4}(81z-18z^3+z^5)$
$2\pi c\displaystyle{\int_{1}^{2}\frac{1}{4}(81z-18z^3+z^5)\,dz =2\pi c(\frac{81}{8}z^2-\frac{9}{8}z^4+\frac{1}{24}z^6)$

Which works out to be $\frac{129}{4}\pi c$.

Thanks
• Sep 1st 2010, 05:28 AM
Ackbeet
Don't forget the c. Otherwise, it looks correct.

[EDIT]: I see you've included the c. Great!