# Math Help - Moment of inertia of a spherical segment

1. ## Moment of inertia of a spherical segment

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

2. 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?

3. Hi
Sorry, yes the sphere is centered at the origin and I need to find the moment of inertia about the z axis.

4. 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?

5. 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

6. Don't forget the c. Otherwise, it looks correct.

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