Calculating rotational inertia of a sphere

[Silimay]

Just how do you calculate the rotational inertia of a sphere?
Assuming the sphere lies at the center of the xyz coordinate system, I divided the sphere into a series of cross-sections of verticle width dz and area pi*y^2. I then multiplied these together and multiplied this by z^2, and multiplied this by density (M/V, or M/(4/3*pi*R^3)), and then tried to integrate with respect to z from -R to R. I wasn't sure whether or not to include z itself in the integration (z=(R^2-Y^2)^(1/2)). I have a feeling I completely messed the entire problem up; however, I'm not sure where. Did I go about doing it in an entirely wrong way? Should I use double integrals (would that be easier)? Do I have to use spherical coordinates or something?
Any help is appreciated.

 

[dextercioby]

Your answer lies here

Hopefully Arildno and Krabs' approach are within your level of understanding...

 

[Silimay]

Thank you thank you thank you! That was REALLY helpful...I've been trying to understand the process of calculating inertia for days...my textbook was of no help. I think I finally understand it.

 

[Andrew Mason]

Just how do you calculate the rotational inertia of a sphere?

Divide the solid sphere into thin disks of thickness dz and mass dm. For the thin disk is $I = \frac{1}{2}MR^2$

The moment of inertia of each disk is
$$dI = \frac{1}{2}x^2dm$$ where $$dm = \rho \pi x^2 dz$$

So $$dI = \frac{1}{2}\rho \pi x^4 dz$$

Then integrate dI from z = -R to R (note: $x^2 = R^2 - z^2$)

That will give you I in terms of $\rho$ which is M/V (where V is the volume of the sphere and M is its mass) so just replace $\rho$ with M/V.

The integration looks a little tough because of the $(R^2 - z^2)^2$ Good luck.

AM