Proving the moment of inertia of a thinwalled hollow sphere

[parsa418]

I was wondering if there is any way to prove the moment of inertia of a thin walled hollow sphere by using y^2 + x^2 = r^2 instead of using the angle method. I want to use a Cartesian coordinate system not a polar coordinate system.

 

[Curl]

First of all y^2 + x^2 = r^2 describes a circle not a sphere.
And yes, you can calculate inertia in Cartesian coordinates by the definition of inertia, you just have to evaluate this surface integral:

8\rho \int_{0}^{r}\int_{0}^{\sqrt{r^2-x^2}}\sqrt{\frac{x^2 r^2}{r^2 - x^2 - y^2}}dy dx

Tell me what you get.

 

[parsa418]

Hi thank you for replying
could you please tell me how you got that integral exactly? as in how did you set it up?

 

[Curl]

use the definition of inertia, distance squared times mass. So integrate x^2 times each surface element. I did one octant and multiplied by 8 due to symmetry.

To integrate you need to know what a surface integral is, otherwise I can't help you much.
http://en.wikipedia.org/wiki/Surface_integral

Also I made a typo, the x^2 should be x^4 if its going to be inside the square root