Proving the moment of inerta of a sphere: can someone help me please?

[vInCEE-j]

I'm having a difficulty in proving the moment of inertia of a sphere.
Using I=p(x^2)dv, we have to show that

I = (3m/r^3) * the integral (from 0 to r) of (x^3)((r^2) - (x^2))^(1/2) dx

where r=radius, x=the axis

I've been trying to prove it and yet no success. if anyone can show me how to do it, i'd greatly appreciate that.
thanks a lot

 

[HallsofIvy]

It would seem more reasonable to me to use spherical coordinates:
$$x= \rhocos(\theta)sin(\phi)$$ and $$dV= \rho^2 sin(\phi)d\rho\d\thetad\phi$$
$$\int\rho x^2dV= p\int_{\phi=0}^\pi\int_{\theta= 0}^{2\pi}\int_{\rho= 0}^R(\rho^3cos^2(\theta)sin^3(\phi)d\rhod\thetad\phi$$

(oops! Thanks, Fermat.)

 

[Fermat]

It would seem more reasonable to me to use spherical coordinates:
$$x= \rho cos(\theta)sin(\phi)$$ and $$dV= \rho^2 sin(\phi)d\rho d\theta d\phi$$
$$\int\rho x^2\ dV= \rho\int_{\phi=0}^\pi\int_{\theta= 0}^{2\pi}\int_{\rho= 0}^R \rho^4cos^2(\theta)sin^3(\phi) d\rho d\theta d\phi$$

Ah, now I can see what it says