# Moment of Inertia of a Torus

## Homework Statement

To calculate the moment of inertia of a solid torus through the z axis(the torus is on the xy plane), using the parallel and perpendicular axis theorem.

## The Attempt at a Solution

Well, first I divided the torus into tiny little disks and calculated the moment of inertia of one of these disks about an axis parallel to the z axis, using the perpendicular theorem. Then by the parallel axis theorem I found that the moment of inertia of the disk about the axis passing through the center of the torus is:
Idisk=dItorus=$dma^{2}/4$+$R^{2}dm$

Then in order to find the moment of inertia of the torus I should integrate that expression, but it doesn't work. I used the relation : $dm/Rd\theta = M/2\pi R$
to substitute for dm. Whats wrong ?
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A torus is given by two radii. You only mention one, R. Which of the two is it? And what's the other one?

A torus is given by two radii. You only mention one, R. Which of the two is it? And what's the other one?
Sorry I didn't write the equations correctly. a is the radius of the little disk and R is the distance from the center of the torus to the center of the little disk, that is : Inner Radius = R-a.

So what does the integral look like?

Itorus=$\int \frac{(a^{2}M)}{8\pi} d\theta$ + $\int \frac{R^{2}M}{2\pi} d\theta$, from 0 to 2$\pi$ and got :

$\frac{(a^{2}M)}{4}$ + $R^{2}M$

Well the substitution was silly because all I had was a bunch of constants and no functions of theta .