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## 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.

## Homework Equations

## 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:

I

_{disk}=dI

_{torus}=[itex]dma^{2}/4[/itex]+[itex]R^{2}dm[/itex]

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 : [itex]dm/Rd\theta = M/2\pi R[/itex]

to substitute for dm. Whats wrong ?

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