Calculating the moment of inertia of a solid sphere

[Rahulrj]

Homework Statement

To calculate moment of inertia of a solid sphere of uniform density[/B]

Homework Equations

$$ I = \int r^2 dm$$

The attempt at a solution

I consider an elemental disk of small thickness $d\theta$
$dm = \frac{M}{4/3 \pi R^3}*\pi R^2\cos^2\theta* Rd\theta$
Therefore $dI = r^2 dm = R^2\cos^2\theta dm$ r is the distance from the axis to the disc which is same as $R\cos\theta$
Then I do the integration $I = \int_\frac{-\pi}{2}^\frac{\pi}{2} dI$

As I am getting a $\cos^4\theta$ I am not able to get the answer $2/5 MR^2$
Can someone tell me if I have gone wrong anywhere in what I have written above?

 

[andrewkirk]

Your first formula implies that $R\,d\theta$ is the height of the incremental disc. It is not. It is the length of the slanted edge of the disc. What do you have to multiply that length by to give the height?

Also, the moment of inertia of the incremental disc is not $r^2\,dm$. Look at this list of moments of inertia to see what it should be.

 

[Rahulrj]

Your first formula implies that $R\,d\theta$ is the height of the incremental disc. It is not. It is the length of the slanted edge of the disc. What do you have to multiply that length by to give the height?

So then instead of $Rd\theta$ it will be $R\cos\theta d\theta$? and yes the MoI of disk is $1/2MR^2$
So now I have to evaluate $\cos^5x$?

 

[andrewkirk]

Yes. You can use multiple angle formulas to do this.
Alternatively, you might prefer to integrate with respect to $z$ instead of $\theta$. That gives an easier integration - but the same result of course.