Moment of inertia of spherical shell

[tsw99]

Homework Statement

Moment of inertia of spherical shell of radius R, mass M along its rotation axis is given by $$\frac{2}{3}MR^{2}$$
I am trying to calculate this

Homework Equations

The Attempt at a Solution

This is my attempt but is unsuccessful,
since the spherical shell is an assembly of rings (of varying radius), and the MI of a ring is
$$I=MR^{2}$$
Hence $$dI=y^{2}dm$$
$$I=\int y^2(2\pi \sigma ydz$$
Using $$y=Rsin\theta$$ and $$z=Rcos\theta$$
I get:
$$I=2 \pi \sigma R^{4} \int sin^{4}\theta d\theta<br /> =2 \pi \sigma R^{4} \frac{3\pi}{8}<br /> =\frac{3\pi MR^{2}}{16}$$
which is incorrect.

Which step I have gone wrong? Thanks

 

[ehild]

The constant surface charge refers to the spherical surface element which is R^2 sinθ dφ dθ in the spherical polar coordinates. After integrating for φ for a ring, it is dA=2πR^2 dθ. You have to multiply this by σ to get dm, and by the square of the distance from the axis, (Rsinθ)^2. So you have only sin^3 in the integrand. ehild