Instead of the usual approach of integrating a bunch of discs, I do it differently. I integrate by subdividing the sphere into a bunch of concentric spheres containing mass on their surface.
The integral becomes:
$\int_0^R r^2 \cdot (4\pi r^2) \cdot \rho \space dr$ which ends up equaling $\frac{3}{5}MR^2$
Inside the integral I'm taking the squared distance from the axis times the surface area of the sphere times the density. The right answer is of course $\frac{2}{5}MR^2$, but why does my approach fail? Conceptually, why does this method of breaking the sphere into smaller shells not work?
Best Answer
The moment of inertia is defined relative to an axis, not a point. Therefore the distance you need is from a mass element in the sphere to the $z$-axis, not to the origin. This is why the disk method is used -- it takes advantage of the symmetry. Your expression uses the distance $r$ to the origin, whereas the distance to the $z$ axis in spherical coordinates is $r\sin \theta$ (where $\theta$ is the angle between the $z$ axis and the point of interest).
The correct expression for a spherical object with arbitrarily varying density in spherical coordinates is
\begin{eqnarray} I &=& \int_0^R {\rm d} r r^2 \int_0^{\pi} {\rm d\theta} \sin \theta \int_0^{2\pi}{\rm d}\phi \rho(r,\theta,\phi)\Delta(r,\theta,\phi)^2, \end{eqnarray} where $\rho(r,\theta,\phi)$ is the density and $\Delta(r,\theta,\phi)=r\sin\theta$ is the distance from a mass element at $r,\theta,\phi$ to the $z$ axis.
For completeness we can work this out for a constant density sphere, $\rho(r,\theta,\phi)=3 M/(4\pi R^3)$ (note that the integrand doesn't depend on $\phi$, so the $\phi$ integral is just $2\pi$) \begin{eqnarray} I &=& 2\pi \frac{3M}{4\pi R^3} \int_0^R {\rm d} r r^2 \int_0^{\pi} {\rm d\theta} \sin \theta\ \left(r \sin\theta\right)^2 \\ &=& \frac{3 M}{2 R^3} \int_0^R {\rm d} r r^4 \int_0^\pi {\rm d}\theta \sin^3 \theta \\ &=& \frac{3 M}{2 R^3} \times \frac{R^5}{5} \times \frac{4}{3} \\ &=& \frac{2}{5} MR^2 \end{eqnarray}