Inertia Tensor of an ellipsoid Given is the following inertia tensor of a certain mass distribution $\rho(\vec{r})$ : 
$$ I_{ij} = \int dV \rho(\vec{r}) \left( \vec{r}^2 \cdot \delta_{ij} - r_ir_j \right) $$
I should compute the inertia tensor for the following ellipsoid: 
$$x^2/a^2 + y^2/b^2 + z^2/c^2 \le 1$$
And I'm allowed to use $\rho(\vec{r}) = \rho_0$, and I should then give the result as a function of the mass $M$. 
It's clear to me how I parameterize the Ellipsoid and how to compute the Jacobian, but I don't get how I should integrate that expression. 
 A: You can use the parametrization
\begin{eqnarray}
x &=& a r \sin\theta\sin\phi \\
y &=& b r \sin\theta\cos\phi \\
z &=& c r \cos\theta
\end{eqnarray}
with $0<r<1$, $0<\theta<\pi$ and $0<\phi<2\pi$. The Jacobian for this choice of coordinates is $abc r^2\sin\theta$, so the integral becomes
\begin{eqnarray}
I_{ij} &=& \rho_0\int{\rm d}r~{\rm d}\theta~{\rm d}\phi ~ abc r^2\sin\theta [\underbrace{r^2\left(\sin^2\theta\left(a^2\cos^2\phi+b^2\sin^2\phi\right)+c^2
\cos^2\theta\right)}_{x^2+y^2+z^2}\delta_{ij} - x_i x_j] \\
&=& \frac{3M}{4\pi} \int{\rm d}r~{\rm d}\theta~{\rm d}\phi ~ r^2\sin\theta [r^2\left(\sin^2\theta\left(a^2\cos^2\phi+b^2\sin^2\phi\right)+c^2
\cos^2\theta\right)\delta_{ij} - x_i x_j]
\end{eqnarray}
where I used $\rho_0 = 3M/(4\pi a b c)$. Now it is a matter of integrating, for example
\begin{eqnarray}
I_{11} &=& \frac{3M}{4\pi} \int{\rm d}r~{\rm d}\theta~{\rm d}\phi ~ r^2\sin\theta [r^2\left(\sin^2\theta\left(a^2\cos^2\phi+b^2\sin^2\phi\right)+c^2
\cos^2\theta\right) - a br^2\sin^2\theta\sin\phi\cos\phi] \\
&\cdots& \\
&=& \frac{M}{5}(b^2 + c^2)
\end{eqnarray}
This is the result you should get
$$
{\bf I } = \frac{M}{5}\begin{pmatrix}
b^2+ c^2 & 0 & 0 \\
0 & a^2+ c^2 & 0 \\
0 & 0 & a^2+ b^2\\
\end{pmatrix}
$$
