Ellipsoid moment of inertia matrix Some background info: torque $\tau$ is defined as $$\tau = I*d\omega$$
Where $I$ is the moment of inertia matrix and $d\omega$ is an object's rotational acceleration. As I understand it, the inertia matrix acts just like mass in that it counteracts the torque (for example, if an object is spinning around the x axis, a big value of $I_{xx}$ means that the object needs more torque around the x axis in order to spin).
However, angular momentum $M$ can be defined as $$M=I * \omega$$
Where $\omega$ is the rotational velocity. So it seems that torque is the time derivative of angular momentum. 
Using these facts, how would I find the moment of inertia matrix for an ellipsoid with uniform density of the form $$\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}≤9$$ with $a≠b≠c≠0$? Would I have to use spherical coordinates somehow? I'm not given any torque or angular velocity information. Any guidance is appreciated.
 A: In usual notation, 
\begin{align*}
  \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} &= 1 \\
  I_{ij} &=
  \iiint \rho(\mathbf{r}) (r^{2} \delta_{ij}-x_{i} x_{j}) d^{3} \, \mathbf{r} \\
  I &= \frac{m}{5}
  \begin{pmatrix}
    b^2+c^2 & 0 & 0 \\
    0 & c^2+a^2 & 0 \\
    0 & 0 & a^2+b^2 \\
  \end{pmatrix}
\end{align*}
In your case: $$I=\frac{3m}{5}
\begin{pmatrix}
    b+c & 0 & 0 \\
    0 & c+a & 0 \\
    0 & 0 & a+b \\
\end{pmatrix}$$
See also the link here.
P.S.:
By symmetry, $i\neq j \implies I_{ij}=0$
Let $x=a X$, $y=b Y$ and $z=c Z$
\begin{align*}
  \iiint_{x^2/a^2+y^2/b^2+z^2/c^2<1} x^{2} dV &=
  abc \iiint_{X^2+Y^2+Z^2<1} a^{2} X^{2} dX dY dZ \\
  &=
  abc\int_{-1}^{1}
  \left[
    \int_{-\sqrt{1-X^2}}^{\sqrt{1-X^2}}
    \left( \int_{-\sqrt{1-X^2-Y^2}}^{\sqrt{1-X^2-Y^2}} dZ \right)  dY
  \right] a^2X^2  dX \\
  &= 
  abc\int_{-1}^{1}
  \left[
    \int_{-\sqrt{1-X^2}}^{\sqrt{1-X^2}}
    2\sqrt{1-X^2-Y^2} dY
  \right] a^2X^2  dX \\
  &= abc\int_{-1}^{1} \pi (1-X^2)  a^2X^2  dX \\
  &= \pi abc \times \frac{4a^2}{15} \\
 \rho &= \frac{3m}{4\pi abc} \\
  I_{11} &= \rho \iiint (y^2+z^2) dV \\
  &= \frac{m}{5} (b^2+c^2)
\end{align*}
Finish the rest by symmetry.
