Tensor of inertia of semi-ellipsoid.

Question

so recently I've been doing a research of rattleback and now I'm stuck on finding the tensor of inertia of it: semi ellipsoid with following equations: ${\begin{cases} \displaystyle {\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}+{\frac {z^{2}}{c^{2}}}<=1. \\ z<= 0\end{cases}}$ While the calculation of centroid is clear, I don't understand how do I do compute the principal axes of inertia: ${\displaystyle \ J_{ij}=\int (\delta _{ij}r^{2}-r_{i}r_{j})\ dm.}$ I have an assumption that it is simply $I=\frac {M}{10} \begin{bmatrix} b^2+c^2 & 0 & 0 \\ 0 & a^2+c^2 & 0 \\ 0 & 0 & a^2+b^2 \end{bmatrix} $ but I can't proof it nor disprove it. Also, I have tried parametrization $\begin{gather} x=a\sin(\theta )\cos(\varphi )\\ y=bsin(\theta)sin(\varphi) \\ z=c\cos(\theta)\end{gather}$ but still did not succeed. It's important to note that the mass is distributed evenly. I would be thankful for any help or explanation.

Answer 1

Your conjecture is correct:

The full ellipsoid has moment of inertia around the $x$-axis $$ I_x=\iiint\limits_{\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\le 1}\rho\,(z^2+y^2)\,dx\,dy\,dz $$ which is known to be $I_x=\frac{M}{5}(b^2+c^2)$ where $M$ is the mass and $\rho$ the uniform density $$\rho=\frac{M}{V}=\frac{M}{\frac{4}{3}abc}\,.$$

If we denote by $I^+_x,I^-_x$ the moments of the two half ellipsoids with $z\ge 0\,,$ resp. $z\le 0$ we know from symmetry that they must be equal and, by the integral formla, add up to $I_x\,.$

The fact that the off diagonal elements $I^\pm_{xy}\,,I^\pm_{xz}\,,I^\pm_{yz}$ are zero is easy to see. For example, $$ I^+_{xy}=\iiint\limits_{\scriptstyle \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\le 1\atop\scriptstyle z\ge 0}-\rho\,xy\,dx\,dy\,dz $$ is zero because the integral $\int_{-d}^dxy\,dx$ is zero and $$ d=a\sqrt{1-\frac{y^2}{b^2}-\frac{z^2}{c^2}}\,. $$