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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.

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    $\begingroup$ Hint: can you show that $I$ will be diagonal? $\endgroup$ – Fabian May 22 '18 at 21:14
  • $\begingroup$ @Fabian as long as only one component of $\vec{r}$ is not equal zero, while the others are, the tensor will be diagonal right ? and now ? $\endgroup$ – Leroy May 22 '18 at 21:20
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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

$$ {\bf I } = \frac{M}{5}\begin{pmatrix} b^2+ c^2 & 0 & 0 \\ 0 & a^2+ c^2 & 0 \\ 0 & 0 & a^2+ b^2\\ \end{pmatrix} $$

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  • $\begingroup$ why did set $\rho_0 = \frac{3M}{4\pi}$ ? and why did you multiply the Jacobian by $x^2 + y^2 + z^2$ ?? $\endgroup$ – Leroy May 22 '18 at 21:53
  • $\begingroup$ @Leroy I assumed the density is constant $\rho = M / V = M / (4\pi a b c / 3)$. I didn't multiply the Jacobian by $x^2+y^2+z^2$, just replaced the expression ${\bf r}^2\delta_{ij} - x_i x_j$ $\endgroup$ – caverac May 22 '18 at 21:57
  • $\begingroup$ Why are all other entries in the matrx $0$ ?? $\endgroup$ – Leroy May 22 '18 at 22:00
  • $\begingroup$ @Leroy Short answer: symmetry. But if you want to convince yourself you will need to calculate the integrals, it is not too complicated, just a bunch of trigonometric functions $\endgroup$ – caverac May 22 '18 at 22:02
  • $\begingroup$ as far as inertia tensors are concerned, can I just assume the $-x_i x_j$ part will always be cancelled out by the integral ? $\endgroup$ – Leroy May 22 '18 at 22:13

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