Moment of inertia of a solid ellipsoid

Points of a solid ellipsoid satisfy the equation,

$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} \le 1$

We can parameterize these points as follows,

$x = a r \sin t \cos s$

$y = b r \sin t \sin s$

$z = c r \cos t$

where $0 \le r \le 1 , 0 \le t \le \pi, 0 \le s \le 2 \pi$ .

The moment of inertia integral is

$I = \rho \displaystyle \int_V (x^2 + y^2) dV$

In order to use the above parameterization, we have to first find the Jacobian $J$ ,

$J = \dfrac{\partial(x, y, z)}{ \partial(r, t, s) }$

$=\begin{bmatrix} a \sin t \cos s && a r \cos t \cos s && a r \sin t (-\sin s) \\ b \sin t \sin s && b r \cos t \sin s && b r \sin t \cos s \\ c \cos t && -c r \sin t && 0 \end{bmatrix}$

Its determinant is, $| J | = a b c r^2 \sin t$

And the moment of inertia integral becomes,

$I = \rho \displaystyle \int_{s = 0}^{2 \pi} \int_{t = 0}^{\pi} \int_{r = 0}^{r = 1} r^2 \sin^2 t (a^2 \cos^2 s + b^2 \sin^2 s ) | J | dr \hspace{4pt} dt \hspace{4pt} ds$

Integrating with respect to $r$ first, the integral reduces to,

$I = \frac{1}{5} a b c \rho \displaystyle \int_{s = 0}^{2 \pi} \int_{t = 0}^{\pi} \sin^3 t ( a^2 \cos^2 s + b^2 \sin^2 s ) dt \hspace{4pt} ds$

Next we'll integrate with respect to $t$ , and our integral reduces to,

$I = \frac{1}{5} abc \rho (\frac{4}{3}) \displaystyle \int_{s = 0}^{2 \pi} (a^2 \cos^2 s + b^2 \sin^2 s) ds$

And finally, by integrating with respect to $s$ , we get,

$I = \frac{4}{15} \pi . abc (a^2 + b^2) \rho$

Now, the volume of the ellipsoid is $V = \frac{4}{3} \pi a b c$ , and its mass is $m$ , so

so $\rho = m/V = \dfrac{3 m}{4 \pi a b c}$ . Substituting this in the expression for $I$ ,

$I = \frac{1}{5} m (a^2 + b^2)$

With $m = 1 , a = 5 , b = 4$ , we get,

$I = \frac{1}{5} (1) (25 + 16 ) = \frac{41}{5} = 8.2$

and for a sphere with $m = 1$ and $R = a = b = 5$

$I = \frac{1}{5} (1) (25 + 25 ) = 10$

so that the sum of the $I$ 's $= 8.2 + 10 = \boxed{18.2}$