Suppose that $\Omega$ is a body of revolution of the function $y=f(x), a\le x \le b$ around the $x$-axis, where $f(x)>0$ is continuous.
How to compute the inertia tensor ${\bf{J}} _{\Omega}$?
After computing ${\bf{J}} _{\Omega}$, how to solve ${\bf{J}} _{\Omega} \dot w={\bf{J}} _{\Omega}w \times w$?
Answer
I will answer the mass moment of inertia tension question. For an infinitesimal clump of mass ${\rm d}m$ located at $\vec{r}$ its effect on the inertia tensor is $${\rm d}{\bf J} = -[\vec{r}\times][\vec{r}\times]{\rm d}m$$
where $[\vec{r}\times]$ is the skew symmetric cross product operator $$ \begin{pmatrix}x\\y\\z \end{pmatrix}\times = \begin{bmatrix}0&-z&y\\z&0&-x\\-y&x&0\end{bmatrix}$$
For a axisymmetric part $\vec{r}(x,y,\theta) = (x,y \cos\theta,y \sin \theta)$ where $x=a\ldots b$, $y=0\ldots f(x)$ and $\theta=0\ldots 2\pi$.
The total mass is $m = \iiint \rho {\rm d}V$ where ${\rm d}V = (y{\rm d}\theta)({\rm d}y)({\rm d}x)$ $$m = \int \limits_a^b \int \limits_0^{f(x)} \int \limits_0^{2\pi} \rho\,y{\rm d}\theta\,{\rm d}y\,{\rm d}x$$ $$\rho = \frac{m}{\pi \int_a^b {f(x)}^2\,{\rm d}x}$$
The inertia tensor is ${\bf J}=\rho \iiint -[\vec{r}\times][\vec{r}\times]\,{\rm d}V$. You can do some math to find that the inner integral is
$$\int \limits_0^{2\pi} -[\vec{r}\times][\vec{r}\times]\,{\rm d}\theta = \begin{bmatrix} 2\pi y^2&0&0\\0&\pi(2 x^2+y^2)&0\\0&0&\pi(2 x^2+y^2)\end{bmatrix} $$
In the end, I get $${\bf J} = \rho\, \begin{bmatrix} \frac{\pi}{2}\int_a^b{f(x)}^4\,{\rm d}x&0&0\\ 0&\frac{\pi}{4}\int_a^b\left({f(x)}^4+4 x^2 {f(x)}^2\right)\,{\rm d}x &0\\0&0&\frac{\pi}{4}\int_a^b\left({f(x)}^4+4 x^2 {f(x)}^2\right)\,{\rm d}x \end{bmatrix}$$
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