Moment of inertia of a n-dimensional sphere

Question

Is it well known that the moment of inertia of a sphere can be calculated starting from its density $\rho=\frac{m}{V}$ where $V$, the volume of the solid is: $V=\frac{4}{3}\pi r^3$. Calling it $I_3$, we get after some calculation: $$I_3=\frac{2}{5}Mr^2$$ Now, the volume of a n-ball is: $$V_n=\frac{\pi^{n/2}}{\Gamma\left(\frac{n}{2}+1\right)}r^n$$ How the moment of inertia of such a n-ball can be calculated, assuming the mass of the ball, is $M$? Thanks in advance.

Answer 1

By definition, the moment of inertial is directly computed by $$\begin{align}I_n&=\int_{V_n(r)}\frac M{V_n(r)}r^2dV\\ &=\frac M{V_n(r)}\int_{-r}^r\int_{S^{n-1}(\sqrt{r^2-z^2})}r^2-z^2dSdz\\ &=\frac M{V_nr^n}\int_{-r}^rS^{n-1}(r^2-z^2)^{\frac{n-1}2}(r^2-z^2)dz\\ &=M\frac {S^{n-1}}{V_nr^n}\int_{-r}^r(r^2-z^2)^{\frac{n+1}2}dz\\ &=\frac{nM}{r^n}\int_{-r}^r(r^2-z^2)^{\frac{n+1}2}dz\end{align}$$ For $n$ is odd, the integrand is a polynomial, which is very easy to integrate. For $n$ is even, it is refer to hypergeometric function and thus has no closed form.