# Moment of inertia of a n-dimensional sphere

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.

RESTATEMENT OF THE PROBLEM

This problem is ill posed. I am correcting it as follows:

QUESTION

For any natural number $n$, given an $n$-ball of radius $R$ and of constant density, how is its moment of inertia about any axis through its center of mass, $I_{c,n}$, related to its mass, $M_n$?

$$I_{c,n} = M_n\,R^2\,\dfrac{ n-1 }{n+2}$$

BACKGROUND

Adapting from , "Generally, when embedded in an ($n$) -dimensional Euclidean space, an ($n$-1)-sphere is the surface or boundary of an ($n$)-dimensional ball.... Thus, the ($n$-1)-sphere would be defined by: $$S^{n-1}=\left\{x\in \mathbb {R} ^{n}:\left\|x\right\|=R\right\}."}$$

Adapting from , in Cartesian coordinates, an expression for moment of inertia of this $n$-ball rotating about a specified axis is $$I_{c,n} = \rho\,\int_{x_n = -\sqrt{R^2 - \sum_{i=n+1}^{n}{x_i^2} }}^{ \sqrt{R^2 - \sum_{i=n+1}^{n}{x_i^2} }}\ \cdots \int_{x_1=-\sqrt{R^2 - \sum_{i=2}^{n}{x_i^2} }}^{ \sqrt{R^2 - \sum_{i=2}^{n}{x_i^2} }} \left\|\vec{r}\right\|^2 \, dx_1\,\ldots\,dx_n.$$ where $\rho$ is the mass density at each point $\vec{r}$ is a vector perpendicular to the axis of rotation and extending from a point on the rotation axis to a point $(x_1, x_2, \ldots, x_n)$ in the solid. Herein, for the sake of argument, let's assume that the $n$-ball is rotating around the $x_n$ axis. We find then that $$\left\|\vec{r}\right\|^2 = \sum_{k=1}^{n-1}x_k^2.$$ Upon rewriting, $$I_{c,n} = \rho \,\int_{x_n = -\sqrt{R^2 - \sum_{i=n+1}^{n}{x_i^2} }}^{ \sqrt{R^2 - \sum_{i=n+1}^{n}{x_i^2} }}\ \cdots \int_{x_1=-\sqrt{R^2 - \sum_{i=2}^{n}{x_i^2} }}^{ \sqrt{R^2 - \sum_{i=2}^{n}{x_i^2} }} \left( \sum\limits_{j=1}^{n-1}x_j^2 \right) \, dx_1\,\ldots\,dx_n.$$ We can now write the ratio of the moment, $I_{c,n}$, to mass $M_n$ as $$\dfrac{I_{c,n}}{M_n} = \dfrac{\int_{x_n = -\sqrt{R^2 - \sum_{i=n+1}^{n}{x_i^2} }}^{ \sqrt{R^2 - \sum_{i=n+1}^{n}{x_i^2} }}\ \cdots \int_{x_1=-\sqrt{R^2 - \sum_{i=2}^{n}{x_i^2} }}^{ \sqrt{R^2 - \sum_{i=2}^{n}{x_i^2} }} \left( \sum\limits_{j=1}^{n-1}x_j^2 \right) \, dx_1\,\ldots\,dx_n}{\int_{x_n = -\sqrt{R^2 - \sum_{i=n+1}^{n}{x_i^2} }}^{ \sqrt{R^2 - \sum_{i=n+1}^{n}{x_i^2} }}\ \cdots \int_{x_1=-\sqrt{R^2 - \sum_{i=2}^{n}{x_i^2} }}^{ \sqrt{R^2 - \sum_{i=2}^{n}{x_i^2} }} \, dx_1\,\ldots\,dx_n}$$

SOLUTION

Again turning to , we can convert this problem to a spherical coordinate system by writing \begin{align} x_1 &= r\,\cos{(\phi_1)} \\ x_2 &= r\,\sin{(\phi_1)}\,\cos{(\phi_2)} \\ x_{n-1} &= r\,\sin{(\phi_1)}\,\ldots \sin{(\phi_{n-2})}\, \cos{(\phi_{n-1})} \\ x_n&= r\,\sin{(\phi_1)}\,\ldots \sin{(\phi_{n-2})} \,\sin{(\phi_{n-1})} \end{align} Please note for future reference, that $$\sum_{k=1}^{n}{x_k^2} = r^2.$$ So $$\sum_{k=1}^{n-1}{x_k^2} = r^2\,\left[1 - \sin^2{(\phi_1)}\,\ldots \sin^2{(\phi_{n-1})}\right].$$

So \begin{align} \dfrac{ I_{c,n}}{M_n} & = \dfrac{ \int_{\phi_{n-1} = 0}^{2\,\pi} \int_{\phi_{n-2} = 0}^{\pi} \cdots \int_{\phi_{1} = 0}^{\pi} \int_{r=0}^{R} r^{n+1}\, \sin^{n-2}{(\phi_1)}\, \sin^{n-3}{(\phi_2)} \cdots\, \sin{(\phi_{n-2})}\,\left[1 - \sin^2{(\phi_1)}\,\ldots \sin^2{(\phi_{n-1})}\right] dr\,d\phi_1\,d\phi_2\cdots d\phi_{n-1} } { \int_{\phi_{n-1} = 0}^{2\,\pi} \int_{\phi_{n-2} = 0}^{\pi} \cdots \int_{\phi_{1} = 0}^{\pi} \int_{r=0}^{R} r^{n-1}\, \sin^{n-2}{(\phi_1)}\, \sin^{n-3}{(\phi_2)} \cdots\, \sin{(\phi_{n-2})}\,dr\,d\phi_1\,d\phi_2\cdots d\phi_{n-1} } \\ & = \dfrac{n\,R^2}{n+2}\dfrac{ \int_{\phi_{n-1} = 0}^{2\,\pi} \int_{\phi_{n-2} = 0}^{\pi} \cdots \int_{\phi_{1} = 0}^{\pi} \, \sin^{n-2}{(\phi_1)}\, \sin^{n-3}{(\phi_2)} \cdots\, \sin{(\phi_{n-2})}\,\left[1 - \sin^2{(\phi_1)}\,\ldots \sin^2{(\phi_{n-1})}\right] \,d\phi_1\,d\phi_2\cdots d\phi_{n-1} } { \int_{\phi_{n-1} = 0}^{2\,\pi} \int_{\phi_{n-2} = 0}^{\pi} \cdots \int_{\phi_{1} = 0}^{\pi} \sin^{n-2}{(\phi_1)}\, \sin^{n-3}{(\phi_2)} \cdots\, \sin{(\phi_{n-2})} \,d\phi_1\,d\phi_2\cdots d\phi_{n-1} } \\ & = \dfrac{n\,R^2}{n+2}\dfrac{ \int_{\phi_{n-1} = 0}^{2\,\pi} \int_{\phi_{n-2} = 0}^{\pi} \cdots \sin^{n-3}{(\phi_2)} \cdots\, \sin{(\phi_{n-2})}\, \int_{\phi_{1} = 0}^{\pi} \, \left[\sin^{n-2}{(\phi_1)} - \sin^{n}{(\phi_1)} \sin^2{(\phi_{2})} \,\ldots \sin^2{(\phi_{n-1})}\right] \,d\phi_1\,d\phi_2\cdots d\phi_{n-1} } { \int_{\phi_{n-1} = 0}^{2\,\pi} \int_{\phi_{n-2} = 0}^{\pi} \cdots \sin^{n-3}{(\phi_2)} \cdots\, \sin{(\phi_{n-2})} \, \int_{\phi_{1} = 0}^{\pi} \sin^{n-2}{(\phi_1)}\, d\phi_1\,d\phi_2\cdots d\phi_{n-1} } \\ & = \dfrac{n\,R^2}{n+2}\dfrac{ \int_{\phi_{n-1} = 0}^{2\,\pi} \int_{\phi_{n-2} = 0}^{\pi} \cdots \sin^{n-3}{(\phi_2)} \sin^{n-4}{(\phi_3)} \cdots\, \sin{(\phi_{n-2})}\, \int_{\phi_{2} = 0}^{\pi} \, \left[1 - \dfrac{n-1}{n}\,\sin^{2}{(\phi_2)}\, \sin^2{(\phi_{3})} \,\ldots \sin^2{(\phi_{n-1})}\right] \,d\phi_2\cdots d\phi_{n-1} } { \int_{\phi_{n-1} = 0}^{2\,\pi} \int_{\phi_{n-2} = 0}^{\pi} \cdots \sin^{n-4}{(\phi_3)} \cdots\, \sin{(\phi_{n-2})} \, \int_{\phi_{2} = 0}^{\pi} \sin^{n-3}{(\phi_2)}\, d\phi_2\cdots d\phi_{n-1} } %%%%% %%%%% \\ & = \dfrac{n\,R^2}{n+2}\dfrac{ \int_{\phi_{n-1} = 0}^{2\,\pi} \int_{\phi_{n-2} = 0}^{\pi} \cdots \sin^{n-4}{(\phi_3)} \cdots\, \sin{(\phi_{n-2})}\, \int_{\phi_{2} = 0}^{\pi} \, \left[\sin^{n-3}{(\phi_2)} - \dfrac{n-1}{n}\,\sin^{n-1}{(\phi_2)}\, \sin^2{(\phi_{3})} \,\ldots \sin^2{(\phi_{n-1})}\right] \,d\phi_2\cdots d\phi_{n-1} } { \int_{\phi_{n-1} = 0}^{2\,\pi} \int_{\phi_{n-2} = 0}^{\pi} \cdots \sin^{n-4}{(\phi_3)} \cdots\, \sin{(\phi_{n-2})} \, \int_{\phi_{2} = 0}^{\pi} \sin^{n-3}{(\phi_2)}\, d\phi_2\cdots d\phi_{n-1} } %%%%% %%%%% \\ & = \dfrac{n\,R^2}{n+2}\dfrac{ \int_{\phi_{n-1} = 0}^{2\,\pi} \int_{\phi_{n-2} = 0}^{\pi} \cdots \sin^{n-5}{(\phi_4)} \cdots\, \sin{(\phi_{n-2})}\, \int_{\phi_{3} = 0}^{\pi} \, \left[\sin^{n-4}{(\phi_3)} - \dfrac{n-2}{n}\,\sin^{n-2}{(\phi_3)}\, \sin^2{(\phi_{4})} \,\ldots \sin^2{(\phi_{n-1})}\right] \, d\phi_3\cdots d\phi_{n-1} } { \int_{\phi_{n-1} = 0}^{2\,\pi} \int_{\phi_{n-2} = 0}^{\pi} \cdots \sin^{n-5}{(\phi_4)} \cdots\, \sin{(\phi_{n-2})} \, \int_{\phi_{3} = 0}^{\pi} \sin^{n-4}{(\phi_3)}\, d\phi_3\cdots d\phi_{n-1} } %%%%% %%%%% \\ & = \dfrac{n\,R^2}{n+2}\dfrac{ \int_{\phi_{n-1} = 0}^{2\,\pi} \left[1 - \dfrac{2}{n}\, \sin^2{(\phi_{n-1})}\right] \, d\phi_{n-1} } { \int_{\phi_{n-1} = 0}^{2\,\pi} \, d\phi_{n-1} } \\ & = \dfrac{n\,R^2}{n+2} \,\left[1 - \dfrac{1}{n} \right] \\ & = \dfrac{n\,R^2}{n+2} \, \dfrac{n-1}{n} \\ & = \dfrac{R^2\,(n-1) }{n+2} \end{align}

Where we have made use of the integral tables (see CRC Encyclopedia, Table of Integrals by Gradshteyn, or similar) to find that for a constant $a$ and a $k\leq n$ \begin{align} & \dfrac{ \int_{\phi_{1} = 0}^{\pi} \, \left[\sin^{n-k}{(\phi_1)} - \sin^{n-k+2}{(\phi_1)} \,a\right] \,d\phi_1 } { \int_{\phi_{1} = 0}^{\pi} \sin^{n-k}{(\phi_1)}\, d\phi_1 } = 1 - \dfrac{n-(k-1)}{n-(k-2)}\,a \end{align}

REFERENCES

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.