# Integral of $\| \vec{x} \|^2$ over unit ball in $\mathbb{R}^n$

I am trying to show that $\displaystyle\int_{\| \vec{x} \| < 1} \| \vec{x} \|^2 dx = \frac{n v_n}{n+2}$, where $n$ is the number of dimensions and $v_n$ is the volume of the n-dimensional unit ball.

It seems like there should be a way to induct, but it gets quite ugly (even with spherical coordinates). It seems like there should be some nice argument, but I don't see it.

An $n$-ball $B_r$ of radius $r>0$ has volume $v_nr^n$. It follows that the area $\omega_n$ of $S^{n-1}$ is given by $$\omega_{n-1}=\lim_{\epsilon\to0+}{1\over\epsilon}{\rm vol}(B_{1+\epsilon}\setminus B_1)=v_n{d\over dr}r^n\biggr|_{r=1}=nv_n.$$ Therefore we obtain, using a "shelling" of $B_1$ by concentric spheres, and Fubini's theorem, $$\int_{B_1}r^2\>{\rm d}(x)=\int_0^1 r^2\> r^{n-1}\omega_{n-1}\>dr={nv_n\over n+2}\ .$$
My idea would be to use spherical coordinates and then use what we know about sphere surface and volume as much as possible. Rewriting your integral to spherical coordinates gives $$\int_{\| \vec{x} \| < 1} \| \vec{x} \|^2 dx = \int_{r<1}r^2 r^{n-1} \mathrm dr \int_{S_{n-1}}\mathrm d \Omega$$ where the second integral stands for the angle part and equals to the surface on a sphere $S_{n-1}$. This part is the same as in the computation of sphere volume.
What now remains is to compare $\int_{r<1} r^{n+1} \mathrm dr$ that we have with $\int_{r<1}r^{n-1} \mathrm dr$ that comes from the volume of a unit ball. This will give the prefactor $\frac{n}{n+2}$.