TITLE: Finding the mean of $x \mapsto |x|^p$ on the unit ball in $\mathbb{R}^n$ without calculating the volume of the unit ball.
QUESTION [0 upvotes]: Find the mean of the function $x\mapsto |x|^p$ on the ball $\{x:|x|<1\}\subset\mathbb{R}^n$ for $p\in(0, \infty)$. (Hint: You do not need the volume of the ball).

I tried doing something similar to the previous exercise (another hint I was given) by considering the set 
$$E=\{(x,z)|0\leq z\leq 1- |x|^p\}\subset \mathbb{R}^n\times\mathbb{R}$$
and then trying to find it's volume in two ways. First integrating $x$ and first integrating $z$ respectively, which gives me that 
$$\int_0^1\int_{|x|<1-z}1dxdz=\int_{|x|<1}\int_0^{1-|x|^p}1dzdx$$
where after doing the first integral on the right side and breaking it up gives
$$\int_{|x|<1}1dx-\int_0^1\int_{|x|<1-z}1dxdz=\int_{|x|<1}|x|^pdx$$
or
$$1-\left(\int_0^1\int_{|x|<1-z}1dxdz\right)\bigg/\left(\int_{|x|<1}1dx\right)=
\left(\int_{|x|<1}|x|^pdx\right)\bigg/\left(\int_{|x|<1}1dx\right)$$
where the right side is the expression I want to find, but I'm not sure how to calculate the left side, at least not without calculating the volumes of the spheres, which leads me to thinking this is not the best way to approach this.

REPLY [1 votes]: The mean of $|x|^p=r^p$ is
$$
\frac{\int_{r\lt1}r^p\mathrm d^nx}{\int_{r\lt1}\mathrm d^nx}=\frac{\int\mathrm d\Omega\int_0^1\mathrm drr^{n-1}r^p}{\int\mathrm d\Omega\int_0^1\mathrm drr^{n-1}}=\frac{\int_0^1\mathrm drr^{n-1}r^p}{\int_0^1\mathrm drr^{n-1}}=\frac n{n+p}\;,
$$
where $\int\mathrm d\Omega$ is the angular integration.
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