Define the standard $n$-ball as usual:
$$B_n(R) = \{x\in\mathbb{R}^n:\|x\|_2\leq R\}$$
How to compute the mean distance to the center of $B_r(R)$?
Specifically I am looking for simple, intuitive ways to approach this problem.
What I tried (not very simple nor very intuitive...)
Divide the $n$-ball into all the (n-1)-spheres inside it (i.e. $S^{n-1}(r) = \{x\in\mathbb{R}^n:\|x\|_2=r\}$ for $0\leq r \leq R$) and denote its sufrace area by $S_{n-1}(r)$
Then one computes
$$\frac{\int_0^R r S_{n-1}(r)dr}{V_n(R)}$$
where $V_n(R)$ is the volume of the $n$-ball.
Simplifying the above expression and using the formulas for the surface area and volume of an $n$-sphere
$$S_{n}(r) = \frac{2\pi^{\frac{n + 1}{2}}}{\Gamma(\frac{n+1}{2})}r^n$$
$$V_n(r)=\frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2}+1)}r^n$$
One arrives at
$$\frac{\int_0^R r \frac{2\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}r^{n-1}dr}{\frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2}+1)}R^n}=\frac{2\Gamma(\frac{n}{2}+1)}{R^n\Gamma(\frac{n}{2})}\int_0^R r^ndr=\frac{2\Gamma(\frac{n}{2}+1)}{\Gamma(\frac{n}{2})}\frac{R}{n+1}=\frac{n}{n+1}R$$
Is there a simpler approach?
If $X$ represents the distance to the origin then $$\mathbb{P}(X\le r)=\cases{0& if $r<0$\cr r^n/R^n& if $0\le r\le R$\cr 1& if $r\ge R$}$$ So, the density function of $X$ is $$f_X(r)=n\frac {r^{n-1}}{R^n}\mathbb{I}_{[0,R]}(r)$$ Hence $$\mathbb{E}(X)=\frac{n}{R^n}\int_0^R r^ndr=\frac{n}{n+1}R$$ Which is the expected distance to the origine.