The following equation is used as an inference but not explained in my solid state physics book (Economou, "The Physics of Solids"). I figure it's a math/geometry problem so I ask here. Can anyone show this?
$$ \langle r^2\rangle = \frac{3}{5} r_o^2 $$
where $r_o$ is the radius of a sphere and $\langle r^2\rangle$ is the average square distance from the center (e.g. for some particle in this volume ($\frac{4}{3}\pi r_o^3$), assuming equal probability distribution).
I would have thought, naively, that $\langle r\rangle = \frac{1}{2}r_o$ thus $\langle r^2\rangle = \frac{1}{4}r_o^2$
If interested this is the context:
The amount of the sphere at a distance of $r$ is $$ 4\pi r^2\,\mathrm{d}r $$ so the average of $r^2$ would be $$ \begin{align} \frac{\int_0^{r_0}4\pi r^2r^2\,\mathrm{d}r}{\int_0^{r_0}4\pi r^2\,\mathrm{d}r} &=\frac{\frac{4\pi}5r_0^5}{\frac{4\pi}3r_0^3}\\[3pt] &=\frac35r_0^2 \end{align} $$