I have found an answer on this site to the question of determining the mean straight-line distance between 2 randomly chosen points in a disc of radius r. (See Average distance between two points in a circular disk) I'm now trying to find an answer to the same question except involving a ball of radius r rather than a disc. Any guidance on this question would be appreciated.
Geometry – Mean Distance Between Two Points in a Sphere
geometryprobability distributionsspheres
Best Answer
(a) Two random points on the unit sphere $S^2$:
Assume the first point at the north pole ($\theta=0$) of $S^2$. Then the distance to a point at latitude $\theta$ is $2\sin{\theta\over2}$. Therefore the mean distance between the north pole and the second point is given by $${1\over 4\pi}\int_0^\pi 2\sin{\theta\over2}\cdot 2\pi\sin\theta\ d\theta={4\over3}\ .$$ $$ $$
(b) Two random points in the unit ball of ${\mathbb R}^3\ $:
Let ${\bf X}$ and ${\bf Y}$ be the two random points. Then $R:=|{\bf X}|$, $\ S:=|{\bf Y}|$, and $\Theta:=\angle({\bf X},{\bf Y})$ are independent random variables with densities $$f_R(r)=3r^2\quad (0\leq r\leq 1)\ ,\qquad f_S(s)=3s^2\quad(0\leq s\leq 1)\ ,$$ and $$f_\Theta(\theta)={1\over2}\sin\theta\quad(0\leq\theta\leq\pi)\ .$$ (Concerning $f_R$ and $f_S$ note that the volume included between $r$ and $r+dr$ is proportional to $r^2$. For $f_\Theta$ you may assume ${\bf X}$ pointing due north. The abstract surface area between $\theta$ and $\theta+d\theta$ is then proportional to $\sin\theta$, as in (a).)
It follows that the mean distance $\delta$ between ${\bf X}$ and ${\bf Y}$ is given by $$\delta=\int_0^1\int_0^1\int_0^\pi \sqrt{r^2+s^2-2rs\cos\theta}\ f_R(r) f_S(s)f_\Theta(\theta) d\theta\ ds\ dr\ .$$ The innermost integral computes to $$\eqalign{{1\over2}\int_0^\pi \sqrt{r^2+s^2-2rs\cos\theta}\ \sin\theta\ d\theta&={1\over 6rs}\bigl(r^2+s^2-2r s\cos\theta\bigr)^{3/2}\Biggr|_0^\pi \cr &={1\over 6rs}\bigl((r+s)^3-|r-s|^3\bigr)\ . \cr}$$ In the sequel we assume $s\leq r$ and compensate this by a factor of $2$. We are then left with $$\delta=\int_0^1\int_0^r 3r s(6r^2 s +2s^3)\ ds\ dr={36\over35}\ .$$ It should not be too difficult to set a similar computation up that is valid for a ball in any ${\mathbb R}^n$, $\ n\geq 2$.