Average arc length between two random points on a unit sphere? I'm trying to find the average arc length between two random points on a unit sphere. The solution I've come up with is rather ugly. Consider a parametric surface:
$$X(u,v)=\sin u\cos v\\Y(u,v)=\cos u\cos v\\Z(u,v)=\sin v$$
(For a sphere, $u\in[0,2\pi]$, $v\in\left[\frac{-\pi}{2},\frac{\pi}{2}\right]$). Then, integrate the distance formula for two points $(x_1,y_1,z_1),(x_2,y_2,z_2)$along all three axes. This turns into a nasty integral, though.
$$F(\ldots)=\int_0^{2\pi}
    \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}
        D(X_1,Y_1,Z_1,X_2,Y_2,Z_2)dv\mbox{ }du$$
$$\int_0^{2\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}F(\ldots)\mbox{ }dv\mbox{ }du$$
(Where $D(\ldots)$ is the arc length formula.) I could plug this into mathematica or something, but I have a feeling this would


*

*take too long to compute

*be too complex

*probably also be wrong


What is a more efficient way to do this? How should I go about solving this problem? 
 A: You can rotate the coordinates so the first point is at the pole and measure $\theta$ from $0$ at that pole to $\pi$ at the other one.  The distance to the other point is then $\theta$.  The probability distribution of $\theta$ is proportional to $\sin \theta$ as that is the radius of the small circle the second point is on.  Your average distance is then 
$$\frac {\int_0^\pi \theta \sin(\theta) d\theta}{\int_0^\pi \sin(\theta) d\theta}=\frac {\pi}{2}$$
We can see this is true by matching up points at $\theta$ with points at $\pi - \theta$  The average of each pair is $\frac \pi 2$
A: For a sphere with unitary radius we have  

The "number" of points having the arc-distance $\alpha$ is measured by the area of the annulus
$\sin \alpha d\alpha$, so
$$
\begin{gathered}
  \overline \alpha   = \frac{1}
{{\underbrace {4\pi }_{\text{tot}\text{.}\,\text{weight}}}}\int_{\alpha  = 0}^{\,\pi } {\overbrace \alpha ^{\text{value}}\,\overbrace {2\pi \sin \alpha d\alpha }^{\text{weight}}\,}  = \frac{1}
{2}\int_{\alpha  = 0}^{\,\pi } {\alpha \,\sin \alpha d\alpha \,}  =  \hfill \\
   = \frac{1}
{2}\left( {\left. { - \alpha \,\cos \alpha } \right|_{\alpha  = 0}^{\,\pi }  - \int_{\alpha  = 0}^{\,\pi } {\cos \alpha d\alpha \,} } \right) = \frac{\pi }
{2} \hfill \\ 
\end{gathered} 
$$
which shall be expected, given the symmetry of the Northern and Southern emisphere.
