Consider a sphere $S$ of radius $a$ centered at the origin. Find the average distance between a point in the sphere to the origin.
We know that the distance $d = \sqrt{x^2+y^2+z^2}$.
If we consider the problem in spherical coordinates, we have a 'formula' which states that the average distance $d_{avg} = \frac{1}{V(S)}\iiint \rho dV$
I think that this is reminiscent of an average density function which I've seen in physics courses, and it is clear that the $\iiint \rho dV$ is equal to the volume of the sphere, but I'm not sure as to why we must integrate over the distance and then divide by the actual volume to calculate the average distance.
I am looking for a way to explain this to my students without presenting the solution as a formula, any insights would be appreciated.
Best Answer
If $\rho$ is the radius, the volume is in fact $\iiint dV$ without the $\rho$. The average distance is then as you have said. This is an example of the general formula for the average value of a variable $X$ over a probability distribution $V$, which is $\bar X= \tfrac {\int X dV}{\int dV}$