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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.

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I would probably explain it by making an analogy with a finite collection of points - i.e. given a collection $\{(x_i,y_i,z_i)\}\subset S$, what is their average distance to the origin? You find all their distances, add them up, and divide by the number of points. Summing up and dividing by the number of points is perfectly analogous to integrating then dividing by the volume. – icurays1 Nov 12 '12 at 21:13
Just a nitpick, but I think you mean a ball, not a sphere. If the points were on the surface of the sphere the average distance would obviously be the radius. – Javier Nov 12 '12 at 21:16
up vote 1 down vote accepted

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 $\bar X= \tfrac {\int X dV}{\int dV}$

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All points are equidistant from the origin ... that is why it is a sphere. Therefore, the average distance of a point on the sphere from its center is the radius.

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In the sphere, not on it. – Arturo don Juan Nov 7 '14 at 1:20
The title of the question says on. – lhf Nov 7 '14 at 1:57

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