What is the average area of a cut on a sphere by a random plane?

[astroboy999]

Homework Statement

Given a sphere of radius r, what is the average area of a cut given by a random plane meeting the sphere?

The Attempt at a Solution

I just need someone to check my answer, and maybe suggest an alternative solution if there is a better one. I assumed that the cut is horizontal, then t in my integral below denotes the distance of the plane from the center of the sphere.

The answer is given by the integral
$$\frac{1}{r} \int_{0}^{r} 2 \pi (r^2 - t^2) dt$$

and it works out to $$\frac{2 \pi r^2}{3}$$

In particular, is there a clever solution that may not use an integral? I also may have made calculation mistakes, so I would be grateful if someone checks the answer for me... thanks!

 

[ystael]

Where did the initial factor of $$2$$ in the integral come from? The area of the disc cut by a plane at height $$t$$ is $$\pi(r^2 - t^2)$$, and you are averaging this over the interval $$t\in[0,r]$$.

 

[astroboy999]

Well, don't I want to average over the interval [-r, r]? But maybe I needed the factor of $$\frac{1}{2r}$$, instead of $$\frac{1}{r}$$. Is that correct?

 

[ystael]

By symmetry, it doesn't matter -- but you must be consistent. You must either integrate from $$-r$$ to $$r$$ and divide by $$2r$$, or integrate from $$0$$ to $$r$$ and divide by $$r$$. Either way, the integrand should be $$\pi(r^2 - t^2)\,dt$$, without a factor of $$2$$.