# Radius probability of random cut hemisphere.

1. Sep 11, 2006

### quasi426

Imagine viewing a hemisphere normal to the equator such that it looks like a circile with the full radius of the hemisphere. Now randomly section or cut the hemisphere in a manner a tomato is sliced. If we only consider the portion of the hemisphere that contains the pole we should be generating a smaller radius then the full radius given we actually cut something off.

My question is the following:
What is the average length of the radius of a randomly cut cross-section?

I did this numerically and obtained 0.7855 of the original radius. What is the analytical approach? Thanks.

2. Sep 11, 2006

### balakrishnan_v

Just do some integration .
You will get
$$\frac{\pi R}{4}$$
where R is the radius of the sphere

Last edited: Sep 12, 2006
3. Sep 12, 2006

### quasi426

Ah I see, I just integrate sqrt(1-x^2) from 0 to R. That's a tough integral though for a guy who hasn't done much calculus in 3 years, I was lazy and used mathematica to obtain the solution. Makes me feel like I would be useless in a deserted island. Thanks for the help.