|Sep11-06, 05:38 PM||#1|
Radius probability of random cut hemisphere.
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.
|Sep11-06, 07:14 PM||#2|
Just do some integration .
You will get
where R is the radius of the sphere
|Sep12-06, 11:46 AM||#3|
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.
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