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Radius probability of random cut hemisphere.

  1. Sep 11, 2006 #1
    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. jcsd
  3. Sep 11, 2006 #2
    Just do some integration .
    You will get
    [tex]\frac{\pi R}{4}[/tex]
    where R is the radius of the sphere
    Last edited: Sep 12, 2006
  4. Sep 12, 2006 #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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