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How can I find an average distance between two points lying inside a circular disk of a certain radius?

I wonder if there is any other way except of using a Monte Carlo method?

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2 Answers

up vote 1 down vote accepted

See the answer to this question. The expected distance is $$ d= {128 r\over 45\pi}. $$

Here is another demonstration of this result.

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Your answer seem to be wrong. Using Monte Carlo simulations (see this script) I got another answer $d=0.72r$, and $\frac{128}{45\pi}\approx0.91$. –  Installero Apr 24 '12 at 6:05
    
You're probably making the mistake decribed here anderswallin.net/2009/05/… –  Gebb Apr 24 '12 at 8:44
    
Exactly! Thank you! –  Installero Apr 26 '12 at 15:12
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Let $D$ be the disk, and let $$M = \iint\limits_{(x_0,y_0)\in D} ~~\iint\limits_{(x,y) \in D}dxdydx_0dy_0.$$

Then the quantity you are looking for should be given by $$ \frac{1}{M}\iint\limits_{(x_0,y_0)\in D} ~~\iint\limits_{(x,y) \in D} \sqrt{(x-x_0)^2+(y-y_0)^2}dxdydx_0dy_0 $$ where a suitable change to polar coordinates in both double integrals would probably be helpful.

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Unfortunately, this method seems to lead me to an unsolvable finite integrals. I would be very thankful if you can help me getting to a final answer. –  Installero Apr 23 '12 at 13:39
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