11 December 2011

A geometric probability problem

Here's a cute problem (from Robert M. Young, Excursions in Calculus, p. 244): "What is the average straight line distance between two points on a sphere of radius 1?"

(Answer to follow.)

If any of my students are reading this: no, this should not be interpreted as a hint to what will be on the final exam.

14 comments:

dfan said...

Fun problem, especially since I'm relearning my calculus and am looking for reasons to use it. I got 64 / 15\pi, but did so much calculation along the way that I assume 1) I did something wrong somewhere and 2) there must be a simpler way to do it. At least the size of the answer is plausible.

dfan said...

(And indeed, I just found the actual solution and I was wrong. Now to find my mistake...)

Rafael said...
This comment has been removed by the author.
Chris Lusto said...

4/pi ?

inverno said...

4/3 only one integral to compute! XD

dr said...

Maybe I'm slow, but I don't get what "average" means in this case. How are the different "rings" weighted? If a particular line of latitude has twice the circumference of another, does it have twice as many infinite points from the given points? The same number?

Rafael said...

I don't know, but 4/3*(pi^2) is a good number. :)

inverno said...

good but not correct in my opinion! its only 4/3 !

Christian Blatter said...

It's unbelievable: This simple problem got 5 different answers (1 deleted) by professional mathematicians! We're talking about the bad effects of pocket calculators here!

dr said...

Oh yes, I am slow. Now I get what "average" means.

Karsten W. said...

+1 for Chris Lusto's 4/pi

inverno said...

i don't think the solution depends on pi.

as i cut pi in the nominator & denominator and in my integral there are only 2 sini (where pi) has an non decimal value!

Jyoti Aneja said...

The answer in general is (4/3)*R. In the given case R = 1, hence it is 4/3.

nish said...

its quite clearly 4/3