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AVT maybe?

What are the expected coordinates of a point chosen uniformly at random inside the upper half \((y>0)\) of the sphere \(x^2+y^2+z^2=r^2\)?

Is it \(\frac{2}{3}\)? Or am I thinking of this incorrectly.

Note by Trevor Arashiro
1 year, 9 months ago

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Using spherical coordinates with the AVT I'm finding that the expected distance from the center to a randomly chosen point is \(\frac{3}{4}r\). For a disc, (or one half of a disc), the expected distance would be \(\frac{2}{3}r\). I'm open to being convinced otherwise. :)

Brian Charlesworth - 1 year, 9 months ago

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I honestly can't argue :3, and you're probably right anyway.

I have no idea how to apply AVT to a 2 dimensional object and much less a 3D solid. The only way I know to apply it is to a function. Do you have to use any form of it other than \(\frac{1}{b-a}\displaystyle \int_a^b f(x)\)?

Trevor Arashiro - 1 year, 9 months ago

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To avoid confusion I'll let the the radius of the hemisphere be \(R\) and the variable distance from the center of the hemisphere be \(r\).

In this case, because of the symmetry with respect to \(\phi\) and \(\theta\), we know that the probability a random point will lie between a distance of \(r\) and \(r + dr\) from the center of the hemisphere is the ratio of the area of the hemispherical shell defined by these two values to the area of the entire hemisphere. We then multiply this probability by \(r\) and integrate from \(0\) to \(R\) to get the expected distance from the center. This comes out to

\(\displaystyle\dfrac{\int_{0}^{R} 2\pi r^{2} * r dr}{\dfrac{2}{3}\pi R^{3}} = \dfrac{3}{4}R\).

If there wasn't the symmetry with respect to \(\phi\) and \(\theta\) then we would require a full-blown triple integral.

Brian Charlesworth - 1 year, 9 months ago

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