AVT maybe?

What are the expected coordinates of a point chosen uniformly at random inside the upper half (y>0)(y>0) of the sphere x2+y2+z2=r2x^2+y^2+z^2=r^2?

Is it 23\frac{2}{3}? Or am I thinking of this incorrectly.

Note by Trevor Arashiro
5 years, 4 months ago

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Comments

Using spherical coordinates with the AVT I'm finding that the expected distance from the center to a randomly chosen point is 34r\frac{3}{4}r. For a disc, (or one half of a disc), the expected distance would be 23r\frac{2}{3}r. I'm open to being convinced otherwise. :)

Brian Charlesworth - 5 years, 4 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 1baabf(x)\frac{1}{b-a}\displaystyle \int_a^b f(x)?

Trevor Arashiro - 5 years, 4 months ago

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To avoid confusion I'll let the the radius of the hemisphere be RR and the variable distance from the center of the hemisphere be rr.

In this case, because of the symmetry with respect to ϕ\phi and θ\theta, we know that the probability a random point will lie between a distance of rr and r+drr + 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 rr and integrate from 00 to RR to get the expected distance from the center. This comes out to

0R2πr2rdr23πR3=34R\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 ϕ\phi and θ\theta then we would require a full-blown triple integral.

Brian Charlesworth - 5 years, 4 months ago
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