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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 43r. For a disc, (or one half of a disc), the expected distance would be 32r. I'm open to being convinced otherwise. :)
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 b−a1∫abf(x)?
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 ϕ and θ, 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
32πR3∫0R2πr2∗rdr=43R.
If there wasn't the symmetry with respect to ϕ and θ then we would require a full-blown triple integral.
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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Using spherical coordinates with the AVT I'm finding that the expected distance from the center to a randomly chosen point is 43r. For a disc, (or one half of a disc), the expected distance would be 32r. I'm open to being convinced otherwise. :)
Log in to reply
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 b−a1∫abf(x)?
Log in to reply
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 ϕ and θ, 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
32πR3∫0R2πr2∗rdr=43R.
If there wasn't the symmetry with respect to ϕ and θ then we would require a full-blown triple integral.