Divergence theorem over a hemisphere

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Homework Help Overview

The discussion revolves around applying the divergence theorem to evaluate an integral over a hemisphere, including considerations for a unit disk at the base. Participants are exploring the relationship between surface integrals and volume integrals in this context.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of the divergence and its implications for the integral over the hemisphere and the disk. There is an exploration of the orientation of normals and how it affects the results.

Discussion Status

There are differing results regarding the final value of the integral, with some participants questioning the provided answer and suggesting that their calculations may indicate a different outcome. The discussion remains open with no clear consensus reached.

Contextual Notes

Participants note potential discrepancies in the provided answer and the calculations they have performed, indicating a need for further verification of their methods and assumptions.

jonwell
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I was told this problem could be done with divergence theorem, instead of as a surface integral, by adding the unit disc on the bottom, doing the calculation, then subtracting it again.

Homework Statement



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Homework Equations



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The Attempt at a Solution



for del . f I get i + j = 2. Which makes the integral equal twice the volume of the hemisphere, or 4/3 pi. Now I'm supposed to subtract the unit disc, but I get pi when I calculate that surface, which leaves me with 1/3 pi. The answer should be 7/6 pi.

Thanks :)
 
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The outward pointing normal to the disk is -k. So F.(-k) is -1 and the contribution from the disk is -pi. The total integral from the divergence theorem is 4pi/3 (as you said). I think that makes the integral over the upper hemisphere 7pi/3, doesn't it? Not 7pi/6?
 
Ya, I got that a couple times too (after I remembered the orientation), however the answer I was provided is 7/6 pi. That could be wrong. I'll re-do the surface integral the other way and see what I come up with I guess.
 
Good idea! Let me know what you get.
 
well, I'm still getting 7pi/3, so I'm going to assume that the two of us combined are smarter than the given answer ;) Thanks for your help!
 

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