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For this question I found in a video:
Find the surface flux through the piece of cylinder with equation $x^2 + y^2 = a^2$ for $0 \leq z \leq h$ in the first octant, with vector field $F = (z,x,y)$.
The answer says that it is $\frac{1}{2} ( ah^2 + a^2 h)$.
However, why can't I use the divergence theorem, with $\mathrm{div}F = 0 + 0 + 0 = 0$, so the triple integral $\iiint \mathrm{div}F dV = 0$?

Twenty-six colours
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1 Answers1

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Because the surface at hand is not the boundary of a domain in $\mathbb{R}^3$.

Amitai Yuval
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  • Thanks. I'm afraid I don't fully understand. Iirc, the only condition I was taught was that the region has to be closed, I'm not sure what you mean by "the boundary of a domain" – Twenty-six colours Jun 02 '17 at 05:25
  • A similar example, with instead of a cylinder, it was replaced with a sphere (in the same octant) and the divergence theorem worked for me. – Twenty-six colours Jun 02 '17 at 05:27
  • @Twenty-sixcolours For the sake of the matter, it's the same thing. And your surface is not closed, in the sense that it has boundary. If something similar worked for you with a piece of a sphere (as opposed to a whole one), then you were lucky. – Amitai Yuval Jun 02 '17 at 05:29
  • Hmm, I'm a bit confused now, do you mean it's not simply closed because it has a boundary? (Since you said "in the sense that it has boundary") and also that it works for a full sphere not for a piece of sphere. Is there an easy way to tell if divergence theorem would work for a solid? I just saw examples on this [url]http://mathinsight.org/divergence_theorem_examples[/url], and it worked for the box (which seems to have 6 boundaries). – Twenty-six colours Jun 02 '17 at 05:37
  • @Twenty-sixcolours This is why I started with "the boundary of a domain"... You say "a solid". Same thing. It works for any solid. However, the piece of cylinder you have in your question is not the boundary of any solid. – Amitai Yuval Jun 02 '17 at 06:09
  • Ah I think I just realised why it's not. It's because the equation technically only sketches out the curved bit right? So in essence, to use divergence theorem, we would have to add the two "rectangular walls" and the two quarter circles (at the top and bottom) to fully close it? – Twenty-six colours Jun 02 '17 at 06:38
  • @Twenty-sixcolours Yes. – Amitai Yuval Jun 02 '17 at 06:55
  • Cheers. So what if instead, I evaluated that closed surface (the one where I added all those walls and quarter circles) with divergence theorem (which equals to zero). Can I then individually (without divergence theorem), evaluate the flux through the "walls" and quarter-circles and do arithmetic to evaluate my original flux integral? – Twenty-six colours Jun 02 '17 at 07:00
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    @Twenty-sixcolours Absolutely. – Amitai Yuval Jun 02 '17 at 10:39