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I was reading Maxwell's relations and came across: $$\oint pdV=\oint TdS\Rightarrow \iint dpdV=\iint dTdS.$$

I know this is straightforward to see since they both represent the surface area, but I've never seen a math theorem on textbooks that indicates $$\oint ydx=\iint dxdy.$$ Is this just a trivial corollary?

Lluvio Liu
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3 Answers3

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This is Stokes' theorem, $$ \int_S \mathrm{d}\omega = \int_{\partial S} \omega$$ for a 1-form $\omega = y\mathrm{d}x$, $S$ a surface and $\partial S$ its boundary.

ACuriousMind
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    While concise, the differential form version of the equation doesn't say much to somebody who doesn't already know it anyway. – Dave Feb 25 '22 at 22:26
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    @Dave Sure, but that's what the Wikipedia link is there for. I don't see reason to replicate detailed explanations of Stokes' theorem in its various forms here when they are already available elsewhere, and I chose the differential form version both because I prefer it and because otherwise I'd just have typed the same equation OP already has in their question again. (for general links, there's a "link rot" argument, but Stokes' theorem is such a basic building block that it's not going to vanish from the internet; also, I certainly don't think this answer should be worth 28 upvotes :P) – ACuriousMind Feb 25 '22 at 22:47
  • It was just an observation; I'm one of the 28. – Dave Feb 25 '22 at 22:52
  • After all the quesiton was "Is there ..." and now "Show me ...". – Florian F Feb 28 '22 at 12:36
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If in Green's formula (special case of Stokes' formula), we replace $L=y$ and $M=0$, we find: $$\int_{\partial S}ydx=-\int_{S}dxdy$$

The Tiler
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Is this just a trivial corollary?

It is clearly not trivial since you are asking about it.

To understand the meaning better you should think about the geometry and specify the region of integration in both cases.

The equality only holds when one integration region is the boundary of the other, not generally.

hft
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