TITLE: Is there a math theorem by which a contour integral is equal to a double integral?
QUESTION [12 upvotes]: 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?

REPLY [38 votes]: 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.

REPLY [8 votes]: If in  Green's formula (special case of Stokes' formula), we replace $L=y$ and $M=0$, we find: $$\int_{\partial S}ydx=-\int_{S}dxdy$$

REPLY [7 votes]: 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.