Why is Green's theorem asymmetric in $x$ and $y$? Green's theorem is
$$\oint_{\partial D} (P\, dx+Q\, dy) = \iint_D dx\,dy \: \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)$$
where one can see that the RHS is asymmetric in $x$ and $y$. Why is this, and what is the physical significance?
I suspect the answer has to do with using a right-handed coordinate system (e.g. one can use Stokes' theorem to show the above, which involves a vector product, which is right-handed), or possibly that the boundary is traversed anti-clockwise, but I'm unable to make any deeper or more accurate statement than that...
 A: The lack of symmetry is only apparent.  Green's formula is a special case of what's known as the (general) Stokes's Theorem, which you should look up in Wikipedia for starters: but think of it like this, if you exchange the coordinates $x$ with $y$ you must also exchange the vector fields components $P$ with $Q$ to be consistent.
A: Both the left side and the right side are asymmetric in $x$ and $y$: The boundary $\partial D$ goes around $D$ in a particular direction and not in the other direction.
A: It is linked with the orientation of space. Denoting $\omega$ the differential form $\,P\mathrm d\mkern1.5mu x+Q\mathrm d\mkern1.5mu y$, Green-Riemann's formula can be written:
$$\int_{\partial D} \omega= \int_D \mathrm d\mkern1.5mu\omega $$
where  $\,\mathrm d\mkern1.5mu\omega$ is the exterior differential of $\omega$, defined by:
\begin{align*}\mathrm d\mkern1.5mu (P\,\mathrm d\mkern1.5mu x+Q\,\mathrm d\mkern1.5mu y)&= \mathrm d\mkern1.5mu P\wedge \mathrm d\mkern1.5mu x+\mathrm d\mkern1.5mu Q\wedge\mathrm d\mkern1.5mu y\\
&=\Bigl(\frac{\partial P}{\partial x}\mathrm d\mkern1.5mu x+ \frac{\partial P}{\partial y}\mathrm d\mkern1.5mu y\Bigr)\wedge\mathrm d\mkern1.5mu x  +\Bigl(\frac{\partial Q}{\partial x}\mathrm d\mkern1.5mu x+ \frac{\partial Q}{\partial y}\mathrm d\mkern1.5mu y\Bigr)\wedge\mathrm d\mkern1.5mu y \\
& = \frac{\partial P}{\partial y}\mathrm d\mkern1.5mu y\wedge\mathrm d\mkern1.5mu x+ \frac{\partial Q}{\partial x}\mathrm d\mkern1.5mu x\wedge\mathrm d\mkern1.5mu y =  \Bigl(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\Bigr)\mathrm d\mkern1.5mu x\wedge\mathrm d\mkern1.5mu y 
\end{align*}
since $\,\mathrm d\mkern1.5mu x\wedge\mathrm d\mkern1.5mu x=\mathrm d\mkern1.5mu y\wedge\mathrm d\mkern1.5mu y=0\,$ and $\,\mathrm d\mkern1.5mu y\wedge\mathrm d\mkern1.5mu x=-\mathrm d\mkern1.5mu x\wedge\mathrm d\mkern1.5mu y$.
