 Signed area as a boundary integral

For a counterclockwise, piecewise smooth simple closed planar curve $C$ enclosing $D$, Green's theorem with $P=-y/2$ and $Q=x/2$ gives $\frac{1}{2}{\oint }_C(xdy-ydx)={\iint }_{D}1dxdy$. The integral is the area of $D$ with a sign determined by the traversal direction. For polar coordinates $x=r\cos \theta$, $y=r\sin \theta$, differentiation gives $xdy-ydx=r^{2}d\theta$, yielding the familiar polar area formula when the curve is traversed once.