Alternative Formulation of Green's Theorem

Let
$C$
be a simple closed curve in the
$xy$
plane and let
$D$
be a simple closed curve enclosed by
$C$
.
The area enclosed by
$C$
is equal to
$A = \int \int_D dx dy \int \int {\partial Q}{\partial x}- {\partial P}{\partial y}dxdy = \int_C P dx - Qdy =$

Assume
$Q=0$
then
$\frac{\partial P}{\partial y}= -1 \rightarrow P=-y$

Hence
$\int \int_D dxdy = - \oint dy$

If
$P=0$
then
$\frac{\partial Q}{\partial y} =1 \rightarrow Q=x$
hence
$A= \int\oint xdy$

These two integrals return the same area so are equal. Hence
$A= \frac{1}{2} \oint_C xdy -ydx$

Example: If
$x=a cos \theta , y= b sin \theta , \: 0 \leq \theta \leq 2 \pi$

Then
$A = \int^{2 \pi}_0 a cos \theta b cos \theta d \theta - b sin \theta (- a sin \theta ) d \theta = \int^{2 \pi}_0 ab ( cos^2 \theta + sin^2 \theta ) d \theta = \pi ab$