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