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