Area Enclosed by a Curve Using Normal

Theorem
If  
\[\mathbf{r} = x \mathbf{i} + y \mathbf{j}\]
  is the position vector of a point on a curve  
\[C\]
  then
\[The \: area \: enclosed \: by \: C = \frac{1}{2} \oint_C \mathbf{r} \cdot \mathbf{n} \: ds \]
  where  
\[\mathbf{n}\]
  is the outer normal to  
\[C\]
  and  
\[s\]
  is the distance along the curve.
Proof
By a version of Greens Theorem,  
\[Area \: enclosed \: by \: C = \frac{1}{2} \oint_C x \: dy - y \: dx\]

The normal to the curve  
\[C\]
  is  
\[\mathbf{n} =\mathbf{T} \times \mathbf{k}= (dx \mathbf{i} +dy \mathbf{j}) \times \mathbf{k} = dy \mathbf{i} - dx \mathbf{j}\]

Hence  
\[\oint_C \mathbf{r} \cdot \mathbf{n} \: ds = \oint_C x \: dy - y \; dx= twice the area enclosed in \]

Hence  
\[The \: area \: enclosed \: by \: C = \frac{1}{2} \oint_C \mathbf{r} \cdot \mathbf{n} \: ds \]