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