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