Suppose we have a closed simple curve written as $(x,y) = F(u)$. To find the area enclosed by this curve, we appeal to Green's theorem:
$$ A = \frac{1}{2}\int_0^{2\pi} [x\frac{\partial y}{\partial u} - y\frac{\partial x}{\partial u}]du $$
According to the paper https://projecteuclid.org/download/pdf_1/euclid.jdg/1214439902 page 7 Lemma 3.1.7, we can reduce the equation above to,
$$A = \frac{-1}{2} \int_0^{2\pi} \langle F, vN\rangle du$$
where $v = \sqrt{(\partial y/\partial u)^2 + (\partial x/\partial u)^2}$ or equivalently, $v = |\partial F/\partial u|$ and $N$ is the (inward pointing) unit normal. However, (embarrassingly) I am not sure how this was done.
I think what I'm trying to prove here is the following statement
$$ -\langle F(u), vN\rangle = -(x,y)\cdot vN = x \frac{\partial y}{\partial u} - y\frac{\partial x}{\partial u} $$
Any help is appreciated.
A vector parallel to the tangent is $\partial F/\partial u = (\partial x/\partial u,\partial y/\partial u)$. Providing that the curve is parametrised anticlockwise, the inward-pointing normal direction is this rotated anticlockwise by $\pi/2$, i.e. $(-\partial y/\partial u,\partial x/\partial u)$. Taking the inner product of this with $F$ gives $$ - x \frac{\partial y}{\partial u} + y \frac{\partial x}{\partial u}, $$ i.e. minus the original integrand. Finally, $N$ is $(-\partial y/\partial u,\partial x/\partial u)/\lvert (-\partial y/\partial u,\partial x/\partial u) \rvert$, so $ (-\partial y/\partial u,\partial x/\partial u) = \nu N $.