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 $.

  • $\begingroup$ Thank you for your help again! $\endgroup$ – Anmol Bhullar May 9 '18 at 18:49

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