Let $S$ be the region enclosed by a piecewise smooth simple closed curve $C$ in the $xy-$plane. Use Green's theorem to show that the area of $S$ is $\frac{1}{2}\int_C xdy-ydx$, where $C$ is oriented anticlockwise.
Attempt: Choose a parametrisation $g(x,y)=(-\frac{1}{2}y,\frac{1}{2}x)$. Then the curl is equal to $1$. So the area is $\int\int 1 dxdy$. I have no idea how to prove the required identity though.
Using the definition of double integral we can conclude that
$\lim_{n\to \infty}\sum_{i=0}^nf(x_i,y_i)\triangle A_i=\iint_Rf(x,y)dxdy \Rightarrow \lim_{n\to \infty}\sum_{i=0}^n \triangle A_i=Area=\iint_Rdxdy$
According to Green`s theorem, we know that
$\iint_R (\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y})dxdy=\frac{1}{2}\int_C Mdy-Ndx $
Let $M=\frac{x}{2}$ and $N=\frac{y}{2}$
$M=\frac{x}{2},N=\frac{y}{2} \Rightarrow\frac{\partial M}{\partial x}=\frac{\partial N}{\partial y}=\frac12$
$\iint_R (\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y})dxdy=\int_C Mdy-Ndx \Rightarrow \iint_R (\frac{1}{2}+\frac{1}{2})dxdy=Area=\frac{1}{2}\int_C xdy-ydx $