Let C be a simple closed curve in a region where Green's Theorem holds. Show that the area of the region is:
\begin{equation} A=\int_{C}xdy=-\int_{C}ydx \end{equation}
Green's theorem for area states that for a simple closed curve, the area will be $A=\frac{1}{2}\int_{C}xdy-ydx$, so where does this equality come from?
On
Stokes' theorem on manifolds: $$\int_M d\omega = \int_{\partial M}\omega$$ where $\omega$ is a k form, and $M$ is a differentiable manifold.
For a closed curve in the plane, the area form is $d\omega=dx\wedge dy$. So the area of $M$ is just the region bounded by $\partial M$ - the closed, simple curve. Therefore for any $\omega$ such that $d\omega = dx\wedge dy$ you can find the area by integrating $\omega$ on the curve $\partial M$, here are some examples: $$Area = \int_M dx\wedge dy = \int_{\partial M}x dy = -\int_{\partial M}ydx = \frac{1}{2}\int_{\partial M}xdy - ydx$$
Let $D$ be the interior of the simple closed curve $\mathcal{C}$. Then we are after $$ A = \iint_D 1\ dxdy$$ We need to find some $f(x,y) = (f_1(x,y),f_2(x,y))$ such that $\frac{\partial f_2}{\partial x} - \frac{\partial f_1}{\partial y} = 1$. Observe that $f(x,y) = (0,x)$ does the trick. Then by Green's Theorem, \begin{align} A &= \iint_D 1\ dxdy\\ &= \iint_D \left(\frac{\partial f_2}{\partial x} - \frac{\partial f_1}{\partial y}\right)\ dxdy\\ &= \int_\mathcal{C} (f_1dx + f_2dy)\\ &= \int_\mathcal{C} x\ dy \end{align} And the other equality is got by defining a different $f(x,y)$ (I'll won't spoil the fun for you there).
EDIT: Let's illustrate this integral on the area of a cirlce of radius $r$. Let $\mathcal{C}$ be the curve parametrized by $\mathbf{r}(t) = (r\cos(t),r\sin(t)), 0 \le t < 2\pi$.
Then, \begin{align} A &= \int_\mathcal{C} x dy \\ &= \int_0^{2\pi} (r\cos(t))\frac{dy}{dt} dt\\ &= r^2 \int_0^{2\pi} \cos(t)\cos(t) dt\\ &= r^2 \int_0^{2\pi} \frac{1}{2}(1 + \cos(2t)) dt\\ &= \frac{1}{2}r^2 \left[t + \frac{1}{2}\sin(2t) \right|_0^{2\pi}\\ &= \pi r^2 \end{align} as expected!