# Area of a simple closed curve

Let C be a simple closed curve in a region where Green's Theorem holds. Show that the area of the region is:

$$$$A=\int_{C}xdy=-\int_{C}ydx$$$$

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?

• Nop. What can be deduced from Green's Theorem is that the are is half that integral: $$A=\frac12\int_Cxdy-ydx$$ – DonAntonio Nov 23 '18 at 0:05
• I edited the question, my mistake – IchVerloren Nov 23 '18 at 0:11

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!
• @IchVerloren $\mathcal{C}$ is an arbitrary simple closed curve, so I've not assumed any particular simple closed curve here. – AlkaKadri Nov 23 '18 at 0:27
• I see it. It seems that you switched the differentials tho. It should be $\int_{C}(f_{2}dy+f_{1}dx)$ right?. For the other equality f(x,y)=(-y,0) works! – IchVerloren Nov 23 '18 at 0:47