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?

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

  • $\begingroup$ Isn't that a particular case? Shouldn't the equality hold for every simple closed curve? $\endgroup$ – IchVerloren Nov 23 '18 at 0:24
  • $\begingroup$ @IchVerloren $\mathcal{C}$ is an arbitrary simple closed curve, so I've not assumed any particular simple closed curve here. $\endgroup$ – AlkaKadri Nov 23 '18 at 0:27
  • $\begingroup$ 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! $\endgroup$ – IchVerloren Nov 23 '18 at 0:47
  • $\begingroup$ @IchVerloren Yes you're absolutely right! I've fixed it now and added an example to show that this works :) $\endgroup$ – AlkaKadri Nov 23 '18 at 1:03

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