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So I'm trying to solve this problem stated like this:

Using Green's Theorem, find the area of the elipse defined by (where $a,b \gt 0$):

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} \leqq 1$$

I'm having trouble doing this,

$$\int_{FrD} \!F \cdot T = {\int\int}_D \left(\frac{df_2}{dx} - \frac{df_1}{dy}\right)$$

Where $F$ is a vector field. My try at solving this was parametize the elipse as polar coords and doing $F$ as $F(x,y) = (1,1)$. Similiar to how to get the area. But I'm pretty sure that's not the correct way.

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This is a standard application, a way to use Green's Theorem to compute areas by doing line integrals.

Let $D$ be the ellipse, and $C$ its boundary $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$. The area you are trying to compute is $$\int\!\!\int_D 1\,dA.$$ According to Green's Theorem, if you write $1 = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$, then this integral equals $$\oint_C(P\,dx + Q\,dy).$$ There are many possibilities for $P$ and $Q$. Pick one. Then use the parametrization of the ellipse \begin{align*} x&=a\cos t\\ y&=b\sin t \end{align*} to compute the line integral.

As you can probably see, the idea of finding $P$ and $Q$ with $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1$ can be used to compute the area of any region enclosed by a simple closed curve. Of course, the line integral may be more complicated than the area computation, but that's another kettle of fish.

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Let $A$ be the area of the region $D$ bounded by the ellipse with equation $$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}=1$$

Let $\partial{D}$ denote the boundary. You can parametrize $\partial{D}$ with counterclockwise orientation, by $$\varphi(t) = (a\cos{t},b\sin{t})$$ Then you have

\begin{align*} A &=\frac{1}{2} \int\limits_{\partial{D}} xdy - ydx \\ &= \frac{1}{2} \int\limits_{0}^{2\pi} (−b \sin(t), a \cos(t)) \cdot (−a \sin(t), b \cos(t) dt \\ &= \frac{1}{2} \int\limits_{0}^{2\pi} (ab \sin^{2}{t} + ab \cos^{2}{t}) \ dt\\ &= \pi \cdot ab \end{align*}

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  • $\begingroup$ Did you understand why this person put that fraction in the integral @Griffinheart ? $\endgroup$ – Neisy Sofía Vadori Jul 18 '18 at 14:04
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It suffices to take $Q =0$ and $P =-y$ then $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=1$ and by Green formula we get, $$ \iint_{\frac{x^2}{a^2}+\frac{y^2}{b^2}\le 1} 1dxdy =\iint_{\frac{x^2}{a^2}+\frac{y^2}{b^2}\le 1}\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}dxdy= -\oint_{\frac{x^2}{a^2}+\frac{y^2}{b^2}= 1} ydx =\\\color{blue}{- \int_0^{2\pi} b\sin td(a\cos t)}= ab\int_0^{2\pi} \sin^2 t dt= ab\int_0^{2\pi} \frac{1-\cos (2t)}{2}dt= \color{red}{ ab\pi}$$ Where we used, $$x= a\cos t~~~and ~~y =b\sin t$$

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