Applying Green's Theorem

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.

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*}