Use Green’s Theorem to find the area of the region enclosed by the ellipse $$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1\qquad a>0\quad b>0$$

Solution We express the ellipse using the parametric equations $x=a\;\cos t$, $y=b\;\sin t$, $0\leq t\leq 2\pi$. Then $dx=-a\;\sin t\,dt$ and $dy=b\;\cos t\,dt$. Now we use (1) to find the area. $$\begin{eqnarray*} A& =&\dfrac{1}{2}\oint_{C}(-y\,dx+x\,dy)=\dfrac{1}{2}\int_{0}^{2\pi }[-b\sin t(-a\sin t\,dt)+a\cos t(b\cos t\,dt)] \\ & =&\dfrac{1}{2}\int_{0}^{2\pi }ab(\sin ^{2}t+\cos ^{2}t)\,dt=\dfrac{1}{2}(ab)\int_{0}^{2\pi }dt=\dfrac{1}{2}ab(2\pi )=\pi ab \end{eqnarray*}$$