Answer
$A=\pi ab$
Work Step by Step
The ellipse is symmetric about the axes, so we find the area in the first quadrant and multiply it by 4.
$A=4\displaystyle \int_{0}^{a}ydx=\quad \left[\begin{array}{ll}
\frac{dx}{d\theta}=-a\sin\theta & \\
x=0\Rightarrow\theta=\frac{\pi}{2}, & x=a\Rightarrow\theta=0
\end{array}\right]$
$=4\displaystyle \int_{\pi/2}^{0}b\sin\theta(-a\sin\theta)d\theta$
$=4ab\displaystyle \int_{\pi/2}^{0}-\sin^{2}\theta d\theta$
$=4ab\displaystyle \int_{0}^{\pi/2}\sin^{2}\theta d\theta\qquad...\left[\begin{array}{l}
\cos 2\theta=1-2\sin^{2}\theta\\
\sin^{2}\theta=(1-\cos 2\theta)/2
\end{array}\right]$
$=4ab\displaystyle \int_{0}^{\pi/2}\frac{1}{2}(1-\cos 2\theta)d\theta$
$=2a[\displaystyle \int_{0}^{\pi/2}d\theta-\int_{0}^{\pi/2}\cos 2\theta d\theta]$
$=2ab[\displaystyle \theta-\frac{1}{2}\sin 2\theta]_{0}^{\pi/2}$
$=2ab[(\displaystyle \frac{\pi}{2}-0)-(0-\frac{1}{2}(0)]$
$=2ab(\displaystyle \frac{\pi}{2})$
$=\pi ab$
