Answer
$Area = \pi~ab$
Work Step by Step
$x = a~cos~\theta$
$dx = -a~sin~\theta~d\theta$
$y = b~sin~\theta$
To find the total area, we can find the area of one quadrant when $x$ goes from $0$ to $a$, and then multiply this area by 4.
$Area = 4\times \int_{0}^{a}y~dx$
When $x = 0,$ then $\theta = \frac{\pi}{2}$
When $x = a,$ then $\theta = 0$
We can substitute to set up a new integral:
$Area = 4\times \int_{0}^{a}y~dx = 4\times \int_{\pi/2}^{0}(b~sin~\theta)(-a~sin~\theta~d\theta)$
$Area = 4\times \int_{\pi/2}^{0}-ab~sin^2~\theta~d\theta$
$Area = 4\times \int_{\pi/2}^{0}-\frac{ab}{2}~(1-cos~2\theta)~d\theta$
$Area = 4\times \int_{\pi/2}^{0}\frac{ab}{2}~(cos~2\theta-1)~d\theta$
$Area = 4\times (\frac{ab}{2})~(\frac{sin~2\theta}{2}-\theta)\Big\vert_{\pi/2}^{0}$
$Area = (2ab)~(\frac{sin~2\theta}{2}-\theta)\Big\vert_{\pi/2}^{0}$
$Area = (2ab)~[(0-0)-(0-\frac{\pi}{2})]$
$Area = (2ab)~(\frac{\pi}{2})$
$Area = \pi~ab$
