The area bounded by the curve  $a^2{x}^2=y^3(2a-y)$ in square units is

The given curve is  $a^2{x}^2=y^3(2a-y)\mathrm{.....}\left(1\right)$
It is symmetrical about y-axis and it cuts the y-axis at the points (0, 0) and 
(0, 2a). The curve does not exist for  $y>2a$ and  $y<0$.
The required area  $=2\times \text{area\hspace{0.17em}}OBA$
$=2{\int }_0^{2a}xdy$
$=2{\int }_0^{2a}\frac{y^{3/2}\sqrt{(2a-y)}}{a}\text{dy,from}\left(1\right).$
$Puttingy=2a{\sin }^2\theta ,weget,$
$=\frac{2}{a}{\int }_0^{\pi /2}{\left(2a\right)}^{3/2}{\sin }^3\theta .\sqrt{\left(2a\right).\cos \theta .4a\sin \theta \cos \theta d\theta }$
$=32a^2.\frac{\mathrm{3.1.1}}{\mathrm{6.4.2}}.\frac{\pi }{2}$ by Wallis formula
Thus, the required area  $=\pi a^2$.