Explanation

(b)
The curve $y^{2}(2a - x) = x^{3}$ is symmetrical about $x$-axis and passes through origin.
$Also,\ \ \frac{x^{3}}{2a - x} < 0\ for\ x > 2a\ and\ x < 0$
So, curve does not lie in $x > 2a\$ and $x < 0$, therefore curves lies wholly on $0 \leq x \leq 2a$
$\therefore Requried\ area = \int_{0}^{2a}{\frac{x^{3/2}}{\sqrt{2a - x}}dx}$
$Put\ \ \ \ x = {2a\sin}^{2}\theta$
$\Longrightarrow \ \ 0\ dx = 4a\sin{\theta\cos{\theta\ d\ \theta}}$
$\therefore\ Requried\ area = \int_{0}^{\frac{\pi}{2}}{{8a}^{2}\sin^{4}\theta\ d\ \theta}$
$= {8a}^{2}\left\lbrack \frac{3}{4} \bullet \frac{1}{2} \bullet \frac{\pi}{2} \right\rbrack$
$= \frac{3{\pi a}^{2}}{2}\ sq\ unit$