Explanation

(b)
Given equation of curve is
$y^{2}(2a - x) = x^{3}$
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Which is symmetrical about $x$-axis and passes through origin
Also, $\frac{x^{3}}{2a - x} < 0$
For $x > 2a$ or $x < 0$
So, curve does not lie in $x > 2a$ and $x < 0$, therefore curve lies wholly on $0 \leq x \leq 2a$
$\therefore$ Required area $= \int_{0}^{2a}{\frac{x^{3/2}}{\sqrt{2a - x}}dx}$
Put $x = 2\ a\sin^{2}\theta$
$\Rightarrow dx = 2a \bullet 2\sin{\theta + \cos{\theta d\theta}}$
$\therefore$ Required area $= \int_{0}^{\pi/2}{8a^{2}\sin^{4}{\theta d\theta}}$
$= 8a^{2}\left\lbrack \frac{3}{4} \bullet \frac{1}{2} \bullet \frac{\pi}{2} \right\rbrack$ (using gamma function)
$= \frac{3\pi a^{2}}{2}\ sq\ unit$