Explanation

(a)
Let $I = \int_{a}^{b}{\sqrt{(x - a)(b - x)}dx}$
Put $x = a\cos^{2}{\theta + bsin^{2}\theta}$
$\Rightarrow dx = - 2a\cos{\theta\sin{\theta + 2b\sin{\theta\cos\theta}}}$
$\Rightarrow dx = \sin 2\theta(b - a)$
$\therefore\ \ I = \int_{0}^{\pi/2}{\sqrt{(b - a)\sin^{2}{\theta(b - a)\cos^{2}\theta}}\ (b - a)}\sin{2\theta d\theta}$
$= \int_{0}^{\pi/2}{\frac{(b - a)^{2}}{2}\sin^{2}{2\theta d\theta}}$
$= \frac{(b - a)^{2}}{2}\int_{0}^{\pi/2}{\left\lbrack \frac{1 - \cos{4\theta}}{2} \right\rbrack d\theta}$
$= \frac{(b - a)^{2}}{4}{\left\lbrack \theta - \frac{\sin{4\theta}}{4} \right\rbrack_{0}}^{\pi/2}$
$= \frac{(b - a)^{2}}{4}\left\lbrack \frac{\pi}{2} - 0 \right\rbrack$
$= \frac{\pi(b - a)^{2}}{8}$