Explanation

(a)
The point intersections of given curves are (2,1) and $( - 2,1)$.
$\therefore Required\ area = 2\int_{0}^{1}{x\ dy + 2\int_{1}^{3}{x\ \ dy}}$
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$= 2\int_{0}^{1}\sqrt{4y}\ dy + 2\int_{1}^{3}\sqrt{6 - 2y}\ dy$
$= 4\left\lbrack \frac{y^{3/2}}{3/2} \right\rbrack_{0}^{1} + 2\left\lbrack \frac{(6 - 2y)^{3/2}}{3/2} \times \frac{1}{- 2} \right\rbrack_{1}^{3}$
$= \frac{8}{3} + \frac{16}{3} = 8\ sq\ units$