Explanation

(b)
Given, curves are $y^{2} = 8x\ \ \ \Longrightarrow y = \sqrt{8x}$
$and\ x^{2} = 8y\ \Longrightarrow y = \frac{x^{2}}{8}$
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The points of intersection of two curves are (0,0),(8,8)
$Now,\ required\ area = \int_{0}^{8}\left( \sqrt{8x} - \frac{x^{2}}{8} \right)dx$
$= \left\lbrack \frac{\sqrt{8}x^{3/2}}{3/2} - \frac{x^{3}}{8.3} \right\rbrack_{0}^{8}$
$= \frac{64}{3}sq\ units.$