Question

The area of the region $S=\left\{(x, y): y^{2} \leq 8 x, y \geq \sqrt{2} x, x \geq 1\right\}$ is

• A. $\frac{13 \sqrt{2}}{6}$
• B. $\frac{11 \sqrt{2}}{6}$
• C. $\frac{5 \sqrt{2}}{6}$
• D. $\frac{19 \sqrt{2}}{6}$

Answer 1

Answer: (B)

$y^{2}=8 x$ ...(1)
$y=\sqrt{2} x$ ...(2)
$y^{2}=2 x^{2}$

$\Rightarrow 8 x=2 x^{2}$
$\Rightarrow x =0\, \&\, 4$
$=\int\limits_{1}^{4} 2 \sqrt{2} \sqrt{x}-\sqrt{2} x d x$
$=2 \sqrt{2}\left(\frac{x^{\frac{3}{2}}}{3 / 2}\right)_{1}^{4}-\sqrt{2}\left(\frac{x^{2}}{2}\right)_{1}^{4}$
$=\frac{4 \sqrt{2}}{3}(8-1)-\frac{\sqrt{2}}{3}(16-1)$
$=\frac{28 \sqrt{2}}{3}-\frac{15 \sqrt{2}}{2}=\frac{11 \sqrt{2}}{6}$