Question

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

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

$\Rightarrow 8x=2x^2$
$\Rightarrow x=0\&4$
$=\underset{1}{\overset{4}{\int }}2\sqrt{2}\sqrt{x}-\sqrt{2}xdx$
$=2\sqrt{2}{\left(\frac{x^{\frac{3}{2}}}{3/2}\right)}_1^4-\sqrt{2}{\left(\frac{x^2}{2}\right)}_1^4$
$=\frac{4\sqrt{2}}{3}(8-1)-\frac{\sqrt{2}}{3}(16-1)$
$=\frac{28\sqrt{2}}{3}-\frac{15\sqrt{2}}{2}=\frac{11\sqrt{2}}{6}$