Question

The area of the region
\left\\{(x,y) : y² ≤ 8x, y ≥ \sqrt2x, x ≥ 1 \right\\}
is

• A. $\frac{13\sqrt2}{6}$
• B. $\frac{11\sqrt2}{6}$
• C. $\frac{5\sqrt2}{6}$
• D. $\frac{19\sqrt2}{6}$

Answer 1

Answer: (B)

$\frac{11\sqrt2}{6}$

Answer 2

The correct answer is (B) : $\frac{11\sqrt2}{6}$

Fig. Area Between Two Curves

Required area
$=\int_{1}^{4} (\sqrt{8x} - \sqrt{2x}) \,dx$
$=\frac{2\sqrt{8}}{3}x^{\frac{3}{2}} - \frac{x^2}{\sqrt{2}} \Bigg|_{1}^{4}$
$= \frac{16\sqrt3}{3} - \frac{16}{\sqrt2} - \frac{2\sqrt8}{3} + \frac{1}{\sqrt2}$
$= \frac{11 \sqrt2}{6}$ $sq.units$