Question

The area enclosed by $y^2=8x$ and $y=\sqrt{2}x$ that lies outside the triangle formed by $y=\sqrt{2}x,x=1,y=2\sqrt{2}$, is cqual to :

• A. $\frac{16\sqrt{2}}{6}$
• B. $\frac{11\sqrt{2}}{6}$
• C. $\frac{13\sqrt{2}}{6}$
• D. $\frac{5\sqrt{2}}{6}$

Answer 1

Answer: (C)

Area of $\Delta ABC=\frac{1}{2}(\sqrt{2})\cdot 1=\frac{\sqrt{2}}{2}$
So required Area $=\underset{0}{\overset{4}{\int }}(\sqrt{8x}-\sqrt{2}x)dx-\frac{\sqrt{2}}{2}$
$=\frac{32\sqrt{2}}{3}-8\sqrt{2}-\frac{\sqrt{2}}{2}=\frac{13\sqrt{2}}{6}$