Question

The area enclosed by $y ^{2}=8 x$ 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 $=\int\limits_{0}^{4}(\sqrt{8 x}-\sqrt{2} x) d x-\frac{\sqrt{2}}{2}$
$=\frac{32 \sqrt{2}}{3}-8 \sqrt{2}-\frac{\sqrt{2}}{2}=\frac{13 \sqrt{2}}{6}$