Question

The area lying in the first quadrant inside the circle $x^2+y^2=12$ and bounded by the parabolas $y^2=4x,x^2=4y$ is:

• A. $2(\frac{\sqrt{2}}{3}+\frac{3}{2}{\sin }^{-1}\frac{1}{3})$
• B. $4(\frac{\sqrt{2}}{3}+\frac{3}{2}{\sin }^{-1}\frac{1}{3})$
• C. $(\frac{\sqrt{2}}{3}+\frac{3}{2}{\sin }^{-1}\frac{1}{3})$
• D. none of these

Answer 1

The correct option is B $4(\frac{\sqrt{2}}{3}+\frac{3}{2}{\sin }^{-1}\frac{1}{3})$

The required area

$={\int }_0^{2}2\sqrt{x}dx+{\int }_2^{2\sqrt{2}}\sqrt{12-x^2}dx-{\int }_0^{2\sqrt{2}}\frac{x^2}{4}dx$

$={\left[\frac{4x^{3/2}}{3}\right]}_0^2+{[\frac{x}{2}\sqrt{12-x^2}+\frac{12}{2}{\sin }^{-1}\frac{x}{2\sqrt{3}}]}_2^{2\sqrt{3}}-\frac{1}{4}{\left[\frac{x^3}{3}\right]}_0^{2\sqrt{2}}$

$=4(\frac{\sqrt{2}}{3}+\frac{3}{2}{\sin }^{-1}\frac{1}{3})$