Question

Area bounded between the curves $y=\sqrt{4-x^2}$ and $y^2=3|x|$ is/are

• A. $\frac{\pi -1}{\sqrt{3}}$
• B. $\frac{2\pi -1}{3\sqrt{3}}$
  
• C. $\frac{2\pi -\sqrt{3}}{3}$
  
• D. $\frac{2\pi -\sqrt{3}}{3\sqrt{3}}$

Answer 1

The correct option is C $\frac{2\pi -\sqrt{3}}{3}$


Required area = $2{\int }_0^1(\sqrt{4-x^2}-\sqrt{3x})dx$
$=2{(\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}si{n}^{-1}\left(\frac{x}{2}\right)-\frac{\sqrt{3}.2x^{\frac{3}{2}}}{3})}_0^1$
$=\frac{2\pi -\sqrt{3}}{3}$