Question

The area of the smaller portion of the circle $x^2+y^2=4$ cut off by the line $x=1$ is

• A. $\frac{4\pi +3\sqrt{3}}{3}$
• B. $\frac{4\pi -3\sqrt{3}}{3}$
• C. $\frac{2\pi -3\sqrt{3}}{3}$
• D. $\frac{4\pi -\sqrt{3}}{3}$

Answer 1

The correct option is B $\frac{4\pi -3\sqrt{3}}{3}$

Required area $=2{\int }_1^2\sqrt{4-x^2}dx={[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-1}\frac{x}{2}]}_1^2$

$=[0+2{\sin }^{-1}\left(1\right)-(\frac{1}{2}\sqrt{3}+2{\sin }^{-1}\left(\frac{1}{2}\right))]$

$=4\left(\frac{\pi }{2}\right)-\sqrt{3}-4\left(\frac{\pi }{6}\right)=\frac{4\pi -3\sqrt{3}}{3}.$