Question

The area of the region between the curve $x^2+y^2=4$ and $x=0,x=1$ is

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

Answer 1

The correct option is A $\sqrt{3}+\frac{2\pi }{3}$

$A=A_1+A_1$

$=2A_1$

$=2{\int }_0^1(\sqrt{4-x^2}-0)dx$

$={2[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-1}\frac{x}{2}]}_0^1$

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

$=2\left[\frac{\sqrt{3}}{2}+2\times \frac{\pi }{6}\right]$

$=\sqrt{3}+2\frac{\pi }{3}$