Question

The area of the smaller part bounded by the semi-circle $y=\sqrt{4-x^2},y=x\sqrt{3}$ and x-axis is

• A. $\frac{\pi }{3}$
• B. $\frac{2\pi }{3}$
• C. $\frac{4\pi }{3}$
• D. none of these

Answer 1

The correct option is B $\frac{2\pi }{3}$

The required area $={\int }_0^1\sqrt{3}xdx+{\int }_1^2\sqrt{4-x^2}dx$

$=\frac{\sqrt{3}}{2}{\left[x^2\right]}_0^1+{[\frac{x}{2}\sqrt{4-x^2}+\frac{2^2}{2}{\sin }^{-1}\frac{x}{2}]}_1^2$

$=\frac{\sqrt{3}}{2}+2.\frac{\pi }{2}-\frac{\sqrt{3}}{2}-2.{\sin }^{-1}\frac{1}{2}=\pi -\frac{\pi }{3}=\frac{2\pi }{3}$