Question

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

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

Answer 1

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

The shaded is the smaller part
$\stackrel{1}{\underset{0}{\int }}\sqrt{3}xdx+\stackrel{2}{\underset{1}{\int }}\sqrt{4-x^2}dx$
$\frac{\sqrt{3}{x}^2}{2}{|{}_0^1+(x/2\sqrt{4-x^2}+4/2{\sin }^{-1}(x/2\left)\right)|}_1^2$
$\frac{\sqrt{3}}{2}+2{\sin }^{-1}\left(1\right)-\frac{\sqrt{3}}{2}-2{\sin }^{-1}\left(1/2\right)$
$2(\pi /2-\pi /6)$
$2\left(\frac{3-1}{6}\right)\pi =\frac{2\pi }{3}squnits.$