Question

Area lying in the first quadrant and bounded by the circle $x^2+y^2=4$, the line $x=\sqrt{3}y$ and x-axis is.

• A. $\pi$ sq units
• B. $\frac{\pi }{2}$ sq units
• C. $\frac{\pi }{3}$ sq units
• D. None of these

Answer 1

The correct option is C $\frac{\pi }{3}$ sq units

Required area

$={\int }_0^1(x_2-x_1)dy$

$={\int }_0^1(\sqrt{4-y^2}-\sqrt{3}y)dy$

$={[\frac{1}{2}y\sqrt{4-y^2}+\frac{1}{2}(4)si{n}^{-1}\frac{y}{2}-\frac{\sqrt{3}{y}^2}{2}]}_0^1$

$=\frac{\sqrt{3}}{2}+si{n}^{-1}\left(\frac{1}{2}\right)-\frac{\sqrt{3}}{2}-2si{n}^{-1}0$

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

Hence, option C is correct.