Question

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

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

Answer 1

Answer: (C)

$\frac{\pi }{3}$

The line and the curve meet at $P(\sqrt{3},1)$ in the $1st$ quadrant.

Draw perpendicular $PM$.

$\therefore A=ar\left(△OPM\right)+{\int }_{\sqrt{3}}^{2}ydx$

Now $x=2\cos \theta ,y=2\sin \theta$

and adjust the limits

$=\frac{1}{2}\sqrt{3}.1+{\int }_{\pi /\theta }^0\left(2\sin \theta \right)(-2\sin \theta )d\theta$

$=\frac{\sqrt{3}}{2}+4{\int }_0^{\pi /6}\frac{(1-\cos 2\theta )}{2}d\theta$

$=\frac{\sqrt{3}}{2}+2{[\theta -\frac{\sin 2\theta }{2}]}_0^{\pi /6}=\frac{\sqrt{3}}{2}+2[\frac{\pi }{6}-\frac{1}{2}\frac{\sqrt{3}}{2}]=\frac{\pi }{3}$