Question

Consider the line $x=\sqrt{3}y$ and the circle $x^2+y^2=4$.

What is the area of the region in the first quadrant enclosed by the x-axis, the line $x=\sqrt{3}y$ and the circle?

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{3}-\frac{\sqrt{3}}{2}$
• D. None of the above

Answer 1

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

$A(\frac{\sqrt{3}}{2},\frac{1}{2});D(0,2);C(2,0)$

Required area $=$ Area enclosed by line $BA$ from $O$ to $A+$ Area enclosed by circle from $A$ to $C$

$⟹A={\int }_0^{\sqrt{3}}\frac{x}{\sqrt{3}}dx+{\int }_{\sqrt{3}}^2\sqrt{4-x^2}⟹A={\left|\frac{x^2}{2\sqrt{3}}\right|}_0^{\sqrt{3}}+{|\frac{x}{2}\sqrt{4-x^2}+\frac{2^2}{2}{\sin }^{-1}(\frac{x}{2}\left)\right|}_{\sqrt{3}}^2⟹A=\frac{\sqrt{3}}{2}+\pi -\frac{\sqrt{3}}{2}-2{\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)⟹A=\pi -\frac{2\pi }{3}=\frac{\pi }{3}$