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 y-axis, the line $x=\sqrt{3}$ and the circle?

• A. $\frac{2\pi }{3}+\frac{\sqrt{3}}{2}$
• B. $\frac{\pi }{2}-\frac{\sqrt{3}}{2}$
• C. $\frac{\pi }{3}-\frac{1}{2}$
• D. None of the above

Answer 1

The correct option is A $\frac{2\pi }{3}+\frac{\sqrt{3}}{2}$

Equation of line $BC$ is $x=\sqrt{3}$

$B(\sqrt{3},1);C(\sqrt{3},-1)$

The equation of circle in first quadrant is $y=\sqrt{4-x^2}$

Area enclosed in first quadrant, $A={\int }_0^{\sqrt{3}}\sqrt{4-x^2}dx{=}_0^{\sqrt{3}}(\frac{x}{2}\sqrt{4-x^2}+\frac{2^2}{2}{sin}^{-1}(\frac{x}{2}\left)\right)=\frac{\sqrt{3}}{2}\left(1\right)+2\left(\frac{\pi }{3}\right)-0=\frac{\sqrt{3}}{2}+\frac{2\pi }{3}$