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?

• Correct answer
  A

  $\frac{\pi}{3}$
• B

  $\frac{\pi}{6}$
• C

  $\frac{\pi}{3}-\frac{\sqrt{3}}{2}$
• D

  None of these

Solution

$\begin{aligned} & x^2+y^2=4 \text { and } x=\sqrt{3} y \\ & P(\sqrt{3}, 1)=\text { First quadrant }\end{aligned}$
The point of intersection of the line and the circle in the first quadrant is $(\sqrt{3}, 1)$. Area of $\triangle \mathrm{OPA}=$ Area of $\triangle \mathrm{OPB}+$ Area of PAB $=\frac{1}{2} \times \mathrm{OB} \times \mathrm{PB}=\frac{1}{2} \times \sqrt{3} \times 1=\frac{\sqrt{3}}{2}$ $\begin{aligned} & \text { Area of PAB } \int_{\sqrt{3}}^2 \mathrm{ydx} \\ & =\int_{\sqrt{3}}^2 \sqrt{4-x^2} d x=\left[\frac{x}{2} \sqrt{4-x^2}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{\sqrt{3}}^2 \\ & {\left[2 \times \frac{\pi}{2}-\frac{\sqrt{3}}{2}[\sqrt{4-3}]-2 \sin ^{-1} \frac{\sqrt{3}}{2}\right]} \\ & =\pi-\frac{\sqrt{3}}{2}-\frac{2 \pi}{3}=\frac{\pi}{3}-\frac{\sqrt{3}}{2} \end{aligned}$ Area enclosed by $x$-axis, the line $=\sqrt{3} . y$, and the circle $x^2+y^2=4$ in the first quadrant $=\frac{\sqrt{3}}{2}+\frac{\pi}{3}-\frac{\sqrt{3}}{2}=\frac{\pi}{3}$

Hence, the correct answer is option: 'A'.