The area (in sq. units) of the region lying in the first quadrant and enclosed by the $X$-axis, the straight line $x-\sqrt{3} y=0$ and the circle $x^2+y^2=4$, is

• Correct answer
  A

  $\frac{\pi}{3}$
• B

  $\frac{2 \pi}{3}$
• C

  $\frac{\pi}{2 \sqrt{3}}$
• D

  $\frac{2 \pi}{3 \sqrt{2}}$

Solution

Equation of given circle
$\begin{aligned} & \Rightarrow x^2+y^2=2^2 \\ & \therefore \text { radius }=2\end{aligned}$
and line $x-\sqrt{3} y=0$
$$ \begin{aligned} (\sqrt{3} y)^2+y^2 & =4 \\ \Rightarrow \quad & 4 y^2=4 \Rightarrow y^2=1 \Rightarrow y= \pm 1 \end{aligned} $$ If $y=+1$ $$ \therefore \quad x=\sqrt{3} $$ If $y=-1$ $$ \therefore \quad x=-\sqrt{3} $$ As point $C$ is in first quadrant $$ \therefore C \text { is }(\sqrt{3}, 1) $$ Area of shaded region $$ \begin{aligned} & =\int_0^{\sqrt{3}} \frac{x}{\sqrt{3}} d x+\int_{\sqrt{3}}^2 \sqrt{4-x^2} d x \\ & =\frac{1}{\sqrt{3}} \int_0^{\sqrt{3}} x d x+\int_{\sqrt{3}}^2 \sqrt{2^2-x^2} d x \end{aligned} $$ $\begin{aligned} & =\frac{1}{\sqrt{3}}\left[\frac{x^2}{2}\right]_0^{\sqrt{3}}+\left\lfloor\frac{x}{2} \sqrt{2^2-x^2}+\frac{2^2}{2} \sin ^{-1} \frac{x}{2}\right]_{\sqrt{3}}^2 \\ & {\left[\because \int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}+c\right\rceil} \\ & =\frac{1}{\sqrt{3}} \cdot \frac{3}{2}+\left\lceil\left(0+2 \sin ^{-1} 1\right)-\left(\frac{\sqrt{3}}{2}\right.\right. \\ & \quad=\frac{\sqrt{3}-3}{2}+\left[2 \times \frac{\pi}{2}-\frac{\sqrt{3}}{2}-2 \frac{\pi}{3}\right\rceil \\ & \quad=\frac{\sqrt{3}}{2}+\pi-\frac{\sqrt{3}}{2}-\frac{2 \pi}{3}=\pi-\frac{2 \pi}{3}=\frac{\pi}{3} \\ & \therefore \text { Required area }=\frac{\pi}{3} \text { sq. units }\end{aligned}$

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