The area (in sq. units) of the region bounded by the circle $x^2+y^2=64$, positive $x$-axis and the line $y=\sqrt{3} x$ is

• A

  $\frac{16 \pi}{3}$
• B

  $\frac{8 \pi}{3}$
• C

  $\frac{64 \pi}{3}$
• Correct answer
  D

  $\frac{32 \pi}{3}$

Solution

$x^2+y^2=64 \Rightarrow y=\sqrt{3} x$ $\begin{aligned} & y^2=3 x^2 \Rightarrow x= \pm 4 \\ & \therefore \quad y=\sqrt{48}\end{aligned}$
Area $=\int_0^4 y d x+\int_4^8 y_2 d x$ $\begin{aligned} & =\int_0^4(\sqrt{3} x) d x+\int_4^8 \sqrt{64-x^2} d x \\ & =\left[\sqrt{3} \frac{x^2}{2}\right]_0^4+\left[\frac{x}{2} \sqrt{64-x^2}+\frac{64}{2} \sin ^{-1}\left(\frac{x}{8}\right)\right]_4^8\end{aligned}$ $\begin{aligned} & =8 \sqrt{3}+\left[32 \frac{\pi}{2}-2 \sqrt{48}-32 \cdot \frac{\pi}{6}\right] \\ & =16 \pi-\frac{16 \pi}{3}=\frac{32 \pi}{3}\end{aligned}$

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