The area of region \( R=\{(x,y):x^{2}+y^{2}\le 16,\ y\le\sqrt{3}x,\ x\ge 0,\ y\ge 0\} \) is: (in Sq. units)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{8\pi}{3} \)
The region is inside the circle of radius 4, above the x-axis and below the line \( y=\sqrt{3}x \). This line makes an angle θ with the x-axis where \( \tan\theta=\sqrt{3} \), so \( \theta=\frac{\pi}{3} \).
So R is the sector of the circle between the angles 0 and \( \frac{\pi}{3} \):
\( \displaystyle \text{Area}=\frac{1}{2}r^{2}\theta \)
\( \displaystyle =\frac{1}{2}\times 16\times\frac{\pi}{3} \)
\( \displaystyle =\frac{8\pi}{3} \)
By integration, the line meets the circle at \( (2,2\sqrt{3}) \), so
\( \displaystyle \text{Area}=\int_{0}^{2}\sqrt{3}x\,dx+\int_{2}^{4}\sqrt{16-x^{2}}\,dx \)
\( \displaystyle =2\sqrt{3}+\left(\frac{8\pi}{3}-2\sqrt{3}\right) \)
\( \displaystyle =\frac{8\pi}{3} \)