Mathematics

Problem

For

0 \le \theta \le \dfrac{\pi}{3}

, a region

R

is bounded by the curve

r = 6\sin\theta

and the origin. What is the exact area of

R

? a)

3\pi + \dfrac{9\sqrt{3}}{4}

3\pi - \dfrac{9\sqrt{3}}{4}

6\pi - \dfrac{9\sqrt{3}}{2}

3\pi - \dfrac{9\sqrt{3}}{2}