## Calculus (3rd Edition)

$area = \frac{9}{4}\left( {\frac{\pi }{3} + \frac{{\sqrt 3 }}{2}} \right) \simeq 4.305$
Using Eq. (2) of Theorem 1 of Section 12.4, the area is $area = \frac{1}{2}\cdot\mathop \smallint \limits_{\pi /3}^{2\pi /3} 9{\sin ^2}\theta {\rm{d}}\theta$ Since ${\sin ^2}\theta = \frac{1}{2}\left( {1 - \cos 2\theta } \right)$, the integral becomes $area = \frac{9}{4}\cdot\mathop \smallint \limits_{\pi /3}^{2\pi /3} \left( {1 - \cos 2\theta } \right){\rm{d}}\theta$ $area = \frac{9}{4}\left( {\theta - \frac{1}{2}\sin 2\theta } \right)|_{\pi /3}^{2\pi /3}$ $area = \frac{9}{4}\left( {\frac{\pi }{3} - \frac{1}{2}\sin \frac{{4\pi }}{3} + \frac{1}{2}\sin \frac{{2\pi }}{3}} \right)$ $area = \frac{9}{4}\left( {\frac{\pi }{3} + \frac{{\sqrt 3 }}{2}} \right) \simeq 4.305$