Answer
$area = \frac{9}{2}$
Work Step by Step
When $\theta=0$, we have $r=3$. When $\theta=3$, we have $r=0$. This corresponds to the shaded region in Figure 15. So, the interval of $\theta$-values is $0 \le \theta \le 3$.
Using Eq. (2) of Theorem 1, the area of the region is
$area = \frac{1}{2}\cdot\mathop \smallint \limits_0^3 {\left( {3 - \theta } \right)^2}{\rm{d}}\theta = \frac{1}{2}\cdot\mathop \smallint \limits_0^3 \left( {9 - 6\theta + {\theta ^2}} \right){\rm{d}}\theta $
$area = \frac{1}{2}\left( {9\theta - 3{\theta ^2} + \frac{1}{3}{\theta ^3}} \right)|_0^3 = \frac{9}{2}$
