Answer
$area = 4\pi $
Work Step by Step
From Figure 4, we see that for the entire circle the limits of integration is 0 and $\pi$.
So, using Eq. (2) of Theorem 1, the area is
$area = \frac{1}{2}\cdot\mathop \smallint \limits_0^\pi {\left( {4\sin \theta } \right)^2}{\rm{d}}\theta = 8\cdot\mathop \smallint \limits_0^\pi {\sin ^2}\theta {\rm{d}}\theta $
Since ${\sin ^2}\theta = \frac{1}{2}\left( {1 - \cos 2\theta } \right)$, the integral becomes
$area = 8\cdot\mathop \smallint \limits_0^\pi \frac{1}{2}\left( {1 - \cos 2\theta } \right){\rm{d}}\theta $
$area = \left( {4\theta - 2\sin 2\theta } \right)|_0^\pi = 4\pi $
