Answer
Area enclosed by the cardioid:
$area = \frac{3}{2}\pi {a^2}$
Work Step by Step
First, we find the limits of integration. When $\theta=0$, we have $r = 2a$, so the cardioid starts at $\left( {2a,0} \right)$. When $\theta=\pi$, we have $r=0$. So, the rectangular coordinates at $\theta=\pi$ is $\left( {0,0} \right)$. This corresponds to the upper-half of the cardioid. Thus, by symmetry, the interval of $\theta$-values of the cardioid is $0 \le \theta \le 2\pi $.
Using Eq. (2) of Theorem 1 of Section 12.4, the area of the cardioid is
$area = \frac{1}{2}\cdot\mathop \smallint \limits_0^{2\pi } {a^2}{\left( {1 + \cos \theta } \right)^2}{\rm{d}}\theta $
$area = \frac{{{a^2}}}{2}\cdot\mathop \smallint \limits_0^{2\pi } \left( {1 + 2\cos \theta + {{\cos }^2}\theta } \right){\rm{d}}\theta $
Since ${\cos ^2}\theta = \frac{1}{2}\left( {1 + \cos 2\theta } \right)$, the integral becomes
$area = \frac{{{a^2}}}{2}\cdot\mathop \smallint \limits_0^{2\pi } \left( {1 + 2\cos \theta + \frac{1}{2}\left( {1 + \cos 2\theta } \right)} \right){\rm{d}}\theta $
$area = \frac{{{a^2}}}{2}\left( {\frac{3}{2}\theta + 2\sin \theta + \frac{1}{4}\sin 2\theta } \right)|_0^{2\pi }$
$area = \frac{{{a^2}}}{2}\left( {3\pi } \right) = \frac{3}{2}\pi {a^2}$
