Answer
Area of the shaded region $ \simeq 8.338$
Work Step by Step
In Exercise 15 for the limaçon $r = 2\cos \theta - 1$ we obtain the area of the inner loop $A$:
$A = \pi - \frac{3}{2}\sqrt 3 \simeq 0.54$.
In Exercise 15 also we see that the entire region of the limaçon corresponds to the interval $0 \le \theta \le 2\pi $.
Next, we compute the entire area of the limaçon:
$area = \frac{1}{2}\cdot\mathop \smallint \limits_0^{2\pi } {\left( {2\cos \theta - 1} \right)^2}{\rm{d}}\theta $
$area = \frac{1}{2}\cdot\mathop \smallint \limits_0^{2\pi } \left( {4{{\cos }^2}\theta - 4\cos \theta + 1} \right){\rm{d}}\theta $
Since ${\cos ^2}\theta = \frac{1}{2}\left( {1 + \cos 2\theta } \right)$, the integral becomes
$area = \frac{1}{2}\cdot\mathop \smallint \limits_0^{2\pi } \left( {2\left( {1 + \cos 2\theta } \right) - 4\cos \theta + 1} \right){\rm{d}}\theta $
$area = \frac{1}{2}\left( {3\theta + \sin 2\theta - 4\sin \theta } \right)|_0^{2\pi } = \frac{1}{2}\left( {6\pi } \right) = 3\pi $
However, the area of the entire limaçon has covered an extra area of the inner loop $A$ in the integration for the interval $0 \le \theta \le 2\pi $. So to find the shaded area of Figure 21, we have to subtract twice the area $A$. Thus,
area of the shaded region $ = 3\pi - 2\cdot\left( {\pi - \frac{3}{2}\sqrt 3 } \right) \simeq 8.338$
