Answer
Area of the inner loop $A \simeq 0.54$
Work Step by Step
Since $r = 2\cos \theta - 1$, the rectangular coordinates of a point on the curve corresponding to $\theta$ is given by
$\left( {x,y} \right) = \left( {r\cos \theta ,r\sin \theta } \right)$,
$\left( {x,y} \right) = \left( {\left( {2\cos \theta - 1} \right)\cos \theta ,\left( {2\cos \theta - 1} \right)\sin \theta } \right)$.
For the interval $0 \le \theta \le 2\pi $, we compute several points in rectangular coordinates corresponding to $\theta = 0,\frac{\pi }{4},\frac{\pi }{3},\frac{\pi }{2},\pi ,\frac{{5\pi }}{4},\frac{{3\pi }}{2},2\pi $ and list them on the following table:
$\begin{array}{*{20}{c}}
\theta \\
0\\
{\frac{\pi }{4}}\\
{\frac{\pi }{3}}\\
{\frac{\pi }{2}}\\
\pi \\
{\frac{{5\pi }}{4}}\\
{\frac{{3\pi }}{2}}\\
{2\pi }
\end{array}\begin{array}{*{20}{c}}
{\left( {x,y} \right)}\\
{\left( {1,0} \right)}\\
{\left( {0.293,0.293} \right)}\\
{\left( {0,0} \right)}\\
{\left( {0, - 1} \right)}\\
{\left( {3,0} \right)}\\
{\left( {1.707,1.707} \right)}\\
{\left( {0,1} \right)}\\
{\left( {1,0} \right)}
\end{array}$
Then we plot the points in rectangular coordinates and sketch the curve by connecting these points. From this figure we see that the upper-half of the inner loop corresponds to the interval $0 \le \theta \le \frac{\pi }{3}$. Also, the entire region of the limaçon corresponds to the interval $0 \le \theta \le 2\pi $. By symmetry, the area of the inner loop is twice the area of the upper-half. Let $A$ denote the area of the inner loop. Using Eq. (2) of Theorem 1, we have
$A = 2\cdot\frac{1}{2}\cdot\mathop \smallint \limits_0^{\pi /3} {\left( {2\cos \theta - 1} \right)^2}{\rm{d}}\theta $
$A = \mathop \smallint \limits_0^{\pi /3} \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
$A = \mathop \smallint \limits_0^{\pi /3} \left( {2\left( {1 + \cos 2\theta } \right) - 4\cos \theta + 1} \right){\rm{d}}\theta $
$A = \left( {3\theta + \sin 2\theta - 4\sin \theta } \right)|_0^{\pi /3}$
$A = \pi + \frac{1}{2}\sqrt 3 - 2\sqrt 3 = \pi - \frac{3}{2}\sqrt 3 \simeq 0.54$
