Answer
area of $A$ $ \simeq 11.1633$
Work Step by Step
First, we find the intersection points of the circle $r=1$ and $r = 4\cos \theta $ by solving the equation: $1 = 4\cos \theta $. So, we have
$\cos \theta = \frac{1}{4}$
The solution is $\theta = \pm {\cos ^{ - 1}}\frac{1}{4}$.
From the figure we see that the region of $A$ is between two polar curves: $r=1$ and $r = 4\cos \theta $. So,
area of $A$ $ = \frac{1}{2}\cdot\mathop \smallint \limits_{ - {{\cos }^{ - 1}}\left( {1/4} \right)}^{{{\cos }^{ - 1}}\left( {1/4} \right)} \left( {{{\left( {4\cos \theta } \right)}^2} - 1} \right){\rm{d}}\theta $
area of $A$ $ = \frac{1}{2}\cdot\mathop \smallint \limits_{ - {{\cos }^{ - 1}}\left( {1/4} \right)}^{{{\cos }^{ - 1}}\left( {1/4} \right)} \left( {16{{\cos }^2}\theta - 1} \right){\rm{d}}\theta $
Since ${\cos ^2}\theta = \frac{1}{2}\left( {1 + \cos 2\theta } \right)$, the integral becomes
area of $A$ $ = \frac{1}{2}\cdot\mathop \smallint \limits_{ - {{\cos }^{ - 1}}\left( {1/4} \right)}^{{{\cos }^{ - 1}}\left( {1/4} \right)} \left( {8 + 8\cos 2\theta - 1} \right){\rm{d}}\theta $
area of $A$ $ = \frac{1}{2}\left( {7\theta + 4\sin 2\theta } \right)|_{ - {{\cos }^{ - 1}}\left( {1/4} \right)}^{{{\cos }^{ - 1}}\left( {1/4} \right)}$
area of $A$ $ = \frac{1}{2}\left( {14{{\cos }^{ - 1}}\frac{1}{4} + 8\sin \left( {2{{\cos }^{ - 1}}\frac{1}{4}} \right)} \right) \simeq 11.1633$
