Answer
The area of the region inside the circle $(x-1)^{2}+y^{2}=1 $, and outside the circle $(x-1)^{2}+y^{2}=1 $ is given by:
$$
\begin{aligned}
2 A(D) &=2 \iint_{D} d A\\
&=2 \int_{0}^{\pi / 3} \int_{1}^{2 \cos \theta} r d r d \theta \\
&=\frac{\pi}{3}+\frac{\sqrt {3}}{2}.
\end{aligned}
$$
Work Step by Step
The region is inside the circle $(x-1)^{2}+y^{2}=1 $, and outside the circle $(x-1)^{2}+y^{2}=1 $
In polar coordinates, the circle is
$$
\begin{aligned}
(x-1)^{2}+y^{2}&=1 \\
(x)^{2} -2x +1+y^{2}&=1 \\
(x)^{2} +y^{2}&=2x \\
& \quad\left[\begin{array}{c}{ \text {since } x^{2} +y^{2} =r^{2}, } \\ { x=r \cos \theta \text { then } }\end{array}\right] \\
& r^{2}= 2r \cos \theta \\
& r= 2 \cos \theta
\end{aligned}
$$
and the circle
$$
\begin{aligned}
x^{2}+y^{2}&=1 \\
& \quad\left[\begin{array}{c}{ \text {since } x^{2} +y^{2} =r^{2}, } \\ { \text { then } }\end{array}\right] \\
& r^{2}= 1\\
\end{aligned}
$$
The curves intersect in the first quadrant when
$$
r= 2 \cos \theta=1 \Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta= \frac{\pi}{3}
$$
The region in the first quadrant is given by
$$
D=\left\{(r, \theta) | 1 \leq r \leq 2 \cos \theta , \quad 0 \leq \theta \leq \frac{\pi}{3} \right\}
$$
Therefore, the area of the region inside the circle $r= 2 \cos \theta $, and outside the circle $ r^{2}= 1 $ is given by:
$$
\begin{aligned}
2 A(D) &=2 \iint_{D} d A\\
&=2 \int_{0}^{\pi / 3} \int_{1}^{2 \cos \theta} r d r d \theta \\
&=2 \int_{0}^{\pi / 3}\left[\frac{1}{2} r^{2}\right]_{r=1}^{r-2} d \theta \\
&=\int_{0}^{\pi / 3}\left(4 \cos ^{2} \theta-1\right) d \theta \\
&=\int_{0}^{\pi / 3}\left[4 \cdot \frac{1}{2}(1+\cos 2 \theta)-1\right] d \theta \\
&=\int_{0}^{\pi / 3}(1+2 \cos 2 \theta) d \theta \\
&=[\theta+\sin 2 \theta]_{0}^{\pi / 3} \\
&=\frac{\pi}{3}+\frac{\sqrt {3}}{2}.
\end{aligned}
$$
