Answer
The area of the region in the first quadrant that lies between the circles $x^{2}+y^{2}=4 $, and $x^{2}+y^{2}=2x $ is given by:
$$
\begin{aligned}
\iint_{D} x d A &=\iint_{x^{2}+y^{2} \leq 4} x d A-\iint_{y \geq 0} x d A \\
& x \geq 0, y \geq 0 \\
&=\int_{0}^{\pi / 2} \int_{0}^{2} r^{2} \cos \theta d r d \theta-\int_{0}^{\pi / 2} \int_{0}^{2 \cos \theta} r^{2} \cos \theta d r d \theta \\
&=\frac{16-3 \pi}{6}.
\end{aligned}
$$
Work Step by Step
The region is in the first quadrant and lies between the circles $x^{2}+y^{2}=4 $, and $x^{2}+y^{2}=2x$.
In polar coordinates, the region is given by:
$$
\begin{aligned}
(x)^{2} +y^{2}&=2x \\
& \quad\left[\begin{array}{c}{ \text {since } x^{2} +y^{2} =r^{2}, } \\ { x=r \cos \theta \text { then } }\end{array}\right] \\
& r^{2}= 2r \cos \theta \\
& r= 2 \cos \theta
\end{aligned}
$$
the region in the first quadrant is given by:
$$
D=\left\{(r, \theta) | 0 \leq r \leq 2 \cos \theta , \quad 0 \leq \theta \leq \frac{\pi}{2} \right\}
$$
and the circle
$$
\begin{aligned}
x^{2}+y^{2}&=4 \\
& \quad\left[\begin{array}{c}{ \text {since } x^{2} +y^{2} =r^{2}, } \\ { \text { then } }\end{array}\right] \\
& r^{2}= 4\\
\end{aligned}
$$
the region in the first quadrant is given by
$$
D=\left\{(r, \theta) | 0 \leq r \leq 2 , \quad 0 \leq \theta \leq \frac{\pi}{2} \right\}
$$
Therefore, the area of the region inside the circle $r^{2}= 2 $, and outside the circle $ r= 2 \cos \theta $, in the first quadrant, is given by:
$$
\begin{aligned}
\iint_{D} x d A &=\iint_{x^{2}+y^{2} \leq 4} x d A-\iint_{y \geq 0} x d A \\
& x \geq 0, y \geq 0 \\
&=\int_{0}^{\pi / 2} \int_{0}^{2} r^{2} \cos \theta d r d \theta-\int_{0}^{\pi / 2} \int_{0}^{2 \cos \theta} r^{2} \cos \theta d r d \theta \\
&=\int_{0}^{\pi / 2} \frac{1}{3}(8 \cos \theta) d \theta-\int_{0}^{\pi / 2} \frac{1}{3}\left(8 \cos ^{4} \theta\right) d \theta \\
&=\frac{8}{3}-\frac{8}{12}\left[\cos ^{3} \theta \sin \theta+\frac{3}{2}(\theta+\sin \theta \cos \theta)\right]_{0}^{\pi / 2} \\
&=\frac{8}{3}-\frac{2}{3}\left[0+\frac{3}{2}\left(\frac{\pi}{2}\right)\right]\\
&=\frac{16-3 \pi}{6}.
\end{aligned}
$$
