2. Use polar coordinates to evaluate the double integral,
\[
\iint_{D} x d A
\]
where $D$ is the region in the first quadrant that lies between the circles $x^{2}+y^{2}=4$ and $x^{2}+y^{2}=2 x$.

Step 1 ：Given the double integral \(\iint_{D} x d A\), where D is the region in the first quadrant that lies between the circles \(x^{2}+y^{2}=4\) and \(x^{2}+y^{2}=2 x\).

Step 2 ：We can rewrite these equations in polar coordinates as \(r^{2}=4\) and \(r^{2}=2 r \cos \theta\), respectively. Solving these equations for r, we get \(r=2\) and \(r=2 \cos \theta\).

Step 3 ：Since the region D is in the first quadrant, \(\theta\) ranges from 0 to \(\pi/2\). The inner radius is given by \(r=2 \cos \theta\) and the outer radius is given by \(r=2\).

Step 4 ：The double integral in polar coordinates is given by \(\iint_{D} x d A = \int_{0}^{\pi/2} \int_{2 \cos \theta}^{2} r \cos \theta \cdot r dr d\theta\), where \(x = r \cos \theta\) and \(dA = r dr d\theta\).

Step 5 ：Calculate the inner integral, we get \(-8\cos(\theta)^{4}/3 + 8\cos(\theta)/3\).

Step 6 ：Calculate the outer integral, we get \(8/3 - \pi/2\).

Step 7 ：Final Answer: The value of the double integral is \(\boxed{\frac{8}{3} - \frac{\pi}{2}}\).