Answer
$\iint_D xdA=\frac{8}{3}-\frac{3\pi}{16}$
Work Step by Step
$x^2+y^2=2x$ is a circle center at $(1,0)$ and radius 1 since it can be rewritten as $(x-1)^2+y^2=1$
Meanwhile, $x^2+y^2=4$ is a circle center at the origin and radius 2.
It describes that the circle $x^2+y^2=4$ is outside the circle $x^2+y^2=2x$.
Convert $x^2+y^2=2x$ in polar coordinates:
$r^2=2r\cos \theta$
$r=\cos\theta$
Then, the region in the first quadrant that lies both circles above is written as the set of points $D=\{(r,\theta)|0\leq \theta \leq\frac{\pi}{2},\cos\theta\leq r\leq 2\}$.
Evaluate the given integral over $D$:
$\iint_D xdA=\int_0^{\pi/2}\int_{\cos\theta}^2r\cos \theta\cdot r dr d\theta$
$=\int_0^{\pi/2}\frac{r^3\cos\theta}{3}]_{\cos\theta}^2d\theta$
$=\int_0^{\pi/2}\frac{8\cos\theta}{3}-\frac{\cos^4\theta}{3}d\theta$
$=\frac{8\sin\theta}{3}-\frac{12\theta+8\sin(2\theta)+\sin(4\theta)}{32}]_0^{\pi/2}$
$=(\frac{8}{3}-\frac{6\pi+0+0}{32})-(0-0)$
$=\frac{8}{3}-\frac{3\pi}{16}$
