For circle one, the equation is simply
$$C_1: r=2$$
For circle two $x^2+y^2=r^2,\quad x=r\cos\theta$, so
$$C_2: r^2=2r\cos\theta \implies r(r-2\cos\theta)=0 \implies r=0,r=2\cos\theta$$
Of these only the second solution is useful (the first just says $C_2$ passes through the origin), so the inner and outer radii for the integral are $2\cos\theta$ and $2$, respectively. And $\theta$ varies from $0$ to $\pi/2$ in the first quadrant.
Since $dA=r\,dr\,d\theta$, we get
$$A=\int_{0}^{\pi/2}{\int_{2\cos\theta}^{2}{r}\,dr\,d\theta}$$
(The original integral for area should have been $A=\iint{dA}$.)
If $I=\iint{x\,dA}$ is required, then from $x=r\cos\theta$
$$I=\int_{0}^{\pi/2}{\int_{2\cos\theta}^{2}{r^2\cos\theta}\,dr\,d\theta}$$
Observe that such integral equals $\pi R^{2}$ (as you have suggested).
That is because the integral of $\sin$ or $\cos$ over a period equals zero.
Consequently, there is no need to evaluate the terms $r^{2}\cos(\theta)$ and $r^{2}\sin(\theta)$.
Hence it remains to calculate the integral:
\begin{align*}
I = \int_{0}^{2\pi}\mathrm{d}\theta\int_{0}^{R}r\mathrm{d}r
\end{align*}
which leads to the desired result.
Best Answer
Let $D_1$ be the region described by $x^2+y^2 \le 4,\, x\ge 0,\, y \ge 0$ and $D_2$ the region $(x-1)^2+y^2 \le 1, \, y \ge 0$. \begin{gather*} \iint_{D}x\, dA = \iint_{D_1}x\, dA -\iint_{D_2}x\, dA =\\[2ex] \int_{0}^{2}\left(\int_{0}^{\sqrt{4-y^2}}x\, dx\right)\, dy - \int_{0}^{1}\left(\int_{1-\sqrt{1-y^2}}^{1+\sqrt{1-y^2}}x\, dx\right)\, dy = \\[2ex] \dfrac{1}{2}\int_{0}^{2}(4-y^2)\, dy - \int_{0}^{1}2\sqrt{1-y^2}\, dy = \dfrac{8}{3}-\dfrac{\pi}{2}. \end{gather*} where the last integral can be interpreted as the area of a half unit circle.