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 $x=r\cos\theta$
$$I=\int_{0}^{\pi/2}{\int_{2\cos\theta}^{2}{r^2\cos\theta}\,dr\,d\theta}$$
Don't try to do this sort of thing by "pure algebra" - always draw the region of integration. If you do this you will see easily that $\theta$ varies from $\pi/4$ to $\pi/2$. So we have
$$I=\int_{\pi/4}^{\pi/2}\int_?^? x^2\,J\,dr\,d\theta$$
where $J$ is the Jacobian. To find the limits for $r$, draw a line on your diagram starting at the origin and heading in the $\theta$ direction (where $\theta$ is between $\pi/4$ and $\pi/2$). You can see that the values of $r$ which are in your region and on this line go from a minimum of $0$ to a maximum on the horizontal line $y=1$. To find the $r$ value on this line we have
$$r\sin\theta=y=1$$
and so $r_{\rm max}=1/\sin\theta$. Hence
$$I=\int_{\pi/4}^{\pi/2}\int_0^{1/\sin\theta} x^2\,J\,dr\,d\theta\ .$$
You should also know that the Jacobian for polar coordinates is $r$ and that $x=r\cos\theta$. Hence
$$I=\int_{\pi/4}^{\pi/2}\int_0^{1/\sin\theta} (r\cos\theta)^2\,r\,dr\,d\theta\ .$$
Best Answer
Sorry, don't know how to insert diagrams in my post so will have to ask you to draw your own.
Draw your circle (the boundary of your region). It has diameter from the origin to $A=(a,0)$. The $\theta$ values in this region go from $-\pi/2$ to $\pi/2$, so that's the first part of your answer. To find the $r$ values, draw a line from the origin making an angle $\theta$ with the $x$-axis, where $-\pi/2<\theta<\pi/2$. Label the intersection of the line and circle $P$, and consider the segment $OP$. The minimum value of $r$ on the segment is obviously $0$. The maximum value of $r$ on the segment depends on the value of $\theta$; you need to find $r_{\rm max}$ in terms of $\theta$.
Hint: consider the triangle $\triangle OPA$.