Integrating the function $f(x,y)=x$ over the area inside the disc $x^2+y^2\le 4$ and outside the disc $(x-1)^2+y^2=1.$ 
Integrate the function $f(x,y)=x$ over the area inside the disc $x^2+y^2\le 4$ and outside the disc $(x-1)^2+y^2=1.$

My attempt:
I chose shifted polar coordinates $$\begin{cases}x=2+r\cos\varphi\\ y=r\sin\varphi.\end{cases}$$ Then $$\begin{aligned}x^2+y^2&=4\\\iff (2+r\cos\varphi)^2+(r\sin\varphi)^2&=4\\\iff r(2\cos\varphi+r)&=0\end{aligned}$$ and $$\begin{aligned}(x-1)^2+y^2&=1\\\iff (1+r\cos\varphi)^2+(r\sin\varphi)^2&=1\\\iff r(4\cos\varphi)&=0\end{aligned}$$
So, I'm integrating over the set $$\left\{(\varphi,r),\frac\pi2\le\varphi\le\frac{3\pi}2, -2\cos\varphi\le r\le -4\cos\varphi\right\}.$$
The integral becomes $$\begin{aligned}\int_{\pi/2}^{3\pi/2}\int_{-2\cos\varphi}^{-4\cos\varphi}r(2+r\cos\varphi)drd\varphi&=\int_{\pi/2}^{3\pi/2}\left(r^2+\frac{r^3}3\cos\varphi\Big|_{-2\cos\varphi}^{-4\cos\varphi}\right)d\varphi\\&=\int_{\pi/2}^{3\pi/2}12\cos^2\varphi-\frac{56}3\cos^4\varphi d\varphi\\&=\int_{\pi/2}^{3\pi/2}\frac{56}3\cos^2\varphi(1-\cos^2\varphi)d\varphi-\int_{\pi/2}^{3\pi/2}\frac{20}3\cos^2\varphi d\varphi\\&=\frac{14}3\int_{\pi/2}^{3\pi/2}(2\sin\varphi\cos\varphi)^2-\frac{10}3\int_{\pi/2}^{3\pi/2}(1+\cos(2\varphi))d\varphi\\&=\frac{14}3\int_{\pi/2}^{3\pi/2}\frac{1-\cos(4\varphi)}2d\varphi-\frac{10}3\left(\varphi+\frac{\sin(2\varphi)}2\Big|_{\pi/2}^{3\pi/2}\right)\\&=\frac{14}6\left(\varphi-\frac{\sin(4\varphi)}4\Big|_{\pi/2}^{3\pi/2}\right)-\frac{10}3\pi\\&=\frac{14}6\pi-\frac{10}3\pi\\&=\frac{-6}6\pi\\&=-\pi\end{aligned}$$
UPDATE:
I've just found my mistake, instead of dividing by $3,$ I divided by $2$ after the first integration in the $r$ variable.

In polar coordinates, the inner circle is $r=2\cos \theta$ and the integral is
$$\int_{-\pi/2}^{\pi/2}\int_0^{2\cos\theta}r\cos\theta \> rdr d\theta=\frac83 \int_{-\pi/2}^{\pi/2}\cos^4\theta\> d\theta= \pi
$$
Per symmetry, the integration over the outer circle vanishes. Thus, the overall result is $-\pi$.
