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.

You can just exploit symmetries without having to do any explicit integral (or have to deal with polar coordinates) to find the answer. Here's an explaination.
Let $D:=\big\{(x,y) \in \mathbb{R}^2 \mid x^2+y^2\le4\big\}$ and $E:=\big\{(x,y) \in \mathbb{R}^2 \mid (x-1)^2+y^2\le1\big\}$. Since $D = \big\{(x,y) \in \mathbb{R}^2 \mid (-x)^2+y^2\le1\big\}$ and $\forall x,y\in \mathbb{R}^2, f(-x,y)=-f(x,y)$, it follows that
\begin{equation*}
\int_Df(x,y)\mathrm{d}x\mathrm{d}y=\int_Df(-x,y)\mathrm{d}x\mathrm{d}y = -\int_Df(x,y)\mathrm{d}x\mathrm{d}y
\end{equation*}
which implies $\int_Df(x,y)\mathrm{d}x\mathrm{d}y=0$.
On the other hand $E\subset D$. It follows that
\begin{equation*}
\int_{D\backslash E}f(x,y)\mathrm{d}x\mathrm{d}y=\int_Df(x,y)\mathrm{d}x\mathrm{d}y-\int_Ef(x,y)\mathrm{d}x\mathrm{d}y = -\int_Ef(x,y)\mathrm{d}x\mathrm{d}y.
\end{equation*}
Now, define $g \colon \mathbb{R}^2\to \mathbb{R}, (x,y) \mapsto x-1 $ and $F:= \big\{(x,y) \in \mathbb{R}^2 \mid x^2+y^2\le1\big\}$. Since $F = \big\{(x,y) \in \mathbb{R}^2 \mid (-x)^2+y^2\le1\big\}$, it follows that:
\begin{equation*}
\int_E g(x,y)\mathrm{d}x\mathrm{d}y= \int_F f(x,y)\mathrm{d}x\mathrm{d}y = \int_F f(-x,y)\mathrm{d}x\mathrm{d}y =-\int_F f(x,y)\mathrm{d}x\mathrm{d}y 
\end{equation*}
which implies $\int_E g(x,y)\mathrm{d}x\mathrm{d}y=0$.
It follows that
\begin{equation*}
\int_Ef(x,y)\mathrm{d}x\mathrm{d}y = \int_E g(x,y)\mathrm{d}x\mathrm{d}y + \int_E 1\mathrm{d}x\mathrm{d}y = 0+ |E|=\pi.
\end{equation*}
In conclusion
\begin{equation*}
\int_{D\backslash E}f(x,y)\mathrm{d}x\mathrm{d}y= -\int_Ef(x,y)\mathrm{d}x\mathrm{d}y = -\pi
\end{equation*}
