<s>[INST] $\iint_A\frac{y}{x^2+y^2}dxdy, \quad A: [\space {(x,y): \space x^2+y^2\ge 4, \quad x^2+y^2\le2 \sqrt{x^2+y^2}+2x}]$ I've been given a "fun" integral and I just want to know if I'm doing it good or I missed something.
$$\iint_A\frac{y}{x^2+y^2}dxdy, \quad A: [\space {(x,y): \space x^2+y^2\ge 4, \quad x^2+y^2\le2 \sqrt{x^2+y^2}+2x}]$$
And this is my solution:
$$
        \begin{matrix}
        x=r \cos \varphi \\
        y=r \sin \varphi \\
        \end{matrix}
$$
$$r\ge2, \quad  \begin{matrix}
        r^2 \le 2r+2r \cos \varphi \\
        r \le 2(1+ \cos \varphi) \\
        \end{matrix}$$
$$x=2 \Rightarrow r \cos \varphi=2 \Rightarrow r=\frac{2}{\cos \varphi}$$
Now I evaluate the integral:
$$
\begin{split}
I
 &= \iint_A \frac{y}{x^2+y^2} dxdy \\
 &= 2\int_0^{\pi/2} d\varphi
       \left(\int_2^{2(1+\cos \varphi)}\frac{\sin \varphi}{r}dr
            +\int_{\frac{2}{\cos \varphi}}^{2(1+\cos \varphi)}
                  \frac{\sin \varphi}{r}dr\right)\\
 &= 2\int_0^{\pi/2}
          \sin \varphi
          \left(\ln|2+2\cos \varphi|
                - \ln2
                + \ln|2+2 \cos|2+2 \cos \varphi|
                - \ln \left| \frac{2}{\cos \varphi} \right|
          \right) d \varphi\\
 &= 2\int_0^{\pi/2}
          \sin \varphi
          (2 \ln|1+\cos \varphi|+\ln|\cos \varphi|)d\varphi\\
 &\text{substitute } \cos \varphi =t \\
 &= -2\int_1^0(2\ln(1+t)+\ln(t))dt=8\ln2-2
\end{split}
$$
Of course, I skipped many trivial calculation steps.
Can anyone check it or maybe comment if I'm doing it wrong?

 [/INST] The integrand is odd function respect to $y$ and the area is symmetric about $x$-axis, so the integral is $0$. On the other hand, you can write
$$\iint_A\frac{y}{x^2+y^2}dxdy=\int_{-\pi/2}^{\pi/2}\int_2^{2(1+\cos\theta)}\dfrac{r\sin\theta}{r^2}r\,dr\,d\theta=0$$
</s>