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?
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 $$\iint_A\frac{y}{x^2+y^2}dxdy=\int_{-\pi/2}^{\pi/2}\int_2^{2(1+\cos\theta)}\dfrac{r\sin\theta}{r^2}r\,dr\,d\theta=0$$