Solving double integrals: iterated integration vs. polar coordinates

I have to solve the following integral

$$ \int_\Omega \frac{xy}{x^2+y^2} dxdy, \;\; \Omega = \{ (x,y)\in \mathbb{R^2}: 1<x^2+y^2<4,x>0,y>0 \}. $$

I rewrited the domain to be simple with respect to the $x$-axis:

$$ \Omega^{'} = \{(x,y) \in \mathbb{R^2}: 1<x<2, \sqrt{1-x^2} < y < \sqrt{4-x^2} \}. $$

Solving now the integral

$$ \int_{1}^{2} \left( \int_{\sqrt{1-x^2}}^{\sqrt{4-x^2}} \frac{xy}{x^2+y^2}dy\right)dx, $$

the result is $\;\frac{3}{4}\mathrm{ln(4)}\;$, which is approximately $1.039$. This is also the result of the integral when solving it with WolframAlpha.

I also used polar coordinates to solve the integral, which became

$$ \int_{1}^{2} \rho \left(\int_{0}^{\frac{\pi}{2}} \mathrm{cos}\theta \; \mathrm{sin}\theta \;d\theta \right) d\rho = \frac{3}{4} = 0.75. $$

My question is: why do the two methods give two different results with such a big difference?

The polar coordinate integral is correct. The cartesian integral is just wrong. The region $\Omega'$ in fact makes no sense: When $x>1$, $\sqrt{1-x^2}$ is undefined. You have to split it up into two integrals, one with $0\le x\le 1$, and the other with $1\le x\le 2$. In the latter, the lower limit on $y$ is $0$. The moral of the story is: Draw pictures.

P.S. I have no idea what you did by hand or with Wolfram Alpha to get a real answer. Well, I have an idea ...