If I would want to integrate the following
$$\int_1^2 \int_1^2\frac{x}{x^2+y^2} \, dx \, dy$$
by doing a polar corrdinates change of variables $x=r\cos\theta$ and $y=r\sin\theta$ I would get
$$\int_\text{?}^\text{?}\int_\text{?}^\text{?}\frac{r\cos\theta}{r^2} r \, dr \, d\theta $$
but what would be my bounderies then? I have trouble finding those bounderies to then integrate.
thank you for your help
On
To use polar coordinates for integrating over a square far from the origin is a bad idea.
By Fubini's theorem
$$ \int_{1}^{2}\int_{1}^{2}\frac{x}{x^2+y^2}\,dx\,dy = \frac{1}{2}\int_{1}^{2}\log\left(\frac{4+y^2}{1+y^2}\right)\,dy = \int_{1}^{2}\left(\arctan\frac{2}{x}-\arctan\frac{1}{x}\right)\,dx$$
and you may pick the one-variable integral you like the most and solve it by integration by parts.
For instance
$$ \int_{1}^{2}\arctan\frac{1}{x}\,dx = \frac{\pi}{2}-\int_{1}^{2}\arctan(x)\,dx = \frac{\pi}{2}-\left[x\arctan x\right]_{1}^{2}+\int_{1}^{2}\frac{x}{1+x^2}\,dx $$
gives
$$ \int_{1}^{2}\arctan\frac{1}{x}\,dx = \frac{\pi}{2}-[x\arctan x]_{1}^{2}+\frac{1}{2}[\log(1+x^2)]_{1}^{2}.$$
At last we have:
$$\boxed{ \iint_{[1,2]^2}\frac{x}{x^2+y^2}\,dx\,dy = \color{red}{\arctan\frac{1}{3}+\log\frac{8\sqrt{2}}{5\sqrt{5}}}}$$
that is very close to one third.
Per comment, polar coordinates are not a good way, but if you really would like to see as you mentioned:
Your area is a square and you need to divide it into two triangles.
For the lower triangle:
$$\theta \in (\arctan \frac{1}{2},\frac{\pi}{4})$$
$r$ is bounded by bottom and right line of the square: $$\frac{1}{\sin \theta} <r < \frac{2}{\cos \theta}$$
Similar for the upper triangle, and I'll leave it to you.