The open set to check for integrability is $U = \{(x,y) \in \mathbb{R}^2 | 0<x^2+y^2<4\}$. I've considered a succesion of sets $U_n = \{(x,y) \in \mathbb{R}^2 | \frac{1}{n^2}<x^2+y^2<4\}$, that comply with $U_n \subset U_{n+1}$ and $U = \bigcup_{n=1}^{\infty}U_n$.
Now, from that, I can obtain that $\lim_{n\to \infty}\int -\ln{n}<\lim_{n \to \infty}\int\ln{\sqrt{x^2+y^2}}<\lim_{n \to \infty}\int \ln{2} \implies -\lim_{n \to \infty}\frac{1}{n}<\lim_{n\to \infty}\int\ln{\sqrt{x^2+y^2}}<\lim_{n\to \infty}n \ln{2} \implies0<\lim_{n\to \infty}\int\ln{\sqrt{x^2+y^2}}<\infty$.
Does this prove something about whether or not the function is integrable? How else can I use exhaustion to prove its integrability?
Thanks a lot!
Switch to polar coordinates. $\int_U ln(\sqrt{x^2+y^2})dxdy=2\pi\int\limits_0^2 ln(r) rdr$$=8\pi(\frac{ln(2)}{2}-\frac{1}{4})$.
To avoid integration, elementary calculus may be used to get bounds on the integrand. Maximum is at $r=2$ and minimum is at $r=e^{-1}$. Bounded continuous function over finite interval is integrable.