I need to compute $\int_0^1\int_{x^2}^x\frac{1}{\sqrt{x^2+y^2}}dydx$.
Switching to polar coordinates, I get $0\leq\theta\leq\pi/4$ (from 0 to 1). Then, for $x^2=y$, $r=\tan(\theta)/cos(\theta)$ since $x=rcos(\theta)$ and $y=rsin(\theta)$. However, I can't figure out how to get the upper extreme of the integral for $r$, since $y=x \iff rcos(\theta)=rsin(\theta)$ and $r$ just cancel. I'm not even sure if the lower extreme of the integral is right.