Let $S$ a surface of revolution with length elements $ds^2 = d \rho^2 + \frac{1}{sin^2 \theta} dt^2$, with $\rho = log(tan(\frac{\theta}{2})$.
Why is it true that any Euclidean circular sector $\Omega = \left\{ (x,y) : y \geq |x|, r_0 \leq \sqrt{x^2 + y^2} \leq r_0 e^{2 \pi} \right\}$ with $r_0 > 0$ is isometric to $S$?