The topologically invariant Euler characteristic of a 2-surface is given by $\chi=\frac{1}{4\pi}\int\sqrt{g}\mathcal{R}$ (where $\mathcal{R}$ is the scalar curvature) and is equivalent to $\chi=2-2g - b$ for a surface with $g$ handles and $b$ boundaries.
For a sphere this is simple, $\sqrt{g}=R^2$ and $\mathcal{R}=\frac{2}{R^2}$, where $R$ is the radius. Therefore, $\chi=2$, as suggested by the second formula above. If we wanted to add a boundary, we could take the usual metric \begin{equation} \mathrm{d}s^2=\mathrm{d}\theta^2 + \sin^2{\theta} \mathrm{d}\phi^2 \end{equation} and limit $\theta$ to $0\leq\theta\leq\theta_0$, with $\theta_0<\pi$. We have added a hole in the sphere, thus adding a boundary, so the Euler charateristic should be 1. By the first formula, however, we get $\chi=(1-\cos^{-1}{\theta_0})$.
Moreover, this should be topologically equivalent to a flat disc, on which we may take a flat metric, so $\mathcal{R}=0$ and $\chi=0$.
Where have I gone wrong?
Many thanks in advance.