I'm trying to verify the Gauss-Bonnet theorem with a triangle on a hyperboloid that can be parametrized by two ways:
$$\varphi(u,v) = (\cosh(u)\cos(v), \cosh(u)\sin(v),\sinh(u))$$
$$\psi(u,v)=(\cos v,\sin v,0)+\sinh(u)(-\sin v, \cos v,1).$$
Now the sides of the triangle are given by the curves:
- $\alpha(u) = \varphi(u,0)$ with $u\in [0,u_1]$
- $\beta(u) = \psi(u,0)$ with $u\in [0,u_1]$
- $\gamma(v)=\varphi(u_1,v)$ with $v\in [0,v_1]$
Now, the tricky part is the integral of the Gaussian curvature $K$. We have to integrate
$$\iint_{R}K(u,v)\sqrt{EG-F}dudv$$
where $R$ is the region in the $uv$ plane which the parametrization maps to the interior of the triangle.
The curves $\alpha$ and $\gamma$ are coordinate curves of the $\varphi$ parametrization, so they map to lines parallel to the axis on the $u-v$ plane. This is the easy part.
The hard part is being $\beta$. I mean, I know that we can invert $\varphi$ as
$$\varphi^{-1}(x,y,z)=(\operatorname{arcsinh}(z),\arctan(y/x))$$
thus we have that since $\beta(u)=(1,\sinh u,\sinh u)$, we obtain
$$\varphi^{-1}(\beta(u))=(u,\arctan(\sinh u))$$
this is the $\beta$ curve on the $uv$ plane.
Now I just don't know how to set up the integral. Furthermore, it all seems wrong.
So, how is the right way to describe this region in the $uv$ plane and how to set up the integral inside the region?