I'm working on the following exercise:
For $r\in \mathbb{R}^{+}$ let the surface $\Sigma_{r}$ be given by \begin{align} \Sigma_r = \left \{ (x,y,z)\in \mathbb{R}^{3}\colon z=\cos \sqrt{x^2+y^2}, x^2+y^2<r^2, x>0,y>0 \right \} \end{align} Determine the value of the integral \begin{align} \int_{\Sigma_r}KdA, \end{align} where $K$ is the Gaussian curvature of $\Sigma_r$.
Perhaps I have to use the fact that: \begin{align} \int \kappa_g(s)ds = 2\pi - \int_{\Sigma_r}KdA \end{align}
Where $\kappa(s)= \left \langle \dot{T}(s),N(s) \right \rangle $. Or maybe \begin{align} \int_{\Sigma_r}KdA = 2\pi \cdot \chi(\Sigma_r) \end{align}
I'm trying to find this expressions although I can't find the parameterization yet. ¿Is this surface homeomorphic to an easier one? Should I try finding $K$? How can I apply the theorem if I really don't know if the surface is compact? I am guided by this question: Exercise in differential geometry using Gauss-Bonnet
I am a bit lost and would appreciate your help very much.
The parameterization is already given in $\Sigma_r,$ it's a graph of the function $z(x,y).$ While let $x=R\cos(\theta),$ $y=R\sin(\theta),$ $z=\cos(R).$ So you can directly compute Gaussian curvature like the answer linked in your question(unaccpeted one).
Or the other way is to use Gauss-Bonnet Thm, which is the accpeted answer in the linked question. As surface $\Sigma_r$ is a graph of the function $z(x,y)$ on a quarter disk with radius $r,$ it's homeomorphic to the quarter disk, and of course disk. So $\chi(\Sigma_r)=1.$ Actually your $\Sigma_r$ is really not compact, but the boundary has no contribution to the integral $\int K dA$. So consider $\overline{\Sigma}_r$ instead, which is a compact one.
If you decide to use Gauss-Bonnet Thm, use symmetry first like the accpeted answer in your link.