# How to calculate $\int_{\Sigma_r}KdA$ of a given surface (Gauss-Bonnet)

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^20,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.

• Gauss-Bonnet theorem ? Nov 15 '21 at 21:05
• @MathiasRousset Yes, I'm supposed to use it but I'm not sure how to do it Nov 15 '21 at 21:27
• Start by writing down the statement of the theorem and figuring out what every single symbol means. Nov 15 '21 at 21:48
• What is the meaning of $N$ here? This is a rather non-standard notation. It is neither the principal normal of the curve nor the surface normal. Nov 16 '21 at 16:54

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.