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.

  • $\begingroup$ Gauss-Bonnet theorem ? $\endgroup$ Nov 15 '21 at 21:05
  • $\begingroup$ @MathiasRousset Yes, I'm supposed to use it but I'm not sure how to do it $\endgroup$
    – didiegop
    Nov 15 '21 at 21:27
  • 1
    $\begingroup$ Start by writing down the statement of the theorem and figuring out what every single symbol means. $\endgroup$ Nov 15 '21 at 21:48
  • $\begingroup$ 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. $\endgroup$ 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.

If you decide to use Gauss-Bonnet Thm, use symmetry first like the accpeted answer in your link.


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