Let $S \subset \mathbb{R}^3$ be the surface given by $x^4+y^4+z^4=1$, and let $K$ be its gaussian curvature. Then what is $\int_S K$?
First of all, I think finding patch seems hard. Should I consider 8 patches given by $(x,y,z)=(\pm \sqrt{\sin u \cos v},\pm \sqrt{\sin u \sin v},\pm \sqrt{cosu})$ where $0<u,v<\pi/2$? But then how should I deal with boudaries of the patches(for example, $u=0, v=0$)?
Best Answer
I guess you are expected to apply the Gauss-Bonnet theorem rather than calculate directly.
But if you really want to integrate $K$, the boundaries of patches can be ignored. The surface is smooth, hence there is no singular component to its curvature. That is, $K$ is a continuous function, and as such integrates to zero over any set of measure zero.
If the surface was $|x|+|y|+|z|=1$ instead, then you would have to worry about the boundaries between triangular faces: in fact, all of the curvature is concentrated there. (This is an octahedron, obviously non-smooth).