I am trying to solve following problem
Compute total Gaussian curvature of a surface $M : x^2+y^4+z^6=1$.
The textbook just says
\begin{align}
\int \int K dM = \int \int K(x) \sqrt{EG-F^2} dudv =\int \int \cos(v) dudv =
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{-\pi}^{\pi} \cos(v) dudv =4\pi
\end{align}
How this computation is possible? I have no idea of computation process.
My trial of parametrization is
\begin{align}
X(u,v) = (\sin(u)\cos(v), \sin^{\frac{1}{2}}(u)\sin^{\frac{1}{2}}(v), \cos^{\frac{1}{3}}(u))
\end{align}
then this is not good. Even this satisfies $x^2+y^4+z^6=1$, the range of $u,v$ is not the same with above textbook integral.
I tried to applied my previous posts obtaining metric or tangent vector in given implicit surface $F(x,y,z)=0$. but apparently, having trouble with tangent vectors due to $z^6$.
Best Answer
The textbook is cheating. The integral of the Gaussian curvature is the area of the image of the Gauss map on the unit sphere. Since this surface is convex, the image is precisely the unit sphere, covered once. This gives you $4\pi$ (and that's the parametric integral they wrote down). Alternatively, you can deduce this immediately from the Gauss-Bonnet Theorem.