Gaussian curvature of a z=f(x,y) function This is not a homework question.
Just wondering if there is a general formula for the gaussian curvature at point (x,y,f(x,y)) in terms of x, y, and f(x,y).
I didn't see any thing like that on the wikipedia article, but maybe it's hidden behind all these letters/symbols which are not actually described.
By the way, I'm studying computer science. For us, a general formula is one that works most of the time. When it doesn't work, we call this an exception, and it requires special handling :). So here of course, the function is assumed everywhere indefinitely continuous and differentiable and bounded inside any box that would fit the RAM - whatever that means - and also in R3, inside a standard classical euclidean space...
Thank you.
 A: There are a few explicit formulas at http://mathworld.wolfram.com/GaussianCurvature.html.
A: If the surface is in Monge form, i.e. written as the graph of a function $z = f(x,y)$ such that $f(0,0) = f_x(0,0) = f_y(0,0) = 0,$ then there is an easy way to calculate the Gauss curvature. You take a rotation about the $z$-axis which kills off the $xy$-term in the quadratic part of $f.$ Then you will be able to write the surface as $z  = \frac{1}{2}(\kappa_1x^2 + \kappa_2y^2) + \cdots,$ where the dots denote terms of order three and above. A rotation is a special orthogonal transformation and does not change the Gauss curvature.
The $\kappa_1$ and $\kappa_2$ will be in terms to the coefficients in the Taylor series of the original expression for $f.$
The numbers $\kappa_1$ and $\kappa_2$ are the principal curvatures of the surface at $(0,0,0).$ Moreover, the $x$-axis and the $y$-axis are the principal directions of the surface at $(0,0,0).$ (Recall that there are always two orthogonal principal directions at each non-umbilic point, i.e. $\kappa_1 \neq \kappa_2,$ of the surface.)The Gauss curvature is given by $K = \kappa_1\kappa_2$ and the mean curvature is given by $H = \frac{1}{2}(\kappa_1 + \kappa_2).$
