Differential Geometry and Its Application - Exercise 3.2.12

Problem:

Show that the Gaussian curvature of the hyperboloid of one sheet $ x(u, v) = (a\cosh u \cos v, b \cosh u \sin v, c \sinh u) $ maybe written in Cartesian coordinate as

$ K = -\frac{1}{a^2b^2c^2[\frac{x^2}{a^4} + \frac{y^2}{b^4} + \frac{z^2}{c^4}]^2} $

Solution:

This code should verify the correctness of the identity. It would be too tedious for manual computation.

clear;
clc;
syms a;
syms b;
syms c;
syms x;
syms y;
z = sqrt(c^2 * (x^2/a^2 + y^2/b^2 - 1));
s = [x; y; z];
sx = diff(s, x);
sy = diff(s, y);
E = simplify(sx.' * sx);
F = simplify(sx.' * sy);
G = simplify(sy.' * sy);
normal = cross(sx, sy);
normal_norm = simplify(sqrt(normal.' * normal));
u = simplify(normal/normal_norm(1));
sxx = simplify(diff(sx, x));
sxy = simplify(diff(sx, y));
syy = simplify(diff(sy, y));
l = simplify(u.' * sxx);
m = simplify(u.' * sxy);
n = simplify(u.' * syy);
K = simplify((l * n - m * m)/(E * G - F * F))

PK = -1/(a^2 * b^2 * c^2 * (x^2 / a^4 + y^2 / b^4 + z^2/c^4)^2);

simplify(K - PK)

Now we get from the program that:
$ K = -\frac{c^2a^6b^6}{(c^2x^2b^4+c^2y^2a^4+a^2b^4x^2+a^4b^2y^2-a^4b^4)^2} $

PK represent the K value we need to prove, and the program output 0 meaning the K we found matches the PK we need to reach!