Ellipsoid gaussian curvature

For an ellipsoid with equation

$\left(\frac{x}{a}\right^2 + \left(\frac{y}{b}\right^2 + \left(\frac{z}{c}\right^2 = =1$

the Gaussian curvature at each point is given by

$K(x,y,z) = \frac{1}{a^2 b^2 c^2 \left(\frac{x^2}{a^4} + \frac{y^2}{b^4} + \frac{z^2}{c^4} \right )^2}$

Now suppose a ≥ b ≥ c > 0. Otherwise relabel the coordinate axes so that this is the case. Then the largest curvature occurs at (±a, 0, 0), and the smallest curvature occurs at (0, 0, ±c).

You could prove this using algebra by manipulating inequalities, or using calculus with Lagrange multipliers.

To see intuitively why this might be true, it helps to exaggerate the shape. First imagine that a is much larger than b or c. Then you have a cigar shape, the greatest curvature as at the two ends. And If you imagine c being much smaller than a and b, you have sort of a pancake shape which is flat on top and bottom.

The maximum curvature is (a/bc)² and the minimum curvature is (c/ab)².

Curvature of an ellipsoid

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