Discussion
The Gaussian curvature of the elliptic cylinder
\[ \begin{aligned} &\{(x, y, z) \in \mathbf{R}^{3} : \\ &\frac{x^{2}}{2} + \frac{y^{2}}{3} = 1\} \end{aligned} \]
is :
Answer: C
The Gaussian curvature is always = 0.
The elliptic cylinder is generated by straight lines parallel to the z-axis through the ellipse \(x^{2}/2 + y^{2}/3 = 1\). Along each ruling the normal curvature is 0, so one principal curvature is 0 everywhere, and \(K = \kappa_{1}\kappa_{2} = 0\).
Equivalently, the cylinder can be unrolled onto a plane without stretching, so it is locally isometric to the plane and by Gauss's Theorema Egregium has the plane's curvature, zero. The other options describe curved surfaces such as ellipsoids (positive) or saddles (negative).
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