Surface of revolution

Example

Let $s \mapsto (r(s), z(s))$, for $s$ in an open interval $I$, be a smooth regular curve in the $xz$-plane. Assume that it is parametrized by arc length and that $r(s) > 0$ for all $s$. Rotating it about the $z$-axis gives the surface of revolution $$x(s, \theta) = (r(s) \cos\theta, \; r(s) \sin\theta, \; z(s)), \qquad s \in I, \ \theta \in \R.$$ The partial derivatives are $x_s = (r' \cos\theta, r' \sin\theta, z')$ and $x_\theta = (-r \sin\theta, r \cos\theta, 0)$. They are orthogonal, $\abs{x_s} = 1$ because $r'^2 + z'^2 = 1$, and $\abs{x_\theta} = r > 0$. So $x$ is an immersion, and if the profile curve is a homeomorphism onto its image then the image of $x$ is a regular surface. The first fundamental form is $ds^2 + r(s)^2 \, d\theta^2$. The meridians ($\theta$ constant) and the parallels ($s$ constant) are lines of curvature, and the Gaussian curvature is $$K = -\frac{r''(s)}{r(s)}.$$