Helicoid

Example

Fix $c > 0$. For $u \in \R$ and $v \in \R$, let $$x(u, v) = (v \cos u, \; v \sin u, \; c u).$$ Its image, the helicoid, is swept out by a horizontal line that turns about the $z$-axis at a constant rate while it moves up. It is a minimal surface.

Check. The partial derivatives are $x_u = (-v \sin u, v \cos u, c)$ and $x_v = (\cos u, \sin u, 0)$. So $E = v^2 + c^2$, $F = 0$ and $G = 1$. The cross product is $x_u \times x_v = (-c \sin u, c \cos u, -v)$, with length $W = \sqrt{v^2 + c^2}$, and $N = (x_u \times x_v)/W$. The second derivatives are $x_{uu} = (-v \cos u, -v \sin u, 0)$, $x_{uv} = (-\sin u, \cos u, 0)$ and $x_{vv} = 0$. So the coefficients of the second fundamental form are $e = 0$, $f = c/W$ and $g = 0$. The mean curvature is $H = \dfrac{eG - 2fF + gE}{2(EG - F^2)} = 0$.