📂Geometry

The Rotational Surface with Zero Curvature

Theorem¹

Let $M$ be the unit speed curve of the rotational surface $\boldsymbol{\alpha}$ and let the Gaussian curvature be $K=0$. Then, $M$ satisfies one of the following conditions.

• It is a part of a cylinder.
• It is a part of a plane.
• It is a part of a cone.

Moreover, these surfaces are locally isometric.

Proof

Let the Gaussian curvature of the rotational surface be $K = 0$. Since the curvature of the rotational surface is $K = -\dfrac{r^{\prime \prime}}{r}$, then $r^{\prime \prime} = 0$. Therefore,

$$r^{\prime \prime}(s) = 0 \implies r^{\prime}(s) = a \implies r(s) = as + b$$

Since $z^{\prime} = \pm\sqrt{1 - (r^{\prime})^{2}}$,

$$z^{\prime} = \pm \sqrt{1 - (r^{\prime})^{2}} = \pm \sqrt{1 - a^{2}}$$

$$\implies z = \pm \sqrt{1-a^{2}}s + d = cs + d$$

Thus, the curve $\boldsymbol{\alpha} = (r, z)$ is as follows.

$$\boldsymbol{\alpha}(s) =\big(r(s), z(s)) = (as + b, cs + d),\quad a,b,c,d \in \mathbb{R}$$

If $a=0$, then $\alpha (s) = (b, cs + d)$ and the resulting rotational surface is part of a cylinder.

If $c=0$, then $\alpha (s) = (as + b, d)$ is a plane, and the resulting rotational surface is part of a plane.

If $a\ne 0, c\ne 0$, then $\alpha (s) = (as + b, cs + d)$ and this line is parallel to neither the $r$ axis nor the $z$ axis. The resulting rotational surface is part of a cone.

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