Parametrization of surface of revolution

Question

A surface of revolution is obtained by picking a curve $C$ on a plane and rotating it around some axis contained on the plane.

Choosing the coordinates in a convenient manner, we can pick $C$ to be contained on the plane $xz$ and chose the axis to be the $z$ axis.

In that case, if $R_z(\theta) : \mathbb{R}^3\to \mathbb{R}^3$ is a rotation around the $z$-axis, the surface of revolution will be

$$S = \{R_z(\theta)\cdot p \in \mathbb{R}^3 : p\in C, \theta\in [0,2\pi]\}.$$

Now, I want to show that $S$ is a regular surface, in the sense that it satisfies Do Carmo's definition:

A regular surface is $S\subset \mathbb{R}^3$ such that for each $p\in S$ there is $V$ open in $S$, $U\subset \mathbb{R}^2$ open in $\mathbb{R}^2$ and $\mathbf{x}: U\to V$ such that

1. $\mathbf{x}$ is differentiable,
2. $\mathbf{x}$ is injective,
3. $\mathbf{x}$ has injective derivative, that is, $d\mathbf{x}_q : \mathbb{R}^2\to \mathbb{R}^3$ is injective for each $q\in U$.

Now, without thinking too much about domains for a while, a natural choice for $\mathbf{x}$ would be as follows: we parametrize $C$ by $\alpha : I\subset \mathbb{R}\to \mathbb{R}^3$ and define $\mathbf{x} : [0,2\pi]\times I\to S$

$$\mathbf{x}(\theta,t)=R_z(\theta)\cdot \alpha(t).$$

This has two problems:

1. $[0,2\pi]\times I$ is not open. Furthermore, $R_z(0)=R_z(2\pi)$, hece we would not have injectivity. This problem would be solved if we replace $[0,2\pi]$ by $(0,2\pi)$ and if $I$ is an open interval.

2. I can't identify the open set $V\subset S$ used with this. I mean, I know that this $\mathbf{x}$ should work, but I don't know how to find open $V$ in $S$ for each $p\in S$ so that there is one $\mathbf{x}$ like that.

I mean, defining $\mathbf{x}$ isn't that hard. What I'm finding quite complicated is finding the correct domains on $\mathbb{R}^2$ and $S$, and then proving inside the domains that $1,2,3$ are satisfied.

How can this be done in this general context?

Answer 1

Your choice of $\mathbf{x}$ is good and your two problems can be fixed quite easily:

1. Restrict to an open set $U \subseteq [0,2\pi] \times I$. For instance, for the point $p = R(\theta_0)\cdot \alpha(t_0) \in S$, you can choose $U = I_1 \times I_2$, where $I_1 \subset [0,2\pi]$ is an open interval containing $\theta_0$ of your choosing, and $I_2 \subset I$ is an open interval containing $t_0$ of your choosing.
2. You don't need to identify $V$ more than saying $V = \mathbf{x}(U)$. If you want something a little more explicit, with the choice of $U$ above, this gives $V = \{R(\theta)\cdot \alpha(t) \colon ~~\theta \in I_1, t\in I_2$}.

In your case, you need to make sure that the curve $C$ itself is a regular curve in order for everything to work. This means that $\alpha$ is differentiable, locally open onto its image and has injective derivative (same conditions as $\mathbf{x}$). This is an assumption that is reasonable to make about your curve.