Show that a surface of revolution can always be parametrized so that $E = E(v), F = 0$ and $G = 1$.

Question

The following exercise is on the Do Carmo's Differential Geometry of Curves and Surfaces in the section about The First Fundamental Form.

Show that a surface of revolution can always be parametrized so that $E = E(v), F = 0$ and $G = 1$.

$\textbf{My attempt:}$

Let be $X(u,v) = \left( f(v) \cos u, f(v) \sin u, g(v) \right)$ parametrization of a surface of revolution generated by through of rotation of a regular plane curve $C$ in $xz$ in turn of $z$ axis which is given by $x = f(v)$ and $z = g(v)$, $f(v) > 0$, then

$X_u = \left( -f(v) \sin u, f(v) \cos u, 0 \right)$,

$X_v = \left( f'(v) \cos u, f'(v) \sin u, g'(v) \right)$,

$E(u,v) = \langle X_u, X_u \rangle = \left[ f(v) \right]^2 = E(v)$,

$F(u,v) = \langle X_u, X_v \rangle = 0$,

$G(u,v) = \langle X_v, X_v \rangle = \left[ f'(v) \right]^2 + \left[ g'(v) \right]^2$

Since always regular curve can be reparametrized by length arc, we can assume that $v = s$, then $\left[ f'(v) \right]^2 + \left[ g'(v) \right]^2 = 1$ and we have to $G = 1$. $\square$

I have two doubts:

1. Are the parameters $u$ and $v$ in the exercise arbitrary or can I choose them in such a way that $E = E(v), F = 0$ and $G = 1$ just as I did?

2. If I can choose the parameters, can I just assume $v = s$ just as I did or I need start with an arbitrary parameter $v$ and exhibit an reparametrization by arc length?

Thanks in advance!

Answer 1

This is fine. All you have to do is let $v$ be the arclength parameter for the curve that's being rotated. There's no need (or capacity) to give an explicit such parametrization, since the curve is arbitrary.