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

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?

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.