Find conditions for a surface of rotation to be regular I am reading Geomtric Structures in Dimension two by Bjørn Jahren, and have some questions about exercise 5.1.4, which says the following:

Let $\alpha(u)=(f(u),g(u)),u\in[a,b]$ be an embedded curve in the $xz$-plane such that $f (u) > 0$ for all $u$. Find a parametrization of the surface of rotation obtained by rotating $\alpha$ around the $z$–axis and find conditions for this to be a regular surface.
Discuss what happens if we remove the condition $f (u) > 0$.

A regular surface is defined by the following:

Definition 5.1.3. Let $S$ be a subset of $\mathbb{R}^3$ which is also a smooth surface, and let $\iota: S \to\mathbb{R}^3$ be the inclusion map. $S$ is called a regular surface if for every local parametrization $x$ of $S$ the Jacobian of the composition $\iota\circ x$ has rank 2 at every point.

A parametrization of this surface is given by $(u,t)\mapsto(f(u)\cos(t),f(u)\sin(t),g(u))$. By the definition of regular surfaces, we need the Jacobian of this parametrizationto be of rank 2. The Jacobian has the columns $(f'(u)\cos(t),f'(u)\sin(t),g'(u))$ and $(-f(u)\sin(t),f(u)\cos(t),0)$. If $g'(u)\neq0$ they are linearly independent if and only if $f(u)\neq 0$, since $\sin$ and $\cos$ are not 0 in the same point. If $g'(u)=0$ linear independence is equivalent to the matrix 
$\begin{bmatrix}
f'(u)\cos(t) & -f(u)\sin(t)\\
f'(u)\sin(t) & f(u)\cos(t)
\end{bmatrix}$
having nonzero determinant, i.e. that $f(u)f'(u)\neq 0$. To conclude, the Jacobian has rank 2 if and only if $g'(u)\neq0\land f(u)\neq0$ or $f(u)\neq0\land f'(u)\neq 0$. Under the condition $f(u)>0$, that is if and only if $\alpha'(u)\neq 0$.
Where do I go from here? I have found conditions given one specific parametrization, but I need to show it for all possible parametrizations. How do I generalize the result? ((EDIT: Maybe this suffices, since I can create an atlas with just this parametrization, by varying the domain?)) What about the points $a$ and $b$ are mapped to? Can I find a parametrization from an open subset of $\mathbb{R}^2$ around these? And what happens if we remove the condition $f(u)>0$, how should I attack that?
Here is a handwritten (i.e. not so readable) and brief solution which may be of some use:

(Available here.)
 A: To prove that a surface is regular, you only need to check that the rank 2 condition is satisfied for all parametrizations in some atlas. What you have done therefore shows that the surface is regular if $\alpha'$ doesn't vanish and $f$ is positive. Since we are assuming that the curve $\alpha$ is embedded, the condition $\alpha' \neq 0$ always holds. Hence, the surface of revolution obtained by rotating $\alpha$ around the $z$-axis is always regular (provided that $f > 0$).

What about the points $a$ and $b$ are mapped to? Can I find a parametrization from an open subset of $\mathbb{R}^2$ around these?

You can't. The curve $\alpha$ is parametrized by a closed interval, so it has a boundary consisting of two points. Rotating $\alpha$ around the $z$-axis therefore produces a surface with a boundary consisting of two circles. In general, you cannot find a parametrization from an open subset of $\mathbb{R}^2$ around points of the boundary of a surface. However, you can find a parametrization defined on an open subset of the upper half-plane $\mathbb{R} \times [0,\infty)$.

And what happens if we remove the condition $f(u)>0$, how should I attack that?

I think the best thing to do here is to look at examples and see what happens. For instance, if we take $\alpha(u) = (u,u)$ with $[a,b] = [0,1]$, then the surface we get is a cone, which is not regular (the vertex is a singular point). I invite you to come up with your own examples; you should be able to convince yourself that zeroes of $f$ generally correspond to singular points on the surface of revolution.
There are some special cases in which the answer is a bit trickier: for instance, what if $f(a) = 0$ or $f(b) = 0$, but $f(u) > 0$ for all $u \in (a,b)$? In this situation, the surface we get can be singular (e.g. with $\alpha(u) = (u,u)$ and $[a,b] = [0,1]$ as above), but it can also be regular (e.g. $\alpha(u) = (\sin(u),\cos(u))$ with $[a,b] = [0,\pi]$).
