= Solution
The coordinate tangent vectors are
$$
\phi_u=(-f\sin u,f\cos u,0),
\qquad
\phi_v=(f'\cos u,f'\sin u,g'),
$$
where $f,f',g'$ are evaluated at $v$. Their <cross product> is
$$
\phi_u\times\phi_v
=\bigl(fg'\cos u,fg'\sin u,-ff'\bigr),
$$
and therefore
$$
\lVert\phi_u\times\phi_v\rVert
=f(v)\sqrt{f'(v)^2+g'(v)^2}>0.
$$
The positivity follows because $f(v)>0$ and the profile is a <regular curve>. Hence $d\phi$ has rank two everywhere, so $\phi$ is an <immersion>. The angle interval $(0,2\pi)$ and the absence of profile self-intersections make $\phi$ injective with a continuous local inverse. It is therefore an <embedded surface parametrization>, and its image is a <regular surface>.
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