A surface of revolution like metric For the metric
$du^2+f(u)^2dv^2$
with $|f'(u)|\le 1$ this is the surface of revolution:
$(f(u)\cos v, f(u)\sin v, \int\sqrt{1-f'(u)^2} du)$.
My question is now for the metric
$du^2+g(u)^2dv^2$
with $g'(u)>1$ is there any way to embed into $\mathbb{R}^3$? If not how do you see it is impossible?
 A: If $f'(u) > 1$, the metric
$$
du^{2} + f(u)^{2}\, dv^{2}
\tag{1}
$$
does not embed into Euclidean $3$-space as a surface of rotation, effectively because your formula for embedding (1) as a surface of rotation is general, and the third coordinate ceases to be real if $f'(u) > 1$.
Geometrically, think of measuring arc length along a meridian (a.k.a., a generator or profile) and asking how much area per unit length the surface has: For a surface of rotation, the maximum occurs when the surface profile is a line perpendicular to the axis of rotation, which (if you check) corresponds to $f'(u) = 1$. (More details can be found, for example, in A Symplectic Look at Surfaces of Rotation, l'Ens. Math 49 (2003) 157-172.)
If $f'(u) > 1$, the resulting metric may embed in Euclidean $3$-space: Think of $f(u) = mu$ for $m > 1$, which embeds as a "saddle cone". Though I don't have a proof, I expect it's difficult to decide in general whether or not (1) embeds globally (isometrically) into Euclidean $3$-space.
A: In general, one cannot do this globally. Notice that the case $f(u)= \cosh u$ corresponds to hyperbolic space and, by Hilbert theorem, this space cannot be regarded as a surface of Euclidean 3-space. (However, locally, hyperobolic space can be regarded as the surface of revolution generated by the tractrix.)
