Are surfaces of revolution with congruent profile curves isometric? Assume that two curves $\alpha,\widetilde{\alpha}\colon I \to \Bbb R^3$ are of the form $\alpha(u) = (f(u),0,g(u))$ and $\widetilde{\alpha}(\widetilde{u}) = (\widetilde{f}(\widetilde{u}),0,\widetilde{g}(\widetilde{u}))$, say, both with unit speed. Assume that $\widetilde{\alpha}$ is obtained from $\alpha$ by a rigid motion in the plane $y=0$. If $${\bf x}(u,v) = (f(u)\cos v, f(u) \sin v, g(u)),$$and $\widetilde{\bf x}$ (similarly defined) are the parametrizations of revolution associated, then are the surfaces of revolution isometric? The metrics in these coordinates are $${\rm d}u^2 + f(u)^2{\rm d}v^2 \quad\mbox{and}\quad {\rm d}\widetilde{u}^2 + \widetilde{f}(\widetilde{u})^2\,{\rm d}\widetilde{v}^2.$$
To gain some insight I tried writing $$\widetilde{\alpha}(\widetilde{u}) = ( f(\widetilde{u})\cos \theta - g(\widetilde{u})\sin \theta,0,f(\widetilde{u})\sin \theta+g(\widetilde{u})\cos \theta)$$for some $\theta$, and this seems to tell me that the "obvious" guess $F$ defined by $F(\mathbf{x}(u,v)) = \widetilde{\bf x}(u,v)$ will not by an isometry. Surely I can compute the Gaussian curvatures as $-f''/f$ and $-\widetilde{f}''/\widetilde{f}$ but this didn't gave me any new insight until now. 
Can someone help me please? For what's it worth, I didn't see this as an exercise in any book, I just thought about it now.
 A: Just to take out this question out of the unanswered queue: plenty of examples were given in the comments.


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*"Consider a "standard" torus in $\Bbb R^3$ with the induced metric. If you take $α$ to be a small segment of the left-side of the circle which generates it by revolution, then the curvature of the revolution is $>0$. If it is on the right-side (which will be a rigid motion of $α$), then the curvature of the revolution is $<0$."

*"Slide a curve farther from the $z$-axis in the $xz$-plane. Then the surfaces will obviously not be isometric, as the radii of the parallels will all be larger in the latter case."

*"Take $α(u)=(1,0,u)$ a vertical straight line segment in the $xz$ plane and $\tilde{α}(\tilde{u})=((\tilde{u}+1)/\sqrt{2},0,(\tilde{u}−1)/\sqrt{2})$ its (say) $π/4$-radian rotation in the same plane. The surface of revolution generated by the first curve is a cylinder, whereas the one generated by the second is a cone. Both are obviously not even conformally equivalent, let alone isometric. You can adjust the interval of definition of $α$ so that $\tilde{u}$ is never equal to $±1$ and the surface remains regular."
