Let $f:\mathbb{R}\to\mathbb{R}$ be a smooth function, and $\text{graf }(f)$ be it's graphic. The surface $S$ generated by rotating $\text{graf }(f)$ around an axis can be parametrized as
\begin{align} \varphi:\mathbb{R}\times J&\to \mathbb{R}^3\\ (u,v)&\mapsto (u,f(u)\sin v,f(u)\cos v) \end{align}
Where $ J =(0,2\pi)$. Since $S$ is a regular surface, it is a 2-dimensional smooth manifold.
Is seems that $S$ induces a differentiable bundle structure over $\mathbb{R}$. The projection is $\pi:S\to\mathbb{R}$ defined by $\pi^{-1}(x) = \varphi(\{x\}\times J)$. Picture it sending the rotated $f(x)$ into $x$.
This projection must be $C^\infty$. Indeed, it is, because $\pi\circ\varphi:(u,v)\mapsto u$ is $C^\infty$.
I wanted to know if this reasoning is correct, that is, the triple $(S,\mathbb{R},\pi)$ is, indeed, a differentiable bundle. If so, is this bundle NOT locally trivial?