I have to decide whether the surface $S$ parametrized by \begin{align*}\mathbf{x}:\mathbb{R}\times \left]-\pi,\pi\right[&\longrightarrow S\\ (u,v)&\longmapsto ((2+\sin u)\cos v,(2+\sin u)\sin v,u),\end{align*} is locally isometric to the helicoid $T$ given by \begin{align*} \mathbf{y}:\mathbb{R}^2&\longrightarrow T\\ (s,t)&\longmapsto (s\cos t,s\sin t,t). \end{align*} Here's my attempt:
I have computed the Gaussian curvature of $S$ and got that it is $$K_S(u)=\frac{\sin u}{(1+\cos^2u)^2(2+\sin u)}.$$ On the other hand, the Gaussian curvature of the helicoid is $$K_T(s)=-\frac{1}{(1+s^2)^2}.$$ If $S$ and $T$ where locally isometric, then they would have the same Gaussian curvature. However, we cannot solve $K_S(u)=K_T(s)$ since $K_S$ is sometimes positive and sometimes negative, while $K_T$ is always negative. Therefore, $S$ and $T$ are not locally isometric.
Is this reasoning correct? Thanks.
PD: I've checked my computations with Mathematica and apparently the curvatures are correct.