Suppose I have a surface in cylindrical coordinates given by $z=f(r,\theta)$. How can I proceed to find the normal unit vector of this surface?
My initial guess was to evaluate it's gradient, which will give me the vector (not normalized):
$$\underline{n}=\left(\frac{\partial f}{\partial r},\frac{1}{r}\frac{\partial f}{\partial \theta},-1\right)$$ And then divide its component by the length $\sqrt{n_r^2+n_z^2}$ to obtain the normal unit vector. However, I've found on the internet that the normal vector of a surface is given by $\frac{\partial \underline{F}}{\partial r}\times\frac{\partial \underline{F}}{\partial\theta}$, where $\underline{F}$ is the regular parametrization of the surface. Taking for the present example $\underline{F}=(r\cos\theta,r\sin\theta,f(r,\theta))$:
$$\underline{n} = \frac{\partial \underline{F}}{\partial r}\times\frac{\partial \underline{F}}{\partial\theta} = \left(\sin\theta\frac{\partial f}{\partial \theta} - r\cos\theta\frac{\partial f}{\partial r} , -r\sin\theta\frac{\partial f}{\partial r} - \cos\theta\frac{\partial f}{\partial \theta},r \right) $$
Which does not look to be the same. Can anyone help me on how can I evaluate this quantity? Thank you in advance.