In cylindrical polar coordinates, $\nabla={\bf e}_{\rho}\frac{\partial}{\partial \rho}+{\bf e}_{\phi}\frac1{\phi}\frac{\partial}{\partial \phi}+{\bf e}_z\frac{\partial}{\partial z}$ and $\frac{\partial {\bf e}_{\phi}}{\partial \phi}={\bf e}_{\phi}, \frac{\partial {\bf e}_{\phi}}{\partial \phi}=−{\bf e}_{\rho}$ ,while all other derivatives of the basis vectors are zero. Derive expressions for $\nabla \cdot {\bf A}$ and $\nabla \times {\bf A}$ where where ${\bf A}=A_{\rho}{\bf e}_{\rho}+A_{\phi}{\bf e}_{\phi}+A_z{\bf e}_z$.
I've just started this question and I began by converting to Cartesian with ${\bf e}_{\rho}=(\cos\phi ,\sin\phi ,0), {\bf e}_{\phi}=(-\sin\phi ,\cos\phi ,0)$ and ${\bf e}_z=(0,0,1)$ and then computing, however this is quickly getting tedious so I was just wondering if, firstly will what I'm doing work out eventually?
Secondly, and this is really the question, is there a better/faster way of doing this? I'm not really asking here for a solution, but I wouldn't mind one either as that should show me what I want to know for future questions.
Any help is appreciated
Thank you
EDIT: I'll include what I started to do here as I doubt it's correct and this should make it clearer:
Coverting to Cartesian I have $\nabla=(\cos\phi \frac{\partial}{\partial\rho}-\sin\phi \cdot \frac1{\rho} \frac{\partial}{\partial \phi},\sin\phi \frac{\partial}{\partial\rho}+\cos\phi \cdot \frac1{\rho} \frac{\partial}{\partial \phi},\frac{\partial}{\partial z})$ and ${\bf A}=(A_{\rho}\cos\phi -A_{\phi}\sin\phi,A_{\rho}\sin\phi -A_{\phi}\cos\phi,A_z)$ and then manually computed from there.
There is no reason to switch to Cartesian coordinates. ${\rm e}_\rho$, ${\rm e}_\phi$, and ${\rm e}_z$ form an orthonormal set. So $\nabla \cdot A=\frac{\partial A_\rho}{\partial \rho}+\frac{1}{\phi}\frac{\partial A_\phi}{\partial \phi}+\frac{\partial A_z}{\partial z}$. You can use the same thing to calculate the cross product. For example ${\rm e}_\rho \times {\rm e}_\phi={\rm e}_z$