I know a lot of alternative methods to the one I describe below, so please DO NOT mention any methods other than the one I have below.
We have an expression for $\nabla\times \mathbf A$ in Cartesian coordinates $$ \left(\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z}\right) \hat{\mathbf x} \\ + \left(\frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x}\right) \hat{\mathbf y} \\ + \left(\frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}\right) \hat{\mathbf z}. $$
I believe I can obtain the formula in cylindrical coordinates, starting from the one above. Here is my plan for computing component $\hat{\boldsymbol\varphi}$:
- Note that $\hat{\boldsymbol\varphi}=-\sin\phi\hat{\mathbf x}+\cos\phi \hat{\mathbf y}$, so to find the $\hat{\boldsymbol\varphi}$ component we just need to evaluate the dot product $$-\sin\phi\left(\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z}\right) +\cos\phi \left(\frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x}\right) $$
- Next, try to change $A_x,A_y$ to $A_\phi,A_r$.
In the second step, I feel things going seriously wrong. I get $$ \frac{\partial A_y}{\partial z}=\frac{\partial (A_r\sin A_\phi)}{\partial z}=\sin A_\phi\frac{\partial A_r}{\partial z}+\cos A_\phi\frac{\partial A_\phi}{\partial z} A_r. $$ This is terrible. I am not expecting to see anything like $\sin A_\phi$ in my final answer, and there is no way I can get rid of it or cancel it out.
What's wrong?