Curl example confusion

Question

For a standard set of $\vec e_\theta, \vec e_r,\vec e_z$, and a differentiable vector field $\vec F = f(z)\vec e_\theta$ show that it's curl is $\frac{f}{r}\vec e_z-f'(z)\vec e_r$. I can show the $\vec e_r$ component but I have no idea how to show the other component. Also I don't have to convert back to cartesian coordinates right? (because right hand basis will preserve the curl?)

Answer 1

This is the section in Wikipedia, where we can change $r \to \rho$ and $\theta \to \phi$.

\begin{align} \operatorname{div} \mathbf A &= \lim_{V\to 0} \frac{\iint_{\partial V} \mathbf A \cdot d\mathbf{S}}{\iiint_V dV} \\ &= \frac{A_\rho(\rho+d\rho)(\rho+d\rho)d\phi dz - A_\rho(\rho)\rho d\phi dz + A_\phi(\phi+d\phi)d\rho dz - A_\phi(\phi)d\rho dz + A_z(z+dz)d\rho (\rho +d\rho/2)d\phi - A_z(z)d\rho (\rho +d\rho/2) d\phi}{\rho d\phi d\rho dz} \\ &= \frac 1 \rho \frac{\partial (\rho A_\rho)}{\partial \rho} + \frac 1 \rho \frac{\partial A_\phi}{\partial \phi} + \frac{\partial A_z}{\partial z} \end{align}

\begin{align} (\operatorname{curl} \mathbf A)_\rho &= \lim_{S^{\perp \boldsymbol{\hat \rho}}\to 0} \frac{\int_{\partial S} \mathbf A \cdot d\mathbf{\ell}}{\iint_{S} dS} \\ &= \frac{A_\phi (z)(\rho+d\rho)d\phi - A_\phi(z+dz)(\rho+d\rho)d\phi + A_z(\phi + d\phi)dz - A_z(\phi)dz}{(\rho+d\rho)d\phi dz} \\ &= -\frac{\partial A_\phi}{\partial z} + \frac{1}{\rho} \frac{\partial A_z}{\partial \phi} \end{align}

\begin{align} (\operatorname{curl} \mathbf A)_\phi &= \lim_{S^{\perp \boldsymbol{\hat \phi}}\to 0} \frac{\int_{\partial S} \mathbf A \cdot d\mathbf{\ell}}{\iint_{S} dS} \\ &= \frac{A_z (\rho)dz - A_z(\rho + d\rho)dz + A_\rho(z+dz)d\rho - A_\rho(z)d\rho}{d\rho dz} \\ &= -\frac{\partial A_z}{\partial \rho} + \frac{\partial A_\rho}{\partial z} \end{align}

\begin{align} (\operatorname{curl} \mathbf A)_z &= \lim_{S^{\perp \boldsymbol{\hat z}}\to 0} \frac{\int_{\partial S} \mathbf A \cdot d\mathbf{\ell}}{\iint_{S} dS} \\ &= \frac{A_\rho(\phi)d\rho - A_\rho(\phi + d\phi)d\rho + A_\phi(\rho + d\rho)(\rho + d\rho)d\phi - A_\phi(\rho)\rho d\phi}{\rho d\rho d\phi} \\ &= -\frac{1}{\rho}\frac{\partial A_\rho}{\partial \phi} + \frac{1}{\rho} \frac{\partial (\rho A_\phi)}{\partial \rho} \end{align}

\begin{align} \operatorname{curl} \mathbf A &= (\operatorname{curl} \mathbf A)_\rho \hat{\boldsymbol \rho} + (\operatorname{curl} \mathbf A)_\phi \hat{\boldsymbol \phi} + (\operatorname{curl} \mathbf A)_z \hat{\boldsymbol z} \\ &= \left(\frac{1}{\rho} \frac{\partial A_z}{\partial \phi} -\frac{\partial A_\phi}{\partial z} \right) \hat{\boldsymbol \rho} + \left(\frac{\partial A_\rho}{\partial z}-\frac{\partial A_z}{\partial \rho} \right) \hat{\boldsymbol \phi} + \frac{1}{\rho}\left(\frac{\partial (\rho A_\phi)}{\partial \rho} - \frac{\partial A_\rho}{\partial \phi} \right) \hat{\boldsymbol z} \end{align}

From this we will retain the term with $-\frac{\partial A_\phi}{ \partial z}$ in the dimension of $\rho$ and also the $\frac{1}{\rho}\frac{\partial (\rho A_\phi)}{\partial \rho}$ term in the dimension of $z$.