and

$C_y\=h_r \frac{y}{r}\+h_{\mathrm{\θ}} \frac{x}{r^2}\=h_r \sin \left(\mathrm{\θ}\right)\+h_{\mathrm{\θ}}\frac{\cos \left(\mathrm{\θ}\right)}{r}$

$B_z\=f_z \frac{y}{r}\+g_z \frac{x}{r}\=f_z \sin \left(\mathrm{\θ}\right)\+g_z \frac{\cos \left(\mathrm{\θ}\right)}{r}$

$C_y-B_z\=h_r \sin \left(\mathrm{\theta }\right)\+h_{\mathrm{\theta }}\frac{\cos \left(\mathrm{\theta }\right)}{r}-f_z \sin \left(\mathrm{\theta }\right)-g_z \frac{\cos \left(\mathrm{\theta }\right)}{r}\equiv u$

$A_z\=f_z \frac{x}{r}-g_z \frac{y}{r}\=f_z \cos \left(\mathrm{\θ}\right)-g_z \sin \left(\mathrm{\θ}\right)$

$C_x\=h_r \frac{x}{r}-h_{\mathrm{\θ}} \frac{y}{r^2}\=h_r \cos \left(\mathrm{\θ}\right)-h_{\mathrm{\θ}} \frac{\sin \left(\mathrm{\θ}\right)}{r}$

$A_z-C_x\=f_z \cos \left(\mathrm{\theta }\right)-g_z \sin \left(\mathrm{\theta }\right)-h_r \cos \left(\mathrm{\theta }\right)\+h_{\mathrm{\theta }} \frac{\sin \left(\mathrm{\theta }\right)}{r}\equiv v$

$B_x\=\frac{r {\left(y f\+x g\right)}_x-\left(y f\+x g\right) \frac{x}{r}}{r^2}\=\frac{y \left(f_r \frac{x}{r}-f_{\mathrm{\θ}} \frac{y}{r^2}\right)\+g\+x \left(g_r \frac{x}{r}-g_{\mathrm{\θ}} \frac{y}{r^2}\right)}{r}-\frac{x y f\+x^{2}g}{r^3}$

$A_y\=\frac{r {\left(x f-y g\right)}_y-\left(x f-y g\right) \frac{y}{r}}{r^2}\=\frac{x \left(f_r \frac{y}{r}\+f_{\mathrm{\θ}} \frac{x}{r^2}\right)-g-y \left(g_r \frac{y}{r}\+g_{\mathrm{\θ}} \frac{x}{r^2}\right)}{r}-\frac{x y f-y^{2}g}{r^3}$

$u \cos \left(\mathrm{\θ}\right)\+v \sin \left(\mathrm{\θ}\right)$

$$\begin{gathered} \=\left[h_r \sin \left(\mathrm{\theta }\right)\+h_{\mathrm{\theta }}\frac{\cos \left(\mathrm{\theta }\right)}{r}-f_z \sin \left(\mathrm{\theta }\right)-g_z \frac{\cos \left(\mathrm{\theta }\right)}{r}\right] \cos \left(\mathrm{\θ}\right) \\ \+\left[f_z \cos \left(\mathrm{\theta }\right)-g_z \sin \left(\mathrm{\theta }\right)-h_r \cos \left(\mathrm{\theta }\right)\+h_{\mathrm{\theta }} \frac{\sin \left(\mathrm{\theta }\right)}{r}\right]\sin \left(\mathrm{\θ}\right) \end{gathered}$$

$\=\frac{h_{\mathrm{\θ}}}{r}-g_z$

$-u \sin \left(\mathrm{\θ}\right)\+v \cos \left(\mathrm{\θ}\right)$

$$\begin{gathered} \=\left[f_z \cos \left(\mathrm{\theta }\right)-g_z \sin \left(\mathrm{\theta }\right)-h_r \cos \left(\mathrm{\theta }\right)\+h_{\mathrm{\theta }} \frac{\sin \left(\mathrm{\theta }\right)}{r}\right] \cos \left(\mathrm{\θ}\right) \\ -\left[h_r \sin \left(\mathrm{\theta }\right)\+h_{\mathrm{\theta }}\frac{\cos \left(\mathrm{\theta }\right)}{r}-f_z \sin \left(\mathrm{\theta }\right)-g_z \frac{\cos \left(\mathrm{\theta }\right)}{r}\right] \sin \left(\mathrm{\θ}\right) \end{gathered}$$

$\=f_z-h_r$

$P≔r\,\mathrm{\θ}\,z\:$$$

$\mathrm{Typesetting}:-\mathrm{Suppress}\left(\left\{f\left(p\right)\,g\left(P\right)\,h\left(P\right)\right\}\right)\:$$$

Loading Student:-VectorCalculus

Define F, a vector field in cylindrical coordinates

$⟨f\,g\,h⟩$ = $\left[\begin{array}{c}f\left(p\right)\\ g\left(r\,\mathrm{\θ}\,z\right)\\ h\left(r\,\mathrm{\θ}\,z\right)\end{array}\right]$$\stackrel{\text{apply coordinates}}{\to }$$\left[\begin{array}{c}f\left(p\right)\\ g\left(r\,\mathrm{\θ}\,z\right)\\ h\left(r\,\mathrm{\θ}\,z\right)\end{array}\right]$$\stackrel{\text{to Vector Field}}{\to }$$\left[\begin{array}{c}f\left(p\right)\\ g\left(r\,\mathrm{\θ}\,z\right)\\ h\left(r\,\mathrm{\θ}\,z\right)\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$F$$$

Convert the cylindrical vector field F to a Cartesian vector field G

$\mathbf{F}$ = $\left[\begin{array}{c}f\left(p\right)\\ g\left(r\,\mathrm{\θ}\,z\right)\\ h\left(r\,\mathrm{\θ}\,z\right)\end{array}\right]$$\stackrel{\text{change coordinates}}{\to }$$\left[\begin{array}{c}\frac{f\left(p\right)x}{\sqrt{x^2\+y^2}}-\frac{g\left(\sqrt{x^2\+y^2}\,\arctan \left(y\,x\right)\,z\right)y}{\sqrt{x^2\+y^2}}\\ \frac{f\left(p\right)y}{\sqrt{x^2\+y^2}}\+\frac{g\left(\sqrt{x^2\+y^2}\,\arctan \left(y\,x\right)\,z\right)x}{\sqrt{x^2\+y^2}}\\ h\left(\sqrt{x^2\+y^2}\,\arctan \left(y\,x\right)\,z\right)\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$G$

Compute the curl of G, which possible since the form of the curl in Cartesian coordinates is known.

Convert this curl from Cartesian to cylindrical coordinates.

$\nabla \times \mathbf{G}$ =

$\left[\begin{array}{c}\frac{{\mathrm{D}}_1\left(h\right)\left(\sqrt{x^2\+y^2}\,\arctan \left(y\,x\right)\,z\right)y}{\sqrt{x^2\+y^2}}\+\frac{{\mathrm{D}}_2\left(h\right)\left(\sqrt{x^2\+y^2}\,\arctan \left(y\,x\right)\,z\right)}{x\left(1\+\frac{y^2}{x^2}\right)}-\frac{{\mathrm{D}}_3\left(g\right)\left(\sqrt{x^2\+y^2}\,\arctan \left(y\,x\right)\,z\right)x}{\sqrt{x^2\+y^2}}\\ -\frac{{\mathrm{D}}_3\left(g\right)\left(\sqrt{x^2\+y^2}\,\arctan \left(y\,x\right)\,z\right)y}{\sqrt{x^2\+y^2}}-\frac{{\mathrm{D}}_1\left(h\right)\left(\sqrt{x^2\+y^2}\,\arctan \left(y\,x\right)\,z\right)x}{\sqrt{x^2\+y^2}}\+\frac{{\mathrm{D}}_2\left(h\right)\left(\sqrt{x^2\+y^2}\,\arctan \left(y\,x\right)\,z\right)y}{x^2\left(1\+\frac{y^2}{x^2}\right)}\\ \frac{\left(\frac{{\mathrm{D}}_1\left(g\right)\left(\sqrt{x^2\+y^2}\,\arctan \left(y\,x\right)\,z\right)x}{\sqrt{x^2\+y^2}}-\frac{{\mathrm{D}}_2\left(g\right)\left(\sqrt{x^2\+y^2}\,\arctan \left(y\,x\right)\,z\right)y}{x^2\left(1\+\frac{y^2}{x^2}\right)}\right)x}{\sqrt{x^2\+y^2}}\+\frac{2g\left(\sqrt{x^2\+y^2}\,\arctan \left(y\,x\right)\,z\right)}{\sqrt{x^2\+y^2}}-\frac{g\left(\sqrt{x^2\+y^2}\,\arctan \left(y\,x\right)\,z\right)x^2}{{\left(x^2\+y^2\right)}^{3\/2}}\+\frac{\left(\frac{{\mathrm{D}}_1\left(g\right)\left(\sqrt{x^2\+y^2}\,\arctan \left(y\,x\right)\,z\right)y}{\sqrt{x^2\+y^2}}\+\frac{{\mathrm{D}}_2\left(g\right)\left(\sqrt{x^2\+y^2}\,\arctan \left(y\,x\right)\,z\right)}{x\left(1\+\frac{y^2}{x^2}\right)}\right)y}{\sqrt{x^2\+y^2}}-\frac{g\left(\sqrt{x^2\+y^2}\,\arctan \left(y\,x\right)\,z\right)y^2}{{\left(x^2\+y^2\right)}^{3\/2}}\end{array}\right]$

$\stackrel{\text{change coordinates}}{\to }$

$\left[\begin{array}{c}\left(\frac{{\mathrm{D}}_1\left(h\right)\left(\sqrt{r^2{\cos \left(\mathrm{\θ}\right)}^2\+r^2{\sin \left(\mathrm{\θ}\right)}^2}\,\arctan \left(r\sin \left(\mathrm{\θ}\right)\,r\cos \left(\mathrm{\θ}\right)\right)\,z\right)r\sin \left(\mathrm{\θ}\right)}{\sqrt{r^2{\cos \left(\mathrm{\θ}\right)}^2\+r^2{\sin \left(\mathrm{\θ}\right)}^2}}\+\frac{{\mathrm{D}}_2}{}\right)\end{array}\right]$