and

$\mathbf{G}$

$\=r^2\cos \left(\mathrm{\theta }\right)\sin \left(\mathrm{\theta }\right) \left[\begin{array}{c}\cos \(\mathrm{\theta }\)\\ -\sin \(\mathrm{\theta }\)\\ 0\end{array}\right]-r \sin \left(\mathrm{\theta }\right) z\left[\begin{array}{c}\sin \(\mathrm{\theta }\)\\ \cos \(\mathrm{\theta }\)\\ 0\end{array}\right] -r \cos \left(\mathrm{\theta }\right) z\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$

$\=\left[\begin{array}{c}{r}^2{\cos }^2\left(\mathrm{\theta }\right)\sin \left(\mathrm{\theta }\right)-r {\sin }^2\left(\mathrm{\theta }\right)z\\ -r^2\cos \left(\mathrm{\theta }\right){\sin }^2\left(\mathrm{\theta }\right)-r\sin \left(\mathrm{\theta }\right)z\cos \left(\mathrm{\theta }\right)\\ -r\cos \left(\mathrm{\theta }\right)z\end{array}\right]$

$\left(h_{\mathrm{\θ}}-{\left(r g\right)}_z\right)\/r$

$\=\sin \left(\mathrm{\theta }\right)\left(r\cos \left(\mathrm{\theta }\right)\+z\right)$

$f_z-h_r$

$\={\cos }^2\left(\mathrm{\theta }\right)r\+\cos \left(\mathrm{\theta }\right)z-r$

$\left({\left(r g\right)}_r-f_{\mathrm{\θ}}\right)\/r$

$\=-r \cos \left(\mathrm{\theta }\right)$

$\nabla \times \mathbf{G}\to$

$\left[\sin \left(\mathrm{\theta }\right)\left(r\cos \left(\mathrm{\theta }\right)\+z\right)\right] {\mathit{e}}_{\mathit{r}}\+\left[{\cos }^2\left(\mathrm{\theta }\right)r\+\cos \left(\mathrm{\theta }\right)z-r\right] {\mathit{e}}_{\mathbf{\θ}}\+\left[-r \cos \left(\mathrm{\theta }\right)\right] {\mathit{e}}_{\mathit{z}}$

$\=\frac{y}{x}\left(x\+z\right)\left(\frac{x}{r} \mathbf{i}\+\frac{y}{r} \mathbf{j}\right)\+\left(\frac{x^2}{r}\+\frac{x z}{r}-r\right) \left(-\frac{y}{r} \mathbf{i}\+\frac{x}{r} \mathbf{j}\right)-x \mathbf{k}$

$\=\frac{\mathbf{i}}{r^2}\left(x^{2}y\+x y z-x^{2}y-x y z\+y r^2\right)\+\frac{\mathbf{j}}{r^2}\left(y^{2}x\+y^{2}z\+x^3\+x^{2}z-x r^2\right)-x \mathbf{k}$

$\=\frac{\mathbf{i}}{r^2}\left(y r^2\right)\+\frac{\mathbf{j}}{r^2}\left(y^{2}x\+r^{2}z\+x^3-x \left(x^2\+y^2\right)\right)-x \mathbf{k}$

$\=y \mathbf{i}\+\frac{\mathbf{j}}{r^2}\left(r^{2}z\right)-x \mathbf{k}$

$\=y \mathbf{i}\+z \mathbf{j}-x \mathbf{k}$

$\=\left[\begin{array}{c}y\\ z\\ -x\end{array}\right]$

Initialize

Loading Student:-VectorCalculus

Define the Cartesian vector field F

$⟨x y\,-y z\,-x z⟩$ = $\left[\begin{array}{c}xy\\ -yz\\ -xz\end{array}\right]$$\stackrel{\text{to Vector Field}}{\to }$$\left[\begin{array}{c}xy\\ -yz\\ -xz\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$F$$$

Obtain the curl of F

$\nabla \times \mathbf{F}$ = $\left[\begin{array}{c}y\\ z\\ -x\end{array}\right]$$$

Alternate calculation of the curl of F

$\mathbf{F}$ = $\left[\begin{array}{c}xy\\ -yz\\ -xz\end{array}\right]$$\stackrel{\text{curl}}{\to }$$\left[\begin{array}{c}y\\ z\\ -x\end{array}\right]$$$

Change F to cylindrical coordinates

$\mathbf{F}$ = $\left[\begin{array}{c}xy\\ -yz\\ -xz\end{array}\right]$$\stackrel{\text{change coordinates}}{\to }$$\left[\begin{array}{c}{r}^2{\cos \left(\mathrm{\theta }\right)}^2\sin \left(\mathrm{\theta }\right)-r{\sin \left(\mathrm{\theta }\right)}^{2}z\\ -r^2\cos \left(\mathrm{\theta }\right){\sin \left(\mathrm{\theta }\right)}^2-r\sin \left(\mathrm{\theta }\right)z\cos \left(\mathrm{\theta }\right)\\ -r\cos \left(\mathrm{\theta }\right)z\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$G$$$

Obtain the curl in spherical coordinates

$\nabla \times \mathbf{G}$ = $\left[\begin{array}{c}\frac{\sin \left(\mathrm{\theta }\right)rz+r^2\sin \left(\mathrm{\theta }\right)\cos \left(\mathrm{\theta }\right)}{r}\\ -r{\sin \left(\mathrm{\theta }\right)}^2+\cos \left(\mathrm{\theta }\right)z\\ \frac{r^2\cos \left(\mathrm{\theta }\right){\sin \left(\mathrm{\theta }\right)}^2+r\sin \left(\mathrm{\theta }\right)z\cos \left(\mathrm{\theta }\right)+r\left(-2{\sin \left(\mathrm{\theta }\right)}^2\cos \left(\mathrm{\theta }\right)r-\sin \left(\mathrm{\theta }\right)\cos \left(\mathrm{\theta }\right)z\right)-r^2{\cos \left(\mathrm{\theta }\right)}^3}{r}\end{array}\right]$$\stackrel{\text{simplify}}{\=}$$\left[\begin{array}{c}\sin \left(\mathrm{\theta }\right)\left(\cos \left(\mathrm{\theta }\right)r+z\right)\\ {\cos \left(\mathrm{\theta }\right)}^{2}r+\cos \left(\mathrm{\theta }\right)z-r\\ -\cos \left(\mathrm{\theta }\right)r\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{curlG}$$$

Change the coordinates in $\nabla \times \mathbf{G}$ back to Cartesian

$\mathrm{curlG}$ = $\left[\begin{array}{c}\sin \left(\mathrm{\theta }\right)\left(\cos \left(\mathrm{\theta }\right)r+z\right)\\ {\cos \left(\mathrm{\theta }\right)}^{2}r+\cos \left(\mathrm{\theta }\right)z-r\\ -\cos \left(\mathrm{\theta }\right)r\end{array}\right]$$\stackrel{\text{change coordinates}}{\to }$$\left[\begin{array}{c}\frac{y\left(x+z\right)x}{x^2+y^2}-\frac{\left(\frac{x^2}{\sqrt{x^2+y^2}}+\frac{xz}{\sqrt{x^2+y^2}}-\sqrt{x^2+y^2}\right)y}{\sqrt{x^2+y^2}}\\ \frac{y^2\left(x+z\right)}{x^2+y^2}+\frac{\left(\frac{x^2}{\sqrt{x^2+y^2}}+\frac{xz}{\sqrt{x^2+y^2}}-\sqrt{x^2+y^2}\right)x}{\sqrt{x^2+y^2}}\\ -x\end{array}\right]$$\stackrel{\text{simplify}}{\=}$$\left[\begin{array}{c}y\\ z\\ -x\end{array}\right]$$$

Initialize

$$\begin{gathered} \mathrm{with}\left(\mathrm{Student}:-\mathrm{VectorCalculus}\right)\: \\ \mathrm{BasisFormat}\left(\mathrm{false}\right)\: \end{gathered}$$

Define the vector field F

$\mathbf{F}≔\mathrm{VectorField}\left(⟨x y\,-y z\,-x z⟩\right)\:$

Obtain the curl of F in Cartesian coordinates

Apply the Curl command.

$\mathrm{Curl}\left(\mathbf{F}\right)$

$\left[\begin{array}{c}y\\ z\\ -x\end{array}\right]$

Change F to cylindrical coordinatesa

$\mathbf{G}≔\mathrm{MapToBasis}\left(\mathbf{F}\,\mathrm{cylindrical}\left[r\,\mathrm{\θ}\,z\right]\right)$

Obtain the curl of G, that is the curl of F in cylindrical coordinates

$\mathbf{curlG}≔\mathrm{simplify}\left(\mathrm{Curl}\left(\mathbf{G}\right)\right)$

Restore Cartesian coordinates in $\nabla \times \mathbf{G}$ 

$\mathrm{simplify}\left(\mathrm{MapToBasis}\left(\mathbf{curlG}\,\mathrm{cartesian}\left[x\,y\,z\right]\right)\right)$