5.104 How to find the Curl of a vector?

First example

restart; 
VectorCalculus:-SetCoordinates( 'cartesian'[x,y,z] ); 
F:=VectorCalculus:-VectorField(<y,-x,0>);
 

\begin{align*} F &= y \mathbf {\bar {e}_{x}}-x \mathbf {\bar {e}_{y}} \end{align*}

And now

VectorCalculus:-Curl(F);
 

\[ -2 \mathbf {\bar {e}_{z}} \]

Second example

restart; 
VectorCalculus:-SetCoordinates( 'cartesian'[x,y,z] ); 
F:=VectorCalculus:-VectorField(<y*z^2,x*z^2+2,2*x*y*z-1>);
 

\begin{align*} F &= y \,z^{2} \mathbf {\bar {e}_{x}}+(x \,z^{2}+2) \mathbf {\bar {e}_{y}}+(2 x y z -1) \mathbf {\bar {e}_{z}} \end{align*}

And now

VectorCalculus:-Curl(F);
 

\[ 0 \]

Since Curl is zero, field is conservative.

Third example, in cylinderical coordinates

restart; 
VectorCalculus:-SetCoordinates( 'cylindrical'[rho,phi,z] ); 
F:=VectorCalculus:-VectorField(<0,-rho,2>);
 

\begin{align*} F &= -\rho \mathbf {\bar {e}_{\phi }}+2 \mathbf {\bar {e}_{z}} \end{align*}

And now

VectorCalculus:-Curl(F);
 

\[ 2 \mathbf {\bar {e}_{z}} \]