The ScalarPotential(v) command computes the scalar potential of the vector field v.  This is a function f such that $\mathrm{Gradient}\left(f\right)=v$.  If a scalar potential does not exist, NULL is returned.

If v is a Vector field, an algebraic expression is returned. If v is a Vector-valued procedure, a procedure is returned.

$\mathrm{with}\left(\mathrm{VectorCalculus}\right)\:$

$\mathrm{SetCoordinates}\left('\mathrm{cartesian}'\left[x,y,z\right]\right)$

${\mathrm{cartesian}}_{x,y,z}$

$v≔\mathrm{VectorField}\left(⟨x\,y\,z⟩\right)$

$v≔\left(x\right){\stackrel{\_}{e}}_x\+\left(y\right){\stackrel{\_}{e}}_y\+\left(z\right){\stackrel{\_}{e}}_z$

$\mathrm{ScalarPotential}\left(v\right)$

$\frac{x^2}{2}+\frac{y^2}{2}+\frac{z^2}{2}$

$v≔\mathrm{VectorField}\left(⟨y\,-x\,0⟩\right)$

$v≔\left(y\right){\stackrel{\_}{e}}_x\+\left(-x\right){\stackrel{\_}{e}}_y\+\left(0\right){\stackrel{\_}{e}}_z$

$\mathrm{ScalarPotential}\left(v\right)$

$\mathrm{ScalarPotential}\left(\left(x\,y\,z\right)\mapsto \frac{⟨x\,y\,z⟩}{x^2+y^2+z^2}\right)$

$\left(x\,y\,z\right)\mapsto \frac{\ln \left(x^2+y^2+z^2\right)}{2}$

$\mathrm{SetCoordinates}\left('\mathrm{spherical}'\left[r,\mathrm{\phi },\mathrm{\theta }\right]\right)$

${\mathrm{spherical}}_{r,\mathrm{\phi },\mathrm{\theta }}$

$v≔\mathrm{VectorField}\left(⟨r\,0\,0⟩\right)$

$v≔\left(r\right){\stackrel{\_}{e}}_r\+\left(0\right){\stackrel{\_}{e}}_{\mathrm{\φ}}\+\left(0\right){\stackrel{\_}{e}}_{\mathrm{\θ}}$

$\mathrm{ScalarPotential}\left(v\right)$

$\frac{r^2}{2}$

$\mathrm{Gradient}\left(\right)$

$\left(r\right){\stackrel{\_}{e}}_r\+\left(0\right){\stackrel{\_}{e}}_{\mathrm{\φ}}\+\left(0\right){\stackrel{\_}{e}}_{\mathrm{\θ}}$