# Find Vector Potential from Vector Field with Div = 0?

This works to compute the scalar potential function of a vector field whose curl is 0 (using the DifferentialForms.m package):

<< DifferentialForms
d[x_]
d[y_]
f = 1/(x^2 + y^2)
omega = d[f]
F[x_, y_] = {omega[[1]]*(omega[[3]][[1]][[1]])*omega[[2]],
omega[[1]]*(omega[[3]][[2]][[1]])*omega[[2]]}
Simplify[Curl[F[x,y], {x,y}]]
F1[x_, y_] = F[x, y][[1]]
F2[x_, y_] = F[x, y][[2]]
phiVec[x_, y_] =
phiVec1[x, y] /.
First[DSolve[Grad[phiVec1[x, y], {x, y}] == {F1[x, y], F2[x, y]},
phiVec1[x, y], {x, y}]]


How could I do something similar to find the vector potential of a vector field whose div is 0, like

<< DifferentialForms
d[q3_]
d[q4_]
d[q5_]
eta1 = (q3*Wedge[d[q4], d[q5]] + q4*Wedge[d[q5], d[q3]] +
q5*Wedge[d[q3], d[q4]])/(q3^2 + q4^2 + q5^2)^(3/2)
Simplify[d[eta1]]
F1[q3_, q4_,
q5_] = {eta1[[
1]]*(eta1[[2]][[1]][[1]]), -(eta1[[1]]*
eta1[[2]][[2]][[1]]), (eta1[[1]]*eta1[[2]][[3]][[1]])}
F11[q3_, q4_, q5_] = F1[q3, q4, q5][[1]]
F12[q3_, q4_, q5_] = F1[q3, q4, q5][[2]]
F13[q3_, q4_, q5_] = F1[q3, q4, q5][[3]]
Simplify[Div[F1[q3, q4, q5], {q3, q4, q5}]]
DSolve[{Curl[{A11[q3, q4, q5], A12[q3, q4, q5], A13[q3, q4, q5]}, {q3,
q4, q5}] == {F11[q3, q4, q5], F12[q3, q4, q5],
F13[q3, q4, q5]}}, {A11[q3, q4, q5], A12[q3, q4, q5],
A13[q3, q4, q5]}, {q3, q4, q5}]


(Also, 1) HomotopyOperator fails due to an improper integral, and 2) a musical isomorphism would be nice in DifferentialForms.m.)