= Solution
To construct a potential for the <exact differential>, first integrate its $dx$ coefficient with respect to $x$:
$$
F=x^3(y^2+1)+g(y,z).
$$
Matching the $dy$ coefficient gives $g_y=-2z^2$, hence $g=-2yz^2+h(z)$. The remaining <partial derivative> is then
$$
F_z=-4yz+h'(z).
$$
Matching the $dz$ coefficient forces $h'=0$. Therefore a potential, valid throughout $\mathbb R^3$, is
$$
\boxed{F(x,y,z)=x^3(y^2+1)-2yz^2+C.}
$$
Differentiating this expression recovers all three coefficients, proving that the <differential form> really is an <exact differential>.
Back to article page
