Past exam of the mathematics course of the University of Cambridge / 2002 / ia / Paper 3 / 3A / b / Solution

To construct a potential for the exact differential, first integrate its $dx$ coefficient with respect to $x$:Matching the $dy$ coefficient gives $g_y=-2z^2$, hence $g=-2y{z}^2+h(z)$. The remaining partial derivative is thenMatching the $dz$ coefficient forces $h^{\prime }=0$. Therefore a potential, valid throughout ${\mathbb{R}}^3$, isDifferentiating this expression recovers all three coefficients, proving that the differential form really is an exact differential.