 Solution

The field is continuously differentiable on all of ${\mathbb{R}}^3$, a simply connected domain. Its vanishing curl therefore implies a global potential of a conservative vector field, so $\mathbf{F}\cdot d\mathbf{r}=d\varphi$ is an exact differential. This global domain condition matters: curl-free fields on domains with holes need not have single-valued potentials.

Integrating ${\varphi }_x=2y{z}^2$ gives $\varphi =2xy{z}^2+g(y,z)$. Matching ${\varphi }_y=2x{z}^2$ forces $g_y=0$, so $g=h(z)$; matching ${\varphi }_z=4xyz$ then forces $h^{\prime }=0$. ConsequentlyThe additive constant does not affect the field or any endpoint difference.