Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2008 / Paper 2 / 15 / a

Integrating the first component in $x$ gives $\varphi =x^{3}y+h(y,z)$. Matching the second requires $h_y=-2y{z}^2$, so $h=-y^2{z}^2+k(z)$. Matching the third gives $k^{\prime }=0$. Thus the field is conservative, with scalar potentialDirect differentiation verifies ${\mathbf{F}}_1=\nabla \varphi$ on all of ${\mathbb{R}}^3$.

The curl is $\nabla \times {\mathbf{F}}_2=(2z,3z^2,0)$, which is not identically zero. A gradient of a smooth scalar potential has zero curl, so ${\mathbf{F}}_2$ is not conservative on ${\mathbb{R}}^3$.