 Solution

Direct calculation of the curl givesSince $\mathbf{a}\ne 0$, the first vector field cannot be a gradient field: the curl of a gradient is zero. The second is a smooth polynomial field on all of the simply connected space ${\mathbb{R}}^3$, so its vanishing curl makes it a conservative vector field. A potential is also constructed explicitly in the next part. Thus ${\mathbf{v}}_2$ is a conservative vector field and ${\mathbf{v}}_1$ is not. A vanishing divergence alone, as for ${\mathbf{v}}_1$, is a different property and does not establish conservativity.