2D test for conservative fields

The cross-partials check: $\mathbf{F}=(P,Q)$ is conservative on a simply connected region exactly when $P_y=Q_x$ everywhere.

Computing two derivatives once and comparing them replaces checking every closed loop by hand: if $\partial P/\partial y$ and $\partial Q/\partial x$ agree everywhere on a simply connected region, the field is guaranteed conservative; if they disagree anywhere, that point is a counterexample — a paddle wheel dropped there will spin, ruling out the whole field. On a region with a hole (not simply connected), $P_y=Q_x$ can hold everywhere $\mathbf{F}$ is defined without $\mathbf{F}$ being conservative.

Examples

For $\mathbf{F}_1=(2xy,\ x^2+3y^2)$ : $P_y=2x=Q_x$ everywhere, so $\mathbf{F}_1$ is conservative.
For $\mathbf{F}_2=(-y,\ x)$ : $P_y=-1 \ne 1=Q_x$ everywhere, so $\mathbf{F}_2$ is NOT conservative.
$\mathbf{F} = \left(\frac{-y}{x^2+y^2},\ \frac{x}{x^2+y^2}\right)$ passes the cross-partials check away from the origin but has nonzero circulation around any loop encircling the missing point, since its domain has a hole.