= Solution
For a continuously differentiable <vector field> $\mathbf F$ on a neighborhood of an oriented, piecewise smooth <surface> $S$, <Stokes theorem> states
$$
\boxed{\oint_{\partial S}\mathbf F\cdot d\mathbf r
=\iint_S(\nabla\times\mathbf F)\cdot\hat{\mathbf n}\,dS.}
$$
Here $\partial S$ is the boundary <curve>, $d\mathbf r$ is its directed tangent element, $\hat{\mathbf n}$ is the chosen <unit normal vector>, $dS$ is the <surface area element>, and $\nabla\times\mathbf F$ is the <curl>. The <right-hand rule> relates the boundary direction to $\hat{\mathbf n}$. Thus <circulation> around the edge equals the <flux> of <curl> through the spanning <surface>. Reversing both orientations changes both signs. One can replace a difficult <line integral> by an easier <surface integral>, or replace one spanning <surface> by another within the region of regularity.
For a bounded <volume> $V$ with piecewise smooth closed boundary and a continuously differentiable <vector field> on a neighborhood of its closure, the <divergence theorem> states
$$
\boxed{\iint_{\partial V}\mathbf F\cdot\hat{\mathbf n}_{\rm out}\,dS
=\iiint_V\nabla\cdot\mathbf F\,dV.}
$$
Here $\hat{\mathbf n}_{\rm out}$ points out of $V$, including into any holes, $dV$ is the <volume element>, and $\nabla\cdot\mathbf F$ is the <divergence>. It equates total outward <flux> to integrated source strength. It converts a closed <surface integral> into a <volume integral>, and explains why internal <fluxes> cancel in a subdivided region. Singularities cannot be included without checking the hypotheses.
For a <conservative force>, the <work> around every closed path is zero. <Stokes theorem> gives zero <flux> of <curl> through every sufficiently small patch. Dividing by its area and shrinking the patch shows that every component of the continuous <curl> vanishes:
$$
\boxed{\oint_C\mathbf F\cdot d\mathbf r=0,\qquad
\nabla\times\mathbf F=0,\qquad \mathbf F=-\nabla U.}
$$
Here $U$ is <potential energy>; conversely, zero <curl> implies a <conservative vector field> on a <simply connected domain>. The <divergence theorem> then gives the compatible relation $\iint_{\partial V}\mathbf F\cdot\hat{\mathbf n}\,dS=-\iiint_V\nabla^2U\,dV$. There is no general zero-<divergence> condition on a <conservative force>: $\mathbf F=-\mathbf r=-\nabla(r^2/2)$ is a concrete counterexample, with $\nabla\cdot\mathbf F=-3$.
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