= Solution
{c}
The sufficiency claim requires a domain qualification. For a continuously differentiable <vector field> on all of $\mathbb R^3$, or more generally on a <simply connected> open domain, the equivalence is valid. On an arbitrary domain, vanishing <curl> is necessary but may fail to be sufficient.
\b[Necessity.] Suppose every closed-curve <line integral> vanishes. Fix a point $p$ in the open domain and a <unit vector> $a$. A sufficiently small planar disk $D_\varepsilon$ of radius $\varepsilon$, centred at $p$ and normal to $a$, lies in the domain. <Stokes theorem> gives
$$
0=\oint_{\partial D_\varepsilon}\mathbf F\cdot d\mathbf r
=\iint_{D_\varepsilon}(\nabla\times\mathbf F)\cdot a\,dS.
$$
Divide by $\pi\varepsilon^2$ and let $\varepsilon\downarrow0$. By <continuity>, the average tends to $(\nabla\times\mathbf F)(p)\cdot a$. Taking the three coordinate directions proves $\boxed{\nabla\times\mathbf F=0}$ everywhere.
\b[Sufficiency on $\mathbb R^3$.] A direct proof constructs a <potential of a conservative vector field>. Put
$$
\phi(x)=\int_0^1\mathbf F(tx)\cdot x\,dt.
$$
Write $F_j$ for the components of $\mathbf F$. Vanishing <curl> says $\partial_iF_j=\partial_jF_i$. Differentiation under the integral, followed by the <chain rule>, gives
$$
\begin{aligned}
\partial_i\phi(x)
&=\int_0^1\left[F_i(tx)+t\sum_{j=1}^3x_j\partial_iF_j(tx)\right]dt\\
&=\int_0^1\left[F_i(tx)+t\frac d{dt}F_i(tx)\right]dt
=\bigl[tF_i(tx)\bigr]_0^1=F_i(x).
\end{aligned}
$$
This is the <radial integral potential for a curl-free field>. The <fundamental theorem for line integrals> now gives zero on every closed curve. On a <simply connected> open domain the analogous argument uses path integrals: every loop can be contracted within the domain, and applying <Stokes theorem> to a piecewise smooth contraction makes its <circulation> zero. Equivalently, whenever a particular loop bounds an oriented spanning surface entirely in the domain, <Stokes theorem> immediately gives zero for that loop.
\b[Why the domain matters.] On $U=\mathbb R^3\setminus\{(0,0,z):z\in\mathbb R\}$, the <smooth> <vector field>
$$
\mathbf F=\left(-\frac y{x^2+y^2},\frac x{x^2+y^2},0\right)
$$
has zero <curl>, but its <line integral> around the unit <circle> in $z=0$ is $2\pi$, as in <1/b/ii>. No spanning surface for this loop can lie entirely in $U$. The literal claim without a condition on the domain therefore has this counterexample.
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