= Solution
For smooth <inviscid flow> of constant <mass density> under a conservative <body force>, <Kelvin's circulation theorem> states that the <circulation> around any closed <material curve> is conserved:
$$
\boxed{\frac{d\Gamma}{dt}=0.}
$$
The constant-<mass density> assumption is the usual homogeneous <incompressible flow> setting here. Merely requiring $\nabla\cdot\mathbf u=0$ in a fluid of varying <mass density> would not by itself make the pressure term below a <gradient>.
Parameterize the <material curve> by a periodic particle label $s\in[0,1]$, writing $\mathbf r_t(s,t)=\mathbf u(\mathbf r(s,t),t)$. Differentiate its <line integral>, keeping $s$ fixed:
$$
\begin{aligned}
\frac{d\Gamma}{dt}
&=\int_0^1\left[\frac{D\mathbf u}{Dt}\cdot\mathbf r_s+\mathbf u\cdot\partial_s\mathbf r_t\right]ds\\
&=\oint_{C(t)}\frac{D\mathbf u}{Dt}\cdot d\mathbf r+\int_0^1\partial_s\left(\frac12|\mathbf u|^2\right)ds\\
&=\oint_{C(t)}\frac{D\mathbf u}{Dt}\cdot d\mathbf r.
\end{aligned}
$$
The second term vanishes by periodicity. If the conservative <body force> per unit <mass> is $-\nabla\Omega$, the <Euler momentum equation> gives
$$
\frac{D\mathbf u}{Dt}=-\frac1\rho\nabla p-\nabla\Omega=-\nabla(p/\rho+\Omega).
$$
Its <line integral> around a <closed curve> is zero, proving the claim. The <velocity field> must remain smooth on the moving curve, which cannot pass through a <line vortex> singularity during this argument. No assumption of <irrotational flow> was needed.
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