= Solution
Orient the planar <material curve> $C(t)$ counterclockwise, and let $\mathbf t$ and $\mathbf n$ be its unit tangent and outward <unit normal vector>. The <circulation> and outward <volume flux> are
$$
\boxed{\Gamma(t)=\oint_{C(t)}\mathbf u\cdot\mathbf t\,ds,\qquad M(t)=\oint_{C(t)}\mathbf u\cdot\mathbf n\,ds.}
$$
For $\mathbf u=(u,v)$ and this orientation, $\mathbf t\,ds=(dx,dy)$ and $\mathbf n\,ds=(dy,-dx)$, giving
$$
\Gamma=\oint_C(u\,dx+v\,dy),\qquad M=\oint_C(u\,dy-v\,dx).
$$
Here $M$ is the instantaneous <flux> of the <velocity field>, rather than transport relative to the moving <material curve>; that relative transport is zero. If the enclosed region contains no source and the <velocity field> is smooth there, <incompressibility> and the planar <divergence theorem> imply $M=0$.
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