= Green theorem circulation examples
{c}
Take the <vector field> $F=(-2y,3x)$. Its scalar <curl> is the constant $3-(-2)=5$, so <Green theorem> predicts that its positively oriented boundary <line integral> is five times the enclosed <area>. This one field makes both a round and a rectangular calculation transparent.
For the <circle> $x^2+y^2=4$, choose the counterclockwise <parametric equations> $\gamma(t)=(2\cos t,2\sin t)$, $0\leq t\leq2\pi$. Direct substitution in the <line integral> gives
$$
\oint_C(-2y\,dx+3x\,dy)
=\int_0^{2\pi}(8\sin^2t+12\cos^2t)\,dt
=20\pi.
$$
The interior <double integral> is shorter: the <disk> has <area> $4\pi$, hence $\iint_D5\,dA=20\pi$. Reversing the <boundary orientation> gives $-20\pi$.
For the <rectangle> $D=[0,3]\times[0,2]$, traverse its bottom, right, top, and left edges in that order. The four <line integrals> are respectively
$$
\int_0^3 0\,dx=0,\qquad
\int_0^2 9\,dy=18,\qquad
\int_3^0(-4)\,dx=12,\qquad
\int_2^0 0\,dy=0.
$$
Their sum is $30$, agreeing with $\iint_D5\,dA=5\cdot3\cdot2$. In both examples the path geometry changes the separate contributions, while constant <curl> makes only <area> matter to the final answer.
\Image[/green-theorem-circle-rectangle-circulation.png]
{title=One vector field, two positively oriented boundaries: scalar curl five gives circulation twenty pi around the radius-two circle and thirty around the three-by-two rectangle}
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