A fluid moving in the plane can do two things relative to a curve: flow along it or flow across it. Along is measured by the tangential component $\mathbf{F} \cdot \mathbf{T}$, which you met in work integrals. Across is measured by the normal component $\mathbf{F} \cdot \mathbf{n}$. Around a closed curve, the first total is the circulation, and the second is the flux, the net rate at which fluid leaves the region inside.
Key idea
For a closed curve $C$ traced counterclockwise and $\mathbf{F} = \langle P, Q \rangle$, with $\mathbf{T}$ the unit tangent and $\mathbf{n}$ the outward unit normal:
$$\text{circulation} = \oint_C \mathbf{F} \cdot \mathbf{T}\,ds = \oint_C P\,dx + Q\,dy, \qquad \text{flux} = \oint_C \mathbf{F} \cdot \mathbf{n}\,ds = \oint_C P\,dy - Q\,dx.$$
For a curve that is not closed, the flux $\int_C P\,dy - Q\,dx$ uses the normal pointing to the right of the direction of travel.
Why it works
$\mathbf{T}\,ds = \langle dx, dy \rangle$. Turning a vector a quarter turn clockwise sends $\langle a, b \rangle$ to $\langle b, -a \rangle$ and keeps its length, so $\mathbf{n}\,ds = \langle dy, -dx \rangle$ is a normal of the same length, pointing to the right of the direction of travel. Walking counterclockwise around a region, the region is on your left, so the right side is the outside. Then $\mathbf{F} \cdot \mathbf{n}\,ds = P\,dy - Q\,dx$.
Worked examples
Worked example: The radial field
Find the circulation and the outward flux of $\mathbf{F} = \langle x, y \rangle$ around the circle of radius $R$ centered at the origin.
- $x = R\cos t$, $y = R\sin t$; $dx = -R\sin t\,dt$, $dy = R\cos t\,dt$.
- Circulation: $$P\,dx + Q\,dy = (-R^2\cos t\sin t + R^2\sin t\cos t)\,dt = 0.$$
- Flux: $$P\,dy - Q\,dx = (R^2\cos^2 t + R^2\sin^2 t)\,dt = R^2\,dt,$$ so the flux is ${}2\pi R^2$.
- The field points straight out: it crosses the circle everywhere and never runs along it.
Answer: circulation ${}0$, flux ${}2\pi R^2$
Worked example: The rotation field
Repeat for $\mathbf{F} = \langle -y, x \rangle$.
- Circulation: $$(-R\sin t)(-R\sin t) + (R\cos t)(R\cos t) = R^2$$ per $dt$: total ${}2\pi R^2$.
- Flux: $$P\,dy - Q\,dx = (-R\sin t)(R\cos t) - (R\cos t)(-R\sin t) = 0.$$
Answer: circulation ${}2\pi R^2$, flux ${}0$
Worked example: Flux across a segment
Find the flux of $\mathbf{F} = \langle x, y + 1 \rangle$ across the segment from $(0, 0)$ to $(2, 0)$, normal to the right of travel.
- $\mathbf{r}(t) = \langle 2t, 0 \rangle$: $dx = 2\,dt$, $dy = 0$, and $y = 0$.
- $$P\,dy - Q\,dx = 0 - (0 + 1)(2)\,dt = -2\,dt,$$ so the flux is $\int_0^1 -2\,dt = -2$.
- Check: traveling right, the right-hand normal points down, $\langle 0, -1 \rangle$. Then $\mathbf{F} \cdot \mathbf{n} = -1$ along a segment of length ${}2$.
Answer: $-2$
Common mistake
Using the tangent for flux. $\oint P\,dx + Q\,dy$ is circulation, not flux. Flux across a curve needs the normal: $\oint P\,dy - Q\,dx$. Mixing them up swaps the answers in the first two examples: the radial field would seem to circulate and the rotation field would seem to flow out.