Green's Theorem: Flux Form

Flux across a closed curve measures net outflow, and divergence is outflow per unit area. Green's theorem has a second form that says what you would hope: the outward flux across the boundary equals the total divergence inside.

Key idea

Green's theorem (flux form). Under the same conditions as the circulation form, with $\mathbf{n}$ the outward unit normal, $$\oint_C \mathbf{F} \cdot \mathbf{n}\,ds = \oint_C P\,dy - Q\,dx = \iint_D \left(\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}\right) dA.$$

Why it works

Apply the circulation form to the field $\mathbf{G} = \langle -Q, P \rangle$ (the field $\mathbf{F}$ turned a quarter turn): $$\oint_C (-Q)\,dx + P\,dy = \iint_D \left(\frac{\partial P}{\partial x} - \frac{\partial(-Q)}{\partial y}\right) dA = \iint_D (P_x + Q_y)\,dA.$$ The left side is exactly the flux integral $\oint P\,dy - Q\,dx$. In words: the flow of $\mathbf{F}$ across $C$ is the flow of the turned field along $C$.

Outward flux of $\langle x^2, y \rangle$ from small cells. Flow that leaves one cell enters its neighbor, so inside edges cancel. Blue cells are sources (divergence ${}2x + 1 > 0$); red cells near $x = -1$ take in more than they give out.

The two forms of Green's theorem match the two derivatives of a plane field. Circulation goes with the scalar curl $Q_x - P_y$; flux goes with the divergence $P_x + Q_y$. A field can have one without the other: the rotation field $\langle -y, x \rangle$ has curl ${}2$ and divergence ${}0$, so it circulates around every circle but has no net flow out of any region, while the radial field $\langle x, y \rangle$ is the reverse. Read physically, the flux form is a conservation law: in a region with no sources or sinks ($\nabla \cdot \mathbf{F} = 0$), whatever flows in across the boundary must flow out somewhere else.

Worked examples

Worked example: The radial field again

Find the outward flux of $\mathbf{F} = \langle x, y \rangle$ across the circle of radius $R$.

$P_x + Q_y = 2$.
$\iint_D 2\,dA = 2\pi R^2$, the same as the direct computation in the lesson on circulation and flux.

Answer: ${}2\pi R^2$

Worked example: A square

Find the outward flux of $\mathbf{F} = \langle x^2, y^2 \rangle$ across the boundary of $[0, 1] \times [0, 1]$.

$P_x + Q_y = 2x + 2y$.
$\int_0^1 \int_0^1 (2x + 2y)\,dy\,dx = 1 + 1 = 2$.
Check: on the right side $\mathbf{n} = \langle 1, 0 \rangle$ and $\mathbf{F} \cdot \mathbf{n} = 1$; on the top $\mathbf{F} \cdot \mathbf{n} = 1$; on the left and bottom $\mathbf{F} \cdot \mathbf{n} = 0$. Total ${}2$.

Answer: ${}2$

Worked example: No net flux

Show that $\mathbf{F} = \langle y^3, x \rangle$ has zero outward flux across every simple closed curve.

$P_x + Q_y = 0 + 0 = 0$ everywhere.
By the flux form the outward flux is $\iint_D 0\,dA = 0$ for any region: whatever flows in also flows out.

Answer: ${}0$

Green's Theorem: Flux Form

Worked examples

Practice

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