Math Core

The flux form of Green's theorem says the net outflow across a closed curve equals the total divergence inside it. The divergence theorem (also called Gauss's theorem) is the same statement one dimension up: the flux of a vector field out of a closed surface equals the triple integral of its divergence over the solid inside. It turns hard flux integrals into easy volume integrals, and it underlies conservation laws in physics.

The theorem

A closed surface is one that encloses a solid, like a sphere, the six faces of a cube, or a cylinder with its top and bottom. Closed surfaces get the outward orientation by default: the unit normal points away from the solid.

The divergence theorem

Let $E$ be a solid region whose boundary $S$ is a closed, piecewise-smooth surface with outward orientation. If $\mathbf{F}$ has continuous partial derivatives on an open region containing $E$ , then

\iint_S \mathbf{F}\cdot d\mathbf{S} = \iiint_E \operatorname{div}\mathbf{F}\,dV.

Why it's true. Think of $\mathbf{F}$ as fluid velocity. The divergence at a point is the outflow per unit volume from a tiny box there. Stack all the tiny boxes that fill $E$ : fluid leaving one box through a shared face enters its neighbor, so interior faces cancel. What's left is the flow through the outer boundary $S$ . Total sources inside equal net flow out.

Worked example: A sphere

Find the outward flux of $\mathbf{F} = \langle x, y, z\rangle$ across the sphere of radius $a$ centered at the origin.

$\operatorname{div}\mathbf{F} = 1 + 1 + 1 = 3$ , so the flux is $3\cdot\text{volume} = 3\cdot\dfrac{4}{3}\pi a^3 = 4\pi a^3$ . This matches the direct surface-integral computation from the previous lessons, with no parametrization needed.

Worked example: A variable divergence

Find the outward flux of $\mathbf{F} = \langle x^3,\ y^3,\ z^3\rangle$ across the unit sphere.

$\operatorname{div}\mathbf{F} = 3x^2 + 3y^2 + 3z^2 = 3\rho^2$ . In spherical coordinates, $dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ :

\iiint_E 3\rho^2\,dV = \int_0^{2\pi}\!\!\int_0^{\pi}\!\!\int_0^1 3\rho^4\sin\phi\,d\rho\,d\phi\,d\theta = \frac{3}{5}\cdot 2\cdot 2\pi = \frac{12\pi}{5}.

Worked example: Six faces at once

Find the outward flux of $\mathbf{F} = \langle x^2,\ yz,\ x + z\rangle$ across the boundary of the unit cube $0 \le x, y, z \le 1$ .

A direct computation would need six surface integrals. Instead, $\operatorname{div}\mathbf{F} = 2x + z + 1$ , and

\iiint_E (2x + z + 1)\,dV = \int_0^1\!\!\int_0^1\!\!\int_0^1 (2x + z + 1)\,dx\,dy\,dz = 1 + \frac{1}{2} + 1 = \frac{5}{2}.

(Each term integrates separately: $\int_0^1 2x\,dx = 1$ and $\int_0^1 z\,dz = \tfrac{1}{2}$ , over a cube of volume $1$ .)

Surfaces that aren't closed

The divergence theorem needs a closed surface. If you want the flux through an open surface, like a hemisphere or a paraboloid cap, you can often close it off with a simple flat piece, apply the theorem to the closed surface, and subtract the flux through the piece you added.

Worked example: Closing a hemisphere

Find the flux of $\mathbf{F} = \langle x,\ y,\ 1\rangle$ outward through the upper unit hemisphere $x^2 + y^2 + z^2 = 1$ , $z \ge 0$ .

Add the disk $D$ : $x^2 + y^2 \le 1$ in the plane $z = 0$ , with outward (downward) normal $-\mathbf{k}$ . Together they bound the solid half-ball $E$ .

Whole closed surface. $\operatorname{div}\mathbf{F} = 1 + 1 + 0 = 2$ , and $E$ has volume $\tfrac{2}{3}\pi$ , so the total outward flux is $\tfrac{4\pi}{3}$ .

Bottom disk. $\mathbf{F}\cdot(-\mathbf{k}) = -1$ , so the flux through $D$ is $-1\cdot\pi = -\pi$ .

Hemisphere. $\dfrac{4\pi}{3} - (-\pi) = \dfrac{7\pi}{3}$ .

Gauss's law

The inverse-square field $\mathbf{E} = \dfrac{q\,\mathbf{x}}{\lvert\mathbf{x}\rvert^3}$ (the electric field of a point charge at the origin, in suitable units) has $\operatorname{div}\mathbf{E} = 0$ everywhere except the origin, where it is undefined. Two cases:

If a closed surface $S$ does not enclose the origin, the divergence theorem applies and the flux is $0$ .
If $S$ does enclose the origin, cut out a tiny sphere around it. The divergence theorem on the region between them shows the flux through $S$ equals the flux through the tiny sphere, which a direct computation gives as $4\pi q$ .

So the flux depends only on the enclosed charge, not on the shape of the surface. That is Gauss's law.

Common mistake

Check two things before using the theorem: the surface must be closed (otherwise close it off and subtract), and $\mathbf{F}$ must be smooth everywhere inside. A field like $\mathbf{x}/\lvert\mathbf{x}\rvert^3$ that blows up inside the surface needs the cut-out argument above.

Tip

If $\operatorname{div}\mathbf{F} = 0$ , the outward flux through every closed surface (enclosing only points where $\mathbf{F}$ is smooth) is $0$ . Spot this before doing any integrals.