The Divergence Theorem, multivariable and vector calculus lesson

Pump air into a balloon and the net amount leaving through the rubber each second must come from somewhere inside. If you know how fast air is being created or squeezed at each point within, you can add that up instead of measuring at the surface. The divergence theorem turns this bookkeeping into an exact equation between a flux integral and a triple integral.

Key terms

Closed oriented boundary

Definition Let $E$ be a bounded solid region in space whose boundary $\partial E$ is a finite union of closed piecewise smooth surfaces. The boundary is oriented by the outward unit normal $\mathbf{n}$ , which at each smooth point is perpendicular to $\partial E$ and points away from $E$ .

In plain words The skin of a solid, with every normal arrow pointing out of the solid.

Example For the unit cube $E = [0, 1]^3$ the boundary has six faces. The face $x = 1$ has outward normal $\mathbf{i}$ , and the face $x = 0$ has outward normal $-\mathbf{i}$ . The same pattern holds for the faces in $y$ and $z$ .

Outward flux

Definition For a vector field $\mathbf{F}$ continuous on $\partial E$ , the outward flux is

\iint_{\partial E} \mathbf{F} \cdot \mathbf{n} \, dS,

computed with the outward normal on every piece and summed over the pieces.

In plain words Net amount leaving the solid per unit time. Positive means more goes out than comes in.

Example For $\mathbf{F} = \langle x^2, y, z \rangle$ on the unit cube, look at the faces in pairs. On $x = 1$ the integrand is $\mathbf{F} \cdot \mathbf{i} = 1$ , and on $x = 0$ it is $-x^2 = 0$ . The $y$ and $z$ faces behave the same way, each pair giving $1$ . The outward flux is $3$ .

Divergence over a solid

Definition For $\mathbf{F} = \langle P, Q, R \rangle$ with continuous first partial derivatives,

\nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z},

and its integral over the solid is $\iiint_E \nabla \cdot \mathbf{F} \, dV$ .

In plain words Divergence is the outflow rate per unit volume at a point. Each small cell of the solid is a little source or sink, and the triple integral adds them all.

Example For $\mathbf{F} = \langle x^2, y, z \rangle$ the divergence is $2x + 2$ , and

\begin{aligned} \int_0^1 \!\! \int_0^1 \!\! \int_0^1 (2x + 2) \, dx \, dy \, dz &= 1 + 2 \\[0.4em] &= 3, \end{aligned}

the same as the outward flux.

The divergence theorem and its regularity assumptions

Definition Let $E$ be a bounded solid region whose boundary $\partial E$ is a finite union of closed piecewise smooth surfaces, oriented by the outward unit normal. Let $\mathbf{F}$ have components with continuous first partial derivatives on an open set containing $E$ and its boundary. Then

\iint_{\partial E} \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_E \nabla \cdot \mathbf{F} \, dV.

Every hypothesis matters. The surface must be the complete boundary of $E$ , every piece must carry the outward normal, and $\mathbf{F}$ must be $C^1$ at every point of $E$ , inside and on the boundary. Drop any one and the equation can fail, as the singular field below shows.

In plain words Outflow through the skin equals the sum of all the sources inside. Neighboring cells exchange flow across shared walls, those exchanges cancel, and only the outer walls remain.

Example For $\mathbf{F} = \langle x^3, y^3, z^3 \rangle$ and the ball of radius $a$ , the divergence is $3(x^2 + y^2 + z^2) = 3\rho^2$ . In spherical coordinates the outward flux is

\begin{aligned} &\int_0^{2\pi} \!\! \int_0^\pi \!\! \int_0^a 3\rho^2 \cdot \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta \\[0.4em] &= 3 \cdot \frac{a^5}{5} \cdot 2 \cdot 2\pi = \frac{12\pi a^5}{5}. \end{aligned}

Cavities

Definition If $E$ has holes inside it, its boundary has several pieces, an outer surface and one inner surface around each cavity. The outward normal of $E$ on an inner surface points away from $E$ , which means into the cavity. The theorem holds with all boundary pieces included, each with this orientation.

In plain words On the wall of a hollow, “out of the solid” means into the hole. Measured with the usual normal of the inner sphere, that flux enters with a minus sign.

Example Let $E$ be the shell $1 \le \rho \le 2$ and $\mathbf{F} = \langle x, y, z \rangle$ , so $\nabla \cdot \mathbf{F} = 3$ .

Piece	$\mathbf{F} \cdot \mathbf{n}$	Area	Flux
Outer sphere $\rho = 2$	$2$	$16\pi$	$32\pi$
Inner sphere $\rho = 1$	$-1$	$4\pi$	$-4\pi$

The total is $28\pi$ , and $3$ times the volume $\frac{4}{3}\pi(8 - 1)$ is also $28\pi$ .

Excluded singularities

Definition Suppose $\mathbf{F}$ is $C^1$ on an open set containing $E$ except at one point $P_0$ in the interior of $E$ . Then the theorem does not apply to $E$ itself. Choose $\varepsilon > 0$ so small that the closed ball $B$ of radius $\varepsilon$ centered at $P_0$ lies in the interior of $E$ , and apply the theorem to the solid $E$ with the open ball removed. Its boundary is $\partial E$ together with the sphere $\partial B$ , and on that sphere the outward normal of the new solid points toward $P_0$ . If $\nabla \cdot \mathbf{F} = 0$ away from $P_0$ , this gives

\iint_{\partial E} \mathbf{F} \cdot \mathbf{n} \, dS = \iint_{\partial B} \mathbf{F} \cdot \mathbf{n}_{\text{out}} \, dS,

where $\mathbf{n}_{\text{out}}$ points away from $P_0$ .

In plain words Cut the bad point out with a tiny sphere. Then the flux through the big surface equals the flux through the tiny one, which is usually easy to compute.

Example Let $\mathbf{F} = \frac{\mathbf{r}}{\lvert \mathbf{r} \rvert^3}$ with $\mathbf{r} = \langle x, y, z \rangle$ . Away from the origin a computation gives $\nabla \cdot \mathbf{F} = 0$ . On a sphere of radius $\varepsilon$ the field has length $1/\varepsilon^2$ and points outward, so the flux is

\frac{1}{\varepsilon^2} \cdot 4\pi \varepsilon^2 = 4\pi.

The cube $[-1, 1]^3$ contains the ball of radius $\varepsilon = \frac{1}{2}$ about the origin in its interior, so its outward flux is $4\pi$ , not $0$ , even though the divergence is zero wherever it is defined. A direct computation agrees, since each of the six faces carries flux $\frac{2\pi}{3}$ by symmetry. If the origin lies outside a solid, the field is $C^1$ on an open set containing the solid and the flux through its boundary is $0$ .

Gauss’s law

Definition A point charge $q$ at the origin produces the electric field

\mathbf{E} = \frac{q}{4\pi\varepsilon_0} \frac{\mathbf{r}}{\lvert \mathbf{r} \rvert^3},

where $\varepsilon_0$ is the permittivity of free space. For any closed piecewise smooth surface $S$ bounding a solid $E$ , with the origin not on $S$ , the outward flux of $\mathbf{E}$ is $q/\varepsilon_0$ if the origin is inside $E$ and $0$ if it is outside. For several point charges, none of them on $S$ , the fields add, and the flux is $Q/\varepsilon_0$ , where $Q$ is the total charge inside $E$ .

In plain words The electric flux through a closed surface counts only the charge inside, no matter the shape of the surface. It is the previous example with a constant in front.

Example A charge of $q = 10^{-9}$ coulombs sits at the origin. In SI units the constant is

\varepsilon_0 \approx 8.854 \times 10^{-12} \ \frac{\text{C}^2}{\text{N} \cdot \text{m}^2},

and the field is measured in newtons per coulomb. Multiplying by area in square meters, the flux out of any closed surface enclosing the charge is

\frac{q}{\varepsilon_0} \approx 112.9 \ \frac{\text{N} \cdot \text{m}^2}{\text{C}}.

Through the surface of the box $[1, 2] \times [0, 1] \times [0, 1]$ it is $0$ , since the charge is outside.

Relationship among integral theorems

Definition The main theorems of vector calculus share one shape. The integral of a derivative over an oriented region equals the integral of the original object over its oriented boundary.

Theorem	Integral over the region	Boundary term
Fundamental theorem	$\int_a^b f'(x)\, dx$	$f(b) - f(a)$
Line integrals	$\int_C \nabla f \cdot d\mathbf{r}$	$f(B) - f(A)$ for $C$ from $A$ to $B$
Green	$\iint_D (Q_x - P_y) \, dA$	$\oint_{\partial D} P\,dx + Q\,dy$
Stokes	$\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$	$\oint_{\partial S} \mathbf{F} \cdot d\mathbf{r}$
Divergence	$\iiint_E \nabla \cdot \mathbf{F} \, dV$	$\iint_{\partial E} \mathbf{F} \cdot d\mathbf{S}$

Each one needs its own orientation of the boundary and a derivative that is continuous on the closed region.

In plain words Adding up a rate of change inside gives the net change measured at the edge. The edge is two points, a closed curve or a closed surface, depending on the dimension.

Example The flat version of the divergence theorem is the flux form of Green’s theorem, with $\mathbf{n}$ the outward normal of a plane region. For $\mathbf{F} = \langle x, y \rangle$ on the unit disk, $\mathbf{F} \cdot \mathbf{n} = 1$ on the circle, so the outward flux is $2\pi$ . The divergence is $2$ , and $2$ times the area $\pi$ is also $2\pi$ .

Check yourself

Find the outward flux of $\mathbf{F} = \langle x^3, y^3, z \rangle$ through the closed surface of the solid cylinder $x^2 + y^2 \le 1$ , $0 \le z \le 2$ .

Show the answer

The divergence is $3x^2 + 3y^2 + 1 = 3r^2 + 1$ , so in cylindrical coordinates

\begin{aligned} &\int_0^2 \!\! \int_0^{2\pi} \!\! \int_0^1 (3r^2 + 1)\, r \, dr \, d\theta \, dz \\[0.4em] &= 2 \cdot 2\pi \cdot \Big(\frac{3}{4} + \frac{1}{2}\Big) = 5\pi. \end{aligned}

Find the upward flux of $\mathbf{F} = \langle x, y, z \rangle$ through the paraboloid $z = 1 - x^2 - y^2$ , $z \ge 0$ , by closing it with the disk at the bottom.

Show the answer

The divergence is $3$ and the solid has volume $\frac{\pi}{2}$ , so the outward flux through the closed surface is $\frac{3\pi}{2}$ . On the bottom disk $z = 0$ with outward normal $-\mathbf{k}$ the integrand is $-z = 0$ . So the flux through the paraboloid is $\frac{3\pi}{2}$ .

For $\mathbf{F} = \frac{\mathbf{r}}{\lvert \mathbf{r} \rvert^3}$ , find the outward flux through the sphere $(x - 3)^2 + y^2 + z^2 = 1$ and through the ellipsoid $\frac{x^2}{4} + y^2 + \frac{z^2}{9} = 1$ .

Show the answer

The first sphere does not enclose the origin, so the field is $C^1$ on the ball it bounds and the flux is $0$ . The ellipsoid encloses the origin, so cutting out a small ball gives $4\pi$ .

The Divergence Theorem

Key terms

Closed oriented boundary

Outward flux

Divergence over a solid

The divergence theorem and its regularity assumptions

Cavities

Excluded singularities

Gauss’s law

Relationship among integral theorems

Check yourself

How this lesson connects