The Divergence Theorem
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 6.8: “The Divergence Theorem” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/6-8-the-divergence-theorem |
| Textbook used in class | Stewart, Calculus, Section 16.9: “The Divergence Theorem” (Examples 1, 2) |
Quick Reference
Divergence Theorem (Gauss’s Theorem): Let $E$ be a simple solid region bounded by a closed, outward-oriented surface $S = \partial E$. If $\mathbf{F}$ has continuous partial derivatives on an open set containing $E$, then $$\oiint_S \mathbf{F}\cdot d\mathbf{S} = \iiint_E \operatorname{div}\mathbf{F}\,dV.$$
Divergence: $$\operatorname{div}\mathbf{F} = \frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}.$$
Motivation
Computing the outward flux through a closed surface directly requires parametrizing every piece of that surface -- top, bottom, sides -- and adding the integrals. The Divergence Theorem replaces all of that with a single triple integral over the interior, which is often far simpler.
Physically, the divergence $\operatorname{div}\mathbf{F}$ measures how much the field spreads out (or converges) at each point. The theorem says the total outward flux through the boundary equals the total amount of spreading inside. A region with no sources or sinks ($\operatorname{div}\mathbf{F} = 0$) has zero net outward flux regardless of the shape of the surface.
Key Concept
The Divergence Theorem is the three-dimensional analogue of the Fundamental Theorem of Calculus. In one dimension: $\int_a^b f'(x)\,dx = f(b)-f(a)$; the derivative inside equals the boundary values outside. Here: $\iiint_E \operatorname{div}\mathbf{F}\,dV = \oiint_{\partial E}\mathbf{F}\cdot d\mathbf{S}$; the divergence inside equals the boundary flux outside.
The theorem requires $S$ to be a closed surface (no boundary of its own). A hemisphere is not closed; the hemisphere plus the disk that caps it is.
Worked Example
Use the Divergence Theorem to find $\oiint_S \mathbf{F}\cdot d\mathbf{S}$ where $\mathbf{F} = \langle x^3, y^3, z^3\rangle$ and $S$ is the sphere $x^2+y^2+z^2 = a^2$.
Divergence: $$\operatorname{div}\mathbf{F} = 3x^2+3y^2+3z^2.$$
Divergence Theorem: $$\oiint_S\mathbf{F}\cdot d\mathbf{S} = \iiint_E 3(x^2+y^2+z^2)\,dV,$$
where $E$ is the ball of radius $a$.
Spherical coordinates ($\rho,\phi,\theta$): $x^2+y^2+z^2 = \rho^2$, $dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta$.
$$= \int_0^{2\pi}\int_0^\pi\int_0^a 3\rho^2\cdot\rho^2\sin\phi\,d\rho\,d\phi\,d\theta = 3\cdot 2\pi\cdot 2\cdot\frac{a^5}{5} = \frac{12\pi a^5}{5}.$$
(Intermediate steps: $\int_0^{2\pi}d\theta = 2\pi$; $\int_0^\pi\sin\phi\,d\phi = 2$; $\int_0^a\rho^4\,d\rho = a^5/5$.)
Doing this directly would require parametrizing the sphere and computing $x^3, y^3, z^3$ in terms of $\phi$ and $\theta$ -- the Divergence Theorem saves the effort.
the Divergence Theorem applies to any surface, open or closed. It applies only to a closed surface $S = \partial E$ that bounds a solid region $E$. An open surface (like a paraboloid cap without a bottom disk) has no enclosed region $E$, so the theorem does not apply. If you try to use it on an open surface, the integral $\iiint_E \operatorname{div}\mathbf{F}\,dV$ has no well-defined region to integrate over.
Common Misconceptions
the Divergence theorem can be applied to an open surface by treating the surface integral as a double integral over the parameter domain.
This is the concept-image-conflicts-definition error. The Divergence theorem requires a closed surface $S = \partial E$ that completely encloses a solid region $E$; without a closed surface there is no region $E$ over which to integrate $\operatorname{div}\mathbf{F}$. An open surface such as the upper hemisphere has a boundary circle, not a solid interior, so the theorem does not directly apply. The correct procedure for an open surface is to add a cap to close it, apply the theorem to the closed surface, then subtract the cap’s contribution.
Leveled Practice
Problem 1. Compute $\oiint_S\mathbf{F}\cdot d\mathbf{S}$ where $\mathbf{F} = \langle x, y, z\rangle$ and $S$ is the unit sphere $x^2+y^2+z^2=1$.
Show answer
$\operatorname{div}\mathbf{F} = 1+1+1 = 3$.
$\iiint_E 3\,dV = 3\cdot\frac{4\pi}{3} = 4\pi$.
Sanity check via direct method: on the unit sphere, $\mathbf{F} = \hat{\mathbf{n}}$, so $\mathbf{F}\cdot\hat{\mathbf{n}} = 1$, and $\oiint_S 1\,dS = 4\pi$. Checks out.