Applications of 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 2, 3) |
Quick Reference
Cap-and-subtract strategy: If $S$ is an open surface, close it by adding a cap $S_{\text{cap}}$. Then: $$\oiint_{S\cup S_{\text{cap}}}\mathbf{F}\cdot d\mathbf{S} = \iiint_E\operatorname{div}\mathbf{F}\,dV.$$ Subtract the cap’s contribution to recover $\iint_S\mathbf{F}\cdot d\mathbf{S}$.
Motivation
Many flux problems involve surfaces that are technically not closed (a paraboloid cap, the top half of a cylinder, etc.). The Divergence Theorem applies only to closed surfaces, so the trick is to close the surface by adding a convenient cap, use the theorem, then subtract the cap’s own flux. The cap is almost always something flat and easy to integrate over.
This mirrors the strategy in Stokes’ Theorem, where you replaced a hard line integral with a surface integral over a convenient surface. Here you are replacing a hard surface integral with a solid integral plus a simple cap.
Key Concept: Physical Meaning of Divergence
For a fluid with velocity field $\mathbf{F}$, $\operatorname{div}\mathbf{F}(P)$ is the net rate of fluid creation (or destruction) per unit volume at point $P$. A source has $\operatorname{div}\mathbf{F} > 0$; a sink has $\operatorname{div}\mathbf{F} < 0$; an incompressible fluid has $\operatorname{div}\mathbf{F} = 0$ everywhere.
The Divergence Theorem then says: the total outward flux (fluid leaving the region per unit time) equals the total source strength inside. If there are no sources or sinks inside $E$, the net outward flux through any closed surface around that region is zero -- whatever enters must leave.
Worked Example
Compute $\iint_S\mathbf{F}\cdot d\mathbf{S}$ where $\mathbf{F} = \langle xy, y^2, yz\rangle$ and $S$ is the surface of the cylinder $x^2+y^2\leq 1$, $0\leq z\leq 1$ (including top and bottom disks), with outward orientation. (Based on Stewart 16.9, Example 2.)
Divergence: $$\operatorname{div}\mathbf{F} = \frac{\partial(xy)}{\partial x}+\frac{\partial(y^2)}{\partial y}+\frac{\partial(yz)}{\partial z} = y + 2y + y = 4y.$$
Region $E$: The solid cylinder $x^2+y^2\leq 1$, $0\leq z\leq 1$.
Triple integral in cylindrical coordinates ($x = r\cos\theta$, $y = r\sin\theta$, $dV = r\,dr\,d\theta\,dz$):
$$\iiint_E 4y\,dV = \int_0^{2\pi}\int_0^1\int_0^1 4r\sin\theta\cdot r\,dz\,dr\,d\theta.$$
The $z$-integral gives a factor of $1$. The $\theta$-integral: $\int_0^{2\pi}\sin\theta\,d\theta = 0$.
$$\iiint_E 4y\,dV = 0.$$
The total outward flux is $0$. Physically, the field $y$ component is antisymmetric: as much fluid flows out the right side as flows in the left.
zero divergence means zero flux through every surface. $\operatorname{div}\mathbf{F} = 0$ at every point inside $E$ implies zero net outward flux through the boundary of $E$. But the flux through a single piece of the boundary (say, the top disk) need not be zero. The theorem equates the flux through the whole closed surface to the total divergence inside. Individual pieces can have nonzero flux as long as they cancel.
Common Misconceptions
if the net outward flux through a closed surface is zero, then $\mathbf{F}$ must be zero inside the region.
This is the concept-image-conflicts-definition error. The Divergence theorem equates zero total flux with zero total divergence inside: $\iiint_E \operatorname{div}\mathbf{F}\,dV = 0$. This means the sources and sinks inside cancel in total, not that $\mathbf{F}$ itself is zero. The field can be large and complex inside the region; positive divergence in one part and negative divergence in another simply cancel when integrated. A rotating (non-divergent) field, for instance, can produce substantial flow everywhere yet have zero net flux through any closed surface.
Leveled Practice
Problem 1. Let $\mathbf{F} = \langle x^2, y^2, z^2\rangle$. Find the outward flux through the cube $[0,1]^3$.
Show answer
$\operatorname{div}\mathbf{F} = 2x+2y+2z$.
$\iiint_E(2x+2y+2z)\,dV = \int_0^1\int_0^1\int_0^1(2x+2y+2z)\,dx\,dy\,dz$.
By symmetry each of $2x$, $2y$, $2z$ integrates to $2\cdot\frac{1}{2}\cdot 1\cdot 1 = 1$, so total $= 3$.