Divergence theorem (Gauss's theorem) — statement, flux-divergence connection: Worked examples

This page is a case zoo. The parent note proved the theorem and showed three examples. Here we hunt down every kind of situation the theorem can hand you — positive and negative divergence, zero divergence, a degenerate (flat) region, a limiting shrink-to-a-point case, a singularity that breaks the rule, a physics word problem, and an exam trap. Each one is worked line by line, and each answers What / Why / What it looks like.

Before we start, one reminder in plain words, so no symbol sneaks in unearned.

The scenario matrix

Every problem you meet lands in one of these cells. The examples below are labelled with the cell they cover.

Cell	What is special	Sign of $\nabla \cdot F$	Example
A	Constant positive divergence (uniform source)	$> 0$	Ex 1
B	Positive divergence, varies with position	$> 0$	Ex 2
C	Negative divergence (net sink / inflow)	$< 0$	Ex 3
D	Exactly zero divergence (solenoidal)	$= 0$	Ex 4
E	Degenerate region — flat/zero-volume	any	Ex 5
F	Limiting case — shrink volume to a point	any	Ex 6
G	Singularity inside — theorem fails naively	undefined at a point	Ex 7
H	Real-world word problem (physics)	$> 0$	Ex 8
I	Exam twist — inward normal / hidden orientation	$> 0$	Ex 9

Nine cells, nine examples. Together they cover positive/negative/zero divergence, degenerate and limiting inputs, the failure mode, an application, and a trap.

Cell A — constant positive divergence

Worked example Ex 1 ·

$F = (2 x, 2 y, 2 z)$ out of the ball of radius $R$ Find the outward flux through the sphere of radius $R$ centred at the origin.

Forecast: the field points straight outward everywhere and grows with distance — guess: a big positive number that scales like $R^{3}$ .

Compute the divergence. $\nabla \cdot F = \frac{\partial ( 2 x )}{\partial x} + \frac{\partial ( 2 y )}{\partial y} + \frac{\partial ( 2 z )}{\partial z} = 2 + 2 + 2 = 6.$ Why this step? The theorem lets us swap the hard surface integral for a volume integral of this number, and here that number is a constant — the easiest possible integrand.
Integrate over the ball. A constant pulls out: $∭_{V} 6 d V = 6 \cdot Vol (V) = 6 \cdot \frac{4}{3} π R^{3} = 8 π R^{3} .$ Why this step? $∭_{V} 1 d V$ is by definition the volume, and the volume of a ball of radius $R$ is $\frac{4}{3} π R^{3}$ .

Answer: $Φ = 8 π R^{3}$ .

Verify: put $R = 1$ : $Φ = 8 π$ . Directly on the unit sphere $\hat{n} = (x, y, z)$ so $F \cdot \hat{n} = 2 (x^{2} + y^{2} + z^{2}) = 2$ , and $\iint_{S} 2 d S = 2 \cdot (4 π) = 8 π$ . ✓ The two sides agree, and the $R^{3}$ growth matches the forecast.

Figure — Divergence theorem (Gauss's theorem) — statement, flux-divergence connection

The figure shows the sphere with the field arrows getting longer as you move outward — that "spreading" is exactly what the positive divergence measures.

Cell B — positive but position-dependent divergence

Worked example Ex 2 ·

$F = (x^{2}, 0, 0)$ over the box $[0, 2] \times [0, 1] \times [0, 1]$ Find the total outward flux.

Forecast: the field only flows in $x$ and speeds up as $x$ grows, so more leaves the far face ( $x = 2$ ) than enters the near face ( $x = 0$ ). Expect a positive number.

Divergence. $\nabla \cdot F = \frac{\partial ( x ^{2} )}{\partial x} + 0 + 0 = 2 x .$ Why this step? Divergence is not constant now — it grows with $x$ . The theorem still applies; we just have a genuine integral to do.
Integrate. $∭_{V} 2 x d V = \int_{0}^{1} \int_{0}^{1} \int_{0}^{2} 2 x d x d y d z .$ Why this step? The box is a product region, so we can integrate one variable at a time.
Do the $x$ -integral first. $\int_{0}^{2} 2 x d x = [x^{2}]_{0}^{2} = 4.$ The remaining $y$ and $z$ integrals each span length $1$ : $4 \cdot 1 \cdot 1 = 4.$ Why this step? Integrating the innermost variable collapses the triple integral into a simple product because nothing depends on $y$ or $z$ .

Answer: $Φ = 4$ .

Verify (surface side). Only the two faces perpendicular to $x$ matter (on the four side faces $\hat{n}$ is perpendicular to $F$ , so $F \cdot \hat{n} = 0$ ).

Far face $x = 2$ : $\hat{n} = (+ 1, 0, 0)$ , $F \cdot \hat{n} = x^{2} = 4$ , area $= 1$ ⟹ $+ 4$ .
Near face $x = 0$ : $\hat{n} = (- 1, 0, 0)$ , $F \cdot \hat{n} = - x^{2} = 0$ , area $= 1$ ⟹ $0$ . Total $= 4$ . ✓ Matches, and the far face dominates as forecast.

Look at the arrows on the two shaded faces: short on the left ( $x = 0$ ), long on the right ( $x = 2$ ). The mismatch is the flux, and the growing arrow length across the box is the positive divergence $2 x$ .

Cell C — negative divergence (a net sink)

Compare with Ex 1's figure: there the arrows fled outward (source, positive flux); here every arrow dives inward through the skin (sink, negative flux). Same shape, opposite sign.

Cell D — zero divergence (solenoidal)

Worked example Ex 4 ·

$F = (y, z, x)$ through any closed surface Find the flux through an arbitrary closed surface $S$ .

Forecast: the components mix coordinates but none of them is a function of its own variable — divergence might vanish, giving $0$ for every region.

Divergence. $\nabla \cdot F = \frac{\partial y}{\partial x} + \frac{\partial z}{\partial y} + \frac{\partial x}{\partial z} = 0 + 0 + 0 = 0.$ Why this step? Each partial derivative is of a variable that doesn't appear in it, so each is zero. The field is divergence-free — solenoidal.
Integrate. $∭_{V} 0 d V = 0$ for any $V$ . Why this step? Integrating zero over anything gives zero — no need to know the shape of $S$ .

Answer: $Φ = 0$ for every closed surface.

Verify: the answer must be $0$ regardless of $V$ ; a solenoidal field creates nothing, so whatever flows in flows back out. ✓ (Compare Ex 3 of the parent, $F = (y, - x, 0)$ — same principle.)

The figure sketches a solenoidal flow through a circle: for every arrow entering (into the region) there is one leaving. The ins and outs balance exactly — that visual cancellation is what $Φ = 0$ looks like.

Cell E — degenerate region (zero volume)

Worked example Ex 5 ·

$F = (x, y, z)$ over a flat region (a disk in a plane) Someone asks: "apply the divergence theorem to the flat disk $z = 0$ , $x^{2} + y^{2} \leq 1$ ." What happens?

Forecast: a disk is 2D — it has area but no volume. If the theorem still speaks, the volume integral should be $0$ , so the flux should be $0$ too.

Check the hypothesis. The divergence theorem needs $V$ to be a solid (3D) region. A flat disk has $Vol = 0$ ; it is a degenerate input. Why this step? The theorem's right-hand side is $∭_{V} (\dots) d V$ . If $V$ has zero volume, that integral is $0$ no matter what $F$ is.
Interpret honestly. For a genuine 3D region the theorem gives flux. For a flat disk there is no "inside," so the correct reading is: the divergence theorem does not apply — you'd instead compute an ordinary surface flux of $F$ through the disk directly (which is generally not zero). Why this step? Degenerate cases are where careless students plug in and get $0$ ; the fix is recognising that $V$ must be a solid.

Answer: the divergence theorem returns $0$ for a zero-volume region, but that number is meaningless as a flux through a one-sided disk — the theorem simply isn't the right tool. Use a direct surface integral.

Verify: the flux of $F = (x, y, z)$ straight through the disk (normal $\hat{n} = (0, 0, 1)$ ) is $\iint_{disk} z d S$ , and on $z = 0$ this is $\iint 0 d S = 0$ here — coincidentally $0$ , but for a general field it would not be. The lesson: match the tool to a real solid. ✓ (the volume-integral $0$ is a hypothesis-failure artefact, not physics.)

Cell F — limiting case (shrink to a point)

Worked example Ex 6 ·

$F = (x^{2}, y^{2}, z^{2})$ over a tiny cube of side $ε$ at the origin Take the box $[0, ε]^{3}$ and look at flux-per-volume as $ε \to 0$ .

Forecast: the parent note says "divergence is flux per box." So $\frac{Φ}{Vol}$ should approach $\nabla \cdot F$ at the origin, which is $2 x + 2 y + 2 z = 0$ there. Guess: the ratio $\to 0$ .

Divergence. $\nabla \cdot F = 2 x + 2 y + 2 z .$ Volume side: $∭_{[0, ε]^{3}} (2 x + 2 y + 2 z) d V .$ Why this step? We want to see the local meaning of divergence emerge from a shrinking region.
Integrate. $\int_{0}^{ε} 2 x d x = ε^{2}$ , and the other two variables each contribute a factor $ε$ : so the $2 x$ term gives $ε^{2} \cdot ε \cdot ε = ε^{4}$ . By symmetry all three terms match: $Φ = 3 ε^{4} .$ Why this step? Each term integrates the same way over the symmetric box.
Divide by the volume $ε^{3}$ and take the limit. $\frac{Φ}{ε ^{3}} = \frac{3 ε ^{4}}{ε ^{3}} = 3 ε ε \to 0 0.$ Why this step? This ratio is the average divergence over the box; its limit is the divergence at the point — here $0$ , matching $2 (0) + 2 (0) + 2 (0)$ .

Answer: $\frac{Φ}{Vol} \to 0 = (\nabla \cdot F)_{origin} .$

Verify: the limit $lim_{ε \to 0} 3 ε = 0$ equals $\nabla \cdot F$ at the origin. ✓ This is the limiting definition of divergence recovered from the theorem.

Three cubes of shrinking side $ε$ nested at the origin: as the box collapses onto the point, the flux-per-volume number written beside each ( $3 ε$ ) marches down toward the divergence value $0$ at the centre.

Cell G — singularity inside (the theorem breaks)

Worked example Ex 7 ·

$F = \frac{r}{∣ r ∣ ^{3}}$ with the origin inside $S$ Here $r = (x, y, z)$ and $∣ r ∣ = x^{2} + y^{2} + z^{2}$ . Find the flux through a sphere of radius $R$ around the origin.

Forecast (the trap): away from the origin this field is divergence-free (you can check), so a naïve student says $∭ 0 d V = 0$ . But the field blows up at the origin — so beware.

Divergence away from the origin. A direct computation gives $\nabla \cdot F = 0$ for every point except $r = 0$ , where $F$ is undefined (division by zero). Why this step? The theorem's hypothesis demands continuous partials throughout $V$ — violated exactly at the enclosed point.
Do the honest surface integral. On the sphere of radius $R$ , $∣ r ∣ = R$ so $F = \frac{r}{R ^{3}}$ , and $\hat{n} = \frac{r}{R}$ . Then $F \cdot \hat{n} = \frac{r \cdot r}{R ^{3} \cdot R} = \frac{R ^{2}}{R ^{4}} = \frac{1}{R ^{2}} .$ Why this step? When the theorem can't be trusted, compute the flux directly on the surface.
Integrate over the sphere. $Φ = \iint_{S} \frac{1}{R ^{2}} d S = \frac{1}{R ^{2}} \cdot (4 π R^{2}) = 4 π .$ Why this step? $F \cdot \hat{n}$ is constant on the sphere, so we just multiply by the sphere's area $4 π R^{2}$ .

Answer: $Φ = 4 π$ — not $0$ , and independent of $R$ .

Verify: $\frac{1}{R ^{2}} \cdot 4 π R^{2} = 4 π$ for any $R$ . ✓ The naïve " $0$ " is wrong precisely because of the enclosed singularity — the fix is to exclude the origin (this is the mathematical heart of Gauss's law).

The picture shows the outer sphere $S$ with the origin (the spike) marked in pink at its centre. Because the field explodes there, we must carve out a tiny inner sphere before the theorem can be trusted — the reason the naïve " $0$ " fails.

Cell H — real-world word problem

Worked example Ex 8 · Heat leaving a warming metal cube

A $1 m$ cube $[0, 1]^{3}$ carries a heat-flux field $F = (3 x, 3 y, 3 z)$ measured in watts per square metre ( $W / m^{2}$ ). How much total heat power leaves the cube's surface?

Forecast: the flux field grows outward from the corner — heat is being generated inside. Expect a positive wattage.

Divergence. $\nabla \cdot F = 3 + 3 + 3 = 9$ (units: $W / m^{3}$ , a heat-source density). Why this step? Divergence of a flux field is the local generation rate per unit volume — the continuity picture.
Integrate over the cube. $∭_{[0, 1]^{3}} 9 d V = 9 \cdot (1 m)^{3} = 9 W .$ Why this step? Constant source density times the volume ( $1 m^{3}$ ) gives total power generated, which by the theorem equals total power radiated out the skin.

Answer: $9 W$ of heat leave the cube.

Verify (units & value). $[W / m^{3}] \times [m^{3}] = [W]$ ✓. Numerically $9 \cdot 1 = 9$ . Surface check: the far $x = 1$ face contributes $F \cdot \hat{n} = 3 x = 3$ over area $1$ = $3 W$ ; the near $x = 0$ face gives $0$ ; same for $y$ and $z$ ⟹ $3 \cdot 3 = 9 W$ . ✓

Cell I — exam twist (inward normal / hidden orientation)

Worked example Ex 9 · The sign trap

An exam asks: " $F = (x, y, z)$ ; compute $\iint_{S} F \cdot \hat{n} d S$ through the unit sphere, where $\hat{n}$ is the inward normal."

Forecast: the divergence theorem is built for the outward normal. Flipping to inward should flip the sign — expect $- 4 π$ , not $+ 4 π$ .

Compute the outward answer first. $\nabla \cdot F = 1 + 1 + 1 = 3$ , so outward flux $= 3 \cdot \frac{4}{3} π (1)^{3} = 4 π .$ Why this step? The theorem only speaks in the outward orientation, so get that number cleanly before doing anything else.
Flip the sign for the inward normal. The inward normal is $- \hat{n}_{out}$ , so $\iint_{S} F \cdot \hat{n}_{in} d S = \iint_{S} F \cdot (- \hat{n}_{out}) d S = - \iint_{S} F \cdot \hat{n}_{out} d S = - 4 π .$ Why this step? Reversing the normal reverses the sign of every dot product $F \cdot \hat{n}$ , hence of the whole integral. This is exactly the trap warned about in the parent's mistakes.

Answer: $Φ_{in} = - 4 π$ (while the outward $Φ_{out} = + 4 π$ ).

Verify: outward $= + 4 π$ (this is Ex 1 with $R = 1$ and half the field: here $\nabla \cdot = 3$ , volume $\frac{4}{3} π$ , product $4 π$ ); negate for inward $\Rightarrow - 4 π$ . ✓ Always read the orientation before touching the theorem.

Left circle: outward normals (blue) poking away, flux $+ 4 π$ . Right circle: the same field but inward normals (pink) poking in — every dot product flips sign, so the flux reads $- 4 π$ . One arrowhead direction is the whole difference.

Coverage check

Recall Did we hit every cell?

Positive constant divergence (A) ::: Ex 1 Position-dependent positive divergence (B) ::: Ex 2 Negative divergence / sink (C) ::: Ex 3 Zero divergence / solenoidal (D) ::: Ex 4 Degenerate zero-volume region (E) ::: Ex 5 Limiting shrink-to-a-point (F) ::: Ex 6 Singularity inside — theorem fails (G) ::: Ex 7 Real-world word problem with units (H) ::: Ex 8 Exam sign/orientation trap (I) ::: Ex 9

Connections

Divergence theorem (Gauss's theorem) — statement, flux-divergence connection — the parent statement and proof.
Flux integrals — the surface side we verified against in every example.
Divergence and curl — the scalar $\nabla \cdot F$ driving each answer.
Gauss's law (electromagnetism) — the singularity case (Ex 7) is its mathematical core.
Continuity equation — the source/sink reading used in Ex 8.
Stokes' theorem · Green's theorem — the circulation cousins (flow along, not across).