Divergence theorem (Gauss's theorem) — statement, flux-divergence connection: Question bank

This page assumes only the vocabulary the parent note built: flux $Φ = \iint_{S} F \cdot \hat{n} d S$ (flow across the skin), divergence $\nabla \cdot F = \partial_{x} P + \partial_{y} Q + \partial_{z} R$ (source strength per unit volume), and the theorem $\iint_{S} F \cdot \hat{n} d S = ∭_{V} (\nabla \cdot F) d V$ . Everything else is defined as it appears.

True or false — justify

The divergence theorem needs

S

to be a closed surface.

True — a "leaky" open surface has no enclosed volume

V

, so the right-hand side

∭_{V}

is undefined; only a boundary

S = \partial V

that fully seals a solid qualifies.

\nabla \cdot F = 0

everywhere inside

V

, the flux out of

S

must be zero.

True provided

F

has no singularity in

V

: then

∭_{V} 0 d V = 0

. If a singularity hides inside, the integrand isn't

0

there and the flux can be nonzero (e.g.

r /∣ r ∣^{3}

gives

4 π

Two different closed surfaces enclosing the same region always give the same flux.

True by definition — "the same region" means the same

V

, and the flux equals

∭_{V} (\nabla \cdot F) d V

, which depends only on

V

and

F

, not on how you describe the skin.

Two closed surfaces enclosing different regions can still give equal flux.

True — if

\nabla \cdot F = 0

in the shell between them, the extra volume contributes zero, so both surfaces report the same flux (this is the trick behind Gauss's law).

A field that is large everywhere on

S

must have large flux.

False — only the component along

\hat{n}

counts. A huge field flowing tangent to the surface (like

(y, - x, 0)

swirling around) contributes nothing to

F \cdot \hat{n}

Positive divergence at a point guarantees positive total flux out of any surface around it.

False — total flux sums divergence over the whole volume; strong sinks (negative divergence) elsewhere in

V

can outweigh one positive point, making net flux negative.

The theorem is just a 3D copy of the fundamental theorem of calculus.

True in spirit — both trade an integral over a region for an evaluation on its boundary. FTC:

\int_{a}^{b} f^{'} = f (b) - f (a)

; Gauss:

∭_{V} \nabla \cdot F = \iint_{\partial V} F \cdot \hat{n}

. The boundary of an interval is two points; the boundary of a solid is its skin.

You may apply the divergence theorem to

F = (1/ x, 0, 0)

on the cube

[0, 1]^{3}

False —

1/ x

and its partial blow up at the face

x = 0

, so

F

lacks continuous first partials throughout

V

; the theorem's hypothesis is violated.

Divergence theorem and Stokes' theorem can be applied to the same integral interchangeably.

False — Gauss converts a flux (

F \cdot \hat{n}

, across) into a volume integral; Stokes converts circulation (

F \cdot d r

, along a curve) into a surface integral of curl. Different dot products, different objects.

Reversing to the inward normal changes only the sign of the flux, not its size.

True —

\hat{n} \to - \hat{n}

flips

F \cdot \hat{n}

everywhere, giving

- ∭_{V} \nabla \cdot F d V

; magnitude is unchanged but the theorem as stated requires the outward choice.

Spot the error

\nabla \cdot F = 0

F

has no field lines crossing

S

Wrong conclusion — divergence-free means net flux is zero (lines in = lines out), not that no lines cross. Many lines can pierce

S

and still balance to zero.

F = (y, - x, 0)

has zero flux, so it's a zero field."

No — the field is nonzero and swirls (it has curl); it is solenoidal (divergence-free), which only forces net flux to zero, not the field itself to vanish.

"Flux out of the unit sphere for

F = (x, y, z)

4 π

, so the flux out of a radius-2 sphere is also

4 π

Error —

\nabla \cdot F = 3

, so flux

= 3 \cdot Vol

. The radius-2 sphere has volume

\frac{4}{3} π (2)^{3}

, giving flux

32 π

, not

4 π

. Constant divergence does not mean constant flux.

"Since divergence is a derivative, the divergence theorem needs

F

continuous but not differentiable."

The requirement is stronger:

F

must have continuous first partial derivatives throughout

V

, because

\nabla \cdot F

is built from those partials and must be integrable.

F = r /∣ r ∣^{3}

has

\nabla \cdot F = 0

, so its flux out of any sphere is

0

The divergence is zero except at the origin, where the field is singular. If the sphere encloses the origin, the theorem doesn't apply directly and the true flux is

4 π

— carve out a tiny sphere around the origin first.

"To use the theorem I must first parameterize all six faces of the cube."

Backwards — the point of the theorem is to avoid the surface work: replace six face integrals with one volume integral

∭_{V} \nabla \cdot F d V

Why questions

Why do interior faces contribute nothing when we glue tiny boxes together?

A shared wall is the outward face of one box and the inward face of its neighbor, so the two flux contributions are equal in size and opposite in sign — they cancel exactly, leaving only the exterior skin.

Why does only the normal component

F \cdot \hat{n}

enter the flux?

Flow parallel to the surface slides along the skin without escaping; only motion perpendicular to

S

actually carries stuff across, and the dot product with

\hat{n}

extracts precisely that perpendicular part.

Why is divergence "flux per unit volume" rather than a total?

It is a local quantity — the limit of (net flux out of a tiny box)/(box volume) as the box shrinks — so integrating it over

V

(multiplying local rates by their volumes and summing) recovers the total flux.

Why does the theorem let us swap a hard surface integral for an easier volume one?

Because both count the same physical thing (net production = net escape); we pick whichever description of the boundary-vs-interior is simpler to integrate, often the volume when

\nabla \cdot F

is a simple polynomial.

Why is the outward orientation, not just "a normal," essential?

The sign of the answer encodes direction of net flow; "outward positive" is the convention that makes source strength (

\nabla \cdot F > 0

) correspond to positive escaping flux. An inward normal reverses this bookkeeping.

Why does the Continuity equation follow naturally from this theorem?

Applying Gauss to the mass-flux field turns "mass leaving through

S

" into "

∭_{V} \nabla \cdot (ρ v) d V

"; equating that to the rate mass drops inside gives the local law

\partial_{t} ρ + \nabla \cdot (ρ v) = 0

once the volume is arbitrary.

Edge cases

What happens if

V

shrinks to a single point?

Both sides

\to 0

, but the ratio flux/volume tends to

\nabla \cdot F

at that point — this limiting case is actually the definition of divergence, the seed the whole theorem grows from.

What if

V

has a hole (a cavity) inside it, like a spherical shell?

The boundary

S = \partial V

is both surfaces — the outer sphere with outward normal and the inner sphere with normal pointing into the cavity (outward from the solid). Flux must be summed over both, each correctly oriented.

What if

\nabla \cdot F

is positive in one part of

V

and negative in another?

The volume integral adds signed contributions, so sources and sinks partially or fully cancel; the flux reports only the net imbalance, which can be positive, negative, or zero.

What if the closed surface

S

passes exactly through a singularity of

F

The theorem is undefined there —

F \cdot \hat{n}

is unbounded on

S

and the hypotheses fail. You must deform

S

to avoid the singular point (or handle it as an improper limit).

What is the flux of a constant field

F = c

out of any closed surface?

Zero —

\nabla \cdot c = 0

, so

∭_{V} 0 d V = 0

. Geometrically, whatever constant flow enters one side leaves the other with no net gain.

What if the region

V

is not simply connected or has an awkward shape?

The theorem still holds as long as

S = \partial V

is a genuine closed boundary and

F

is smooth on

V

; we simply subdivide into "box-friendly" pieces whose shared interior faces cancel, exactly as in the derivation.

Recall One-line survival kit

Ask three questions before applying Gauss: (1) Is $S$ closed and outward-oriented? (2) Is $F$ smooth everywhere inside — no hidden singularities? (3) Am I computing flux (across, use divergence) or circulation (along, use Stokes' theorem / Green's theorem)? If any answer is "no", fix it before trusting the equation.

Connections

Flux integrals — the "across" side these traps keep testing.
Divergence and curl — divergence vs. curl underlies the flux-vs-circulation confusion.
Stokes' theorem · Green's theorem — the circulation cousins.
Gauss's law (electromagnetism) · Continuity equation — where "same flux, different surface" and the local law come from.