The divergence theorem

Theorem

The divergence theorem

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Statement

Let \( V \subseteq \mathbb{R}^3 \) be a bounded region whose boundary \( \partial V = S \) is a piecewise-smooth, closed, orientable surface, oriented by the outward unit normal \( \mathbf{n} \). Let \( \mathbf{F} = (F_1, F_2, F_3) \) be a vector field of class \( C^1 \) on an open set \( U \) with \( \overline{V} \subseteq U \). Then \[ \iiint_V (\nabla \cdot \mathbf{F})\, dV \;=\; \oiint_S \mathbf{F} \cdot \mathbf{n}\, dS, \] where \( \nabla \cdot \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \) is the divergence of \( \mathbf{F} \).

Why it matters

The divergence theorem (Gauss's theorem, Ostrogradsky's theorem) is the three-dimensional capstone of the fundamental theorem of calculus: it converts a volume integral of a local differential quantity (divergence, a limit of flux per unit volume) into a boundary integral of the flux itself. It is the mechanism by which local conservation laws — mass, charge, energy, probability — are derived from global balance statements, and conversely by which continuum field equations (Maxwell's equations, the continuity equation, the Navier–Stokes equations) are extracted from integral physical principles.

Structurally it sits alongside Green's theorem and the classical Stokes theorem as an instance of the generalized Stokes theorem for differential forms, \( \int_M d\omega = \int_{\partial M} \omega \); understanding its proof is the natural route into that unification.

Hypotheses

\( \mathbf{F} \) is \( C^1 \) on a neighbourhood of \( \overline{V} \). Counterexample: on \( \mathbb{R}^3 \setminus \{0\} \) take \( \mathbf{F} = \mathbf{r}/|\mathbf{r}|^3 \), which is smooth away from the origin with \( \nabla\cdot\mathbf{F}=0 \) there. If \( V \) is the unit ball, \( \mathbf{F} \) fails to be \( C^1 \) (indeed undefined) at the interior point \( 0 \in V \), and the theorem fails: the surface integral over the unit sphere gives \( 4\pi \), while the "volume integral of \( 0 \)" would wrongly suggest \( 0 \). The singularity at the interior point is precisely what the hypothesis rules out.

\( V \) is bounded. Counterexample: take \( V = \{ x \gt 0\} \subset \mathbb{R}^3 \) (a half-space) and \( \mathbf{F} = (1,0,0) \). Then \( \nabla\cdot\mathbf{F}=0 \) everywhere, so the volume integral is \( 0 \), but the flux through the bounding plane \( x=0 \) (outward normal \( (-1,0,0) \)) is \( \iint_{x=0} (-1)\,dS \), an infinite (divergent) integral — the two sides are not even both finite, let alone equal.

\( \partial V \) is piecewise smooth and \( V \) has finite volume with a well-defined outward normal a.e. on \( \partial V \). Counterexample: let \( V \) be the region under the graph of the Weierstrass-type nowhere-differentiable function, or more simply a domain whose boundary is a fractal curve rotated into a surface of infinite \((2\text{-dimensional Hausdorff})\) area confined to finite volume. Surface measure and the very notion of "outward normal" break down almost everywhere, so \( \oiint_S \mathbf{F}\cdot\mathbf{n}\,dS \) is not even well-defined as a classical surface integral.

\( S \) is coherently oriented (outward-pointing normal, consistent across pieces). Counterexample: if one face of a cube is given the inward normal by mistake while the rest keep the outward normal, the computed "flux" differs from the true flux by twice the flux through that one face, and the identity fails numerically even though every other hypothesis holds.

Proof

We give the classical proof by decomposing \( V \) into elementary regions — regions simultaneously projectable onto each coordinate plane — proving the componentwise identity on each, and gluing.

It suffices to prove the three scalar identities \[ \iiint_V \frac{\partial F_i}{\partial x_i}\,dV = \oiint_S F_i\, n_i\, dS \qquad (i=1,2,3) \] separately and sum them.

Both sides of the full statement are sums of these three terms by linearity of the integral and of \( \nabla\cdot\mathbf{F} = \sum_i \partial F_i/\partial x_i \) and \( \mathbf{F}\cdot\mathbf{n} = \sum_i F_i n_i \). A

We prove the \( i=3 \) case, \( \displaystyle\iiint_V \frac{\partial F_3}{\partial z}\,dV = \oiint_S F_3\, n_3\, dS \), first for a \(z\)-simple region: \[ V = \{ (x,y,z) : (x,y) \in D,\ \phi_1(x,y) \le z \le \phi_2(x,y) \} \] where \( D \subset \mathbb{R}^2 \) is a bounded region with piecewise-smooth boundary and \( \phi_1 \le \phi_2 \) are \( C^1 \) on \( D \).

Definition of a \(z\)-simple (vertically simple, "type I") solid region — the building block for the elementary-region argument. B

\[ \iiint_V \frac{\partial F_3}{\partial z}\, dV = \iint_D \left( \int_{\phi_1(x,y)}^{\phi_2(x,y)} \frac{\partial F_3}{\partial z}(x,y,z)\, dz \right) dA = \iint_D \Big[ F_3(x,y,\phi_2(x,y)) - F_3(x,y,\phi_1(x,y)) \Big] dA. \]

Fubini's theorem reduces the triple integral to an iterated integral (legitimate since the integrand is continuous on the compact region); the inner integral is evaluated by the fundamental theorem of calculus in \( z \), using that \( F_3 \) is \( C^1 \). B

The boundary \( S=\partial V \) splits into the top cap \( S_2 = \{z=\phi_2(x,y)\} \), bottom cap \( S_1=\{z=\phi_1(x,y)\} \), and the (possibly empty) vertical lateral wall \( S_L \) lying over \( \partial D \), where the outward normal has \( n_3=0 \).

Geometric decomposition of the boundary of a \(z\)-simple region; the lateral wall is ruled by lines parallel to the \(z\)-axis, so its normal is horizontal there. A

On \( S_2 \), parametrized by \( (x,y)\mapsto (x,y,\phi_2(x,y)) \), the outward (upward) normal satisfies \( n_3\, dS = +\,dA \); on \( S_1 \), the outward (downward) normal satisfies \( n_3\, dS = -\,dA \). Hence \[ \oiint_S F_3 n_3\, dS = \iint_{S_2} F_3 n_3\,dS + \iint_{S_1} F_3 n_3\, dS + \underbrace{\iint_{S_L} F_3 n_3\, dS}_{=0} = \iint_D F_3(x,y,\phi_2) \, dA - \iint_D F_3(x,y,\phi_1)\, dA. \]

Standard fact for a graph surface \( z=\phi(x,y) \): the upward unit normal is \( \mathbf{n} = \dfrac{(-\phi_x,-\phi_y,1)}{\sqrt{1+\phi_x^2+\phi_y^2}} \) and the surface-area element is \( dS = \sqrt{1+\phi_x^2+\phi_y^2}\,dA \), so \( n_3\,dS = dA \); the outward normal on the lower cap is the downward one, giving the sign flip. The lateral term vanishes because \( n_3=0 \) there. C

Comparing Step 3 and Step 5: \[ \iiint_V \frac{\partial F_3}{\partial z}\, dV = \oiint_S F_3\, n_3\, dS \] for every \(z\)-simple region \(V\).

Direct comparison of the two displayed expressions, which are identical. B

By the analogous argument with \(x\)-simple and \(y\)-simple regions, \[ \iiint_V \frac{\partial F_1}{\partial x}\,dV = \oiint_S F_1 n_1\, dS, \qquad \iiint_V \frac{\partial F_2}{\partial y}\,dV = \oiint_S F_2 n_2\, dS \] hold on any region \(V\) that is simultaneously \(x\)-, \(y\)-, and \(z\)-simple ("elementary" / "normal" region — e.g. any convex region, or any region cut out by graphs over all three coordinate planes, such as a ball, ellipsoid, or box).

Repeats Steps 2–6 with the roles of \(x,y,z\) permuted; a convex bounded region with \(C^1\) or piecewise-\(C^1\) boundary is automatically simple in all three directions. A

Summing the three identities of Steps 6–7 (via Step 1) gives the divergence theorem for any elementary region \(V\): \[ \iiint_V \nabla\cdot\mathbf{F}\, dV = \oiint_{\partial V} \mathbf{F}\cdot\mathbf{n}\, dS. \]

Linearity, as in Step 1. A

For a general bounded region \(V\) with piecewise-smooth boundary, partition \( V = \bigcup_{k=1}^N V_k \) into finitely many elementary regions \(V_k\) with pairwise-disjoint interiors, meeting along shared internal faces \(\Sigma_{jk} = \partial V_j \cap \partial V_k\).

Such a partition exists for any region bounded by a piecewise-smooth surface (e.g. by slicing with finitely many coordinate-aligned or boundary-adapted cutting surfaces); this is the 3-dimensional analogue of subdividing a plane region into vertically/horizontally simple pieces in the proof of Green's theorem. C

\[ \sum_{k=1}^N \iiint_{V_k} \nabla\cdot\mathbf{F}\, dV = \iiint_V \nabla\cdot\mathbf{F}\, dV, \qquad \sum_{k=1}^N \oiint_{\partial V_k} \mathbf{F}\cdot\mathbf{n}_k\, dS = \oiint_{\partial V} \mathbf{F}\cdot\mathbf{n}\, dS, \] because on each internal shared face \( \Sigma_{jk} \), the outward normal of \( V_j \) is exactly opposite to the outward normal of \( V_k \), so the two flux contributions \( \iint_{\Sigma_{jk}} \mathbf{F}\cdot\mathbf{n}_j\,dS \) and \( \iint_{\Sigma_{jk}} \mathbf{F}\cdot\mathbf{n}_k\, dS \) are negatives of each other and cancel in the sum, leaving only the integral over the pieces of \(\partial V_k\) lying on the true outer boundary \( \partial V \).

Additivity of the volume integral over the partition (Step 9); additivity of the surface integral together with the "interior cancellation" lemma: adjacent elementary pieces induce opposite orientations on their shared internal face. C

Combining Step 8 (applied to each \(V_k\)) with Step 10: \[ \iiint_V \nabla\cdot\mathbf{F}\, dV = \sum_k \iiint_{V_k}\nabla\cdot\mathbf{F}\,dV = \sum_k \oiint_{\partial V_k} \mathbf{F}\cdot\mathbf{n}_k\, dS = \oiint_{\partial V}\mathbf{F}\cdot\mathbf{n}\, dS, \] establishing the theorem for general \(V\). \( \blacksquare \)

Chains the three preceding steps. B

Result

\[ \iiint_V \nabla\cdot\mathbf{F}\; dV = \oiint_{\partial V} \mathbf{F}\cdot\mathbf{n}\; dS \]

Reading. The total "outflow-producing" activity of \(\mathbf{F}\) summed throughout the volume equals the net flow of \(\mathbf{F}\) escaping through the boundary — sources and sinks inside \(V\) are exactly what the boundary flux measures.

Scope. Applies to any bounded \( V\subset\mathbb{R}^3 \) with piecewise-smooth, coherently outward-oriented boundary, for any vector field \(C^1\) on a neighbourhood of \(\overline V\). Generalizes verbatim to \(\mathbb{R}^n\) (flux through an \((n-1)\)-dimensional boundary equals the volume integral of divergence) and is the \(n=3\) instance of the generalized Stokes theorem \( \int_V d\omega = \int_{\partial V}\omega \) for the \(2\)-form \(\omega = \iota_{\mathbf F} (dx\wedge dy\wedge dz)\).

Corollaries & converses

Solenoidal fields have zero net flux: if \( \nabla\cdot\mathbf{F}=0 \) throughout \(V\), then \( \oiint_{\partial V}\mathbf{F}\cdot\mathbf{n}\,dS = 0 \) for every such \(V\) — this is why magnetic flux through any closed surface vanishes (\( \nabla\cdot\mathbf{B}=0 \)).
Green's identities: taking \( \mathbf{F} = u\nabla v \) with \( u,v \in C^2 \) yields Green's first identity \( \iiint_V (u\nabla^2 v + \nabla u\cdot\nabla v)\,dV = \oiint_S u\,\partial v/\partial n\, dS \); antisymmetrizing gives Green's second identity.
Local converse (pointwise characterization): if \( \oiint_{\partial B_r(p)} \mathbf{F}\cdot\mathbf{n}\, dS = o(r^3) \) fails to vanish suitably, one shows \( \nabla\cdot\mathbf{F}(p) = \lim_{r\to 0} \frac{1}{\text{vol}(B_r(p))}\oiint_{\partial B_r(p)}\mathbf{F}\cdot\mathbf{n}\,dS \); this recovers divergence from flux and is often taken as the coordinate-free definition of divergence.
Global converse fails: the identity holding for one particular pair \((V,\mathbf F)\) does not imply \( \nabla\cdot\mathbf{F}\equiv 0\) or anything about \(\mathbf F\) off \(V\); the theorem is an identity for each admissible \((V,\mathbf F)\), not a biconditional characterizing \(\mathbf F\).

Fails without

Drop \(C^1\)-on-a-neighbourhood-of-\(\overline V\): \( \mathbf F=\mathbf r/|\mathbf r|^3 \) on the unit ball \(V\) (singular at the interior point \(0\)) gives \( \oiint_S \mathbf F\cdot\mathbf n\,dS = 4\pi \neq 0 = \) the (improper, and actually divergent as a Lebesgue integral of \(0\) plus a delta-like concentration) naive volume side; correctly, \(\nabla\cdot\mathbf F = 4\pi\delta_0\) as a distribution, and the classical theorem simply does not apply.
Drop boundedness of \(V\): the half-space example in Hypotheses shows a finite divergence integral (here \(0\)) can be paired with an infinite or ill-defined flux integral, so the two sides cannot be compared, let alone equated.
Drop coherent outward orientation: reversing the normal on part of \(S\) (e.g. treating an inner boundary component of an annular shell with the wrong sign) changes the right side by twice the flux through that component while leaving the left side untouched, breaking equality.

Common errors

Forgetting that \(S\) must be closed: applying the divergence theorem to an open surface (like a hemisphere alone, without its disk base) and expecting the flux to equal a volume integral — there is no enclosed volume, so the theorem does not apply at all.
Using an inward-pointing normal (common when a surface is naturally parametrized "into" the region) without negating the result, silently flipping the sign of the answer.
Applying the theorem to a field with a singularity strictly inside \(V\) (e.g. inverse-square fields at the origin, or \(1/r\) potentials) without excising a small ball around the singularity first — the "\(\nabla\cdot\mathbf F=0\) almost everywhere" trap.
Computing \(\nabla\cdot\mathbf F\) in a curvilinear coordinate system using the Cartesian formula \(\partial F_1/\partial x + \cdots\) applied naively to non-Cartesian components, instead of the correct curvilinear divergence formula (e.g. in spherical coordinates).
Confusing the divergence theorem with Stokes' theorem (curl and a bounding curve vs. divergence and a bounding surface) and mismatching the dimension of the boundary object with the differential operator used.

Discussion

The theorem was discovered piecemeal: special cases appear in Lagrange's work on fluid mechanics (1762), Gauss used it in 1813 in the context of gravitational attraction, and Mikhail Ostrogradsky gave the first general proof in 1826 (hence "Ostrogradsky's theorem" in much of the Russian and French literature) while studying heat flow. Its physical content predates its formal statement: any local conservation law "rate of accumulation = – net outflow" becomes, via the divergence theorem, the continuity equation \( \partial\rho/\partial t + \nabla\cdot\mathbf{J} = 0 \), since the theorem is exactly the tool that converts "flux through every closed surface obeys a global balance" into "a pointwise differential equation holds everywhere."

The proof strategy given here — verify on elementary building blocks, then glue via boundary cancellation — is the template shared by Green's theorem in the plane and the classical Stokes theorem in space, and it is no accident: all three, together with the fundamental theorem of calculus itself, are the low-dimensional shadows of the single generalized Stokes theorem for differential forms on oriented manifolds with boundary, \( \int_M d\omega = \int_{\partial M}\omega \). The divergence theorem corresponds to integrating the exterior derivative of a 2-form (the "flux form" associated to \(\mathbf F\)) over a 3-manifold.

The cancellation-on-shared-faces argument (Step 10) is the discrete/finite version of a much more general phenomenon: it is exactly what makes the boundary operator on chains satisfy \( \partial\circ\partial = 0 \) in simplicial/singular homology, and the same mechanism underlies the divergence theorem's role as the archetype for defining weak (distributional) divergence via integration by parts — \( \int_V u\,\nabla\cdot\mathbf F\,dV = -\int_V \nabla u\cdot\mathbf F\, dV + \int_{\partial V} u\,\mathbf F\cdot\mathbf n\, dS \) — the starting point of the weak formulation used throughout PDE theory and finite element methods.

Common misconception. Students often believe the theorem requires \(V\) to be convex or "nice" like a ball; in fact any bounded region admitting a decomposition into finitely many simple pieces works — this includes regions with holes (as long as all boundary components, inner and outer, are given consistent outward orientation) and quite irregular polyhedral or piecewise-smooth shapes.

Worked examples

Compute \( \oiint_S \mathbf F\cdot\mathbf n\, dS \) for \( \mathbf F(x,y,z) = (x,y,z) \) and \(S\) the sphere of radius \(R\) centred at the origin, outward normal.

Setup: direct surface computation would require parametrizing the sphere; the divergence theorem replaces this with a volume integral. A

\[ \nabla\cdot\mathbf F = \frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial z}{\partial z} = 1+1+1 = 3. \]

Direct differentiation. A

\[ \iiint_V \nabla\cdot\mathbf F\,dV = \iiint_V 3\, dV = 3\cdot \text{vol}(V) = 3\cdot \frac{4}{3}\pi R^3 = 4\pi R^3. \]

\(V\) is the solid ball, an elementary (convex) region, so the divergence theorem applies as proved above; volume of a ball is a standard fact. A

Check by direct computation: on \(S\), \( \mathbf n = \mathbf r/R \), so \( \mathbf F\cdot\mathbf n = (x,y,z)\cdot(x,y,z)/R = R^2/R = R \) (constant on \(S\)), giving \( \oiint_S \mathbf F\cdot\mathbf n\, dS = R\cdot 4\pi R^2 = 4\pi R^3 \), matching Step 3.

Direct evaluation of the flux integral using the geometry of the sphere, confirming the theorem. B

\[ \oiint_S \mathbf F\cdot\mathbf n\, dS = 4\pi R^3 \]

Reading. The radial field \(\mathbf F=\mathbf r\) has constant divergence \(3\) everywhere, so the flux through any sphere is exactly \(3\) times the enclosed volume.

Compute \( \oiint_S \mathbf F\cdot\mathbf n\, dS \) for \( \mathbf F(x,y,z) = (x^2, y^2, z^2) \) and \(S\) the boundary of the unit cube \( [0,1]^3 \), outward normal.

Direct computation would require six separate face integrals; the divergence theorem collapses this to one volume integral. A

\[ \nabla\cdot\mathbf F = 2x+2y+2z. \]

Direct differentiation. A

\[ \iiint_{[0,1]^3} (2x+2y+2z)\,dV = 2\int_0^1\!\!\int_0^1\!\!\int_0^1 (x+y+z)\,dx\,dy\,dz. \]

The cube is elementary (indeed a box), so the divergence theorem applies; rewrite as an iterated integral by Fubini. A

\[ \int_0^1\!\!\int_0^1\!\!\int_0^1 x\,dx\,dy\,dz = \tfrac12,\quad\text{and by symmetry the } y \text{ and } z \text{ integrals each equal } \tfrac12, \] so the triple integral of \(x+y+z\) is \( \tfrac12+\tfrac12+\tfrac12 = \tfrac32 \), giving \( 2\cdot\tfrac32 = 3 \).

Elementary integration of \(x\) over \([0,1]\) gives \(1/2\); the \(y,z\) factors integrate to \(1\) each; symmetry of the cube under permuting coordinates gives the other two terms for free. A

Spot check on the face \(x=1\) (outward normal \((1,0,0)\)): contribution \( \iint_{[0,1]^2} 1^2\,dy\,dz = 1 \); on the opposite face \(x=0\) (outward normal \((-1,0,0)\)): contribution \( \iint -0^2\, dy\,dz = 0 \). By the \(x\leftrightarrow y \leftrightarrow z\) symmetry of \(\mathbf F\), the \(y\)-faces and \(z\)-faces contribute \(1+0\) each too, totalling \(3\), matching Step 4.

Direct face-by-face flux computation, confirming the theorem's result via the definition of surface flux for a graph aligned with a coordinate plane. B

\[ \oiint_S \mathbf F\cdot\mathbf n\, dS = 3 \]

Reading. Even though \(\mathbf F\) is not radial, the divergence theorem still reduces a six-face flux computation to a single, symmetric volume integral.

Problems

Compute \( \oiint_S \mathbf F\cdot\mathbf n\, dS \) where \( \mathbf F=(0,0,z) \) and \(S\) is the boundary of the cylinder \( x^2+y^2\le a^2,\ 0\le z\le h \).
Solution
\( \nabla\cdot\mathbf F = 1 \), so by the divergence theorem the flux equals \( \text{vol}(V) = \pi a^2 h \). (Direct check: the flux is \(0\) through the curved side since \(F_3=z\) contributes only through \(n_3\) which is \(0\) there; through the top \(z=h\), \(n=(0,0,1)\), contribution \( h\cdot\pi a^2 \); through the bottom \(z=0\), \(n=(0,0,-1)\), contribution \(0\). Total \(\pi a^2 h\), matching.)
Let \( \mathbf F = (x,y,z)/|(x,y,z)|^3 \) on \( \mathbb{R}^3\setminus\{0\} \). Show directly that \( \nabla\cdot\mathbf F = 0 \) away from the origin, and explain why the divergence theorem does not give \( \oiint_S \mathbf F\cdot\mathbf n\,dS = 0 \) when \(S\) is a sphere centred at the origin.
Solution
With \(r=|(x,y,z)|\), \(F_1=x r^{-3}\), so \( \partial F_1/\partial x = r^{-3} + x\cdot(-3r^{-4})(x/r) = r^{-3}-3x^2 r^{-5}\); summing the analogous \(y,z\) terms gives \( 3r^{-3} - 3r^{-5}(x^2+y^2+z^2) = 3r^{-3}-3r^{-3}=0 \) for \(r\neq 0\). The divergence theorem requires \(\mathbf F\) to be \(C^1\) on a neighbourhood of the closed ball \(\overline V\), including the interior point \(0\); since \(\mathbf F\) is undefined (blows up) at \(0\in V\), the hypothesis fails and the theorem simply does not apply — indeed direct computation gives \( \oiint_S \mathbf F\cdot\mathbf n\,dS = 4\pi \neq 0 \).
Use the divergence theorem to evaluate \( \oiint_S \mathbf F\cdot\mathbf n\, dS \) for \( \mathbf F = (2x, -y, z) \) over the surface of the box \( [0,2]\times[0,1]\times[0,3] \).
Solution
\( \nabla\cdot\mathbf F = 2-1+1=2 \), constant. Volume of the box is \(2\cdot1\cdot3=6\), so the flux is \(2\cdot 6 = 12\).
A region \(V\) has a spherical cavity: \(V = B_2(0)\setminus \overline{B_1(0)}\) (the shell between radii 1 and 2), and \( \mathbf F=(x,y,z) \). Using that \(\partial V\) has two components (outer sphere \(r=2\) with outward normal pointing away from origin, inner sphere \(r=1\) with outward normal pointing toward the origin, since it points out of \(V\)), verify the divergence theorem.
Solution
\( \nabla\cdot\mathbf F = 3 \), so \( \iiint_V 3\,dV = 3\left(\tfrac43\pi 2^3 - \tfrac43\pi 1^3\right) = 3\cdot\tfrac43\pi(8-1)=28\pi \). On the outer sphere, \( \mathbf F\cdot\mathbf n = r = 2 \), giving \(2\cdot 4\pi(2)^2=32\pi\). On the inner sphere the outward-from-\(V\) normal is \(-\mathbf r/r\), so \( \mathbf F\cdot\mathbf n = -r=-1 \), giving \((-1)\cdot 4\pi(1)^2=-4\pi\). Total flux \(=32\pi-4\pi=28\pi\), matching the volume integral.
Using the divergence theorem, prove that for any closed surface \(S=\partial V\) and any constant vector \( \mathbf c \), \( \oiint_S (\mathbf c\times \mathbf r)\cdot \mathbf n\, dS = 0 \), where \( \mathbf r=(x,y,z) \).
Solution
Write \( \mathbf F = \mathbf c\times\mathbf r \) with \( \mathbf c=(c_1,c_2,c_3) \) constant. Then \( \mathbf F = (c_2 z - c_3 y,\ c_3 x - c_1 z,\ c_1 y - c_2 x) \), so \[ \nabla\cdot\mathbf F = \frac{\partial}{\partial x}(c_2 z-c_3y) + \frac{\partial}{\partial y}(c_3x-c_1z)+\frac{\partial}{\partial z}(c_1y-c_2x) = 0+0+0=0 \] identically, since each component of \(\mathbf F\) does not depend on its own differentiating variable (each term involves only the other two coordinates, and \(\mathbf c\) is constant). By the divergence theorem, \( \oiint_S \mathbf F\cdot\mathbf n\,dS = \iiint_V \nabla\cdot\mathbf F\, dV = \iiint_V 0\,dV=0 \) for every admissible \(V\).