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Choose $\mathbf{F}=\nabla \psi$ in the preceding product rule and apply the divergence theorem. Since $\nabla \cdot \nabla \psi =\Delta \psi$, this gives the first of Green's identities:Here ${\partial }_n\psi =\mathbf{n}\cdot \nabla \psi$ is the outward normal derivative. Interchange $\varphi$ and $\psi$, then subtract. The gradient dot products cancel, yielding the second of Green's identities:For example, $\varphi ,\psi \in C^2(\overline{V})$ with the needed extensions suffice for these applications of the divergence theorem.

For a continuously differentiable vector field $\mathbf{F}$ on a neighborhood of the closure of $V$, the divergence theorem statesHere $\mathbf{n}$ is the outward unit normal vector; the right side is the outward flux through the boundary.

For any bounded volume $V$ with piecewise smooth closed boundary and outward unit normal vector $\mathbf{n}$, Gauss's law in integral form isThus the outward electric flux equals the enclosed electric charge divided by the vacuum permittivity. To prove it, apply the divergence theorem to the electric field and use its given divergence:The classical proof assumes the requisite differentiability of the electric field. For piecewise smooth charge densities, one may apply the divergence theorem piece by piece and cancel the internal fluxes.

Apply the divergence theorem to $\varphi \mathbf{c}$, using ii. Because the vector $\mathbf{c}$ is constant, it can be taken outside both integrals:This holds for every $\mathbf{c}$, in particular each of the three Cartesian basis vectors. Equality of the three components proves the gradient volume-to-boundary identityIt suffices that $\varphi$ be a continuously differentiable function on a neighbourhood of the closure of $V$.

For a bounded region $V\subset {\mathbb{R}}^3$ with piecewise smooth boundary, suppose $\mathbf{F}$ is a continuously differentiable vector field on an open neighbourhood of its closure. The divergence theorem, with outward unit normal vector $\mathbf{n}$, isThe surface integral measures total outward flux. Edges and isolated vertices have zero surface area and do not affect the integral.

Pair each point $\mathbf{x}$ of the square's boundary with $-\mathbf{x}$. The vector field obeys $\mathbf{f}(-\mathbf{x})=\mathbf{f}(\mathbf{x})$, whereas the outward unit normal vector changes sign: $\mathbf{n}(-\mathbf{x})=-\mathbf{n}(\mathbf{x})$. Corresponding arc length elements are equal. Thus the two contributions to the flux cancel, proving without evaluating any boundary integral thatThe same cancellation appears in the divergence, which is an odd function of $x$ over a region symmetric under $x\mapsto -x$. To justify using the planar divergence theorem despite the lack of differentiability at the origin, remove a disk of radius $\epsilon <1$. The vector field is continuously differentiable on the remaining region, its divergence has zero double integral by this symmetry, and the magnitude of the inner flux is at most $2\pi {\epsilon }^2$, since $\Vert \mathbf{f}\Vert \le \epsilon$ there. Letting $\epsilon \to 0$ recovers the result. This is flux around a removable point singularity.

Let $\Omega \subset {\mathbb{R}}^3$ be a bounded region with a piecewise smooth closed boundary, and let $\mathbf{F}$ be a continuously differentiable vector field on an open neighbourhood of its closure. With $\mathbf{n}$ the outward unit normal vector, the divergence theorem statesThe boundary surface integral is the total outward flux. All boundary components are included; on the boundary of a cavity, the outward unit normal vector points into the cavity.

For any fixed subregion $\Omega$, with no internal heat generation, conservation of energy says that the rate of stored thermal energy equals the incoming heat flux:The minus sign is needed because $\mathbf{q}\cdot \mathbf{n}>0$ denotes outgoing heat flux. Substitute Fourier's law $\mathbf{q}=-k\nabla T$ and apply the divergence theorem:Here differentiation under the integral uses a fixed region and sufficient regularity of $T$. Since the identity holds for every $\Omega$, continuity of the integrands gives the local heat equationFor spatially varying thermal conductivity, the product rule gives $\nabla \cdot (k\nabla T)=k{\nabla }^{2}T+\nabla k\cdot \nabla T$; the second term cannot generally be discarded.

Integrate the steady state equation over $R$ and use the divergence theorem:The zero heat flux condition makes the boundary term vanish. But $\rho >0$ and $S>0$ throughout a region of positive volume, soa contradiction. Thus no steady-state solution exists. This is the insulated steady heat balance: positive internal heat generation continually increases the stored thermal energy when no heat can escape.

Suppose $T^{(1)}$ and $T^{(2)}$ are two sufficiently regular steady state solutions with the same prescribed temperature on the boundary. Their difference $w=T^{(1)}-T^{(2)}$ satisfiesThe identical PDE source terms cancel. Apply the product rule and the divergence theorem to $wk\nabla w$; this is the weighted form of Green's first identity:Since $k>0$, the nonnegative integrand must vanish, so $\nabla w=0$. Hence $w$ is constant on each connected component of $R$, and its zero Dirichlet boundary condition makes each constant zero. Therefore $\boxed{T^{(1)}=T^{(2)}}$: the solution, if it exists, is unique. This proves Dirichlet uniqueness for variable-conductivity heat conduction; it does not require a constant thermal conductivity.

For any closed surface enclosing a region $V$, Gauss's law for gravity states that the outward flux of the gravitational field isHere $G$ is the gravitational constant and $M_V$ is the enclosed mass; the negative sign expresses attractive gravity.

Use the gravitational potential sign conventions specified by this paper: $\mathbf{g}=\nabla \phi$. Thus $\phi$ is the negative of the usual Newtonian gravitational potential $\Phi$ defined by $\mathbf{g}=-\nabla \Phi$. For example, setting the value at infinity to zero gives $\phi =GM/r$ for a point mass. The divergence theorem now givesSince $V$ is arbitrary, continuity of the integrands implies the local Poisson's equationConversely, integrating this Poisson's equation over any $V$ and applying the divergence theorem recovers Gauss's law for gravity. At a discontinuity of the mass density, the local equation is interpreted piecewise, or globally in the sense of distributions.

On the unit sphere, the outward unit normal vector is $\mathbf{n}=\mathbf{r}$ and $\mathbf{u}=\mathbf{r}$. Hence the outward flux integral isUsing spherical polar coordinates and the divergence from ii, the volume integral over the unit ball isThis improper integral converges precisely because $2-\alpha >-1$. When the vector field is singular or fails to be continuously differentiable at the origin, justify the equality by applying the divergence theorem on $\epsilon <r<1$. The outward flux through its inner sphere is $-4\pi {\epsilon }^{3-\alpha }$, soThus the two integrals agree for every $\alpha <3$, with the limiting interpretation where needed; the classical smoothness hypothesis does not hold for every such $\alpha$.

Let $V$ be a bounded region with piecewise smooth closed boundary $\partial V$, and let $\mathbf{F}$ be a continuously differentiable vector field on an open neighbourhood of its closure. The divergence theorem stateswhere $\mathbf{n}$ is the outward unit normal vector. Thus the volume integral of the divergence equals the outward boundary flux integral. For a region containing a hole, the outward unit normal vector on its inner boundary points into that hole.

For $\alpha =3$, the divergence vanishes at every point with $r>0$, so its ordinary volume integral, ignoring the undefined single point, is zero. Nevertheless, the outward flux integral through the unit sphere is still $4\pi$:The hypotheses of the divergence theorem fail at the origin. Indeed $|\mathbf{u}|=r^{-2}$ there, so no continuous extension is possible. On the punctured region $\epsilon <r<1$, the divergence theorem remains valid: the outer flux is $4\pi$ and the inner flux is $-4\pi$. Unlike iii, this inner contribution does not tend to zero. The discrepancy therefore comes from the persistent flux at the singular point, rather than a failure of the divergence theorem under its hypotheses.

The trigonometric identityshows that only angular Fourier modes $1$ and $3$ are required. Differentiating the harmonic expansion on a disk at $r=1$ multiplies the coefficient of each nonconstant Fourier mode by $n$. Thus $B_1=3/4$ and $3B_3=-1/4$, givingEach term is a regular harmonic function, and the radial derivative isThe Neumann compatibility condition for Laplace's equation holds becauseas is necessary by the divergence theorem applied to the gradient of a harmonic function.

The additive constant is undetermined. This Neumann boundary condition is unchanged by $T\mapsto T+C$, so the paper's blanket uniqueness assertion requires a normalization in this case. The uniqueness of the Neumann problem gives uniqueness up to a constant; prescribing $T(0)=0$, or zero mean temperature, selects the displayed answer with $C=0$.

Take $v=|\mathbf{x}{|}^2$ in Green's second identity. Since ${\nabla }^{2}u=0$, ${\nabla }^{2}v=6$, $v=1$ on the unit sphere, and ${\partial }_{n}v=2$ there,The final term vanishes by the divergence theorem, because it equals ${\iiint }_B{\nabla }^{2}udV=0$. ThereforeWe assume sufficient boundary regularity for the stated boundary values and integrals, or use the corresponding trace/limit formulation. This is the equality of volume and surface averages for a harmonic function, since the unit ball and sphere have volume $4\pi /3$ and area $4\pi$.

Apply the divergence theorem to $\mathbf{F}=u\nabla v-v\nabla u$. The product rule givesbecause the two gradient inner products cancel. Also $\mathbf{F}\cdot \mathbf{n}=u{\partial }_{n}v-v{\partial }_{n}u$. ThusThis is Green's second identity, under the usual regularity assumptions permitting these derivatives and boundary integrals.

For a bounded region $R\subseteq {\mathbb{R}}^3$ with piecewise smooth boundary and a continuously differentiable vector field $\mathbf{F}$ on a neighborhood of its closure, the divergence theorem sayswhere $\mathbf{n}$ is the outward unit normal vector. The left side integrates divergence over the volume; the right side is the outward flux.

Parametrize the surface of revolution byAn oriented vector surface area element isUse the vector field $\mathbf{F}=(x/2,y/2,0)$, whose divergence is one. If this parametrization has the outward orientation on an enclosing surface, the divergence theorem givesThe second equality is integration by parts, using the endpoint values. For example this outward convention applies when the positive-$t$ branch lies above the negative-$t$ branch, $f(t)>f(-t)$ for $0<t<\pi /2$.

The stated $C^1$ regularity and endpoint conditions alone do not guarantee an embedded enclosing surface: branches can cross or coincide. Without an appropriate orientation assumption the flux formula is an algebraic signed volume. If “inside” is defined as the region between the two graph branches for each radius, its nonnegative geometric volume is insteadThis explicitly handles a reversal or change in the order of the two branches.

For each of the three constant Cartesian unit vectors ${\mathbf{e}}_j$, apply the divergence theorem to the vector field $\varphi {\mathbf{e}}_j$. Its divergence is ${\partial }_j\varphi$, soThese are precisely the three components ofThus the volume integral of the gradient is the boundary surface integral weighted by the vector surface element $d\mathbf{S}=\mathbf{n}dS$. The regularity needed here is that $\varphi$ be continuously differentiable on a neighbourhood of the closure of $R$.