Divergence theorem

Let $V$ be a bounded region in Euclidean space with piecewise smooth boundary and outward unit normal vector $n$. For a continuously differentiable vector field $F$ on a neighborhood of its closure,The total outward flux equals the accumulated divergence inside. Contributions across shared internal faces cancel when a region is divided into small pieces.

For $F=(x,y,z)$ on a ball of radius $R$, the divergence is $3$, giving $3(4\pi R^3/3)=4\pi R^3$. On the boundary, $F\cdot n=R$, so the surface integral is also $R(4\pi R^2)=\boxed{4\pi R^3}$. This agreement turns the derivative into a geometric flux measurement. The required regularity matters: a singular field at the center cannot be included without checking its contribution separately.

A smooth embedded invariant two-torus in three-dimensional Euclidean phase space bounds a solid torus of positive volume. The vector field is tangent to its boundary, so its normal flux is zero. If the divergence is strictly negative throughout the enclosed volume, the divergence theorem gives a strictly negative flux, a contradiction. Thus a globally negative-divergence three-dimensional vector field has no such quasiperiodic invariant torus. This argument is stronger than merely noting contraction of transported phase volume; an attractor's own zero volume is not a contradiction.

For the outward spherical patch $r=R$, $y\le 0$, $z\le 0$, and the polynomial vector field $F=(y{x}^2,x{y}^2-3z^4,x^3+y^2)$, close the quarter-ball with its two planar half-discs. The divergence is $4xy$, whose volume integral vanishes by symmetry in $x$. The caps have fluxes $-3\pi R^6/16$ and $\pi R^4/8$, obtained from half-disc fourth and second moments. The divergence theorem gives the displayed spherical flux. All pieces use normals pointing out of the same closed volume; the caps are auxiliary and must be subtracted.

A continuous extension of a vector field across one point need not be differentiable, so it does not automatically meet the classical hypotheses of the divergence theorem. Instead apply the divergence theorem after deleting a ball $B_{\epsilon }$ around that point. If $\Vert \mathbf{F}\Vert \le C{\epsilon }^{\alpha }$ on its boundary in $d$ dimensions, the magnitude of the inner flux is at most a constant times ${\epsilon }^{d-1+\alpha }$. When $d-1+\alpha >0$ this tends to zero. If the divergence is also integrable, passing to the limit recovers the outer flux from the volume integral. For example, a planar vector field with $\Vert \mathbf{F}\Vert \le r$ has an inner flux of order ${\epsilon }^2$, even when its extension is not differentiable at the origin.

For a continuously differentiable vector field $v$ on a bounded sufficiently regular region $V$, apply the divergence theorem to $a\times v$ for an arbitrary constant vector $a$. The vector calculus identity $\nabla \cdot (a\times v)=-a\cdot (\nabla \times v)$ and the cyclic scalar triple product identity show that the dot products of both sides of the displayed formula with $a$ agree. Since $a$ is arbitrary, the vector integrals agree. Here $n$ is the outward unit normal vector. Reversing the order on the boundary gives the equivalent positive-sign formula $\int n\times vdS=\int \nabla \times vdV$.

To evaluate outward flux through the lateral surface of a finite right circular cone, add its planar cap and apply the divergence theorem to the enclosed solid. Subtract the cap flux, whose outward normal points away from the solid. For a cone $r=az$ with $0\le z\le h$, the volume element is $rdrd\phi dz$ and the cap has radius $ah$. The cone's vertex has no unique normal, but is a set of area zero; one can remove a small tip and take a limit when the field is regular there.

Consider the vector field $\mathbf{F}=(p(x),q(x),r(z))$ in the open unit cube, where $p$ and $r$ extend to continuously differentiable functions on $[0,1]$, and $q$ is continuous on $(0,1]$ with ${\int }_0^1|q(x)|dx<\infty$. Even if $q$ is unbounded as $x\downarrow 0$, the normal component on an $x$-face has the finite trace $\pm p(x)$. The two $y$-faces have finite improper integral fluxes $\pm {\int }_0^{1}q(x)dx$ and therefore cancel. The remaining face contributions giveTo justify this use of the divergence theorem, first apply it to $[\epsilon ,1]\times [0,1{]}^2$, on which the field is regular. Its divergence is $p^{\prime }(x)+r^{\prime }(z)$, independent of $q$. Its outward flux is $p(1)-p(\epsilon )+(1-\epsilon )[r(1)-r(0)]$. Taking $\epsilon \downarrow 0$ yields the displayed answer, and absolute convergence makes the two $y$-face limits finite separately. For example, $q(x)=1/[x(1+(\ln x{)}^2)]$ diverges at zero butThis argument defines the normal traces and the face integrals explicitly; it does not assert that the whole vector field has a finite value on the singular face.

For the upper curved half of $x^2/a^2+y^2/b^2+z^2/c^2=1$, take the vector field $(x{y}^2,y^2,C)$ and outward unit normal. Its divergence is $y^2+2y$. The odd term integrates to zero over the half-ellipsoid. Scaling a unit half-ball gives $\int y^{2}dV=2\pi a{b}^{3}c/15$. The equatorial closing disk has downward flux $-C\pi ab$. The divergence theorem therefore gives the displayed curved-surface flux. The constant vertical component is not detected by the divergence, but is recovered by accounting for the closing disk.

A vector field singular on an axis need not satisfy the classical divergence theorem on the original closed region. Remove a thin tube $\rho <\epsilon$, apply the theorem to the remaining smooth region, then examine the added cylindrical boundary. If its flux tends to zero and the remaining volume integral converges, the improper boundary flux equals the improper integral of the divergence. For $\mathbf{F}=(\log {\rho }^2,\log {\rho }^2,z/\rho )$ with $\rho =\sqrt{x^2+y^2}$, the cylindrical unit normal has zero $z$ component, so the added flux is $O(\epsilon |\log \epsilon |)$ on a bounded axial interval. The divergence $(2\cos \varphi +2\sin \varphi +1)/\rho$ is locally integrable under the cylindrical coordinates measure $\rho d\rho d\varphi dz$. Hence no finite line-source correction survives, despite the field itself being unbounded.

For a continuously differentiable scalar $\Omega$ on a bounded region $V$ with outward boundary normal, ${\int }_V\nabla \Omega dV={\int }_{\partial V}\Omega \mathbf{n}dS$. Apply the divergence theorem to $\mathbf{c}\Omega$ for every constant vector $\mathbf{c}$; equality of all scalar projections gives the vector identity. The theorem applies to piecewise smooth boundaries, with singular points removed by a limiting argument when necessary.

For a continuously differentiable scalar-valued function $f$ on a bounded sufficiently regular region, apply the divergence theorem to $fk$, where $k$ is any constant vector. Since $\nabla \cdot (fk)=k\cdot \nabla f$, the dot products of the two displayed vector integrals with every $k$ agree. The vector integrals are therefore equal. This is the scalar-gradient analogue of the usual flux identity.

For a bounded sufficiently regular region, the divergence theorem applied to the vector field $x_i{e}_j$ gives the displayed identity. Contracting with constant vectors gives ${\int }_{\partial \Omega }(b\cdot x)(c\cdot n)dS=|\Omega |b\cdot c$. Contracting with the Kronecker delta or the Levi-Civita symbol respectively gives $\int x\cdot ndS=3|\Omega |$ and $\int x\times ndS=0$ in three dimensions.

For $X\in C^1({\mathbb{R}}^n;{\mathbb{R}}^n)$, integration over $B_R$ gives $|\kappa ||B_R|\le \Vert X{\Vert }_{\infty }|\partial B_R|$, so $|\kappa |\le n\Vert X{\Vert }_{\infty }/R$. Let $R\to \infty$. In particular, the bounded flux $Du/\sqrt{1+|Du{|}^2}$ of an entire constant-mean-curvature graph forces its divergence constant to be zero, independently of any bound on $Du$.

On an oriented Riemannian manifold with boundary, the divergence of a Riemannian vector field obeys the displayed integral identity for smooth compactly supported vector fields $X$, or for arbitrary smooth $X$ if the manifold is compact. Here $N$ is the outward unit normal vector and the boundary has the outward-normal-first boundary orientation. Its induced Riemannian volume form is $j^{\ast }{\iota }_N{\omega }_g$. Splitting $X$ into its normal and tangential components gives $j^{\ast }{\iota }_X{\omega }_g=g(X,N){\omega }_{\tilde{g}}$, since a top form vanishes on $n$ boundary-tangent arguments. Apply the Stokes theorem to $d{\iota }_X{\omega }_g=(\mathrm{div}X){\omega }_g$.

Without a support or decay condition, flux at infinity can invalidate the result even if the integrals converge. On $[0,\infty )$ with Riemannian metric $d{x}^2$, $X=(\arctan x){\partial }_x$ has boundary flux zero but integral of divergence $\pi /2$.