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One sufficient version of the divergence theorem is the following. Let $R$ be a bounded region in ${\mathbb{R}}^3$ with a closed, piecewise smooth surface as its boundary, and let $\mathbf{F}$ be a continuously differentiable function with values in ${\mathbb{R}}^3$ on an open neighbourhood of the closure of $R$. With $\mathbf{n}$ the outward unit normal vector,The boundary surface integral is the total outward flux integral. If the boundary has several components, each has the outward orientation relative to the material of $R$; inner components therefore point into the holes. Edges and isolated vertices may be omitted when assigning the unit normal vector, since they have zero surface area.

Use cylindrical coordinates $(\rho ,\theta ,z)$. The right circular cone is described by $0\le z\le 1$ and $0\le \rho \le a(1-z)$, and the divergence of $\mathbf{F}=(0,0,z^2)$ is $2z$. Direct integration gives

To check the boundary flux integral independently, first consider the base $z=0$. Its outward unit normal vector is $-\mathbf{k}$, but $\mathbf{F}=0$ there, so its flux is zero. Parametrize the lateral surface byThe cross productpoints outward: its horizontal component points away from the axis of the cone. This gives the vector surface element without separately normalizing it. Consequently the lateral flux integral isAdding the zero base contribution agrees with the volume integral, verifying the divergence theorem. The apex has zero surface area; the vector field is smooth there, so no additional flux is hidden at that point.

Take the outward orientation of a surface. The divergence is $\nabla \cdot \mathbf{F}=y^2+z^2+x^2$, so the divergence theorem converts the closed surface flux into an integral over the box:Each term separates; for example the $x^2$ term is $(2a^3/3)(2b)(2c)$. Summing givesHere $a,b,c>0$ are the box half-lengths. Reversing orientation of a surface would negate the result.

Differentiate the ellipsoidal parametrization. Its outward vector surface element isFor $a,b,c>0$, its dot product with $\mathbf{r}$ simplifies to $abc\sin \theta$. Thus the requested direct surface integral isAs a check, the divergence of $\mathbf{r}$ is three, and the ellipsoid has volume $4\pi abc/3$, so the divergence theorem gives the same value.

Write $r=|\mathbf{r}|$ and $q=\mathbf{p}\cdot \mathbf{r}$. Away from the origin, $\nabla q=\mathbf{p}$ and $\nabla r^{-m}=-m\mathbf{r}/r^{m+2}$. The product rule givesConsequently the divergence of $\mathbf{E}=3q\mathbf{r}{r}^{-5}-\mathbf{p}{r}^{-3}$ is zero. For any closed surface bounding a region that excludes the origin, the field is smooth throughout that region, and the divergence theorem provesFor the infinite cylinder, use $\mathbf{r}=(a\cos \varphi ,a\sin \varphi ,z)$ and outward unit normal $(\cos \varphi ,\sin \varphi ,0)$. With $\mathbf{p}=(0,0,1)$, $q=z$, $dS=ad\varphi dz$, and $\mathbf{E}\cdot \mathbf{n}=3az/(a^2+z^2{)}^{5/2}$. Hence the direct surface flux isThe integrand is odd in $z$ and absolutely integrable. This calculation treats the unbounded cylindrical side directly, without applying the bounded-region divergence theorem across the field's singularity.

Compute the curl component by component from the given vector field:For example, $B_x={\partial }_y{A}_z-{\partial }_z{A}_y=0-(-2xz)$ and $B_z={\partial }_x{A}_y-{\partial }_y{A}_x=(1-z^2)-(z^2-1)$. Its divergence isas expected because the divergence of a curl is zero.

Take $C$ counterclockwise when viewed from positive $z$, the disc $D$ upward oriented, and the hemisphere $H$ outward oriented. These choices give both surfaces the same positive boundary orientation. Parametrize $C$ by $r(t)=(\cos t,\sin t,0)$, $0\le t\le 2\pi$. On it, $A=(-\sin t,\cos t,0)$ and $dr=(-\sin t,\cos t,0)dt$, so the line integral isOn $D$, use polar coordinates $(x,y)=(r\cos \varphi ,r\sin \varphi )$ and $d\mathbf{S}=(0,0,1)rdrd\varphi$. Since $B_z=2$ there, the surface flux isOn the unit hemisphere, the outward unit normal vector is $n=(x,y,z)$. Since $x^2+y^2=1-z^2$, we have $B\cdot n=4z(1-z^2)$. In spherical polar coordinates, $z=\cos \theta$ and $dS=\sin \theta d\theta d\varphi$, giving the explicit surface integralIndeed the inner integral is $[{\sin }^4\theta {]}_0^{\pi /2}=1$.

These equalities illustrate Stokes theorem twice: $B=\nabla \times A$ and either upward $D$ or outward $H$ spans the oriented circle $C$, so each surface flux equals the same line integral. For the divergence theorem, close the upper half-ball by adjoining the downward oriented base, namely $-D$. Its outward surface flux isThe minus sign on the base is required because its outward normal vector points down. Reversing a specified orientation reverses that integral's sign.

For a bounded volume $V$ with piecewise smooth closed boundary $\partial V$, oriented by the outward unit normal vector, and a continuously differentiable vector field on a neighborhood of $\overline{V}$, the divergence theorem statesFor an oriented piecewise smooth surface $S$ with piecewise smooth boundary $C=\partial S$, and a continuously differentiable vector field on a neighborhood of $S$, Stokes theorem statesThe boundary orientation follows the right-hand rule relative to the chosen normal vector. Reversing that orientation reverses both sides of the latter identity.

The divergence is $\nabla \cdot F=2xyz+2xyz+2xyz=6xyz$. With outward orientation, the divergence theorem converts the surface flux into a volume integral:Directly, the three faces lying in coordinate planes contribute zero, and each opposite face contributes $1/4$, confirming the sign and normalization.

Let $\Omega$ be the interior of the tetrahedron, and take any constant vector field $\mathbf{q}$. Its divergence is zero, so the divergence theorem givesChoosing $\mathbf{q}$ to be each Cartesian basis vector proves ${\sum }_i{\mathbf{V}}_i=0$. This is the general fact that vector areas of a closed polyhedron sum to zero, specialized to four faces. The outward orientations are essential: the scalar face areas are all positive and do not sum to zero.

Let $V\subset {\mathbb{R}}^3$ be a bounded region with a piecewise smooth closed boundary $S=\partial V$, and let $\mathbf{F}$ have continuous first partial derivatives on a neighborhood of its closure. The divergence theorem stateswhere $\mathbf{n}$ is the outward unit normal vector. The right side is the outward flux; reversing the orientation reverses its sign.

On the radius-$a$ sphere, the outward unit normal vector is $(x,y,z)/a$, so $\mathbf{F}\cdot \mathbf{n}=(x^2+2y^2+3z^2)/a$. For a direct surface integral, put $x=a\sin \theta \cos \phi$, $y=a\sin \theta \sin \phi$, $z=a\cos \theta$ and $dS=a^2\sin \theta d\theta d\phi$. The integral isThe azimuthal integrals are $\pi ,\pi ,2\pi$, while ${\int }_0^{\pi }{\sin }^3\theta d\theta =4/3$ and ${\int }_0^{\pi }{\cos }^2\theta \sin \theta d\theta =2/3$. ThusThe flux of a linear field through a sphere also follows from the equality of the three spherical second moments, without using the divergence theorem.

The divergence is $\nabla \cdot \mathbf{F}=1+2+3=6$. Since the volume enclosed by a sphere is $4\pi a^3/3$,This agrees with the independently computed outward surface integral, as the divergence theorem requires.

For a bounded region $V$ with piecewise smooth closed boundary $S$, outward unit normal vector $\mathbf{n}$ and continuously differentiable vector field $\mathbf{F}$, the divergence theorem states ${\int }_S\mathbf{F}\cdot \mathbf{n}dS={\int }_V\nabla \cdot \mathbf{F}dV$. Here $dS$ is scalar surface area, $dV$ is volume, and $d\mathbf{S}=\mathbf{n}dS$ is the oriented area element.

For $\mathbf{F}=\mathbf{r}$, its divergence is three. The ellipsoid has semiaxes $a,a,b$, so its volume is $4\pi a^{2}b/3$, obtained by scaling the unit ball. Consequently

Since $\mathbf{a}$ is constant, componentwise differentiation gives $\nabla \cdot (\mathbf{a}\varphi )={\sum }_i{a}_i{\partial }_i\varphi =\mathbf{a}\cdot \nabla \varphi$. Applying the divergence theorem to this vector field givesThis holds for every constant vector $\mathbf{a}$, so the vector integrals agree. This is the vector gradient form of the divergence theorem. With $\varphi =z$, the gradient is the constant unit vector ${\mathbf{e}}_z$. Therefore

The identity $\nabla \cdot (\mathbf{a}\times \mathbf{v})=\mathbf{v}\cdot (\nabla \times \mathbf{a})-\mathbf{a}\cdot (\nabla \times \mathbf{v})$ reduces to $-\mathbf{a}\cdot (\nabla \times \mathbf{v})$ when $\mathbf{a}$ is constant. In components, this follows by differentiating ${\epsilon }_{ijk}{a}_j{v}_k$ and using the antisymmetry of the Levi-Civita symbol. The divergence theorem and the cyclic scalar triple product identity now giveSince $\mathbf{a}$ is arbitrary,The order of the cross product determines the minus sign, as in the vector curl form of the divergence theorem.

The divergence is $\nabla \cdot F=-1+0+1=0$. Take the hemisphere's outward orientation, and close it with the unit disk $D$ at $z=0$, whose outward normal is $-e_z$. The divergence theorem then gives ${\int }_{S}F\cdot dS+{\int }_{D}F\cdot dS=0$. On the disk, $F\cdot (-e_z)=-x^2$, and polar coordinates giveConsequently the curved hemisphere has fluxThe reversed hemisphere orientation would reverse the sign.

The divergence isFor any closed piecewise smooth surface bounding a volume, use the divergence theorem with the outward orientation:Reversing the orientation still gives zero. The field is smooth everywhere, so there is no hidden singular source inside the surface.

Take the sphere's outward normal. Since $F=k(x,y,z)$ with constant $k$, its divergence is $3k$. The divergence theorem givesThe translated centre does not enter the answer, because the divergence is constant. Equivalently the constant field produced by translating coordinates has zero net flux through a closed sphere. The inward orientation would negate the result.

Integrate the normal component on each face directly, with outward normals. Each unlisted pair of coordinates ranges independently from zero to one:For example, on $x=1$ the integral is ${\int }_0^1{\int }_0^{1}2(y+z)dydz=2$. Adding all six contributions givesThis calculation uses the six explicit surface integrals, independently of the divergence theorem.

For a sufficiently smooth vector field on a volume $V$ and its closed boundary $S$, the divergence theorem stateswith the unit normal directed outwards. HereHence the net volume flux through every closed surface in the liquid is zero. This verifies incompressible flow directly. It does not require the velocity to vanish pointwise on an arbitrary closed surface; inward and outward contributions can cancel.