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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 $R$.

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.

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.

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.

The divergence is the sum of the three diagonal partial derivatives:

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

In Cartesian coordinates, the divergence is the sum of the three component derivatives. Therefore

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.

In Cartesian coordinates, the divergence uses the matching coordinate for each component:Each of these three partial derivatives vanishes.

Writing $r^2=x^2+y^2+z^2$, the gradient is $\mathbf{v}=2e^{r^2}(x,y,z)$. Its divergence, equivalently the Laplacian of $\varphi$, isOn the unit sphere this scalar is the constant $10e$. The lower hemisphere has surface area $2\pi$, soThis is a scalar surface integral of the divergence, not the electric- or vector-field flux integral. There is no normal component to insert.

In Cartesian coordinates, calculate the divergence by differentiating each component with respect to its matching coordinate:For the curl, take the alternating cross-derivatives:

Assume the components of $\mathbf{F}$ have continuous second partial derivatives. Expanding the divergence of its curl givesEach positive term cancels a negative term because the corresponding mixed derivatives commute. Consequently