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Away from the origin, the product rule and $\nabla r=\mathbf{r}/r$ giveThe field is undefined at $r=0$. It has outward flux $4\pi$ through any sphere centred there, so its global distributional divergence is $4\pi {\delta }^{(3)}(\mathbf{r})$, not the zero distribution. The pointwise result must not be extended through the singularity when applying the divergence theorem.

Let $D$ be the disk $x^2+y^2\le 3$. The upper boundary is the graph $z=3-x^2-y^2$, with outward vector vector surface element $(2x,2y,1)dxdy$. Thus its flux isThe lower disk has outward normal vector $-{\mathbf{e}}_z$ and zero flux, since $F_z=0$. The total outward surface integral is therefore $9\pi$.

Independently, $\nabla \cdot \mathbf{F}=2$ and the enclosed volume isConsequently ${\iiint }_V\nabla \cdot \mathbf{F}dV=2V=\boxed{9\pi }$, agreeing with the divergence theorem. Both the orientation of the cap and inclusion of the closing disk have been accounted for.

This is a scalar surface integral, not an oriented flux. Split the whole closed surface into the bottom disc, curved side and tilted top; shared edge curves have zero surface area.

On the bottom, $z=0$ and the integrand is $y$. Its integral over the symmetric disc $x^2+y^2\le 3$ vanishes by oddness:On the top, $z=4-y$, so $y+z=4$. The graph has $z_x=0$, $z_y=-1$, giving $dS=\sqrt{1+z_x^2+z_y^2}dxdy=\sqrt{2}dxdy$. HenceFor the side, let $R=\sqrt{3}$ and parametrize by $(R\cos \varphi ,R\sin \varphi ,z)$, with $0\le z\le 4-R\sin \varphi$. The surface area element is $Rdzd\varphi$. ThusAdding all three pieces givesThe tilted top remains above the bottom everywhere because $4-y\ge 4-\sqrt{3}>0$. The calculation is an instance of a scalar surface integral over a cylinder with one tilted cap; reversing a surface orientation would not change any of its terms.

Take the normal vector to point outward from the box. The divergence theorem changes the closed-surface flux into a volume integral. The divergence isThe first term integrates to zero in $x$, since ${\int }_{-1}^2\pi \cos (\pi x)dx=\sin 2\pi -\sin (-\pi )=0$. ThereforeHenceA direct face check gives zero from the two $x$ faces, $45/2$ from the pair of $y$ faces, and $45$ from the pair of $z$ faces. This also checks the outward-orientation signs.

For a bounded volume $V$ with outward-oriented piecewise smooth boundary $S$, the divergence theorem statesHere $G$ is a continuously differentiable vector field on a neighborhood of $V$, $n$ the outward unit normal vector, and $dV$ the volume element. The left side is its outward flux.

For a twice continuously differentiable field $F$, equality of mixed partial derivatives givesApply the divergence theorem to $G=\nabla \times F$ to obtainThe regularity hypotheses justify differentiating and integrating the curl; a field singularity cannot simply be ignored in a volume argument.

On an outward-oriented sphere of radius $R>0$, the electric field is normal with constant magnitude $A/R^2$. Thus its flux isIt is independent of $R$. There is no contradiction with solution: the classical divergence theorem cannot be applied to a full ball while pretending that $E$ is smooth at its center. On a punctured ball the outer flux and the inward-normal inner flux cancel, in agreement with zero ordinary divergence there.

The three-dimensional Dirac delta distribution ${\delta }^{(3)}(r)$ is supported at the origin and satisfies the sampling identityfor smooth compactly supported test functions $\phi$. It is not an ordinary function with a finite value at zero; its total integral is one, and translating it moves the point at which the test function is sampled.

The globally valid source equation isFor a direct weak verification, excise a ball of radius $\epsilon$ and integrate $-E\cdot \nabla \phi$ outside it. Since the ordinary divergence vanishes there, the result is $A{\int }_{S^2}\phi (\epsilon \hat{r})d\Omega \to 4\pi A\phi (0)$. This is precisely the displayed distributional divergence and restores the divergence theorem with the point charge included.

Use the outward normal on the spherical patch. It is the quarter of the spherical surface lying below both the horizontal plane and the plane $y=0$, with both signs of $x$ included. Close it by the two half-discs $D_y$ in $y=0,z\le 0$ and $D_z$ in $z=0,y\le 0$.

The divergence of the polynomial vector field is $\nabla \cdot F=4xy$. Its volume integral over the enclosed quarter-ball vanishes by oddness in $x$. Thus the divergence theorem gives ${\Phi }_S+{\Phi }_{D_y}+{\Phi }_{D_z}=0$.

On $D_y$, the outward normal is $+\hat{y}$ and $F_y=-3z^4$. On $D_z$, it is $+\hat{z}$ and $F_z=x^3+y^2$; the odd term integrates to zero. The needed half-disc moments for radius $R$ areandAt $R=4$, the cap fluxes are $-768\pi$ and $32\pi$, respectively. Therefore the flux through a spherical quadrant by planar caps isThe source does not explicitly specify a normal for this part; the boxed answer adopts the usual outward spherical orientation. Choosing the inward normal reverses its sign.

For this vector field,The divergence theorem and spherical coordinates therefore give the outward fluxThe $a^5$ dependence also follows by combining the cubic field magnitude with the quadratic scaling of surface area.

Assume $a>0$ and choose the outward orientation. On the sphere, $\mathbf{n}=(x,y,z)/a$, so $\mathbf{F}\cdot \mathbf{n}=(x^2+y^2+2z^2)/a=(a^2+z^2)/a$. In spherical coordinates, $z=a\cos \theta$ and $dS=a^2\sin \theta d\theta d\varphi$. The direct surface integral isIndependently, $\nabla \cdot \mathbf{F}=1+1+2=4$. Its volume integral over the ball isThe two calculations agree exactly with the divergence theorem. The expression requested is a scalar flux of a vector field, namely the dot product with the directed surface area element.

Let $\mathbf{F}=(F_x,F_y,F_z)$ be continuously differentiable near $P$. Take a small axis-aligned rectangular volume around $P$, with side lengths $\Delta x,\Delta y,\Delta z$ shrinking with bounded aspect ratios. The outward flux through its two $x$ faces is, to leading order,where $\Delta V=\Delta x\Delta y\Delta z$. The analogous face pairs give $\partial F_y/\partial y$ and $\partial F_z/\partial z$. Dividing total flux by $\Delta V$ and taking the limit givesThe divergence is thus the local outward flux per unit volume; the symbols on the right are Cartesian partial derivatives.

For a bounded volume $V$ with piecewise smooth closed boundary $S$, subdivide into small cells. Each internal face occurs twice with opposite normal vectors, so its flux cancels. The remaining sum is the exterior surface integral; the cell sums of $(\nabla \cdot \mathbf{F})\Delta V$ converge to the volume integral. Assuming continuous first partial derivatives on a neighborhood of the closure of $V$, this proves the divergence theorem:The unit normal vector $\mathbf{n}$ points out of $V$. Multiple boundary components are allowed; at a cavity wall it points into the cavity.

Choose the outward normal vector on the side of the elliptic cylinder. Parametrize it bySince ${\mathbf{r}}_{\theta }\times {\mathbf{r}}_z=(\frac{1}{2}\cos \theta ,\sin \theta ,0)$ points outward, the parameter rectangle has its bottom edge traversed with increasing $\theta$, its right cut edge with increasing $z$, its top edge with decreasing $\theta$, and its left cut edge with decreasing $z$. After identification, the two cut edges are the same physical curve with opposite directions. Their line integrals cancel for a single-valued vector field. Viewed from above, the bottom rim is counterclockwise and the top rim clockwise.

To apply Stokes theorem to a flux rather than a curl integral, first find a vector potential:This vector potential is smooth everywhere. On either horizontal rim, $d\mathbf{r}$ is horizontal whereas $\mathbf{A}$ is vertical, giving $\mathbf{A}\cdot d\mathbf{r}=0$. For a cut at ${\theta }_0=\pi /4$, $xy=1/4$, so the upward cut contribution is ${\int }_{-1}^1{z}^2/4dz=1/6$ and the downward one is $-1/6$. ConsequentlyA direct check uses the oriented surface element above: $\mathbf{F}\cdot ({\mathbf{r}}_{\theta }\times {\mathbf{r}}_z)=\frac{1}{2}{z}^2\cos 2\theta$, whose integral over a full turn vanishes. Reversing the chosen orientation still gives zero.

For a continuously differentiable vector field $\mathbf{F}$ on a neighborhood of an oriented, piecewise smooth surface $S$, Stokes theorem statesHere $\partial S$ is the boundary curve, $d\mathbf{r}$ is its directed tangent element, $\hat{\mathbf{n}}$ is the chosen unit normal vector, $dS$ is the surface area element, and $\nabla \times \mathbf{F}$ is the curl. The right-hand rule relates the boundary direction to $\hat{\mathbf{n}}$. Thus circulation around the edge equals the flux of curl through the spanning surface. Reversing both orientations changes both signs. One can replace a difficult line integral by an easier surface integral, or replace one spanning surface by another within the region of regularity.

For a bounded volume $V$ with piecewise smooth closed boundary and a continuously differentiable vector field on a neighborhood of its closure, the divergence theorem statesHere ${\hat{\mathbf{n}}}_{\mathrm{out}}$ points out of $V$, including into any holes, $dV$ is the volume element, and $\nabla \cdot \mathbf{F}$ is the divergence. It equates total outward flux to integrated source strength. It converts a closed surface integral into a volume integral, and explains why internal fluxes cancel in a subdivided region. Singularities cannot be included without checking the hypotheses.

For a conservative force, the work around every closed path is zero. Stokes theorem gives zero flux of curl through every sufficiently small patch. Dividing by its area and shrinking the patch shows that every component of the continuous curl vanishes:Here $U$ is potential energy; conversely, zero curl implies a conservative vector field on a simply connected domain. The divergence theorem then gives the compatible relation ${\iint }_{\partial V}\mathbf{F}\cdot \hat{\mathbf{n}}dS=-{\iiint }_V{\nabla }^{2}UdV$. There is no general zero-divergence condition on a conservative force: $\mathbf{F}=-\mathbf{r}=-\nabla (r^2/2)$ is a concrete counterexample, with $\nabla \cdot \mathbf{F}=-3$.

Take $0<r_1<r_2$, so the shell excludes the possible singularity at the origin. The divergence and curl of a radial vector field formula in spherical coordinates givesWith volume element $dV=r^2\sin \theta drd\theta d\varphi$, direct integration givesFor the independent boundary calculation, the outer sphere has outward unit normal vector $+\hat{\mathbf{r}}$ and the inner sphere has outward unit normal vector $-\hat{\mathbf{r}}$. Their fluxes areTheir sum equals the direct volume integral, verifying the divergence theorem. The restriction $\beta \ne 2$ permits the displayed power-antiderivative step. At $\beta =2$, treated separately, both the shell volume integral and net boundary flux are zero; the excluded origin must still not be filled in.

The divergence of $\mathbf{F}=\nabla z$ is the LaplacianIts area integral over the square isFor the boundary flux, ${\int }_0^{2}A(s)ds=-4$. On $x=0$, $\mathbf{F}\cdot \mathbf{n}=4A(y)$, giving $-16$; on $x=2$ it is $8A(y)$, giving $-32$. The corresponding horizontal sides give $-16$ and $-32$. The total is again $-96$, verifying the divergence theorem with outward normals. With a counterclockwise boundary, this is the same orientation as $Mdy-Ndx$.

Choose the normal pointing away from the origin, equivalently the outward normal of the first-octant tetrahedron. Parametrize the face by $\mathbf{r}=(x,y,2-x/2-y/2)$ on $D=\{x,y\ge 0:x+y\le 4\}$. Its oriented area element isThus the surface integral isThe divergence theorem checks this result: $\nabla \cdot \mathbf{F}=1$, the tetrahedron's volume is $4\cdot 4\cdot 2/6=16/3$, and each coordinate-plane face has zero flux. The opposite orientation gives $-16/3$; a signed flux requires an orientation.

In this two-pool calcium model, $c$ is the cytosolic calcium ion concentration, and $c_s$ measures calcium in an intracellular store, such as the endoplasmic reticulum. The variables in these balances are expressed using compatible effective volumes, so internal transfers enter their two equations with equal and opposite signs. If actual compartment concentrations use different reference volumes, the corresponding volume ratio must first be included in their balances.

The flux $J_{+}$ pumps calcium from the cytosol into the store, $J_{-}$ releases it back, and $k_s{c}_s$ is passive store leakage back to the cytosol. The external input $r$ supplies cytosolic calcium, while $kc$ exports it from the biological cell. Thus $k,k_s$ have units of inverse time and all five fluxes have units of concentration per time. Their directions can be sketched asAdding the two ordinary differential equations verifies the total calcium balanceInternal uptake, release and leakage redistribute calcium; only the exchanges across the cell membrane change this total. In particular any equilibrium must have $c=r/k$ when $k>0$.

Choose positive reference concentrations $C,C_s$ and a reference flux $J_0$, and defineOverdots now mean derivatives with respect to the dimensionless time $\tau$. The store balance becomesThe total balance becomes $\dot{u}+(C_s/C)\dot{v}=r/(kC)-u$. ThereforeandThe reference quantities are scale choices, not additional kinetic variables; typically the response functions supply natural saturation concentrations and a maximal uptake flux. The identity $\delta /\epsilon =k_s/k$ checks the nondimensionalization. A small $\epsilon$ means store exchange is fast compared with external cytosolic removal.

At equilibrium $Q_s=M$, so $Q=Q_{\mathrm{in}}+Ms$ for an upstream boundary flux $Q_{\mathrm{in}}$. HenceWithout a specified upstream condition this integration constant must be retained. On a region of positive area, the characteristic speed is $Q^{\prime }(A)=c{A}^m>0$, increasing downstream along the equilibrium. Small disturbances move downstream, spread in space as this speed increases, and decrease in area amplitude in the manner derived next. They are transported away from a fixed location; the model has no diffusion giving instantaneous smoothing. Finite-amplitude compressive disturbances can steepen into shocks because the flux is convex, $Q^{\prime \prime }(A)=cm{A}^{m-1}>0$, so the small-disturbance picture is not a global no-shock assertion.

Taylor expansion gives $Q(A_{\mathrm{eq}}+a)=Q(A_{\mathrm{eq}})+c{A}_{\mathrm{eq}}^{m}a+O(a^2)$. Subtract the equilibrium equation and write $V(s)=c{A}_{\mathrm{eq}}^m(s)$ to obtain $a_t+(Va{)}_s=0$. Set $b=Va$. Since $V$ is time independent, $b_t+V{b}_s=0$. The travel coordinate for a conservative transport perturbation $\xi (s)={\int }_0^{s}V(r{)}^{-1}dr$ transforms this into $b_t+b_{\xi }=0$, and thereforeThe conserved quantity along a characteristic is the flux perturbation $Va$, explaining the diminishing area amplitude as $V$ grows.

Differentiate the equilibrium relation to get $A_{\mathrm{eq}}^{\prime }=M/(c{A}_{\mathrm{eq}}^m)=M/V$. Thus the arrival time is explicitlyFor zero upstream flux the integral remains finite, giving the limiting formula with its second term zero. However $V(0)=0$ then makes the displayed amplitude formula singular at the head; a uniformly small arbitrary area disturbance imposed exactly there is outside this linearization. A positive upstream flux avoids that degeneracy.

For quasistatic bed stability with suspended sediment, in the leading approximation, $(hu{)}_x=0$ and ${\eta }_x=0$. For spatially varying perturbations at fixed unit water flux and fixed surface level, write $u=1+U$, $h=1+H$, $c=1+C$, $s=s_{\ast }+S_1$. The constraints give $H=-U$ and $S_1=-H=U$. Linearizing the suspended and bed equations yieldsFor a Fourier mode $e^{\sigma t+i\kappa x}$, the first relation gives $C=4U/(1+i\kappa )$; substitution into the second givesThus every nonzero spatial frequency has negative growth rate, independently of $\beta$. The parameter changes propagation, not damping. The spatially uniform mode has $\sigma =0$ and is neutral, representing a change among uniform reference states compatible with the leading constraints. Hence there is no growing linear mode in this approximation; the claim is stability, not strict decay of every homogeneous perturbation. Restoring the small discarded terms or changing boundary conditions requires a separate calculation.