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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.

A scalar surface integral integrates a scalar field against the unsigned surface area element. For a regular parametrization $\mathbf{r}(u,v)$, it is $\int f(\mathbf{r}(u,v))\Vert {\mathbf{r}}_u\times {\mathbf{r}}_v\Vert dudv$. Reversing orientation does not change it. For example, the scalar surface integral of the constant one is the surface area; a signed flux instead integrates the normal component of a vector field.

Let $S$ be the full boundary of $x^2+y^2\le R^2$, $0\le z\le H-\alpha y$, with $H>|\alpha |R$. The bottom contribution is zero by symmetry. On the top, the integrand is the constant $H$ and the surface area element is $\sqrt{1+{\alpha }^2}dxdy$. On the side, set $y=R\sin \varphi$ and integrate in $z$ first: the result is $R[H^2-{\alpha }^2{R}^2{\sin }^2\varphi ]/2$. Integrating from zero to $2\pi$ gives the displayed scalar surface integral. No orientation signs enter; a scalar area integral is different from a flux.

Let $S$ be the whole boundary of $x^2+y^2\le R^2$, $a\le z-\alpha x\le b$, where $R>0$ and $a<b$. On the curved cylinder, the parametrization $(R\cos \theta ,R\sin \theta ,\alpha R\cos \theta +s)$ has surface area element $Rd\theta ds$ and $\alpha x-z=-s$. Each planar cap has area $\pi R^2\sqrt{1+{\alpha }^2}$. Summing the side and both caps gives the scalar surface integralThe surface area element is unsigned; cap orientation does not change the signs of the scalar integrand. This separates the geometric area factor from the field values and avoids mistaking a scalar surface integral for a flux.

A solid torus of major radius $A$ and minor radius $R<A$, centred on the horizontal plane, has volume $2{\pi }^{2}A{R}^2$. In circular torus coordinates, $z=r\cos \alpha$. Integrating $z^2$ with the Jacobian determinant $r(A+r\sin \alpha )$ gives ${\int }_V{z}^{2}dV=({\pi }^2/2)A{R}^4$; the part odd in $\sin \alpha$ vanishes. Thus the mean squared height is $R^2/4$. The identity is useful for moments of inertia and flux calculations using the divergence theorem.

A stationary solution of an evolution differential equation is independent of time. Substituting ${\partial }_{t}u=0$ leaves the spatial boundary value problem that determines its profile. For an ordinary differential equation this is an equilibrium; for a partial differential equation it may vary in space. A stationary profile need not have zero flux, although zero flux is forced in a one-dimensional conserved system with reflecting endpoints.

For a smooth tangent vector field $v$ on a compact orientable surface with smooth boundary and outward conormal of a surface boundary, the integral of its surface divergence is its boundary flux. Extend the unit normal vector to a unit field $m$. The curl of a cross product gives $m\cdot \nabla \times (m\times v)=\nabla \cdot v-m_i{m}_j{\partial }_j{v}_i$ on the surface: $m\cdot v=0$ there and $m\cdot {\partial }_{j}m=0$ by unit length. Apply Stokes theorem and use $(m\times v)\cdot d\mathbf{r}=v\cdot (d\mathbf{r}\times m)$ to obtain the formula.

For $a_t+(V(s)a{)}_s=0$ with $V>0$, the flux perturbation $b=Va$ solves $b_t+V{b}_s=0$. The travel coordinate transforms it into unit-speed transport. Thus the flux perturbation is constant on characteristics while area amplitude changes as $1/V$. In a supplied power-law river, $A_{\mathrm{eq}}^{\prime }=M/V$, giving travel time $[A_{\mathrm{eq}}(l)-A_{\mathrm{eq}}(0)]/M$. At a zero-speed head, this time can remain integrable even though the amplitude formula is not uniform.

For serial stages with survival functions $S_i(a)=\exp [-{\int }_0^a{\Gamma }_i(s)ds]$, set $H_i(r)={\int }_0^{\infty }{\Gamma }_i(a)S_i(a)e^{-ra}da$. If completion of stage two produces $m$ new stage-one individuals, a common exponential mode requires $m{H}_1(r)H_2(r)=1$. Indeed its age profiles are ${\varphi }_i(a)={\varphi }_i(0)e^{-ra}{S}_i(a)$, while the incoming fluxes give ${\varphi }_2(0)=H_1{\varphi }_1(0)$ and ${\varphi }_1(0)=m{H}_2{\varphi }_2(0)$. Cancelling nonzero amplitudes proves the equation. Under ordinary positive residence-time assumptions a positive root exists exactly when $m{H}_1(0)H_2(0)>1$. The normalized profiles are $e^{-ra}{S}_i(a)/\int e^{-rs}{S}_i(s)ds$.

Two identical uniform circular discs of angular diameter $\alpha$ and equal flux have $g_{\star }(Y)=2J_1(x)/x$, $x=\pi \alpha Y/\lambda$, with limiting value one at $Y=0$. Multiplication by $\cos (\pi \beta Y/\lambda )$ gives their signed binary coherence. Its first single-disc envelope zero is $Y=1.21967\lambda /\alpha$, while the binary modulation has a separate scale $\lambda /\beta$. Limb darkening and other nonuniform profiles change the envelope, not the elementary Fourier shift factor.

Vector calculus studies differentiation and integration of scalar fields and vector fields. Three operators answer different local questions: the gradient describes change of a scalar value, the divergence describes net outward flow, and the curl describes local circulation. Their associated integral theorems turn local derivatives into measurements along paths and across boundaries.

The difference between divergence and curl is especially visible in two smooth vector fields on three-dimensional Euclidean space:At each point they have the same Euclidean norm $\sqrt{x^2+y^2}$, but their directions differ. For $F$, $\nabla \cdot F=2$ and $\nabla \times F=0$. For $G$, $\nabla \cdot G=0$ and $\nabla \times G=(0,0,2)$. One expands radially in each horizontal plane; the other describes rigid rotation.

Around the counterclockwise unit circle in the plane $z=0$, $F$ is normal to the boundary and has zero tangential component. Its outward planar flux integral is $2\pi$, while its line integral around the circle is zero. For $G$ the values are reversed: outward planar flux is zero, and circulation is $2\pi$. The divergence and curl predict exactly which boundary measurement survives.

Why is this useful? A velocity field can move material through a boundary or circulate it within a region. Conservation of mass connects its divergence with density changes through the continuity equation; its curl is vorticity. In electromagnetism, the Maxwell equations use the same operators to connect electric fields, magnetic fields, and their sources. The common mathematical language lets the same integral calculations work across different physical models. ^[ref].

The geometric framework behind these operators is developed in differential forms. The Generalized Stokes theorem unifies the fundamental theorem for line integrals, Green theorem, Stokes theorem, and divergence theorem. In coordinates their formulas look different, but each relates accumulation over a region to a measurement on its boundary.

Wilson's Vector Analysis, published in 1901, was based on J. Willard Gibbs's lectures at Yale. Wilson's preface explains that examples from geometry, mechanics, and physics are interspersed with the mathematical development, so readers encounter applications as soon as they have the tools to understand them. That is a useful way to approach vector calculus: learn each operation through a measurement it makes possible. ^[ref].

A vector area element combines the scalar surface area element with a chosen unit normal vector. Reversing the order of the two parameter directions reverses its sign. Flux through an oriented surface is the integral of the dot product of a vector field with this vector element. For a graph $z=f(x,y)$, the upward choice is $(-f_x,-f_y,1)dxdy$.