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Because the magnetic field obeys $\nabla \cdot \mathbf{B}=0$, it can be represented locally as the curl of a magnetic vector potential:This automatically enforces the absence of magnetic monopoles. It is also a global representation throughout ordinary all-space problems without monopole singularities. It is not unique: the gauge transformationleaves $\mathbf{B}$ unchanged because the curl of a gradient vanishes. In a time-dependent problem,makes the same transformation leave both physical fields unchanged and automatically satisfies Faraday's law.

In vacuum magnetostatics, choose the Coulomb gauge $\nabla \cdot \mathbf{A}=0$. Substitution into Ampère's law and the identity $\nabla \times (\nabla \times \mathbf{A})=\nabla (\nabla \cdot \mathbf{A})-{\nabla }^2\mathbf{A}$ givesfor a localized steady electric current density and a potential vanishing at infinity. Taking the curl recovers the Biot-Savart law. The magnetic vector potential converts a constrained vector field problem into componentwise Poisson equations once a gauge is chosen.

For a uniform magnetic field, $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{r}$ is a convenient choice. For a localized magnetic dipole moment $\mathbf{m}$, the leading far-field potential isAnother useful relation is Stokes theorem, which relates a closed-path integral to the magnetic flux through a spanning surface:The units of $\mathbf{A}$ are $\mathrm{T}\mathrm{m}$, equivalently $\mathrm{Wb}{\mathrm{m}}^{-1}$.

For a fixed contour, Faraday's law of induction isThe second equation is the differential Maxwell equation, related to the first by Stokes theorem. For a moving wire, the relevant motional electromotive force is $\mathcal{E}=\oint (\mathbf{E}+{\mathbf{v}}_{\mathrm{wire}}\times \mathbf{B})\cdot d\mathbf{l}=-d{\Phi }_B/dt$. Thus a static magnetic field can induce current in a moving circuit even when $\nabla \times \mathbf{E}=0$.

For translational magnetic damping of a square wire loop, while the upper horizontal leg remains in the field, a downward speed $v$ reduces the flux at rate $Bav$. Hence $|\mathcal{E}|=Bav$, $|I|=Bav/R$, and the upward magnetic braking force is $B^2{a}^{2}v/R$. With negligible electrical self-inductance,Let the wire cross-sectional area be $S$, mass density ${\rho }_m$, and resistivity ${\rho }_e$. Since $m=4aS{\rho }_m$ and $R=4a{\rho }_e/S$,These use the printed copper values and $B=1\mathrm{T}$. The wire dimensions cancel. The terminal-speed approximation is restricted to the interval before the upper leg leaves the field, as assumed in the question.

For rotational magnetic damping of a square wire loop, consider rotation about the central $x$ axis in a uniform field, the flux is $B{a}^2\cos \theta$. The induced current produces torque $-B^2{a}^4\dot{\theta }{\sin }^2\theta /R$. A thin square loop has moment of inertia ${\mathcal{I}}_x=m{a}^2/6$: the two legs parallel to $x$ contribute $m{a}^2/8$, and the other two $m{a}^2/24$. Averaging ${\sin }^2\theta$ over a rotation givesIf “decay time” means rotational kinetic energy rather than angular speed, $K\propto {\omega }^2$ gives ${\tau }_{\mathrm{rot},\mathrm{K}}={\tau }_t/6=0.400512\mathrm{ms}$. This distinction, and the assumed separation between rotation period and slow decay, are essential. The electrical and mechanical time scales in magnetic braking must be separated: electrical self-inductance is neglected in both calculations.

For a consistently oriented surface and boundary, Ampère's law isStokes theorem gives the left side as ${\int }_S(\nabla \times \mathbf{B})\cdot \mathbf{n}dS$, so the given curl equation proves the law.

On a circle of radius $r$ around the axis, the assumed azimuthal field is tangent and constant in magnitude. Thus the circulation is $2\pi r{B}_{\theta }(r)$. For $r\le a$, the enclosed electric current is $I{r}^2/a^2$; for $r\ge a$, it is $I$. ThereforeThe field is continuous at $r=a$ and vanishes regularly at the axis. Its direction follows the right-hand rule for positive electric current along ${\mathbf{e}}_z$. The radial component of the curl equation also gives ${\partial }_z{B}_{\theta }=0$, so the expression is valid at every height; no unmentioned axial dependence remains.

The four dimensional equations express mass conservation, momentum balance with pressure, gravity and viscous stress, heat advection and conduction, and a linear thermal-expansion equation of state. For heating from below, suitable plate conditions are $\mathbf{u}=0$ and $T=T_0+\Delta T$ at $z=0$, and $\mathbf{u}=0$, $T=T_0$ at $z=d$. No slip and thermal insulation on the sidewall provide one suitable lateral completion; prescribed lateral temperatures would be another specified thermal setup.

Let $\kappa =k/({\rho }_0{c}_p)$ be thermal diffusivity and $\nu =\mu /{\rho }_0$ kinematic viscosity. Use $d$, $d^2/\kappa$, $\kappa /d$ and $\Delta T$ as length, time, velocity and temperature scales. Subtract the static reference pressure $p_H$ with $\nabla p_H=-{\rho }_{0}g\mathbf{k}$ and scale its remainder by $\mu \kappa /d^2$. ThenDensity becomes ${\rho }_0(1-BT)$. Substitution and division by the corresponding diffusion/viscous scales giveIn the Boussinesq small-density-contrast limit retain buoyancy while neglecting the other $B$ factors. At infinite Prandtl number, after any fast initial adjustment, the equations areThe velocity is instantaneously constrained by a viscous elliptic balance; the temperature still evolves.

Take the divergence of the vector Poisson equation. Derivatives commute, and divergence of a curl is zero, so $\Delta (\nabla \cdot \mathbf{A})=0$. Its boundary value is zero, hence the maximum principle gives $\nabla \cdot \mathbf{A}=0$ throughout $D$. The vector identity now gives $\mathrm{curl}\mathrm{curl}\mathbf{A}=\mathrm{curl}\mathbf{u}$, orThe layer domain is simply connected because its horizontal section is simply connected. For a curl-free field on such a domain, define $\chi (x)$ as its line integral from a fixed base point to $x$. Contractibility of loops and Stokes theorem make the integral path independent, and differentiation gives its gradient equal to the field. Thus $\mathrm{curl}\mathbf{A}-\mathbf{u}=\nabla \chi$.

Incompressibility then gives $\Delta \chi =0$, while the prescribed normal curl gives ${\partial }_n\chi =0$. Green's identity yieldsHence $\chi$ is constant on the connected domain and $\boxed{\mathrm{curl}\mathbf{A}=\mathbf{u}}$. The extra condition $\mathbf{A}\cdot \mathbf{n}=0$ belongs to the stated potential boundary problem; the reconstruction argument above uses the divergence and normal-curl conditions. On a multiply connected domain a curl-free field need not be a globally single-valued gradient.

Finally $\mathrm{curl}(T\mathbf{k})=(T_y,-T_x,0)$. Taking the curl of the infinite-Prandtl momentum balance and using $\mathrm{curl}\mathbf{u}=-\Delta \mathbf{A}$ givesThe derivation is an instance of velocity reconstruction from a mixed-boundary vector Poisson problem.

On a nonempty compact Kähler manifold of complex dimension $n$, the Kähler form is closed and ${\int }_X{\omega }^n>0$. If ${\omega }^k=d\eta$ for $1\le k\le n$, then ${\omega }^n=d(\eta \wedge {\omega }^{n-k})$; Stokes theorem contradicts the positive integral. Thus the de Rham cohomology is nonzero in every even degree through $2n$.

For static electric fields and magnetic fields, Ampère's law and $\mathbf{E}=-\nabla \varphi$ give $\mathcal{P}=\mathbf{E}\times \mathbf{H}=\varphi {\mathbf{J}}_f-\nabla \times (\varphi \mathbf{H})$. Stokes theorem on an oriented separating surface then giveswhere each perfect conductor crossing it has constant electric potential ${\varphi }_k$ and signed electric current $I_k$. The line term is absent for a closed surface and must vanish in the limiting sense at infinity for a noncompact surface. Under that condition the incoming Poynting flux equals terminal power. If the signed electric currents sum to zero, the result is independent of the reference for electric potential.

On a compact oriented smooth manifold $N$ of dimension four without boundary, set $Q([\alpha ],[\beta ])={\int }_N\alpha \wedge \beta$ for closed real two-forms. The Stokes theorem makes the expression independent of representatives, and the even degrees make it symmetric. It is the real de Rham cohomology version of the intersection form. The Hodge decomposition theorem and Hodge star operator express it as $Q(h,k)=(h,\ast k{)}_{L^2}$ on harmonic representatives.

Two homotopy strips with the same endpoint paths and boundary on Lagrangian submanifolds glue to a relative $2$-cycle. If that cycle is zero in $H_2(M,L_0\cup L_1)$, its symplectic area vanishes: apply Stokes theorem, using $d\omega =0$ and the zero restriction of $\omega$ to each Lagrangian. Thus relative homology controls the ambiguity in a Lagrangian path-area functional.

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 $\pi /2$.

For a rigidly rotating prescribed vacuum path, use spatial coordinates attached to the apparatus and laboratory time $t$. At a path point $\mathbf{r}$, the local apparatus velocity is $\mathbf{v}=\mathbf{\Omega }\times \mathbf{r}$. A light displacement has instantaneous laboratory displacement $d\mathbf{r}+\mathbf{v}dt$, up to the common rotation of the spatial axes. The null condition is thereforeWrite $d\ell =|d\mathbf{r}|$. To first order in $|\mathbf{v}|/c$, the positive root isReversing the path reverses the dot product term but leaves the leading path-length term unchanged. Subtract the two return times and use the scalar triple product identity:This is a line integral version of the Sagnac area law; the vector area identity also follows from Stokes theorem. For a mirror-defined polygon the actual ray paths move slightly as the mirrors rotate; the stationary optical path at zero rotation gives the same first-order result. The prescribed-path calculation is an idealization, not an exact finite-speed formula for every moving-mirror layout.

The dot product explains the geometry: only the projection of the oriented vector area along the angular velocity contributes. A planar loop whose normal vector is perpendicular to $\mathbf{\Omega }$ has zero leading delay. For repeated traversals with the same orientation, the signed vector areas add; oppositely oriented lobes can cancel even when the total unsigned area is large. Equal perimeter alone does not imply equal response. A constant translational velocity contributes nothing to this leading closed-path line integral, since ${\oint }_{C}d\mathbf{r}=0$.

Suppose two oriented surfaces have the same oriented boundary and $S_1\cup (-S_2)$ bounds a volume inside a domain where the vector field $\mathbf{F}$ is smooth and solenoidal. The divergence theorem gives zero flux through their closed union, proving the displayed equality. If a global vector potential $\mathbf{F}=\nabla \times \mathbf{G}$ exists on the relevant surfaces, the Stokes theorem proves the equality directly from their common boundary. The domain condition matters: a singularity inside the joining volume can obstruct the divergence argument.

For a smooth map between oriented unit spheres, integrate the pullback of the target area form. Localizing a normalized top-degree form near a regular value, and using Stokes theorem to discard exact-form differences, identifies this integral with the degree as a sum of local degrees.

For an upward-oriented band of the circular elliptic paraboloid between radii $r_0$ and $R$, the vector field $B=(-y^3,x^3,z^3)$ has vertical curl $3(x^2+y^2)$. Its surface integral is $3\pi (R^4-r_0^4)/2$. Stokes theorem gives the same value from the counterclockwise outer circle minus the counterclockwise inner circle. Equivalently the induced inner boundary direction is clockwise. Forgetting the inner boundary or its reversed orientation changes the answer.

Stokes theorem applies to an oriented surface with several boundary components. Cutting a cylindrical side along one generator produces a parameter rectangle: the two copies of the cut are traversed in opposite directions and their line integrals cancel for a single-valued smooth vector field. The upper and lower rims retain opposite directions when viewed from the same side. This proves why cylindrical topology does not require a modification of Stokes theorem; it requires inclusion of the entire oriented boundary.

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 compact connected oriented boundaryless $n$-dimensional smooth manifold, choose a Riemannian metric. Harmonic functions are constant because $⟨f,\Delta f⟩=\Vert df{\Vert }^2$, and Hodge star commutes with the Hodge Laplacian. Thus the Hodge star operator identifies the constant functions with the harmonic differential forms of top degree, exactly $\mathbb{R}{\omega }_g$. The Hodge decomposition theorem identifies these with the top de Rham cohomology. The metric volume form has nonzero class by Stokes theorem. The harmonic-representative theorem is stated in Denis Auroux's Hodge theory lecture.

The radial vector field $B=e_r/r^2$ on punctured three-dimensional Euclidean space has flux $4\pi$ through every centered sphere. It has no vector potential smooth on an entire enclosing sphere: the Stokes theorem on a closed surface would force zero flux. Two overlapping patches suffice. In spherical coordinates take $A_{+}=(1-\cos \theta )e_{\varphi }/(r\sin \theta )$, regular away from the negative $z$ axis, and $A_{-}=-(1+\cos \theta )e_{\varphi }/(r\sin \theta )$, regular away from the positive $z$ axis. Their curls both equal $B$.

The Cartesian forms $A_{+}=(-y,x,0)/[r(r+z)]$ and $A_{-}=(y,-x,0)/[r(r-z)]$ show that the first is smooth at the north pole and the second at the south pole. On the overlap their difference is $A_{+}-A_{-}=2\nabla \varphi$, where the azimuth $\varphi$ is defined locally. Its multivaluedness around the axis explains the circulation difference $4\pi$. Splitting the sphere along a latitude, applying the Stokes theorem with $A_{+}$ on the north and $A_{-}$ on the south, and using their opposite boundary orientations recovers the total flux. A coordinate singularity in the tangential spherical basis does not by itself make a vanishing smooth vector field singular at a pole.

The vector area of an oriented surface is the integral of its unit normal vector times its area element. If its oriented boundary is a closed curve $C$, Stokes theorem givesConsequently it depends only on the oriented boundary, not on the spanning surface. For a planar surface it is signed area times a unit normal vector. A current loop has magnetic dipole moment $\mathbf{m}=I\mathbf{S}$; the scalar area of a nonplanar spanning surface is not interchangeable with this vector.

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].

The graph $z=g(x,y)$ has upward-oriented vector area element $d\mathbf{S}=(-g_x,-g_y,1)dxdy$ and scalar area element $dS=\sqrt{1+g_x^2+g_y^2}dxdy$. These follow from the cross product of the parametrization derivatives $(1,0,g_x)$ and $(0,1,g_y)$. The vector element is used in flux surface integrals and Stokes theorem; reversing orientation changes its sign.