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Let $S$ be an oriented smooth surface with piecewise smooth boundary, and let $F$ be a continuously differentiable vector field on a neighborhood of $S$. With the compatible boundary direction determined by the right-hand rule,For $G=(-y,x,0)$ on the horizontal unit disc with upward normal, the left side is $2$ times its area, giving $2\pi$. The counterclockwise boundary has tangential field of magnitude one and circumference $2\pi$, giving the same line integral. Reversing the surface normal reverses the boundary direction and both signs. The theorem replaces a circulation calculation with a surface integral of the curl, whichever is easier.

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.

A transverse electric mode of a rectangular waveguide has $E_z=0$ while its axial magnetic field is generally nonzero. For time dependence $e^{-i\omega t}$ and axial dependence $e^{i\beta z}$, set $\Delta =(\omega /c{)}^2-{\beta }^2$. The Maxwell equations reconstruct the transverse components from $B_z$:The axial complex amplitude satisfies a transverse Helmholtz equation, $({\partial }_x^2+{\partial }_y^2+\Delta )B_z=0$. Zero tangential electric field at each perfect conductor wall becomes a Neumann boundary condition on $B_z$. Separation of variables givesThe index $m$ counts half-periods across $a$, and $n$ across $b$. Either index can vanish, allowing ${\mathrm{TE}}_{10}$; both cannot vanish. Indeed a spatially constant axial oscillating magnetic field would give nonzero changing magnetic flux through the cross-section, whereas the line integral of tangential electric field around its metal boundary is zero. The integral form of Faraday's law excludes this ${\mathrm{TE}}_{00}$ candidate. Dividing the reconstruction formulas by $\Delta =0$ would be invalid and cannot produce the separate transverse electromagnetic mode.

For $m=1,n=0$, choose $B_{\ast }=-i(\pi /a)E_0/\omega$. The formulas recover exactly the $E_y$, $B_x$ and $B_z$ in rectangular electromagnetic waveguide. In particular, the transverse magnetic field accompanies $B_z$ and ensures zero divergence; an oscillating longitudinal magnetic field on its own is not the guided solution.

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].

For an oriented parametric curve $\mathbf{r}(t)$ and scalar field $V$, define the componentwise line integralIts value is a vector and reversing orientation changes its sign. It differs from the scalar arc length integral ${\int }_{C}Vds=\int V(\mathbf{r}(t))|{\mathbf{r}}^{\prime }(t)|dt$, which is independent of orientation. A request to integrate a scalar field along a curve should specify which differential is intended.

For the counterclockwise circle $\mathbf{r}(\varphi )=(R\cos \varphi ,R\sin \varphi ,z_0)$, suppose the restricted field has Fourier series $V(\varphi )=A_0+{\sum }_{n\ge 1}(A_n\cos n\varphi +B_n\sin n\varphi )$. Orthogonality givesThe oriented vector integral selects the first angular harmonics, whereas the scalar arc-length integral selects the mean.

Use a velocity potential to reconstruct an irrotational flow from one scalar field. A velocity potential $\varphi (\mathbf{x},t)$ satisfiesIt has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$ and is unchanged as a description of the velocity field when any function of time alone is added. For twice continuously differentiable $\varphi$, the identity curl of a gradient proves zero vorticity. In incompressible flow, taking the divergence also gives ${\nabla }^2\varphi =0$, the Laplace equation. This turns a three-component velocity field into a single harmonic function with appropriate boundary conditions. ^[ref]

For uniform flow $\mathbf{u}=(U,0,0)$, take $\varphi =Ux$. With $U=2\mathrm{m}{\mathrm{s}}^{-1}$, the change from $x=0$ to $x=3\mathrm{m}$ is $\Delta \varphi =6{\mathrm{m}}^2{\mathrm{s}}^{-1}$. By the fundamental theorem for line integrals, the line integral of velocity between these points equals that change along any path in the domain; a closed curve gives zero circulation.

For the potential line vortex, choose a local angle branch in polar coordinates and set $\varphi =K\theta$. Then $u_r={\partial }_r\varphi =0$ and $u_{\theta }=r^{-1}{\partial }_{\theta }\varphi =K/r$. One counterclockwise turn changes the continued velocity potential by $2\pi K$, although the velocity field returns to its original value. With $K=1{\mathrm{m}}^2{\mathrm{s}}^{-1}$, this change is $2\pi {\mathrm{m}}^2{\mathrm{s}}^{-1}$, exactly the circulation. A full annulus therefore admits local angle branches and a multivalued velocity potential, but no global single-valued velocity potential when $K\ne 0$. Cutting the annulus along a radial line allows a single-valued angle branch on the remaining domain. The obstruction is path independence, not a failure of the local gradient formula.

Work measures the energy transferred by a force as its point of application moves. For a point mass following a path $C$, add the contribution of the force along each small displacement:This line integral uses the actual path, and the dot product selects the component of force along the motion. A large force perpendicular to every small displacement does zero work; holding something still also does zero mechanical work on it, even if holding it feels tiring. The SI unit is $\mathrm{J}=\mathrm{N}\mathrm{m}$. ^[ref]

The usefulness of work is that it connects a description in terms of forces and distances to one in terms of energy. The work-energy theorem gives the change in kinetic energy from the total net work, often without solving for the entire motion. The work of one selected force equals that change only when the other forces contribute zero total work.

For a fixed force $\mathbf{F}$ and endpoint displacement $\Delta \mathbf{r}$, the line integral reduces to a dot product:Here $\theta$ is the angle between the force and the displacement. For straight motion, $d$ is also the distance traveled. A $3\mathrm{N}$ force acting along a $4\mathrm{m}$ displacement does $12\mathrm{J}$ of work. Turning the same force perpendicular to the displacement gives $\cos (\pi /2)=0$ and zero work; reversing it gives $\cos \pi =-1$ and $-12\mathrm{J}$. At $\theta =\pi /3$, the work is $6\mathrm{J}$. These signs describe energy transferred to or from the moving body. ^[ref]

For a point mass of constant mass $m$ observed in an inertial frame, the net work along its actual path equals its change in kinetic energy:The middle expression is a line integral; the dot product selects the component of force along the displacement. Positive net work increases speed, negative net work decreases it, and zero net work leaves the endpoint speeds equal. Equal endpoint speeds do not imply equal velocities: a perpendicular force can continually change direction while doing zero work. ^[ref]

The bridge to Newton's second law is short: take the dot product of $m\dot{\mathbf{v}}={\mathbf{F}}_{\mathrm{net}}$ with velocity. The chain rule gives $d(m|\mathbf{v}{|}^2/2)/dt={\mathbf{F}}_{\mathrm{net}}\cdot \mathbf{v}$, equating the rate of change of kinetic energy to instantaneous power. Integrating gives the theorem. No conservative force assumption entered this argument; friction also contributes to the net work.

For the harmonic oscillator with spring stiffness $k=2N/m$, release a mass $m=0.5\mathrm{kg}$ from rest at $x=0.3\mathrm{m}$. The spring exerts $F=-kx$. Its work before the first passage through $x=0$ isThus $m{v}^2/2=0.09\mathrm{J}$ gives speed $0.6m/s$ there. The velocity is negative on this passage because the position is decreasing. The energy calculation alone determines its magnitude, so the direction still requires information about the motion.

For a time-independent potential energy $V$, the conservative force contributes work $-\Delta V$. Separating all other work as $W_{\mathrm{other}}$ gives $\Delta (T+V)=W_{\mathrm{other}}$. Consequently conservation of mechanical energy requires zero net other work; the work-energy theorem itself does not require that conservation. With damping, mechanical energy decreases even though the theorem remains valid. ^[ref]