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A tensor derivation is a real-linear operation preserving every tensor type, obeying the tensor-product Leibniz rule, and commuting with every tensor contraction. A derivation $D$ of smooth functions and a real-linear operator on vector fields satisfying $D(fY)=fDY+(Df)Y$ extend uniquely to a tensor derivation. On a differential one-form it must satisfyThis expression is linear over smooth functions in $Y$. In a local frame $D{e}_a=A^b{}_a{e}_b$, the dual rule is $D{ϵ}^a=-A^a{}_b{ϵ}^b$. Apply the product rule to every coefficient and frame factor to define the extension; these dual signs cancel under contraction. Frame changes agree by differentiating the inverse matrix. Cutoffs prove locality and hence uniqueness from local expansions.

For a vector field $V$, the tensorial second derivative is ${\nabla }^{2}V(X,Y)={\nabla }_X{\nabla }_{Y}V-{\nabla }_{{\nabla }_{X}Y}V$. With $R(X,Y)V={\nabla }_X{\nabla }_{Y}V-{\nabla }_Y{\nabla }_{X}V-{\nabla }_{[X,Y]}V$ and geometric torsion tensor $\mathcal{T}$, antisymmetrization givesThis follows by subtracting the two definitions and using ${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y]+\mathcal{T}(X,Y)$. By contrast, the directional operator commutator subtracts no derivative-index connection and obeys $[{\nabla }_X,{\nabla }_Y]V=R(X,Y)V+{\nabla }_{[X,Y]}V$. On scalars the directional commutator is $[X,Y]f$, while the tensorial Hessian commutator is $-\mathcal{T}(X,Y)f$. Distinguishing these objects prevents an apparent contradiction when a scalar Hessian has an antisymmetric part despite commuting coordinate vector fields.

A multilinear operation on vector fields is tensorial when it is linear over smooth functions in every argument. Its value at a point then depends only on the argument vectors at that point. To see this, use a smooth cutoff function to reduce to local fields, expand them in a local frame, and apply linearity over smooth functions to their coefficients. A smoothly valued tensorial operation therefore defines a tensor field. The curvature of an affine connection is tensorial, whereas a covariant derivative differentiates a scalar coefficient in its second argument and is not tensorial there.

The product of three circles is a compact smooth three-manifold. Quotient coordinates allow constant commuting vector fields. A rank-two plane field can have dense immersed leaves, as in a dense immersed cylinder in a three-dimensional torus.

If $z(t)$ solves an autonomous differential equation, $w(t)=z(-t)$ solves the equation with negated vector field. The trajectories and equilibrium points stay in place but their orientations reverse; stable and unstable directions interchange. A saddle equilibrium remains a saddle, and a clockwise centre equilibrium becomes counterclockwise.

A smooth action of the compact abelian Lie group ${\mathbb{R}}^n/(2\pi \mathbb{Z}{)}^n$. Commuting vector fields with a common periodic normalization give such an action. In action-angle variables, the torus acts by translation of the angles while leaving the actions fixed.

For the local flow of a smooth compactly supported vector field and any scalar BV space function $w$, the change of variables formula and polar decomposition of a vector measure giveThe two cofactor matrices are $I\pm tB+O(t^2)$ uniformly. On $|{\sigma }_w|=1$, the linear terms in their Euclidean norm expansions cancel. Integration bounds the remainder by $C{t}^2|Dw|(\Omega )$. The identity includes all components of the decomposition of a BV derivative; it does not require smoothness of $w$ or its jump interfaces.

For a sufficiently small analytic periodic vector field $w$, a near-identity angular diffeomorphism $\chi$ and a constant vector $a$ solve $D\chi \omega =\omega +w\circ \chi +a$. They satisfy $\chi -\mathrm{id}=O(w)$, $a=O(w)$; if $⟨w⟩=0$, then $a=O(w^2)$. For the defect $e={\mathcal{D}}_{\omega }\chi -\omega -w\circ \chi -a$, set $A=D\chi$, choose $\Delta a=⟨A^{-1}{⟩}^{-1}⟨A^{-1}e⟩$, and solve ${\mathcal{D}}_{\omega }v=A^{-1}(\Delta a-e)$ with zero mean. Updating $\chi$ by $Av$ leaves a quadratic defect $(De)v-[w(\chi +Av)-w(\chi )-Dw(\chi )Av]$. The torus small-divisor estimate and geometrically decreasing strip losses yield a convergent analytic iteration.

As a quadrature rule, the trapezoidal rule replaces an integrand by the linear interpolation joining its two endpoint values:For nonnegative endpoint values, this is the area of a trapezoid; in general it is a signed area. The quadrature nodes are $a,b$, each with quadrature weight $(b-a)/2$. Unlike the midpoint quadrature rule, it uses endpoint samples, which can be reused between adjacent panels.

Every affine function $f(x)=\alpha x+\beta$ agrees with its linear interpolation, so the trapezoidal rule integrates it exactly. On $[0,2]$, $f(x)=2x+1$ gives $T(f)=(2/2)(1+5)=6=I(f)$. On $[0,1]$, the polynomial $f(x)=x^2$ instead gives $T(f)=1/2$, exceeding $I(f)=1/3$ by $1/6$. Its degree of precision of a quadrature rule is exactly one.

If $f\in C^2[a,b]$, its signed approximation error satisfiesA convex function lies below its endpoint chord, so the trapezoidal rule overestimates its integral; a concave function gives the reverse inequality. The error formula assumes that the second derivative exists and is a continuous function throughout the closed interval.^[ref] A short derivation is the identityobtained by two integrations by parts. The nonnegative Peano kernel has integral $(b-a{)}^3/12$, and the weighted mean value theorem for integrals gives the formula.

The trapezoidal rule for an ordinary differential equation averages vector fields at the old and new states and is A-stable.

This time-stepping use applies the same trapezoidal rule to the integral of the derivative in an ordinary differential equation. That connection explains its place under implicit Runge-Kutta methods; its quadrature behavior and the nonlinear issue in trapezoidal rule fails B-stability concern different uses of the same endpoint averaging.

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.

For smooth vector fields $X,Y$, the local flows commute wherever both compositions are defined exactly when $[X,Y]=0$. The derivative of the pullback of $Y$ by the $X$ flow is the pullback of $[X,Y]$. A zero bracket therefore makes $Y$ invariant under that flow; uniqueness of solutions to its ordinary differential equation gives commutation. Conversely commutation implies invariance, whose derivative gives the zero bracket. Neither field need have a complete flow.

For the faces of a closed polyhedron, take outward unit normal vectors. Applying the divergence theorem to an arbitrary constant vector field $c$ givesSince $c$ is arbitrary, the vector areas sum to zero. This is an oriented statement; the positive scalar surface areas do not cancel. The same proof works for every piecewise smooth closed boundary to which the divergence theorem applies.

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].

In elliptic cylindrical coordinates, write a vector field as $A=A_u{e}_u+A_v{e}_v+A_z{e}_z$ using the local orthonormal unit vectors. The scale factors of orthogonal coordinates are $(h,h,1)$, so the divergence and curl areApplying the divergence to the gradient of a scalar function gives $\Delta \varphi =({\varphi }_{uu}+{\varphi }_{vv})/h^2+{\varphi }_{zz}$. These formulas hold where $h>0$. They use physical components along unit vectors, rather than components along the unnormalized coordinate tangent vectors.

For a continuously differentiable vector field $v$ on a bounded sufficiently regular region $V$, apply the divergence theorem to $a\times v$ for an arbitrary constant vector $a$. The vector calculus identity $\nabla \cdot (a\times v)=-a\cdot (\nabla \times v)$ and the cyclic scalar triple product identity show that the dot products of both sides of the displayed formula with $a$ agree. Since $a$ is arbitrary, the vector integrals agree. Here $n$ is the outward unit normal vector. Reversing the order on the boundary gives the equivalent positive-sign formula $\int n\times vdS=\int \nabla \times vdV$.

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 $(-f_x,-f_y,1)dxdy$.

For an isotropic, integrable background phase-space distribution function, rotational symmetry makes the integral radial in velocity space. Its divergence with respect to $\mathbf{V}$ is $4\pi f(V)$ by the Dirac delta identity for the divergence of the inverse-square vector field. Integrating over a velocity-space sphere and using the divergence theorem proves the displayed identity. Exterior velocity shells have zero contribution, just as in the spherical shell theorem. This explains why the leading-log Chandrasekhar dynamical friction integral contains only particles slower than the test body.

A volume integral integrates a scalar or componentwise vector field over a three-dimensional region with its volume form. In spherical polar coordinates the Euclidean volume form is $r^2\sin \theta drd\theta d\varphi$. The divergence theorem relates a volume integral of a divergence to a boundary flux integral.

A vector field preserves a chosen volume form $\omega$ when its local flow pulls $\omega$ back to itself. Infinitesimally this is ${\mathcal{L}}_X\omega =0$, using the Lie derivative. The condition depends on the chosen volume form.

The divergence defined through weak derivatives: pair a locally integrable vector field $J$ with a smooth compactly supported test function $\phi$ using the displayed integration-by-parts identity. A piecewise smooth field can have a surface contribution where its normal component jumps. Thus zero classical divergence on each side does not imply zero weak divergence across the interface.