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Equality of line integrals on two selected paths is a necessary test for a conservative vector field, but it is not sufficient. The paths may both miss the component responsible for nonzero curl.

For an explicit example, take $W(x,y,z)=(z-y,x-z,y-x{)}^T$. Its curl is the nonzero constant $2(1,1,1{)}^T$, so it is not a gradient field on ${\mathbb{R}}^3$. Nevertheless, on the plane $x=y$, the field equals $(z-x,x-z,0{)}^T$. Every tangent vector to that plane has equal first and second components, giving zero dot product with $W$. Thus every line integral of $W$ along a path confined to that plane is zero, regardless of endpoints or the path's shape.

In particular, a broken straight path and a genuinely curved path in that plane can have matching line integrals while revealing nothing about global path independence. To establish a conservative vector field, construct a vector-field potential, or verify the appropriate global criterion such as zero curl on a simply connected domain.

A differential form is a smooth field of alternating multilinear maps that can be integrated over an appropriately oriented manifold of matching dimension. A one-form measures a directed displacement; a two-form measures an oriented area element. The point is to express what is being accumulated along a curve or over a surface without tying the answer to one set of coordinates. This makes line integrals, surface integrals, and the integral theorems of vector calculus parts of one construction.

For a concrete one-form, take $\alpha =xdy$ in the plane. Travel from $(0,0)$ to $(1,1)$ along the bottom and then the right edge of the unit square. The horizontal leg contributes zero because $dy=0$; the vertical leg has $x=1$ and contributes ${\int }_0^{1}dy=1$. Taking the left and then the top edge instead gives zero: the vertical leg has $x=0$, and the horizontal leg again has $dy=0$. The two paths have the same endpoints but different integrals.

Now replace $\alpha$ with the exact differential $\beta =d(xy)=ydx+xdy$. Both paths give $1$, because integrating an exact differential along a curve gives the endpoint difference of its potential. Here that difference is $xy{|}_{(1,1)}-xy{|}_{(0,0)}=1$. This is the fundamental theorem of calculus applied to a function along a parametric curve, and is the basic reason conservative vector fields have path-independent work.

The exterior derivative records the local obstruction in the first example: $d(xdy)=dx\wedge dy$. The Generalized Stokes theorem turns the difference between the two path integrals into the integral of this two-form over the square, whose oriented area is $1$. Interior edges cancel when adjoining pieces are added; only the outer boundary survives. That cancellation connects a local calculation with a global measurement.

More precisely, a degree-$k$ differential form assigns to each point $p$ a multilinear scalar-valued function of $k$ tangent vectors at $p$. Interchanging two arguments changes its sign. A one-form is therefore a covector field, while a two-form vanishes on a pair of parallel vectors. Degree zero simply means a smooth function. The space ${\Omega }^k(M)$ is a vector space; the wedge product of differential forms combines degrees, and the exterior derivative raises degree by one.

The usefulness goes beyond bookkeeping. Pullbacks of a differential form supply the right transformation law automatically under a coordinate change. The Generalized Stokes theorem unifies the fundamental theorem of calculus, Green theorem, Stokes theorem, and divergence theorem. Meanwhile, the distinction between closed differential forms and exact differential forms detects holes through de Rham cohomology. Reyer Sjamaar develops these connections in sections 5.4 and 9.3 of his university notes. ^[ref].

In electrostatics, the electric potential $\Phi$ tells us the interaction potential energy per unit test electric charge in the electric field of prescribed sources. Its most directly useful quantity is an electric potential difference: how much work per unit electric charge an external agent must supply to move that test electric charge slowly between two points. This replaces the bookkeeping of a vector force along a path with the difference of two scalar values. The test electric charge must be small enough not to rearrange the sources appreciably.

For motion from $A$ to $B$, write $\Delta \Phi =\Phi (B)-\Phi (A)$. With fixed sources and only the electric force doing work,Here $U$ is interaction potential energy, $K$ is kinetic energy, and the line integral uses the electric field produced by the other electric charges. For slow transport with unchanged kinetic energy, the external work is $W_{\mathrm{ext}}=\Delta U$, so $\Delta \Phi =W_{\mathrm{ext}}/q$ for nonzero $q$. This external work has the opposite sign to the work done by the electric field. The SI unit is the volt,A volt measures energy per unit electric charge, not energy itself. For example, crossing an electric potential difference of $+5\mathrm{V}$ changes the potential energy of $q=+2\mathrm{nC}$ by $+10\mathrm{nJ}$, but changes that of $q=-2\mathrm{nC}$ by $-10\mathrm{nJ}$. Both test electric charges encounter the same electric potential difference; their opposite signs reverse the energy change.^[ref]

The electrostatic potential is a scalar function of position. The electric field is a conservative vector field, and all three components follow from the negative gradient of the electric potential:The electric field points toward decreasing electric potential. A positive test electric charge feels electric force in that direction; a negative test electric charge feels electric force in the opposite direction, since $\mathbf{F}=q\mathbf{E}$. A freely accelerated negative test electric charge therefore moves toward higher electric potential while lowering its potential energy $U=q\Phi$.^[ref]

Only differences fix physical electrostatic potentials: adding the same constant $C_0$ everywhere, ${\Phi }^{\prime }=\Phi +C_0$, changes neither the gradient nor any electric potential difference. Choosing an electrical reference potential is this elementary gauge transformation. Setting zero at infinity is convenient for finitely many localized point charges; it is not available for an ideal uniform electric field extending to infinity. There we can choose a finite reference point. Assigning zero at ground is a convention, whereas physically grounding an electrical conductor also allows exchange of electric charge to impose a boundary value. Zero electric potential at a point never by itself implies zero electric field.^[ref]

For a fixed point charge $Q$ at the origin in vacuum, choose zero at infinity. Coulomb's law giveswhere ${\epsilon }_0$ is the vacuum permittivity. Use the stipulated approximation $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$ and $Q=+2\mathrm{nC}$. At $r=0.20\mathrm{m}$,A test electric charge $q=-3\mathrm{nC}$ has $U=-2.697\times 10^{-7}\mathrm{J}$ and radial electric force $F_r=-1.3485\times 10^{-6}\mathrm{N}$, directed inward. Moving it inward from $r_A=0.20\mathrm{m}$ to $r_B=0.10\mathrm{m}$ raises the electric potential from $89.9$ to $179.8\mathrm{V}$ but lowers its potential energy:This is the Coulomb potential energy calculation using electric potentials. The source's singular self-field at $r=0$ is excluded; when finding the force on a source point charge, use only the electric field of the other sources.^[ref]

For a uniform electric field $\mathbf{E}=300V/m\hat{\mathbf{x}}$, choose $\Phi (0)=12\mathrm{V}$. ThenMoving from $x_A=0$ to $x_B=0.04\mathrm{m}$ gives $\Delta \Phi =-12\mathrm{V}$. For $q=+2\mathrm{nC}$ the change in potential energy is $-24\mathrm{nJ}$ and the electric field does $+24\mathrm{nJ}$ of work. For $q=-2\mathrm{nC}$, the same prescribed displacement gives $\Delta U=+24\mathrm{nJ}$: the external agent must supply that work for slow transport. This uniform electric field approximates the interior of a parallel-plate capacitor sufficiently far from the edges; the fringing field of a capacitor limits the approximation.^[ref]

An equipotential is a level set $\Phi =\text{constant}$. On a smooth equipotential with nonzero gradient, every tangent displacement has $d\Phi =\nabla \Phi \cdot d\mathbf{l}=0$, so the electric field is normal to the equipotential. Thus moving a test electric charge along an equipotential requires no net electrical work. Around a point charge, the equipotentials are spheres; the circles in the diagram are their sections, not an assumption of two-dimensional Coulomb's law. A uniform electric field has parallel plane equipotentials. For contours separated by the same small electric potential difference, smaller normal separation means a stronger electric field, approximately $|\mathbf{E}|\approx |\Delta \Phi |/\Delta n$. At a point where $\nabla \Phi =0$, the electric field is zero and has no direction to call normal.^[ref]

The superposition principle adds electric potentials as signed scalars. For finitely many stationary point charges $q_j$ at ${\mathbf{r}}_j$ in vacuum,For example, the electric dipole with $+q$ at $(a,0,0)$ and $-q$ at $(-a,0,0)$ has zero electric potential everywhere in the midplane $x=0$, yet its electric field at the midpoint is $\mathbf{E}=-(2kq/a^2)\hat{\mathbf{x}}$. With $q=2\mathrm{nC}$ and $a=0.10\mathrm{m}$, this is $-3596N/C\hat{\mathbf{x}}$. At $(2a,0,0)$, the same model gives $\Phi =2kq/(3a)\approx 119.867\mathrm{V}$ and $E_x=8kq/(9a^2)\approx 1598.222N/C$. Cancellation of electric potentials is quite different from cancellation of vector electric fields.

A motivating measurement is conductive-paper electric potential mapping. A documented teaching experiment uses slightly conductive paper on an insulating board, two circular electrodes in highly conducting ink, a steady supply, and a digital voltmeter. Connect the supply to the two electrodes, keep the voltmeter common lead on the negative electrode, and use its other lead as a movable probe. Adjust the supply using the measured electric potential difference between electrodes; the published procedure uses $10\mathrm{V}$. The observable is the voltmeter reading $\Phi (P)-\Phi (\mathrm{COM})$ at the probe position $P$. Locate several points reading $2\mathrm{V}$, record their positions on a separate map, and join them into an equipotential; repeat at other readings. Lift the probe between contacts so it does not damage the paper. This procedure makes the electrical reference potential concrete: every reading compares two contacts.^[ref]

From neighboring readings infer an electric field component by a finite difference, $E_x\approx -[\Phi (x+h)-\Phi (x)]/h$. Reconstruct two in-plane components with displacements along two independent directions. Alternatively, draw the electric field normal to mapped equipotentials, toward lower electric potential. The second diagram illustrates the same measurement principle with bar electrodes spanning a uniform sheet, insulating top and bottom edges, and a synthetic model $\Phi (x)=(50V/m)x$. The model readings $4.0\mathrm{V}$ at $x=0.08\mathrm{m}$ and $5.0\mathrm{V}$ at $x=0.10\mathrm{m}$ imply $E_x=-50V/m$. These values are calculated illustrations, not reported experimental data. The documented circular-electrode experiment similarly estimates the electric field from readings at successive centimetre marks.^[ref]

Useful controls include checking the voltmeter zero, checking that the electric potential difference along each painted electrode is negligible, and repeating reference readings to detect supply drift. Reversing the supply should reverse the signs relative to the same reference contact. Contact resistance, leakage electric currents, and touching the sheet can distort the map. A voltmeter has finite input electrical resistance and draws a small electric current; voltmeter loading changes the quantity being measured if that electrical resistance is insufficiently large relative to the sheet's effective source electrical resistance.^[ref][ref]

The paper carries a steady electric current, so this is a planar analogue rather than a direct measurement of isolated point charges in three-dimensional vacuum. For constant electrical conductivity, Ohm's law and steady charge conservation give $\mathbf{j}=-\sigma \nabla \Phi$ and $\nabla \cdot \mathbf{j}=0$, hence the two-dimensional Laplace equation away from electrodes. Insulating sheet edges impose zero normal electric current, while finite electrodes fix boundary values. This derivation explains both the analogy and its limits: inhomogeneous paper, finite boundaries, and finite electrodes can change the contours. One should not fit the three-dimensional point charge formula $kQ/r$ to a planar map without testing that model. Small probe spacing resolves spatial variation better, but amplifies uncertainty in the inferred finite difference. The published laboratory instructions explicitly ask students to assess whether their proposed field fit is appropriate.^[ref][ref]

The scalar description has a crucial limit. Faraday's law of induction for a stationary closed path $\Gamma$ sayswhere ${\Phi }_B$ is magnetic flux. Nonzero changing magnetic flux gives nonzero circulation, incompatible with a single-valued electric potential alone. More generally, a scalar electric potential and a magnetic vector potential together giveConsequently, in the presence of induction, two voltmeter lead routes between the same contacts can produce different readings if the enclosed magnetic flux changes. This follows by subtracting the two path line integrals; it is a deduction from Faraday's law of induction, not a claimed outcome of the paper experiment. The steady setup and negligible induction justify interpreting its readings as electric potential differences. Merely introducing a scalar electric potential that depends on time cannot supply a nonzero curl of the electric field.^[ref]

A short mathematical justification connects the statements above. Along a smooth path, the chain rule gives $d\Phi =\nabla \Phi \cdot d\mathbf{l}$. Integrating and using $\mathbf{E}=-\nabla \Phi$ yields the endpoint formula by the fundamental theorem for line integrals, proving path independence. Conversely, vanishing circulation around every closed path lets us define $\Phi (P)=-{\int }_{P_0}^P\mathbf{E}\cdot d\mathbf{l}$ independently of the route, with $\Phi (P_0)=0$. Zero curl is a local test; for a continuously differentiable curl-free vector field, a simply connected open region suffices for a global potential. The explicit finite-source formula already gives a global electric potential away from its source points.^[ref]

A force field specifies a force vector as a function of position and, when necessary, time, for a specified test body. It is a vector field. A time-independent conservative vector field has $\mathbf{F}=-\nabla V$, where $V$ is the body's potential energy; its work along a path equals the decrease in $V$. Position-dependent forces can have nonzero curl, so a scalar potential need not exist.

For a continuously differentiable function $\Phi$ and a piecewise smooth parametric curve $\gamma$ from $a$ to $b$,The chain rule converts the integrand into $d(\Phi (\gamma (t)))/dt$, then the fundamental theorem of calculus supplies the endpoint difference. This is why a conservative vector field has a path independent line integral. For $\Phi (x,y)=x^2+y^2$, every path from $(0,0)$ to $(1,2)$ gives $5$, provided it lies in the domain where the potential is defined.

A differential one-form on a specified domain is inexact if it is not the differential of a single-valued scalar potential there. For $Pdx+Qdy$, unequal cross derivatives $P_y\ne Q_x$ prove inexactness locally. Equality is sufficient on a simply connected plane domain under the usual smoothness hypotheses, but holes can obstruct a global potential. The form $42ydx+xdy$ has cross derivatives $42$ and $1$ and counterclockwise unit-square integral $-41$, so it is inexact. A conservative vector field has an exact differential as its work form.

If $\psi (A_1,A_2)$ is a conservative vector field, equality of the mixed derivatives of its potential gives ${\partial }_y(\psi A_1)={\partial }_x(\psi A_2)$. Expanding proves the displayed necessary equation. For continuously differentiable data on the whole plane, the vanishing planar curl is also sufficient for a potential. A useful integrating factor is nonzero on the region under consideration.

Parametrize the straight path by $\mathbf{r}(t)=(t,t,t)$ for $0\le t\le 1$. Then $\dot{\mathbf{r}}=(1,1,1)$ and $\mathbf{A}=(3t^2+6t,-14t^2,20t^3)$. ThusThis differs from the preceding path's value. Indeed, $\nabla \times \mathbf{A}=(14y,-20z^2,-6)$ is not zero, so the field is not a conservative vector field and its integral need not depend only on endpoints.

Construct a global vector-field potential rather than relying only on vanishing curl:Thus $\mathbf{F}$ is a conservative vector field. The endpoints of the given parametric curve are $\mathbf{r}(0)=(3,0,0)$ and $\mathbf{r}(\pi /2)=(0,2,1)$. The fundamental theorem for line integrals givesThe intervening turns of the curve have no effect because this line integral is path independent.

The field is the gradient of a globally defined polynomial:It is therefore a conservative vector field on all of ${\mathbb{R}}^3$. The fundamental theorem for line integrals givesfor every closed piecewise smooth loop. Constructing the global vector-field potential avoids relying on a zero-curl test without checking the topology of the domain.

The field is the gradient of $\Phi =x^2-3y^2+4z^2$ on all of ${\mathbb{R}}^3$. It is therefore a conservative vector field, and the fundamental theorem for line integrals gives ${\int }_{C}F\cdot dr=\Phi (r_{\mathrm{end}})-\Phi (r_{\mathrm{start}})$. The endpoints are $(3,0,0)$ and $(0,2,-1)$, soThe complicated intermediate path does not affect the work. Directly, $\nabla \times F=0$, consistent with the globally defined potential function.

A conservative vector field can be written as $F=\nabla \Phi$ for a single-valued scalar potential $\Phi$. Its line integral between two points depends only on the endpoints, and its integral around every closed curve is zero. A necessary local test for a smooth planar field $F=(P,Q)$ is $P_y=Q_x$. On a simply connected domain such as the whole plane, this is also sufficient.

HereAn explicit scalar potential is even more decisive:Indeed ${\Phi }_x=2x{e}^{xy}+x^{2}y{e}^{xy}$ and ${\Phi }_y=x^3{e}^{xy}+2y$. Thus the field is conservative on ${\mathbb{R}}^2$.

An exact differential is a differential one-form equal to $d\Phi$ on its domain; an inexact differential is not the differential of a single-valued potential there. For example, $d(xy)=ydx+xdy$ is exact. For the supplied form $\eta =42ydx+xdy$, the cross derivatives areThey disagree, so $\eta$ is not even closed and hence cannot be exact. Equivalently, its counterclockwise line integral around the unit square is $1-42=-41\ne 0$. ThereforeThe connection is $F\cdot dr=d\Phi$: a vector field is conservative precisely when its work differential one-form is exact. Locally vanishing curl alone need not guarantee global exactness on a domain with holes.

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 $\nabla \cdot \mathbf{F}=-3$.

Since $\mathbf{a}$ is constant, direct differentiation gives $(\mathbf{r}\cdot \nabla )\mathbf{a}=0$, $(\mathbf{a}\cdot \nabla )\mathbf{r}=\mathbf{a}$, $\nabla \cdot \mathbf{r}=3$ and $\nabla \cdot \mathbf{a}=0$. The stated right-hand side is therefore $2\mathbf{a}$. Independently, index notation givesThus $\boxed{\nabla \times (\mathbf{a}\times \mathbf{r})=2\mathbf{a}}$. A conservative vector field has zero curl, so the field is not conservative unless $\mathbf{a}=0$. For a closed planar circuit, its circulation is $2\mathbf{a}\cdot \mathbf{A}$ by Stokes theorem; the zero-vector exception is conservative with a constant potential.

A conservative vector field is the gradient of a single-valued differentiable scalar potential. Its line integral is path independent and equals the potential difference. If $\psi \mathbf{A}=\nabla F$ with a twice continuously differentiable potential, equality of mixed partial derivatives givesThe product rule then gives ${\psi }_y{A}_1+\psi A_{1,y}={\psi }_x{A}_2+\psi A_{2,x}$. Rearranging proves the integrating-factor equation for a planar vector field:This argument does not require division by $\psi$, so it remains valid even at its zeros. A nonvanishing integrating factor is normally sought when one wants to recover useful potential curves for the original field.

For a continuously differentiable vector field on all of ${\mathbb{R}}^3$, the necessary and sufficient condition for a conservative vector field is $\nabla \times \mathbf{F}=0$. Necessity follows from equality of mixed partial derivatives of a potential; sufficiency uses the ${\mathbb{R}}^3$ being a simply connected space. On a general domain the topology cannot be omitted.

Here the relevant mixed partial derivatives areThus the curl vanishes. Integrating the first component in $x$ gives $\Phi =x^2\cos y-2x{z}^3+A(y,z)$. Matching the second component gives $A_y=3+2y{e}^z$, hence $A=3y+y^2{e}^z+B(z)$. Matching the last component forces $B^{\prime }=0$. A potential of a conservative vector field with the convention $\mathbf{F}=\nabla \Phi$ is consequentlyIf a physical potential is defined instead through $\mathbf{F}=-\nabla V$, it is $V=-\Phi$.

A conservative vector field on ${\mathbb{R}}^n$ is a vector field $\mathbf{V}=\nabla F$ for a globally defined smooth potential of a conservative vector field $F$. The chain rule givesso its line integral is path independent.

Green theorem states that, for a bounded planar region $D$ with piecewise smooth boundary and continuously differentiable $P,Q$ on a neighbourhood of its closure,The boundary is positively oriented: the region lies on the left while it is traversed, so outer components run counterclockwise and hole boundaries clockwise.

For the proposed potential of a conservative vector field, the fundamental theorem of calculus and differentiation under the integral sign giveSince $Q_x=P_y$, the last integral equals $P(x,y)-P(x,0)$, including negative $y$ with the usual signed-integral convention. Hence the proposed potential is global andThe fact that the vector field is defined on all of ${\mathbb{R}}^2$ ensures both straight integration segments stay inside its domain; a hole in the domain could obstruct a global potential.

Integrating the oscillatory term together with the constant components gives a potential of a conservative vector fieldDirect partial derivatives recover both components of $\mathbf{V}$, proving it is a conservative vector field. Its line integral is therefore path independent. At an integer endpoint $(m,n)$, the sine term vanishes, so for any of the specified pathsAny other potential has the same gradient, so differs from $F$ by a constant on the connected plane. In particular every potential changes by $1$ under $x\mapsto x+1$ and by $2$ under $y\mapsto y+1$. Equivalently the integral from $(0,0)$ to $(1,0)$ is $1$, whereas a unit-periodic potential would make that difference zero. There is no potential periodic with period one in both coordinate directions. This is an example of how periods obstruct a periodic potential: a periodic vector field need not have a periodic potential.

Part (c) givesThe half-space $x_1>0$ is a simply connected domain, so the Poincaré lemma makes this curl-free field a conservative vector field: it is the gradient of a single-valued scalar potential.

Using $\tan u+\cot u=2/\sin (2u)$,The spherical-coordinate formula from part (b) shows that a potential must satisfy $f_{\varphi }=2$ and may be independent of $r,\theta$. On $x_1>0$ the azimuthal angle has the single-valued branchHence one solution, up to an additive constant, is

The Stokes theorem states that for an oriented smooth surface $S$ with positively oriented boundary $\partial S$,A field is conservative vector field if its line integral is path-independent, equivalently if every closed line integral vanishes. If $F=\nabla \varphi$, the fundamental theorem for line integrals givesThis is zero for every closed curve, so every globally defined gradient field is conservative.