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On the first segment, parameterize ${\mathbf{r}}_1(t)=(t,t,0{)}^T$, $0\le t\le 1$. The dot products with ${\mathbf{r}}_1^{\prime }=(1,1,0{)}^T$ areTheir line integrals are one and $\sin 1$, respectively.

On the second segment take ${\mathbf{r}}_2(t)=(1,1,t{)}^T$, with tangent vector $(0,0,1{)}^T$. The corresponding integrands are $a-2+2t$ and $c$, whose integrals are $a-1$ and $c$. Adding the segments yields

For $\mathbf{F}$, all three curl components vanish for every $a$. More constructively,Hence $\mathbf{F}$ is conservative for every real $a$. Its endpoint vector-field potential difference is $a$, checking both line integrals.

For the other vector field, direct partial differentiation givesIt therefore has a vector-field potential exactly when $b=0$, with no restriction on $c$:The value of $a$ is irrelevant to $\mathbf{G}$. Its rotational part for general $b$ is $b(1,1,1{)}^T\times \mathbf{r}$. Both selected paths lie in $x=y$, and this rotational part has zero dot product with every tangent vector to that plane. That explains why these paths cannot detect its nonzero curl when $b\ne 0$.

On the circular arc, $z=0$ and $dz=0$, so the line integral reduces toThis exact differential evaluation applies to the specified arc; for example its parametric equation is $(x,y,z)=(1-\cos \theta ,\sin \theta ,0)$, $0\le \theta \le \pi /2$. On the vertical segment $x=y=1$, only $dz$ contributes, giving ${\int }_0^1(\lambda +1)dz=\lambda +1$. The whole path therefore givesTo answer the final conservativeness requirement, compute the curl:It vanishes identically exactly whenOn the simply connected whole space this condition is sufficient for a conservative vector field. Explicitly a vector-field potential isBoth path calculations then give $13/6$, equal to $\varphi (1,1,1)-\varphi (0,0,0)$ by the fundamental theorem for line integrals. Equality of just these two path values would only force $\mu =2\lambda +1$; it is the curl calculation that establishes full path independence.

Parameterize the straight line by $\mathbf{r}(t)=(t,t,t{)}^T$, $0\le t\le 1$. Then ${\mathbf{r}}^{\prime }=(1,1,1{)}^T$ and the line integral becomes

For the parametric curve, ${\mathbf{r}}^{\prime }(t)=(2t,3t^2{)}^T$ andThe line integral is therefore

Using Cartesian coordinates, the Levi-Civita symbol and the Einstein summation convention, the constant-vector assumption givesRecognizing the divergence and directional derivative provesHere the contractions use the Kronecker delta.

For an oriented piecewise smooth surface $S$ with boundary $C$, Stokes theorem stateswhere the orientations obey the right-hand rule and $\mathbf{Q}$ is continuously differentiable near $S$. Apply it to $\mathbf{Q}=\mathbf{F}\times \mathbf{a}$. The scalar triple product givesThe last integrand is $\mathbf{a}\cdot [(d\mathbf{S}\times \nabla )\times \mathbf{F}]$, as the component formulashows. Since $\mathbf{a}$ is arbitrary, the cross-product form of Stokes theorem follows:The derivatives in the surface expression act on $\mathbf{F}$, not on the surface element.

For the specified square, the path is clockwise when viewed from positive $z$, so the compatible unit normal is $-{\mathbf{e}}_z$. On the first and fourth edges, $d\mathbf{l}$ and $\mathbf{r}$ are parallel, giving zero cross product. On the second edge, $y=1$ and $x$ runs from $0$ to $1$, so $d\mathbf{l}\times \mathbf{r}={\mathbf{e}}_{z}dx$ contributes ${\mathbf{e}}_z$. On the third edge, $x=1$ and $y$ runs from $1$ to $0$, so the contribution is $-{\mathbf{e}}_z{\int }_1^{0}dy={\mathbf{e}}_z$. Thus the line integral is $2{\mathbf{e}}_z$.

For an independent surface integral, ${\partial }_i{F}_j={\delta }_{ij}$ and $\nabla \cdot \mathbf{F}=3$ for the position vector field. Its integrand is therefore $d\mathbf{S}-3d\mathbf{S}=-2d\mathbf{S}$. The square has unit area and downward orientation, giving

Use the geometric cross-section shown in the source: the outer cylinder has radius $2a$ and centre $O$, and the circular cavity has radius $a$ and centre $C=a\hat{\mathbf{x}}$. Let the electric current be in the $+z$ direction. The conducting cross-sectional area is $3\pi a^2$, so the uniform electric current density is $\mathbf{J}=I\hat{\mathbf{z}}/(3\pi a^2)$.

To obtain the exterior magnetic field of a wire with an eccentric cylindrical cavity, apply the superposition principle to a full cylinder of electric current density $\mathbf{J}$ centred at $O$ and a second cylinder of electric current density $-\mathbf{J}$ centred at $C$. Their electric currents are $4I/3$ and $-I/3$. This reproduces the actual electric current density everywhere, including zero in the cavity. Outside a long circular cylinder, Ampère's law gives $\mathbf{B}={\mu }_0{I}_{\mathrm{cyl}}\hat{\mathbf{z}}\times \mathbf{r}/(2\pi r^2)$. Every point outside the outer boundary is outside both component cylinders. Consequently, writing $\mathbf{r}=x\hat{\mathbf{x}}+y\hat{\mathbf{y}}$,In Cartesian coordinates, the same magnetic flux density isThe cavity breaks rotational symmetry. Thus the magnetic flux density is generally neither constant in magnitude nor tangential on a circle centred at $O$; replacing its line integral by $2\pi rB$ would be unjustified. Its line integral is nevertheless ${\mu }_{0}I$, as Ampère's law requires. At distances $r\gg a$, the leading term becomes ${\mu }_{0}I\hat{\mathbf{\varphi }}/(2\pi r)$, with corrections of order ${\mu }_{0}Ia/r^2$.

Stokes theorem converts circulation around an oriented boundary into curl flux through a spanning surface. For a sufficiently smooth vector field on an oriented piecewise smooth surface $S$,The boundary orientation follows the right-hand rule relative to $\mathbf{n}$. The field must be smooth on the surface, not merely on its boundary; a singularity crossing the surface cannot be ignored. For a fixed loop, applying the theorem to Faraday's differential law gives the integral flux law of induction.

The planar form is Green theorem. For a region $D$ with positively oriented boundary and sufficiently smooth $P,Q$,The outer boundary is counterclockwise and any holes have clockwise boundaries. For instance $P=-y/2,Q=x/2$ makes the area of $D$ equal to half the line integral of $xdy-ydx$; the unit circle gives area $\pi$.

A closely related use of “Green's theorems” in field theory is Green's identities, obtained by applying the divergence theorem to $\varphi \nabla \psi$:and, after subtracting the expression with $\varphi ,\psi$ exchanged,These relate interior sources to boundary data in electrostatics and other boundary-value problems. For example, a harmonic potential difference vanishing on a Dirichlet boundary condition has zero integral of its squared gradient by the first identity, establishing uniqueness. Both the planar circulation theorem and the field identities are included here to make the convention clear.

Let the approximately conserved magnetic flux be $\Phi$. ThenThe length $L$ shown in the original diagram is the untapered core arc; the two tapers are separate lengths $s$ each. In the untapered core, flux conservation givesFor one taper use distance $u$ from its narrow end. The straight taper in the diagram has local radiusand magnetic field intensityIts contribution to the line integral in the magnetostatic Ampère law is found explicitly:Adding the gap, the untapered arc and both tapers givesThereforeThis uses a linear radius profile, approximately uniform magnetic flux density over each section, a linear constant magnetic permeability, and negligible leakage and magnetic fringing. For a non-linear taper profile $a(u)$ the honest generalization isThus an arbitrary smooth taper with the same end radii and length need not give the printed formula; the straight-sided geometry supplies the additional information.

A conservative vector field has path-independent work: the integral ${\int }_A^B\mathbf{F}\cdot d\mathbf{r}$ depends only on its endpoints. Equivalently, the line integral around every closed path in its domain vanishes. One can then define a single-valued potential energy $U$ byThe arbitrary additive constant in $U$ has no effect on the force. A useful zero is often chosen at infinity for a localized attractive interaction.

A differentiable conservative force has $\nabla \times \mathbf{F}=0$ because the curl of a gradient vanishes. Conversely, a smooth curl-free field on a simply connected domain admits a global scalar potential. The domain condition matters: on the punctured plane, $\mathbf{F}=(C/r)\hat{\mathbf{\theta }}$ has zero curl away from the missing origin but closed-loop work $2\pi C$ around it. It is not globally conservative there. The existence of a local potential does not establish a single-valued potential throughout a domain with holes.

The work-energy theorem gives $\Delta T=W$ for kinetic energy $T$. If all forces are conservative and $U$ has no explicit time dependence, thenThis is conservation of mechanical energy. In one dimension, an equilibrium occurs at $dU/dx=0$; a local minimum with $U^{\prime \prime }>0$ is stable, while a maximum with $U^{\prime \prime }<0$ is unstable. Expanding about a minimum gives $U\simeq U_0+U^{\prime \prime }(x_0)(x-x_0{)}^2/2$, producing simple harmonic motion with ${\omega }^2=U^{\prime \prime }(x_0)/m$.

Examples include the spring force $F_x=-kx$ with $U=k{x}^2/2$, the near-surface gravitational force $-mg\hat{\mathbf{z}}$ with $U=mgz$, and the central Newtonian gravitational field with $U=-GMm/r$. An electrostatic force on a fixed electric charge $q$ has $U=q\Phi$, where $\Phi$ is the electric potential.

Sliding friction provides a contrasting physical test: traversing a closed path still dissipates mechanical energy, since the frictional force opposes the local motion on both outward and return legs. Its work depends on path length and cannot be represented by a position-only $U$. Total energy can remain conserved when the resulting internal energy is included. Also, even if $\mathbf{F}=-\nabla U(\mathbf{r},t)$ at each instant, explicit time dependence gives $d(T+U)/dt=\partial U/\partial t$; a static conservative-field argument alone does not guarantee constant mechanical energy in that situation.

For steady electric currents in vacuum, Ampère's circuital law isChoose an oriented closed contour and a spanning surface; the positive enclosed electric current crosses that surface according to the right-hand rule for the contour. First use the source symmetry to establish the direction of $\mathbf{B}$ and which positions have the same magnitude. Then choose the contour so its useful segments are parallel to $\mathbf{B}$ with constant magnitude, and the remaining segments are perpendicular to $\mathbf{B}$ or have vanishing field contribution. The line integral reduces to the unknown magnitude times a known length or sum of lengths.

For example, cylindrical symmetry around an infinite wire allows a circular contour, giving $B(2\pi r)={\mu }_{0}I$. A uniform current sheet instead calls for a rectangle straddling it. The law is always valid in this magnetostatic setting, but by itself does not determine the local magnetic field from an arbitrary enclosed current without these extra symmetry arguments. Here ${\mu }_0$ is the vacuum permeability.

For a twice continuously differentiable scalar potential, mixed second partial derivatives commute. Taking the spatial curl of the equation from part (i) givesFor example, its first component is $-{\partial }_2{\partial }_3\xi +{\partial }_3{\partial }_2\xi =0$. The curl acts on the spatial representation of the material acceleration, not directly on its expression in reference coordinates.

This is a necessary local condition. It is sufficient locally, and sufficient globally on a simply connected domain under the usual smoothness assumptions. On a domain with holes, a global scalar potential additionally requires every closed-loop line integral of the material acceleration to vanish. The homogeneous motion in part (c) admits an explicit global scalar potential, so this topological qualification causes no obstruction there.

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.

A line integral is path dependent when two admissible paths with the same starting and ending points give different values. This disproves path independence and, for a vector field, shows that it is not a conservative vector field on the relevant domain.

A line integral is path independent when its value depends only on the two endpoints. On a connected domain, this holds for the line integral of every exact differential.

For a vector-field line integral, this condition is equivalent to the field being a conservative vector field.

If a periodic vector field has a periodic potential of a conservative vector field, the line integral between points separated by a period vector must vanish, since it is the potential difference. A periodic gradient field need not satisfy this: the constant field $(1,2)$ on the plane has potential $x+2y$, but its integrals over the two unit translations are $1$ and $2$. On the periodic quotient these translation integrals are nonzero periods, obstructing a single-valued potential there.

Potential energy assigns energy to a system's arrangement, so a change of arrangement accounts for the work done by a conservative force. For a time-independent potential energy $V(\mathbf{r})$,The gradient points toward increasing potential energy; the force points toward decreasing potential energy. Only differences matter: adding a constant to $V$ changes neither the force nor its work. This endpoint rule follows by substituting $-\nabla V$ into the line integral for work. ^[ref]

Near Earth's surface, stipulate constant gravitational acceleration $g=9.81m/{\mathrm{s}}^2$ and measure height $h$ upward from a chosen zero. Then gravitational potential energy is $V=mgh$. Raising a $2\mathrm{kg}$ mass by $0.5\mathrm{m}$ increases its gravitational potential energy by $9.81\mathrm{J}$, while gravity does $-9.81\mathrm{J}$ of work. These are computed model values, not measurements of a particular lift. An ideal spring obeying Hooke's law instead has $V=k{x}^2/2$, with displacement $x$ from its relaxed state and spring constant $k$: $k=100N/m$ and $x=0.2\mathrm{m}$ give $V=2\mathrm{J}$ relative to $x=0$. ^[ref]

If $h:[\alpha ,\beta ]\to [a,b]$ is an increasing piecewise smooth change of parameter, with endpoints mapped to endpoints, the chain rule gives $(\gamma \circ h{)}^{\prime }=({\gamma }^{\prime }\circ h)h^{\prime }$. Change of variables in the defining line integral proves the displayed identity. A decreasing change of parameter reverses the endpoints and the sign. The result concerns the same traversal with the same multiplicity; retracing a loop additional times changes the integral in general.

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 ${\oint }_{C}d\mathbf{r}=0$.

Signed swept area counts the area traversed by a position vector, positively or negatively according to its direction of turning. For a continuously differentiable planar trajectory, its accumulated value relative to $t_0$ is the line integralA short segment contributes the oriented triangle area $\frac{1}{2}\mathbf{n}\cdot (\mathbf{r}\times d\mathbf{r})$; taking its time derivative gives areal velocity. Repeated turns count repeated area; a reversal subtracts it. Thus this quantity need not equal the area of the union of all previously swept regions. The origin matters: the relevant origin for Kepler's second law is the attracting focus of the ellipse, rather than its geometric centre.