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Archimedes' principle states that buoyancy equals the downward gravitational weight, or weight, of the displaced fluid. In a resting fluid of constant mass density ${\rho }_f$ under uniform gravitational acceleration $g$, its magnitude isHere $V_{\mathrm{displaced}}$ is the volume from which the body excludes the fluid: the whole external volume for a fully immersed solid body, and just the immersed volume for a partly immersed one. The body's own mass does not enter this buoyancy formula. ^[ref]

A short proof outline for a fully immersed body is to imagine replacing it by exactly the fluid it displaces. The surrounding hydrostatic pressure is unchanged. That replacement fluid is at rest, so its upward pressure force must balance its downward force from gravity. The original body receives the same inward pressure forces on the same closed surface, hence the same upward buoyancy.

Equivalently, let $\Omega$ be the displaced region, $\partial \Omega$ its closed surface and ${\mathbf{e}}_z$ the upward unit direction. The divergence theorem, applied component by component to the surface integral, and hydrostatic pressure giveThis also explains why horizontal forces cancel even for an irregular shape. A spatially constant surface pressure contributes zero to the closed-surface integral. If the mass density varies with depth, replace ${\rho }_f{V}_{\mathrm{displaced}}$ by the mass ${\int }_{\Omega }{\rho }_{f}dV$ of the displaced fluid.

The closed-surface argument assumes that the fluid applies pressure over the entire boundary of a fully immersed body. A body sealing against a tank bottom may have no fluid beneath it; then the lower-face pressure force used in the proof is absent, and a supporting contact force must be treated separately. For partial immersion, the usual buoyancy formula assumes the exposed surface receives the common atmospheric pressure, neglects the much smaller buoyancy from the air, and neglects surface tension.

For the $0.10\mathrm{m}$ cube in the stipulated water, full immersion displaces $0.001{\mathrm{m}}^3$, so $F_b=1000\times 9.81\times 0.001=9.81\mathrm{N}$. If its uniform mass density is $600\mathrm{kg}{\mathrm{m}}^{-3}$, its mass is $m=600\times 0.001=0.600\mathrm{kg}$ and its downward force from gravity is $mg=5.886\mathrm{N}$. To hold it fully immersed at rest requires an additional downward forceOn release from rest it initially moves upward; its eventual floating equilibrium has a smaller displaced volume. A same-size cube of uniform mass density $1200\mathrm{kg}{\mathrm{m}}^{-3}$ has mass $1.200\mathrm{kg}$ and downward force from gravity $11.772\mathrm{N}$. Full immersion still gives $9.81\mathrm{N}$ of buoyancy, leaving an initial net downward force of $11.772-9.81=1.962\mathrm{N}$. Subsequent motion also involves drag.

An integral over the boundary of a region, such as a line integral around a planar region or a surface integral over the boundary of a three-dimensional region. Green's identities convert certain volume integrals into boundary integrals.

Charge conservation says that the total signed electric charge of an isolated collection stays constant. More generally, a collection closed to electric charge exchange has constant total electric charge, even if energy crosses its boundary. This is the charge bookkeeping relevant to a closed system or an isolated system; isolation from energy exchange is not required. Positive and negative electric charge count with opposite signs. Creating or removing an electron and a Positron together is compatible with charge conservation, because their combined electric charge is zero. ^[ref].

Imagine a fixed box initially containing electric charge $Q(0)=10\mu \mathrm{C}$. Stipulate incoming conventional current $I_{\mathrm{in}}=3\mu \mathrm{A}$ and outgoing conventional current $I_{\mathrm{out}}=1\mu \mathrm{A}$ throughout $0\le t\le 2\mathrm{s}$, with no other boundary transport. Since one ampere is one coulomb per second, the stored electric charge grows at $2\mu C/s$:If instead the outgoing electric current also equals $3\mu \mathrm{A}$, the stored electric charge remains $10\mu \mathrm{C}$ while electric charge passes through. These are stipulated transport examples; constant terminal electric currents alone do not guarantee a time-independent interior charge density.

Locally, charge conservation is the charge continuity equation, an instance of the general continuity equation:Here $\rho (\mathbf{r},t)$ is signed volume charge density and $\mathbf{J}(\mathbf{r},t)$ is current density. A positive divergence of current density means net outward transport from a small neighborhood, so its charge density decreases. For a fixed control volume $V$ with closed surface $\partial V$ and outward unit normal $\mathbf{n}$, the equivalent integrated balance isThe surface integral is signed: incoming positive electric charge contributes negatively to outward electric current, and outgoing positive electric charge contributes positively. Integrate the charge continuity equation over fixed $V$ and apply the divergence theorem to obtain this balance. Zero net boundary transport gives constant $Q_V$; the global and local statements describe the same conservation law at different scales. ^[ref].

In a steady transport state, charge density and current density are time-independent, so the charge continuity equation reduces to $\nabla \cdot \mathbf{J}=0$. At an electrical circuit junction, choose each terminal electric current $I_k$ positive outward and assume the terminals account for all boundary transport. ThenKirchhoff's current law, ${\sum }_k{I}_k=0$, follows when accumulation of electric charge at the junction is negligible. During storage, retain the same balance with its nonzero derivative; charge conservation continues to hold. This makes charge conservation useful for checking electrical circuit models and for tracking the charging of a capacitor. ^[ref].

The beautiful consistency with Maxwell equations is that the electric field and magnetic field already enforce this transport balance. For sufficiently smooth fields in vacuum SI, take the divergence of the Ampère-Maxwell equation, use that the divergence of a curl is zero, and substitute Gauss's law, $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$. Constant vacuum permittivity ${ϵ}_0$ givesThe displacement current term supplies the part needed when electric charge accumulates. Historically, Maxwell added that term to extend the steady-current equation; see the existing charge conservation from Maxwell equations for this connection. ^[ref].

Use circulation to summarize how much a velocity field runs along an oriented closed curve $C$, such as a circle around a fluid vortex. Its definition is the line integralwhere $d\mathbf{l}$ is tangent to $C$ in the chosen direction. Reversing that direction reverses the sign of circulation. Its units are ${\mathrm{m}}^2{\mathrm{s}}^{-1}$, because velocity is multiplied by a length. Counterclockwise circulation in the horizontal plane is positive when the chosen unit normal vector is $+{\mathbf{e}}_z$, according to the right-hand rule.

For solid-body rotation $\mathbf{u}=(-\Omega y,\Omega x,0)$, a counterclockwise circle of radius $R$ has constant tangential velocity $\Omega R$. Multiplying by its circumference givesFor the potential line vortex $\mathbf{u}=K{\mathbf{e}}_{\theta }/r$, the same circle instead gives ${\Gamma }_C=(K/R)(2\pi R)=2\pi K$, independent of $R$. Stipulate $\Omega =1{\mathrm{s}}^{-1}$, $R=1\mathrm{m}$ and $K=1{\mathrm{m}}^2{\mathrm{s}}^{-1}$: both velocity fields have ${\Gamma }_C=2\pi {\mathrm{m}}^2{\mathrm{s}}^{-1}\approx 6.283185{\mathrm{m}}^2{\mathrm{s}}^{-1}$, although their vorticities in the fluid are $2{\mathbf{e}}_z{\mathrm{s}}^{-1}$ and zero respectively. On a circle of radius $2\mathrm{m}$, their circulations are $8\pi {\mathrm{m}}^2{\mathrm{s}}^{-1}$ and $2\pi {\mathrm{m}}^2{\mathrm{s}}^{-1}$. The comparison plot in vorticity shows this difference in radius dependence.

The bridge between circulation and vorticity is Stokes theorem:Here the velocity field must be continuously differentiable on a neighborhood of the entire spanning surface $S$ inside the fluid, and its boundary orientation must match the unit normal vector $\mathbf{n}$. For solid-body rotation, the disk is admissible and its surface integral is $2\Omega \pi R^2$. For the potential line vortex, the origin is singular and excluded from the fluid: a disk crossing it is not admissible. On an annulus, Stokes theorem instead gives ${\Gamma }_{R_{\mathrm{out}}}-{\Gamma }_{R_{\mathrm{in}}}=0$, with both circulations here measured counterclockwise; the induced inner boundary direction is clockwise. Nonzero circulation therefore does not imply nonzero local curl in the available fluid. ^[ref]

This is useful because Stokes theorem lets a local vorticity measurement determine a loop's circulation when a suitable surface exists, while Kelvin's circulation theorem tracks circulation along a moving material curve. The beauty is the precise link between a local derivative and a global line integral, with the domain deciding which conclusions are valid.

The current density $\mathbf{J}(\mathbf{r},t)$ describes the amount and direction of local electric charge transport per area per time. Its SI unit is $A/{\mathrm{m}}^2$. Through a small area element with unit normal $\mathbf{n}$, the contribution to electric current is $dI=\mathbf{J}\cdot \mathbf{n}dA$: the dot product selects transport through the surface. A current density tangent to the surface contributes zero. If current density is uniform and perpendicular to a wire's transverse slice, its magnitude $j=|\mathbf{J}|$ satisfiesHere $A$ is the wire cross-sectional area. For nonuniform current density, $I/A$ is only the average normal component, and the surface integral is needed. ^[ref].

Consider two separate uniform-flow cases, each carrying electric current $I=2\mathrm{A}$. A wire cross-sectional area of $1\mathrm{m}{\mathrm{m}}^2=10^{-6}{\mathrm{m}}^2$ gives $j=2\times 10^{6}A/{\mathrm{m}}^2$; doubling the wire cross-sectional area to $2\mathrm{m}{\mathrm{m}}^2$ gives $j=10^{6}A/{\mathrm{m}}^2$. Equal electric current can therefore mean different local transport intensity. This distinction matters for Joule heating, whose local power per volume depends on current density and electric field.

For a toy electron transport model, stipulate mobile electron number density $n=8\times 10^{28}{\mathrm{m}}^{-3}$ and drift velocity ${\mathbf{v}}_d=-10^{-4}m/s\hat{\mathbf{x}}$, with the positive unit normal $\hat{\mathbf{x}}$. Each electron has electric charge $-e$, where the elementary charge $e=1.602176634\times 10^{-19}\mathrm{C}$ is exact in SI. These are defined model inputs, rather than measured parameters of a particular material. ^[ref] The current density and its magnitude areFor wire cross-sectional area $A=1\mathrm{m}{\mathrm{m}}^2$ and constant transport over time $10\mathrm{s}$, this givesThe two minus signs make conventional current point opposite to the electron drift velocity. A small drift velocity can carry an appreciable electric current because many charge carriers contribute; it does not specify the propagation speed of an electrical disturbance.

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].

The time derivative of the electric displacement field enters the macroscopic Ampère-Maxwell equation alongside free electric current density. In vacuum, ${\mathbf{J}}_D={ϵ}_0{\partial }_t\mathbf{E}$. It is not transport of free charges through the vacuum gap of a capacitor; its surface integral nevertheless supplies the term needed for compatibility with the continuity equation.

Let $V$ be a bounded region in Euclidean space with piecewise smooth boundary and outward unit normal vector $n$. For a continuously differentiable vector field $F$ on a neighborhood of its closure,The total outward flux equals the accumulated divergence inside. Contributions across shared internal faces cancel when a region is divided into small pieces.

For $F=(x,y,z)$ on a ball of radius $R$, the divergence is $3$, giving $3(4\pi R^3/3)=4\pi R^3$. On the boundary, $F\cdot n=R$, so the surface integral is also $R(4\pi R^2)=\boxed{4\pi R^3}$. This agreement turns the derivative into a geometric flux measurement. The required regularity matters: a singular field at the center cannot be included without checking its contribution separately.

An electric current measures how quickly signed electric charge crosses a chosen fixed oriented surface $S$. Choose its unit normal $\mathbf{n}$; positive electric charge crossing along $\mathbf{n}$ counts positively, while crossing against it counts negatively. If $Q_{\mathrm{cross}}(t)$ is the accumulated signed electric charge transported through $S$, thenHere $\mathbf{J}$ is the current density; the surface integral collects its local normal contributions. Reversing the unit normal reverses $I_S$. The electric current is a signed scalar associated with the whole chosen oriented surface, whereas current density is a vector at each point. The SI unit is the ampere, one coulomb per second. ^[ref]; ^[ref].

For a constant electric current of $2\mathrm{A}$ over a time interval of $3\mathrm{s}$, the transported electric charge is $6\mathrm{C}$. This need not be electric charge accumulating in the wire: in steady transport, electric charge enters and leaves. For a closed oriented surface with outward unit normal, electric charge conservation instead gives $I_{\mathrm{out}}=-d{Q}_{\mathrm{inside}}/dt$. Keeping transported and enclosed electric charge distinct connects electrical circuit bookkeeping to the local continuity equation.

Electric flux measures how much electric field crosses a chosen oriented surface, with a sign distinguishing the two crossing directions. A strong electric field running along a surface contributes nothing to its electric flux: only the normal component counts. This lets Gauss's law turn a potentially complicated electric field into one number recording the net enclosed electric charge.

For a surface $S$ with chosen unit normal $\mathbf{n}$, the electric flux is the surface integralHere $dA$ is positive surface area, while $d\mathbf{S}$ is the vector area element. The dot product makes electric flux a signed scalar, rather than a vector like the electric field. Its SI unit is $\mathrm{N}{\mathrm{m}}^2/C$: electric field measured using the newton per coulomb, multiplied by surface area in square metres. Reversing the orientation of a surface replaces $\mathbf{n}$ by $-\mathbf{n}$ and reverses the electric flux. Calling it electric flux does not imply that electric charge is moving through the surface. ^[ref].

For an open patch in a plane, with surface area $A$ and a constant electric field, the unit normal is constant and the surface integral simplifies toThe angle $\theta$ is between the electric field and the chosen unit normal, not between the electric field and the plane. The cosine describes the signed projected surface area seen by the electric field. Thus electric flux is positive for $0\le \theta <90^{\circ }$, zero at $90^{\circ }$, and negative for $90^{\circ }<\theta \le 180^{\circ }$. ^[ref].

For example, take $E=100N/C$ and $A=0.02{\mathrm{m}}^2$, so $EA=2\mathrm{N}{\mathrm{m}}^2/C$. Three choices of angle giveKeeping the $60^{\circ }$ patch fixed and reversing its unit normal gives $120^{\circ }$ and changes $+1$ to $-1\mathrm{N}{\mathrm{m}}^2/C$. The electric field and physical patch have not changed. If the electric field is instead tangent to the patch, its dot product with the unit normal vanishes: electric flux is zero even though $E=100N/C$.

For a closed surface enclosing a volume $V$, choose the outward unit normal. Gauss's law givesHere ${\epsilon }_0$ is the vacuum permittivity, and $Q_{\mathrm{enclosed}}$ is the signed total electric charge inside. The electric field in the surface integral includes all sources, including external electric charges. Assume no ideal point charge lies on the integration surface. Zero enclosed electric charge implies zero total electric flux; it does not imply zero electric field everywhere on the closed surface or inside it. Positive and negative contributions to the surface integral can cancel. ^[ref].

To see the cancellation with sources inside, place point charges $+q$ at $(-a,0,0)$ and $-q$ at $(a,0,0)$, where $q>0$, $a=0.30\mathrm{m}$, and both lie inside the sphere of radius $R=1\mathrm{m}$ centered at the origin. This separated pair has the electric dipole geometry. Its enclosed electric charge is $q-q=0$, so its total outward electric flux is zero. At the point $P=(0,0,R)$ on the sphere, Coulomb's law and the superposition principle give the individual electric fieldsThe $z$ components cancel and the positive $x$ components add:For the stipulated approximation $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$ and $q=1\mathrm{nC}$, this is about $4.740N/C$ along positive $x$. At $P$, the outward unit normal is $\hat{\mathbf{z}}$, so this nonzero electric field is tangent to the sphere. At other points of the sphere, its normal component can be positive or negative. The source-field formula applies away from the two point charges, where the electric field has zero divergence; it is singular at the sources. ^[ref].

The diagram shows streamlines of the three-dimensional electric field in the section $y=0$. The circle is a cross-section of the sphere, not the integration surface: Gauss's law uses the entire three-dimensional surface integral. A two-dimensional line integral of outward field around that circle would be a different quantity, with units $\mathrm{N}m/C$ rather than the electric flux units $\mathrm{N}{\mathrm{m}}^2/C$.

Faraday's law of induction relates a changing magnetic field to circulation of the electric field. For a fixed oriented surface $S$ with boundary $C=\partial S$, choose the positive direction around $C$ by the right-hand rule: the fingers follow $C$ while the thumb follows the chosen unit normal. The fixed-loop electromagnetic electromotive force isThe final derivative of magnetic flux is an ordinary time derivative, with the surface held fixed. Reversing both the unit normal and the boundary direction reverses both signs. The loop may be an imagined path: the field law does not require a wire or an electric current. ^[ref].

For a computed example, a fixed single-turn loop has area $A=0.020{\mathrm{m}}^2$ and out-of-page unit normal. Prescribe an external uniform magnetic field in that direction with $B(t)=0.10\mathrm{T}+(0.30T/s)t$ for $0\le t\le 1.0\mathrm{s}$. ThenIn a quasistatic resistive model with electrical resistance $R=2.0\Omega$, neglect the loop's self-inductance and its own contribution to changing magnetic flux. Ohm's law givesThe magnetic flux rises from $0.0020\mathrm{Wb}$ to $0.0080\mathrm{Wb}$ while the induced electromotive force stays constant: it measures the rate of change, rather than the amount of magnetic flux. The negative sign means clockwise electric current; its magnetic field through the loop points into the page, as explained by Lenz's law. The toy model describes the interval of steady ramping, excluding switching transients at its ends.

For a moving conducting loop $C(t)$, the electromagnetic electromotive force instead includes the Lorentz force contribution from the local velocity $\mathbf{v}$ of each conductor segment:This form assumes a thin-wire model with a well-defined closed contour and spanning oriented surface that move compatibly, a magnetic field satisfying Gauss's law for magnetism, and conductor speeds small compared with the speed of light. Ideal sliding contacts such as those in motional electromotive force can be treated piecewise. Arbitrary changes of electrical connections require checking the actual contour and conductor velocities; differentiating the flux through an arbitrary chosen moving path is insufficient. The velocity here describes conductor motion, not the charge carriers' drift along the wire. ^[ref].

A short derivation of the fixed-loop integral statement applies Stokes theorem to the local curl equation and differentiates under the fixed surface integral. For the moving loop, the magnetic flux transport for a moving circuit adds the swept-area term $-{\oint }_C(\mathbf{v}\times \mathbf{B})\cdot d\mathbf{l}$ to $d{\Phi }_B/dt$. Combining these two identities gives the moving-loop statement. This derives the integral forms from the local field law; it is not a proof of that empirical law from mathematics alone.

The useful unity is geometric: changing the magnetic field, the enclosed area, or the orientation can change magnetic flux and hence produce an electromotive force. A coil's multiple turns are accounted for by magnetic flux linkage, rather than by multiplying a single-turn flux without checking which flux each turn receives. ^[ref].

Faraday's report read on 24 November 1831 describes opposite momentary responses when a neighboring circuit is connected and disconnected, with no continuing response while its steady electric current is maintained. These are concrete experimental roots of the changing-flux idea. The report predates the modern Maxwell equations notation used above. ^[ref].

A Gaussian surface is a closed surface used with the integral form of Gauss's law to relate electric flux to enclosed electric charge. It need not coincide with the surface of any charged body. Choosing it to match a symmetry can make the normal component of the electric field constant on a portion of it, allowing that component to be taken outside a surface integral. A concentric sphere works for an isolated spherically symmetric charge distribution; an arbitrarily placed sphere generally does not.

Zero enclosed electric charge implies zero total electric flux. Inferring zero electric field additionally needs information such as the spherical symmetry used in the electric field of a uniformly charged spherical shell. The imagined boundary does not itself block or redirect the electric field.

Gauss's law relates the electric field on a closed surface to the net electric charge inside it. In SI units, with outward unit normal $\mathbf{n}$,Here $\rho$ describes a volume charge density, and ${\epsilon }_0$ is the vacuum permittivity; for ideal point or surface charges, use the enclosed electric charge directly rather than an ordinary volume-density integral. The left side is electric flux: the surface integral of the outward normal component of the electric field. Field directed outward contributes positively; field directed inward contributes negatively. The equality uses net signed electric charge, so positive and negative enclosed charges can cancel. The Gaussian surface is an imagined accounting boundary, not necessarily a material object. See OpenStax, explaining Gauss's law.

Where the electric field and charge density are smooth, the local statement is $\nabla \cdot \mathbf{E}=\rho /{\epsilon }_0$. The divergence theorem connects this divergence to the surface integral. This is the satisfying connection between local sources and global accounting: a surface can register how much net electric charge it encloses without revealing where that charge is located within it. A point charge requires care with the singular field at its position; the integral formulation remains useful with a boundary that avoids the charge.

The law fixes total electric flux, rather than every value of the electric field. In an uncharged region, a uniform electric field $\mathbf{E}=(E_0,0,0)$ crosses a cube of side $\ell$. Its left and right faces contribute $-E_0{\ell }^2$ and $+E_0{\ell }^2$; the other four faces contribute zero. Thus the net electric flux is zero while the electric field is nonzero everywhere. Charges outside a Gaussian surface can change the field on it without changing its net electric flux. Conversely, fields cannot generally be reconstructed from enclosed electric charge alone. Electrostatics also requires the electric field to be the negative gradient of an electric potential, with appropriate boundary information. This distinction is emphasized in Feynman lectures II.5, application of Gauss' law.

Spherical symmetry supplies the extra information needed for a particularly simple calculation. An isolated spherically symmetric charge distribution in vacuum produces a radial electric field, constant in magnitude on each concentric sphere. Its electric flux is therefore $4\pi r^2{E}_r(r)$, givingOutside the distribution, doubling $r$ quadruples the surface area and divides the radial electric field by four, while the electric flux stays $Q/{\epsilon }_0$. The inverse-square dependence is an area-dilution result. The electric field of a uniformly charged ball and electric field of a uniformly charged spherical shell show that the same total charge can give different interior fields. These symmetry arguments and their limits are developed in OpenStax, applying Gauss's law.

Lift is the component of the fluid force perpendicular to the far-field flow relative to the body. It need not point vertically: a tilted flow changes the lift direction, and a banked wing can produce a sideways component. In a planar problem choose a unit vector ${\mathbf{e}}_D$ along the incoming relative velocity ${\mathbf{U}}_{\infty }$ and a perpendicular unit vector ${\mathbf{e}}_L$ pointing towards the chosen positive-lift side. ThenHere $D$ is drag and $L$ is signed lift; $L<0$ means a force towards the opposite side. In three dimensions the plane perpendicular to ${\mathbf{U}}_{\infty }$ also admits a separate side-force direction, so a lift axis must be specified. Lift is also distinct from buoyancy, which can exist without a relative flow. ^[ref]

The force comes from adding local surface tractions, rather than assigning a single pressure to each side. Let $\Sigma$ be the wetted body surface and $\mathbf{n}$ its unit normal vector pointing out of the body into the fluid. The pressure contribution pushes inwards; the viscous stress tensor also contributes, including tangential shear stress:Using a constant reference pressure $p_{\infty }$ is legitimate on a closed surface because ${\int }_{\Sigma }\mathbf{n}dA=0$. This surface integral makes the direction explicit: a surface-normal force is generally neither pure lift nor pure drag. Pressure and skin shear stress both enter the resultant, although the pressure term usually dominates lift on an airfoil. ^[ref]

The lift coefficient isolates a useful scale. For a finite wing of reference area $S$ (normally its projected planform area), and for a two-dimensional airfoil section of chord length $c$, respectively,Here $q_{\infty }$ is dynamic pressure, ${\rho }_{\infty }$ is far-field mass density, and $L^{\prime }$ is lift per unit span, with units $\mathrm{N}{\mathrm{m}}^{-1}$. The signs of $C_L$ and $c_l$ follow the chosen lift axis. For stipulated values ${\rho }_{\infty }=1.20\mathrm{kg}{\mathrm{m}}^{-3}$, $U_{\infty }=20\mathrm{m}{\mathrm{s}}^{-1}$, $S=0.50{\mathrm{m}}^2$ and $C_L=0.60$, $q_{\infty }=240\mathrm{Pa}$ and $L=72\mathrm{N}$. Doubling $U_{\infty }$ gives four times the lift only if $C_L$ stays fixed: changes in Reynolds number, Mach number, surface condition or angle of attack can change it. ^[ref]

A motivating experiment is to compare two independently observed quantities: the lift registered by a force balance, and the lift inferred from surface pressures. A low-speed wind tunnel holds the model fixed while air passes it at adjustable speed; a calibrated balance measures forces, while small flush holes on both surfaces connect by tubing to differential-pressure sensors referenced to $p_{\infty }$. Iowa State's GA(W)-1 teaching model has $43$ taps around a $101\mathrm{mm}$ chord; its instructions give their coordinates and close the contour by repeating the first tap. A classroom comparison can combine this pressure-panel method with a balance on the same model; this is an experiment design, not a claim that those instructions report a combined measurement. ^{[ref] [ref]}

Before turning on the wind, apply known positive and negative loads to the balance through its resolving centre; use $mg$ for the known load from a mass. Record all balance channels during loading and unloading, including response to drag and torque loads, to detect cross coupling, hysteresis and zero drift. Record a wind-off tare at each model orientation so its weight does not become an aerodynamic lift reading. Map tunnel flow direction and uniformity, rather than using the fan setting as a speed measurement. Covert's MIT guide describes these calibration, tare and flow-survey requirements. ^[ref]

For each pressure channel, apply zero and several known pressure differences spanning the expected range; determine the voltage-to-pressure relation and recheck it after a run. Inspect tubing for leaks and blocked taps, and wait for readings to settle before averaging: tubing can delay a pressure signal. This calibration principle is used in NASA's electronically scanned pressure systems, where several applied calibration pressures determine each sensor's response and repeated calibration tracks drift. A simple pressure standard can instead use a liquid-column height difference and the hydrostatic pressure law. ^[ref]

Measure far-field static and total pressure outside the model's fluid wake; in low-speed incompressible flow their difference gives $q_{\infty }$ by the Bernoulli equation, so $U_{\infty }=\sqrt{2q_{\infty }/{\rho }_{\infty }}$. Record temperature and ambient pressure to determine mass density using the ideal gas law. At fixed $q_{\infty }$, step the angle of attack, repeat selected settings, and sweep back down to check reproducibility. Record balance lift, each tap's mean pressure, and fluctuation sizes. NASA describes pressure probes connected to differential transducers and emphasizes recording their reference pressure. ^[ref]

Convert each tap to a pressure coefficient $C_{p,i}=(p_i-p_{\infty })/q_{\infty }$. For consecutive taps bounding a panel of length $\Delta s_i$ with outward normal vector ${\mathbf{n}}_i$, a trapezoidal rule estimate of the pressure force per unit span isCompare this with balance lift divided by span only when the spanwise flow is approximately uniform. Agreement tests the pressure-resultant picture; disagreement calls for checking unresolved leading-edge pressure variations, skin shear stress, support loads and three-dimensional end effects before rejecting it. A force balance senses the total load, while taps alone reconstruct the pressure part. Tunnel walls and model blockage alter the effective flow and require corrections; end plates or a model extending between walls approximate two-dimensional flow but do not guarantee it. Record the Reynolds number and Mach number with the result, and do not interpret tunnel lift coefficients as universal wing data. ^{[ref] [ref]}

Magnetic flux measures the signed amount of magnetic field crossing an oriented surface $S$. Choose a unit normal $\hat{\mathbf{n}}$ and use the oriented surface element $d\mathbf{S}=\hat{\mathbf{n}}dA$. The defining surface integral isFor a uniform magnetic field across a flat surface of area $A$, the dot product reduces this towhere $\theta$ is the angle between the magnetic field and the chosen unit normal, rather than the plane of the surface. The SI unit is the weber, $1\mathrm{Wb}=1\mathrm{T}{\mathrm{m}}^2$. ^[ref].

For a stipulated magnetic field $B=2\mathrm{T}$ and area $A=3{\mathrm{m}}^2$, the magnetic flux is $6\mathrm{Wb}$ at $\theta =0$, $3\mathrm{Wb}$ at $\theta =\pi /3$, and zero at $\theta =\pi /2$. In the last case the magnetic field lies in the plane of the surface, so it does not cross it. Reversing the chosen unit normal reverses the sign of magnetic flux without changing the physical magnetic field; at $\theta =\pi$ the same example gives $-6\mathrm{Wb}$.

The geometric idea is projected area: tilting a flat surface reduces the area seen perpendicular to the magnetic field by the factor $\cos \theta$. This explains how rotation alone can change magnetic flux even when the magnetic field is constant; Faraday's law of induction turns such changes into an electromotive force. ^[ref].

For a regular parametric surface $\mathbf{r}:U\to X$, the displayed surface integral counts each image point as many times as its preimages in $U$, apart from sets of zero surface area. It equals the ordinary integral over the image when the parametrization is injective except on such sets. A periodic parametrization integrated over more than one period can give a different answer from integration over its image set. For example, $(u,v)\mapsto (\cos u,\sin u,v)$ with $0\le u\le 4\pi$ and $0\le v\le 1$ gives parameter area $4\pi$, whereas the image circular cylinder has surface area $2\pi$.

Orientability asks whether local choices of handedness can agree everywhere on a manifold. On a smooth manifold, an orientation of a smooth manifold chooses an orientation of a vector space for each tangent space, continuously with position. An ordered basis describes the same handedness as another one when the change-of-basis determinant is positive. An oriented atlas implements this condition through positive Jacobian determinants of the smooth transition maps.

A paper strip makes the obstruction visible. Join its short ends without twisting: the resulting untwisted band admits a consistent choice of unit normal. Add a half-turn before joining: on the Möbius band, following the core circle of a Möbius band reverses that choice when you first return to the same point. A local decision about handedness has encountered a global contradiction. The normal reversal around the Möbius core makes this precise for an explicit model.

Both rectangles have coordinates $(u,v)\in [0,1]\times [-1,1]$. The untwisted band identifies $(0,v)$ with $(1,v)$; the Möbius band identifies $(0,v)$ with $(1,-v)$. Across a seam the corresponding local coordinate changes preserve or reverse $v$, giving determinants $+1$ and $-1$. This sign explains the handedness change; the drawing also tracks the two boundary circles of the untwisted band and the single boundary circle of the Möbius band.

For a smooth topological surface embedded in ${\mathbb{R}}^3$, choosing an orientation of a surface is equivalent to choosing a continuous unit normal. This picture is specific to surfaces in three-dimensional space: the general definition uses compatible local orientations, without requiring an embedding. Orientability is a topological invariant, so a homeomorphism cannot turn a Möbius band into a untwisted band. The sphere and torus are orientable surfaces; the Klein bottle is not. ^[ref]

The distinction matters when a sign has geometric meaning. A signed flux surface integral needs an orientation of a surface, and Stokes theorem relates that choice to the outward-normal-first boundary orientation. Ordinary unsigned surface area does not require orientability. Thus a Möbius band still has a perfectly meaningful area even though it has no continuous global unit normal.

A connected orientable manifold of positive dimension has two possible orientations: reversing every local choice gives the other one. If a manifold has several connected components, each component can be oriented independently when it is orientable. For topological surfaces, a useful obstruction criterion is the presence of a Möbius band: a surface is nonorientable exactly when it contains such a band. ^[ref]

The domain of a function $U$ used by a parametrization $\mathbf{r}$ is its parameter domain. A parametrized surface may require several such domains, and duplicated images must be accounted for in a surface integral.

Integrate the surface charge density using the surface integral in spherical coordinates:Since $r_0>a$, the elementary integral isSubstitution yields the total induced electric chargeThe equality with the image charge also follows by applying Gauss's law to the exterior electric field: the physical and fictitious descriptions give the same electric flux through a sphere enclosing the conductor but excluding the physical point charge.

Write the constant spherical harmonic $Y_{00}=C$. Its orthonormality gives $4\pi |C{|}^2=1$, independently of its phase convention. Orthogonality to $Y_{00}$ givesOnly the $\ell =m=0$ term contributes to the surface integral of the surface charge density. Multiplying by the surface-area factor $a^2$ givesThus the spherical harmonic calculation reproduces the total induced electric charge: