Line and surface integrals

Choose what is being accumulated before choosing a formula

A thin wire with density needs mass per unit length multiplied by tiny lengths. A force doing work needs its tangential component multiplied by oriented displacement. A coated sheet needs mass per unit area multiplied by tiny surface areas. A flow through a sheet needs its normal component multiplied by oriented area. These are four related questions with different weights and different orientation behaviour.

The curves and surfaces live in physical input space. Their parameters are coordinates used to describe them, not additional physical directions. The preparation is dot/cross products 121, derivatives and Jacobians 131, multiple integrals 133, and fields 134. A general surface chart has a 3-by-2 derivative; its area factor cannot be the determinant of that nonsquare matrix.

We start with continuous integrands and piecewise C¹ regular curves, and C¹ regular surface patches over compact regular parameter regions. Regular means nonzero curve velocity or two independent surface tangent vectors, apart from explicitly handled boundaries or exceptional sets. The integrals below are finite in this compact continuous setting. Singular or unbounded cases require additional convergence work.

The exact examples initially use dimensionless coordinates. For a physical wire, sheet, force or flux, supply the stated length, density and time scales before interpreting a numerical answer as kilograms, joules or a flow rate.

A scalar curve integral uses arclength

Let r:[a,b]→R³ describe a regular curve C once, with continuous scalar f on C. Tiny path pieces contribute approximately f(r(t*)) Δs. Since ds=‖r′(t)‖dt, the scalar line integral is ∫C f ds=∫ₐᵇ f(r(t))‖r′(t)‖dt. Here ds is nonnegative arclength, not an oriented displacement component.

With f=1, the integral is curve length. With nonnegative linear mass density λ, it is mass. A signed scalar f can produce a negative scalar line integral; orientation independence does not mean every scalar integral is nonnegative. Additivity applies across finitely many smooth pieces, including a path with corners.

For the straight wire r(t)=(t,2t,2t), 0≤t≤1, the speed is 3. With λ=1+x, the density along it is 1+t and mass is ∫₀¹3(1+t)dt=9/2. Omitting speed would give 3/2, corresponding to the parameter interval rather than the physical wire.

Its first x-moment is ∫C xλ ds=3∫₀¹t(1+t)dt=5/2, hence x̄=5/9. Since every point has y=2x and z=2x, the other centroid coordinates are ȳ=z̄=10/9. The normalising mass must be finite and positive; signed integrals are not automatically valid mass-centre denominators.

Changing speed must not change the integral over the same wire

Suppose t=h(s) is a C¹ one-to-one parameter change with nonzero derivative on the interior and maps one parameter interval onto the other. Then r̃(s)=r(h(s)) has speed ‖r′(h(s))‖|h′(s)|. The absolute value makes the scalar arclength integral unchanged for either increasing or decreasing h. A piecewise version covers ordinary finite subdivisions.

For the same wire, use r̃(s)=(s/2,s,s), 0≤s≤2. Its speed is 3/2 and density 1+s/2. The mass becomes (3/2)∫₀²(1+s/2)ds=9/2, as before. Reversing the original path with r(1−s), 0≤s≤1, also leaves the mass unchanged.

A parameterisation that traces the wire twice counts both traversals. For nonnegative density it doubles the path integral, so it cannot represent the mass of the geometric wire counted once. At a self-intersection, finitely many crossing points have zero length and are harmless by themselves; repeated positive-length arcs change multiplicity.

A parameterisation with isolated zero speed at an endpoint may still be handled by a limit or a regular subdivision. Do not use the regular reparameterisation theorem as if its nonzero-derivative hypothesis had been checked there. Choose a regular parameterisation when one is available, and state the treatment of any exceptional point.

A vector line integral uses a dot product with the tangent displacement

For a continuous vector field F along an oriented piecewise C¹ path r, the work integral is ∫C F·dr=∫ₐᵇ F(r(t))·r′(t)dt. Equivalently it is ∫C F·T ds, where T is the unit tangent in the chosen direction. The dot product selects the tangential component, including its sign.

For F=(P,Q,R), one may write ∫C P dx+Q dy+R dz. The differentials are oriented component changes: dx=x′dt and similarly for y,z. There is no extra speed factor after using F·r′dt. Multiplying that expression by ‖r′‖ again would count the parameter scaling twice.

An orientation-preserving reparameterisation leaves work unchanged. Reversing the physical traversal changes r′ to the opposite tangent and reverses work’s sign. Travelling out and back along the same curve gives zero net work for a single-valued field evaluated at the same positions, while the scalar arclength integral of a positive density adds on both legs.

Work can be zero along a nontrivial path if the force is always perpendicular to its tangent. In physical units, newtons dotted with metres give joules. The mathematical formula also applies to other vector fields, but its value should be called physical work only when the field and units have that meaning.

A circulation example makes reversal visible

For F=(−y,x,0) and the counterclockwise unit circle r(t)=(cos t,sin t,0), 0≤t≤2π, F(r(t))=r′(t). Their dot product is 1, so work is 2π. Reversing the circle gives −2π. The curve length stays 2π under either orientation.

The endpoints alone need not determine work. From (0,0) to (1,1) in this planar field, moving first along x and then along y gives work 1. Moving first along y and then along x gives −1. The straight diagonal r(t)=(t,t), 0≤t≤1, gives zero because (−t,t)·(1,1)=0.

Each calculation names the entire oriented path. Comparing two parameterisations of the same traversal is different from comparing two different paths with the same endpoints. Neither an endpoint sketch nor a numerical arrow field establishes path independence.

A spatial example is the helix r(t)=(2cos t,2sin t,t), 0≤t≤2π. Its speed is √5 and length 2π√5. For F=(−y,x,1), the work integrand is 4+1=5, so work is 10π. The speed is not 5; the tangent’s squared norm happens to enter the dot product because F(r)=r′.

The chain rule supplies a genuine path-independence proof

If φ is C¹ on an open set containing a piecewise C¹ path C from A to B, then ∫C grad φ·dr=φ(B)−φ(A). On each smooth piece the integrand is d[φ(r(t))]/dt; integrate and telescope the endpoint terms. The potential must be a single-valued function on the path’s domain, not a multivalued angle notation.

For φ=x²y+y³+xz from 134, the gradient is (2xy+z,x²+3y²,x). Work from (0,0,0) to (1,2,3) is 2+8+3=13 along every admissible path in R³. On the straight path r(t)=(t,2t,3t), the direct integrand is 30t²+6t and its integral is 13, independently confirming the endpoint calculation.

On a path-connected open domain with continuous F, if every piecewise smooth path integral depends only on endpoints, fix A and define φ(x) by work from A to x. Comparing x with x+heⱼ along a short coordinate segment gives [φ(x+heⱼ)−φ(x)]/h→Fⱼ(x), by continuity. Thus φ is C¹ with grad φ=F. This explains the equivalence between a potential and path independence in that setting.

Zero curl alone is a local derivative test. The punctured-plane field (−y/r²,x/r²) has nonzero closed-circle work and no global single-valued potential there, despite zero curl away from the origin. Lesson 134’s domain hypotheses still apply; writing “curl=0” is not permission to evaluate every work integral by unspecified endpoints.

A surface patch needs the cross product of its tangent vectors

Let r:D⊂R²→R³ describe a regular patch S once, with parameters u,v. Its tangent vectors are r_u and r_v. A small parameter rectangle maps to a near-parallelogram whose area is approximately ‖r_u×r_v‖ Δu Δv. Thus dS=‖r_u×r_v‖du dv, while the oriented area vector is n dS=(r_u×r_v)du dv for the normal chosen by that parameter order.

Regularity requires r_u×r_v≠0 in the patch interior. The unit normal is n=(r_u×r_v)/‖r_u×r_v‖. The cross product is usually not a unit normal: it already includes the area stretching. Keeping the unit normal, scalar area factor and oriented area vector distinct prevents an extra factor from appearing in flux calculations.

In matrix terms, Dr has two columns r_u,r_v. The Gram matrix (Dr)ᵀDr is 2-by-2, and its determinant is ‖r_u‖²‖r_v‖²−(r_u·r_v)²=‖r_u×r_v‖². The nonnegative square root is the area scale. Equivalently it is the product of the two nonzero singular values when the patch has rank two, connecting surface area with the singular value decomposition in lesson 128.

For a graph r(x,y)=(x,y,g(x,y)), the cross product r_x×r_y=(−gₓ,−gᵧ,1) points upward. Hence dS=√(1+gₓ²+gᵧ²)dx dy. The xy region is a projection of the surface, not generally a piece of equal area.

Scalar surface integrals weight actual surface area

For continuous scalar f on a regular patch S, define ∬S f dS=∬D f(r(u,v))‖r_u×r_v‖du dv. With f=1 this is surface area; with a nonnegative surface density σ it is mass. Reversing the surface orientation leaves this scalar integral unchanged because the cross-product norm is unchanged.

For the graph z=x+2y over the unit square, r_x×r_y=(−1,−2,1), whose norm is √6. Its area is √6. With density σ=1+x, mass is √6∫₀¹∫₀¹(1+x)dy dx=3√6/2. Using the projected area alone would miss the slope-dependent factor.

The x first moment is √6∫₀¹x(1+x)dx=5√6/6, so x̄=5/9. Symmetry of the density in y gives ȳ=1/2. Because z=x+2y on the sheet, z̄=5/9+1=14/9. A constant surface factor cancels in these centroid ratios, but it does not disappear from the physical mass.

For the paraboloid z=x²+y² above the unit disk, dS=√(1+4x²+4y²)dx dy. Polar coordinates in the projection give area 2π∫₀¹r√(1+4r²)dr=(π/6)(5√5−1). The factor r comes from the planar polar change of variables; the square root comes from the sloping surface. They have different geometric origins and both are needed.

Flux selects a normal component and keeps its sign

For a continuous field F and a consistently oriented regular surface S, flux is ∬S F·n dS. With the normal induced by a chart, this becomes ∬D F(r(u,v))·(r_u×r_v)du dv. Reverse the chosen normal and the result changes sign. Positive flux means net flow in the chosen normal direction; the convention must be stated before interpreting the sign.

For the upward graph z=g(x,y), flux is ∬D F(x,y,g)·(−gₓ,−gᵧ,1)dx dy. One may instead use the upward unit normal and dS=√(1+gₓ²+gᵧ²)dx dy; the normalisation cancels. Multiplying the unnormalised vector (−gₓ,−gᵧ,1) by that square root again is incorrect.

On the upward plane z=x+2y over [0,1]², take F=(x,y,1). The flux integrand is −x−2y+1 and total flux is −1/2. The negative answer means that net flux is opposite the chosen upward-pointing tilted normal, even though the field’s vertical component is positive everywhere. Downward orientation gives +1/2.

For the upward paraboloid z=x²+y² over the unit disk and F=(0,0,z), the flux integrand in projected coordinates is x²+y². Its integral is π/2. This is different from the scalar surface integral of z, which would include √(1+4r²). The same surface and same scalar z can appear in different physical questions with different weights.

The chart changes while the chosen geometric integral stays fixed

Suppose (u,v)=ψ(s,t) is a C¹ one-to-one regular change of parameter coordinates mapping the new parameter domain onto the original domain D. By the chain rule, the new tangent cross product is r̃_s×r̃_t=det(Dψ)(r_u×r_v) at the corresponding point. Its norm has |det(Dψ)|, so ordinary planar change of variables 133 proves scalar surface-integral invariance.

For the plane patch r(u,v)=(u,v,u+2v), 0≤u,v≤1, use u=2s,v=t on 0≤s≤1/2,0≤t≤1. The new cross product is (−2,−4,2), of norm 2√6. Integrating over a parameter rectangle of area 1/2 still gives surface area √6. A different parameter speed or scale must not change the geometric area.

If det(Dψ)<0, the induced cross-product normal reverses. To describe the same chosen oriented surface, compensate that chart-induced sign when using its normal. If instead the physical normal itself is reversed, flux changes sign. Renaming parameters alone does not reverse a previously fixed physical orientation.

Overlapping charts need an explicit rule to avoid counting shared positive-area pieces twice: use a suitable disjoint patch decomposition up to area-zero boundaries, or weighted chart methods in a more advanced treatment. A chart that covers a surface twice counts multiplicity; an unsigned surface integral then doubles unless the parameter domain is restricted. Signed flux can additionally cancel if different traversals induce opposite normals.

Spherical coordinates have seams and rank loss at the poles

For a sphere of radius R>0, set r(φ,θ)=(Rsinφ cosθ,Rsinφ sinθ,Rcosφ), with 0<φ<π and 0<θ<2π. Here φ is colatitude from the positive z-axis and θ is azimuth. In this parameter order, r_φ×r_θ=R²sinφ e_r points outward. The scalar area factor is R²sinφ.

This open chart misses a meridian and the poles. The repeated meridian at θ=0 and θ=2π is a seam, and r_θ vanishes at the poles. These are area-zero exceptions on the smooth geometric sphere; one can use additional regular charts, or first integrate truncated patches and take limits with continuous bounded integrands. The chart’s rank loss is not a singularity of the sphere itself.

Its area is ∫₀²π∫₀π R²sinφ dφ dθ=4πR². For F(x)=x, the outward normal component on the sphere is R, giving outward flux 4πR³. For F(x)=x/‖x‖³, that component is 1/R², giving flux 4π. These follow directly from the surface integral; no divergence theorem is needed to obtain them.

The second field is undefined inside at the origin. It is continuous on the sphere, so its surface integral is valid, but a later volume theorem applied to the entire solid ball would need regularity inside as well. This distinction preserves the singular-source warning from 134.

Choose a continuous side when a flux problem needs one

A regular orientable surface admits a continuous unit normal field. For a graph, “upward” is available globally. For a closed surface bounding a regular solid, the outward convention points away from the solid. An open patch may have two choices without a distinguished physical one; specify the intended side.

A Möbius strip has scalar surface area and can carry scalar mass density, but it admits no continuous global choice of normal. The ordinary global oriented flux integral ∬F·n dS therefore cannot be assigned by pretending one normal remains consistent around the entire strip. Local patch calculations do not remove that obstruction.

A curve’s direction and a surface’s normal become linked when an integral theorem relates a boundary curve to the surface. Viewed from the positive normal side, the compatible boundary traversal is counterclockwise for a simple planar patch. Lesson 136 states the full boundary-orientation conventions and hypotheses for Green’s and Stokes’s theorems.

At this stage, compute the required curve or surface integral directly and label its orientation. Keeping that direct calculation available provides a useful independent check when a later theorem offers a shorter route.

The geometric factor carries dimensions and meaning

For a wire, density kg m⁻¹ times ds in metres gives kilograms. For a sheet, density kg m⁻² times dS in square metres also gives kilograms. A work integrand uses force in newtons and oriented displacement in metres, giving joules. A velocity flux in m s⁻¹ through an area in m² gives m³ s⁻¹; a mass flux density in kg m⁻² s⁻¹ gives kg s⁻¹.

Parameter dimensions may differ from spatial dimensions. If a dimensionless surface parameter maps into metres, each tangent derivative carries metres and their cross product carries square metres. If a coordinate parameter itself has length units, the tangent derivative and its parameter differential must be interpreted together. Units are an additional check, not a substitute for correct geometry.

Before evaluation, state the path or patch, its parameter domain, whether it covers once, the regularity, the scalar or vector quantity, and the physical orientation if required. Then form the speed, cross-product norm or oriented cross product, simplify, and integrate using the actual bounds. Finally test a reversal or a simpler equivalent parameterisation.

Numerical quadrature and plotted meshes can check a finite calculation and reveal a missing scale factor. They do not prove chart injectivity, eliminate a singularity, resolve a nonorientable surface, or justify an improper limit. Preserve the analytic domain and orientation arguments alongside any numerical or interactive result.

Worked examples

Try it yourself

1. What factor converts dt into ds for r(t)=(t,2t,2t)?

   Show the answer and reasoning

   ‖r′‖=3.

   The tangent is (1,2,2); its norm is √9.

2. Must a scalar line integral be nonnegative?

   Show the answer and reasoning

   No.

   Its arclength weight is nonnegative, but a signed integrand can be negative. A mass density is separately required to be nonnegative.

3. Does reversing a wire’s parameter direction change its mass integral?

   Show the answer and reasoning

   No, for the same wire counted once.

   The speed contains an absolute parameter derivative, so orientation does not change scalar arclength accumulation.

4. What does a parameterisation tracing the same positive-density wire twice do to its path integral?

   Show the answer and reasoning

   It counts both traversals and doubles the value.

   It is not a once-counted geometric mass integral unless multiplicity is corrected or the parameter domain restricted.

5. After writing F(r(t))·r′(t)dt, should one multiply by ‖r′(t)‖ again?

   Show the answer and reasoning

   No.

   The tangent displacement already supplies the parameter scaling. An extra speed factor changes the integral.

6. What happens to work when the same curve is traversed in reverse?

   Show the answer and reasoning

   Its sign reverses.

   The oriented tangent displacement reverses while the field values at the same points do not.

7. Find the length of (2cos t,2sin t,t), 0≤t≤2π.

   Show the answer and reasoning

   2π√5.

   The speed is √(4sin²t+4cos²t+1)=√5.

8. What is work of a single-valued C¹ potential’s gradient around a closed piecewise C¹ path in its domain?

   Show the answer and reasoning

   0.

   The endpoint potential values are equal. The path and potential must lie in the stated domain.

9. Can zero curl on the punctured plane alone justify an endpoint-only work calculation?

   Show the answer and reasoning

   No.

   The angle-gradient field has nonzero circulation around the missing origin and no global single-valued potential.

10. What condition makes a C¹ surface chart regular?

    Show the answer and reasoning

    r_u×r_v≠0, equivalently rank Dr=2.

    The two tangent vectors must be linearly independent; otherwise the ordinary local area construction degenerates.

11. May the area factor of a 3-by-2 surface derivative be written as det Dr?

    Show the answer and reasoning

    No.

    That determinant is undefined. Use ‖r_u×r_v‖ or √det((Dr)ᵀDr).

12. For z=x+2y, what is the upward unit normal?

    Show the answer and reasoning

    (−1,−2,1)/√6.

    The cross product (−1,−2,1) has norm √6; it is an area vector before normalising.

13. What is upward flux of the constant field (0,0,1) through z=x+2y over the unit square?

    Show the answer and reasoning

    1.

    Dotting with the graph area vector gives 1 throughout. Its tilted area √6 is not the vertical flux.

14. If upward flux through a sheet is −1/2, what is downward flux?

    Show the answer and reasoning

    +1/2.

    The scalar surface area stays the same while the chosen normal reverses.

15. For sphere parameters ordered (φ,θ), with φ colatitude and θ azimuth, what is the outward area vector?

    Show the answer and reasoning

    R²sinφ e_r dφ dθ for 0<φ<π.

    The tangent order r_φ×r_θ gives the outward direction. Swapping parameter order reverses the induced normal.

16. Does a zero spherical-chart tangent at a pole mean the geometric sphere has a singular surface there?

    Show the answer and reasoning

    No.

    The chosen chart degenerates. Other regular charts cover the poles, or continuous bounded integrals can use area-zero exception limits.

17. May a Möbius strip have scalar surface area while lacking a globally continuous normal for ordinary oriented flux?

    Show the answer and reasoning

    Yes.

    Scalar area is independent of orientation; a global signed normal field is obstructed by nonorientability.

18. What are the units of velocity flux through a surface when velocity is m s⁻¹?

    Show the answer and reasoning

    m³ s⁻¹.

    Normal velocity times area m² gives a volume flow rate, not a volume or a mass without further factors.

What to carry forward

• Scalar curve accumulation uses speed; work uses the oriented tangent displacement.
• Scalar surface accumulation uses a cross-product norm; flux uses an oriented cross product or unit normal times area.
• Scalar integrals are orientation-independent for the same geometric object counted once. Work and flux reverse sign with physical orientation.
• Correct parameter changes include both their derivative scale and their transformed domain; repeated coverage changes multiplicity.
• A single-valued potential proves endpoint work by the chain rule. Zero curl alone does not supply a global potential on every domain.
• The area scale of a surface chart is √det((Dr)ᵀDr), connecting tangent geometry to Gram matrices and singular values.
• A chart seam or coordinate rank loss may be harmless for a smooth surface when explicitly covered or treated by justified limits.
• A singularity inside a surface does not prevent a well-defined surface integral, but it can invalidate a later volume-theorem application.
• Units, regularity, injectivity, orientation and actual bounds are part of the calculation, not optional annotations.
• Direct curve and surface calculations prepare independent checks for Green, Stokes and divergence theorems 136.