The fundamental theorem of calculus, Green’s theorem and the divergence theorem share one structure: a change measured inside a region is related to a measurement on its oriented boundary.
Differential forms make that structure explicit. Instead of treating line integrals, surface integrals and volume integrals as separate collections of formulas, we build objects that already know how many tangent directions they accept, how orientation affects their sign, and how they change under a parametrisation.
The reward is not merely shorter notation. A form keeps the mathematical meaning attached to a calculation. It distinguishes signed circulation from unsigned length, a flux integral from surface area, and a locally defined potential from one that exists across an entire domain.
This guide follows Tangent, Cotangent and Tensor Fields. A differential form is a special alternating covariant tensor field. The earlier Smooth Manifolds and Maps guide supplies the coordinate framework, while Curves and Surfaces supplies concrete parametrised examples.
Level and prerequisites: undergraduate differential geometry and advanced enrichment. You should know partial differentiation, the chain rule, determinants, line and double integrals, and the difference between vectors and covectors. We use real smooth forms. School-level study routes remain separate in the BTT Mathematics Hub.
Reading route: what a form measures → wedge products → exterior differentiation → pullbacks → integration → Stokes’ theorem → closed and exact forms → practice → worked solutions.
1. A form has a degree and a job
A differential k-form assigns to each point an alternating multilinear function of k tangent vectors at that point. Its degree k is the number of vector inputs. A 0-form is simply a smooth scalar function. A 1-form accepts one vector; a 2-form accepts two; a 3-form accepts three.
Alternating means that exchanging two inputs changes the sign. If two inputs agree, the value is zero. More generally, linearly dependent inputs give zero. This is exactly the behaviour needed for oriented area and volume: collapsing two directions onto one should destroy the corresponding area.
In coordinates on R³, a 1-form looks like P dx+Q dy+R dz. A 2-form can be written A dx∧dy+B dx∧dz+C dy∧dz. A 3-form has the form f dx∧dy∧dz. Their coefficients may vary smoothly from point to point.
The covectors dx,dy,dz extract coordinate components of tangent vectors. The wedge symbol ∧ will organise alternating products. These are linear and multilinear operations, not mysterious infinitesimal numbers that must be manipulated without definitions.
The coordinate and alternating-tensor descriptions are reconciled in Lorenzo Sadun’s Notes on Differential Forms. The examples below use both viewpoints: components for calculation, vector inputs for checking meaning.
2. Why there are only certain independent components
On an n-dimensional manifold, a basis for k-forms at a point consists of dxⁱ¹∧⋯∧dxⁱᵏ with strictly increasing indices i₁<⋯<iₖ. There are n choose k such basis elements. Repeated indices give zero, and changing their order only changes a sign.
Thus a 2-form in three dimensions has three independent components, not nine. A 2-form in four dimensions has six. A form of degree greater than n is zero because any collection of more than n tangent vectors is linearly dependent.
This degree count provides a useful early check. Pulling a 2-form back to a one-dimensional curve gives zero. To integrate along a curve in the ordinary differential-form sense, the integrand must be a 1-form. A scalar function can also be integrated against arc length, but arc length supplies additional metric information rather than turning the scalar into a 1-form by itself.
3. The wedge product records orientation
For 1-forms α and β, define their wedge product by (α∧β)(v,w)=α(v)β(w)−α(w)β(v). This is the determinant of the two measurements of the two input vectors.
In the plane, (dx∧dy)(v,w)=vₓwᵧ−vᵧwₓ. It measures signed coordinate area. Interchanging v and w reverses the sign. If the vectors are parallel, the determinant vanishes.
For example, let v=(1,2) and w=(3,−1). Then (dx∧dy)(v,w)=−1−6=−7. The negative sign records orientation relative to the ordered Cartesian basis. Taking an absolute value would preserve the area magnitude but discard the orientation information that Stokes’ theorem needs.
Worked expansion: keep the signs visible
Take α=2dx+dy and β=dx−3dy. Distributing the wedge product gives 2dx∧dx−6dx∧dy+dy∧dx−3dy∧dy. The repeated-coordinate terms vanish, and dy∧dx=−dx∧dy. Therefore α∧β=−7dx∧dy.
The calculation resembles ordinary expansion but uses different multiplication rules. Replacing ∧ with ordinary commutative multiplication would erase the result’s geometric meaning.
4. Higher-degree signs are controlled by degree
If α has degree p and β has degree q, then α∧β has degree p+q and α∧β=(−1)ᵖᑫβ∧α. Moving p one-form factors past q others creates pq sign changes.
Two 1-forms anticommute. Two 2-forms commute, because (−1)⁴=1. Consequently, the square of an odd-degree real form is zero, but the square of an even-degree form need not be zero.
On R⁴ with coordinates x,y,z,w, let β=dx∧dy+dz∧dw. The two self-products vanish because they repeat coordinates. The two cross-products agree, so β∧β=2dx∧dy∧dz∧dw, which is not zero.
This example prevents an overgeneralisation from α∧α=0 for 1-forms. The degree, not merely the repeated name of the form, decides the sign rule. Dimension also matters: a 4-form would vanish on a two-dimensional manifold even if the same algebraic pattern is nonzero on R⁴.
5. Exterior differentiation raises degree
The exterior derivative d takes k-forms to (k+1)-forms. For a function f, it is the ordinary differential df=Σfᵢdxⁱ. In coordinates, differentiate each coefficient and wedge its differential onto the coordinate-form factors already present.
For α=P dx+Q dy in the plane, dα=dP∧dx+dQ∧dy. Expanding gives Pᵧdy∧dx+Qₓdx∧dy, since the dx∧dx and dy∧dy terms vanish. Thus dα=(Qₓ−Pᵧ)dx∧dy.
For α=−y dx+xdy, the derivative is 2dx∧dy. The factor two is not optional: both coefficient derivatives contribute with the same final sign.
For α=xyz dx on R³, dα=(yz dx+xz dy+xy dz)∧dx=−xz dx∧dy−xy dx∧dz. This calculation is a useful way to practise reordering signs without relying on a memorised curl formula.
The elementary exterior-derivative calculations and their vector-calculus interpretations are developed in Donu Arapura’s Introduction to Differential Forms.
6. The product rule and the equation d²=0
Exterior differentiation obeys a graded product rule: d(α∧β)=dα∧β+(−1)ᵖα∧dβ when α has degree p. The sign records moving a one-form derivative past the p factors of α.
Its other central property is d(dω)=0 for every smooth form ω. For a smooth function f on the plane, d(df)=(fᵧₓ−fₓᵧ)dx∧dy=0. In higher degrees, the same symmetry of mixed partial derivatives cancels against wedge antisymmetry.
The statement does not mean that every form has zero derivative. The form −y dx+xdy has derivative 2dx∧dy. It means that applying d a second time to a derivative gives zero.
In vector-calculus language, familiar identities such as curl of a gradient equals zero and divergence of a curl equals zero are manifestations of this structure, once the relevant Euclidean identifications between forms and vector fields are specified.
7. Pullbacks bring a form onto the parameter space
Let F:M→N be smooth and let ω be a k-form on N. Its pullback F*ω is a k-form on M. At p, it is defined by feeding the pushed-forward tangent vectors into ω: (F*ω)ₚ(v₁,…,vₖ)=ωꜰ₍ₚ₎(dFₚv₁,…,dFₚvₖ).
The direction reverses. Points travel from M to N, but forms on N become forms on M. This is exactly what an integration parametrisation requires: measurements on a geometric curve or surface must be expressed on the interval or plane region where ordinary integrals are calculated.
Computationally, substitute F into coefficient functions and replace every target differential by the differential of the corresponding component of F. No inverse map is required. A parametrisation need not be a diffeomorphism onto the entire target for a pullback to make sense.
Pullback preserves wedge products and commutes with d: F*(α∧β)=F*α∧F*β and F*(dω)=d(F*ω). For composable maps, (G∘F)*=F*∘G*. These rules are developed in Sadun’s section on pullbacks.
8. The polar Jacobian emerges from the wedge product
Take F(r,θ)=(r cosθ,r sinθ). Then F*dx=cosθ dr−r sinθ dθ and F*dy=sinθ dr+r cosθ dθ.
Wedge the two expressions. The repeated dr and repeated dθ terms vanish. The cross-terms give rcos²θ dr∧dθ−rsin²θ dθ∧dr. Reordering the second term yields F*(dx∧dy)=rdr∧dθ.
The polar scale factor r has appeared automatically. It is the determinant of the derivative of F. More generally, the pullback of a top-degree coordinate form by a map between equal-dimensional spaces multiplies by the signed Jacobian determinant.
Use a polar chart with r>0 and an angular range that avoids a seam when claiming a one-to-one coordinate system. For disc integrals, the centre and seam can be handled through appropriate charts or by the usual measure-zero argument. They should not be ignored when making a global chart claim.
A reversal shows why the sign matters. For F(u,v)=(v,u), F*(dx∧dy)=dv∧du=−du∧dv. Ordinary unsigned area uses the absolute Jacobian. Oriented forms retain its sign.
9. A line integral evaluates a form on the velocity
For a smooth curve γ:[a,b]→M and a 1-form α, define ∫γ α=∫ₐᵇγ*α. In plane coordinates, this becomes ∫ₐᵇ[P(γ(t))x′(t)+Q(γ(t))y′(t)]dt.
The integrand is α evaluated on the curve’s velocity. There is no absolute-value speed factor unless the specific problem supplies one through a different construction. Reversing the path reverses the integral’s sign, while an orientation-preserving reparametrisation leaves it unchanged.
For piecewise smooth curves, calculate each smooth segment with its orientation and add the results. A path that retraces a segment in the opposite direction contributes the opposite signed amount on that segment.
Worked example: a parabola
Let γ(t)=(t,t²), 0≤t≤1, and α=−y dx+xdy. Substitute x=t,y=t²,dx=dt,dy=2t dt. Then γ*α=−t²dt+2t²dt=t²dt. The integral is 1/3.
This is not the arc length of the parabola. Arc length would integrate √(1+4t²). The two integrals answer different questions about the same path.
Worked example: one circle, two distinct measurements
On γ(t)=(R cos t,R sin t), 0≤t≤2π, the same form pulls back to R²dt. Its integral is 2πR². The curve’s length is 2πR. One scales quadratically with radius and the other linearly.
Changing the direction of traversal changes 2πR² to −2πR², but leaves the length positive. Scaling and orientation checks reveal whether the correct integral has been chosen.
10. Exact forms recover the fundamental theorem immediately
If α=df for a globally defined smooth function f, then γ*α=d(f∘γ). The ordinary fundamental theorem gives ∫γ df=f(γ(b))−f(γ(a)).
The integral depends only on the ordered endpoints, not on the intervening path. Every closed loop has integral zero for an exact 1-form. This provides an efficient obstruction: one closed loop with a nonzero integral proves that a form is not globally exact on the domain containing that loop.
For f=x²y+sin x+y³, the differential is (2xy+cos x)dx+(x²+3y²)dy. Its integral along any piecewise smooth path from (0,0) to (1,1) is 2+sin1. The path can be straight, curved or broken into segments; exactness accounts for all of them at once.
The word “globally” matters. A potential defined only on a small neighbourhood does not automatically justify endpoint-only integration along a path that leaves that neighbourhood.
11. Surface integration pulls back a 2-form
For an oriented surface patch X:D⊂R²→M and a 2-form η, integrate X*η over D with the corresponding orientation. The pullback evaluates η on the two tangent directions Xᵤ and Xᵥ and produces a scalar coefficient multiplying du∧dv.
In standard oriented Euclidean R³, the vector field F=(P,Q,R) corresponds to the flux 2-form η=Pdy∧dz+Qdz∧dx+Rdx∧dy. Evaluating it on two vectors gives the dot product of F with their cross product.
Consequently X*η=[F(X)·(Xᵤ×Xᵥ)]du∧dv. This is the usual flux integrand with the normal and area factor combined. A separate unit normal is unnecessary if the cross product is already being used.
The conversion between flux and 2-form integration is illustrated in Arapura’s surface-integration treatment. It uses the Euclidean metric and orientation to associate a vector field with a form; the integration of the form itself is the more general operation.
Worked example: a saddle patch with negative flux
Take X(u,v)=(u,v,uv), 0≤u,v≤1, oriented upward. Its tangent cross product is (−v,−u,1). For the radial vector field F=(x,y,z), the dot product is (u,v,uv)·(−v,−u,1)=−uv.
Therefore the flux is ∫₀¹∫₀¹−uv du dv=−1/4. The negative sign is meaningful: the selected oriented normal has a negative dot product with the field over the patch interior.
The same result follows by substituting dx=du,dy=dv,dz=vdu+udv into η=xdy∧dz+ydz∧dx+zdx∧dy. The three contributions are −uv,−uv and +uv. Their sum is −uv. Direct pullback and vector flux agree.
12. Orientation is part of the integral’s input
On an oriented k-dimensional manifold, a k-form can be integrated after suitable integrability conditions are established. Reversing orientation changes the integral’s sign. The form and orientation together specify the signed measurement.
A compact manifold and a smooth form provide a convenient setting. On a noncompact manifold, compact support is a useful sufficient condition; other cases require appropriate convergence assumptions.
When one chart does not cover the relevant region, a partition of unity splits the form into locally supported pieces. Each piece is integrated in coordinates, and the results are added. Compatible orientation and the change-of-variables rule make the result independent of this auxiliary choice.
Ordinary unsigned area is different. A nonorientable surface can still have an area density supplied by a Riemannian metric, even though it has no nowhere-zero global ordinary top-degree form. “Not orientable” does not mean “has no measurable area”. It means the signed top-form construction cannot use one globally consistent orientation.
13. Stokes’ theorem is the common boundary law
Let M be a compact oriented smooth n-manifold with boundary, and let ω be a smooth (n−1)-form on M, smooth up to the boundary. Then
∫M dω = ∫∂M ω.
On the right, ω is restricted to the boundary by pullback along its inclusion. The boundary uses the orientation induced from M. The degrees now match: dω has degree n for integration over M, while ω has degree n−1 for integration over its boundary.
The boundary orientation uses the outward-first convention: an outward transverse vector followed by a positively oriented boundary basis gives a positively oriented basis of M. For an upward-oriented plane disc, this makes the boundary anticlockwise when viewed from above.
This manifold formulation appears in Sadun’s Stokes’ theorem section. Rectangles, polyhedra and other piecewise smooth regions require a suitable version accounting for corners, such as Brian Conrad’s Stokes’ Theorem with Corners. A theorem stated for smooth boundary should not silently be applied to arbitrary singular sets.
14. Why the boundary remains after local cancellation
In a simple rectangular coordinate region, Stokes reduces to applying the one-variable fundamental theorem along coordinate directions and combining the endpoint terms with the correct signs.
When a region is divided into adjoining pieces, the shared edges or faces carry opposite induced orientations from the two sides. Their contributions cancel. Only the outside boundary remains.
This is the geometric mechanism behind the formula. It is not a proof that every imaginable region satisfies it; smoothness, orientation and integrability make the local calculations and assembly legitimate. The proof strategy explains why losing a boundary component or reversing one orientation changes the answer.
In dimension one, M=[a,b] with its increasing orientation has positively oriented endpoint b and negatively oriented endpoint a. Taking ω=f, a 0-form, gives ∫ₐᵇdf=f(b)−f(a). The familiar subtraction is the one-dimensional boundary orientation.
15. Green’s theorem is the planar case
On an oriented plane region D, put ω=Pdx+Qdy. Since dω=(Qₓ−Pᵧ)dx∧dy, Stokes gives the boundary integral of Pdx+Qdy as the integral of Qₓ−Pᵧ over D, provided the boundary and coefficient regularity conditions hold.
For a region with a hole, the outer boundary is anticlockwise and the inner boundary clockwise under the standard plane orientation. Treating both components as anticlockwise adds where the theorem requires subtraction.
Worked example: a radius-two disc
Let ω=(x²−y)dx+(x+y²)dy and let D be x²+y²≤4. Here Qₓ=1 and Pᵧ=−1, so dω=2dx∧dy. Therefore the positively oriented boundary integral is 2 times the disc area, or 8π.
A direct parametrisation provides an independent check. Set x=2cos t,y=2sin t. The contribution from −y dx+xdy is 4dt, integrating to 8π. The remaining terms x²dx+y²dy are d(x³/3+y³/3), whose integral around the closed circle is zero.
The two calculations separate the part carrying circulation from the exact part. This is more informative than expanding every trigonometric term and hoping cancellations eventually appear.
16. Curved surfaces obey the same rule
Take the cap S parametrised by X(u,v)=(u,v,1−u²−v²) over the unit disc, oriented upward. Its boundary is the unit circle in the plane z=0.
For ω=−y dx+xdy, we again have dω=2dx∧dy. Pulling back by X gives 2du∧dv, so ∫S dω=2π. Along the positively oriented boundary, x=cos t,y=sin t,z=0, the form becomes dt and its integral is also 2π.
The surface is curved, but the answer is not twice its surface area. The form 2dx∧dy measures signed projected area in this example. The actual area element of the cap would be √(1+4u²+4v²)du dv, a different integrand.
This distinction matters whenever a form is written in ambient coordinates. A familiar pair of symbols dx and dy does not automatically mean intrinsic area on every surface.
17. The divergence theorem is the three-dimensional case
For F=(P,Q,R), use the flux form η=Pdy∧dz+Qdz∧dx+Rdx∧dy. Exterior differentiation gives dη=(Pₓ+Qᵧ+R_z)dx∧dy∧dz.
Stokes on a suitable oriented three-dimensional region says that the integral of this divergence over the volume equals the outward flux through its boundary. The ordinary divergence theorem is therefore the same boundary formula in a different degree.
For F=(x,y,z), the divergence is 3. On a ball of radius R, the volume calculation gives flux 3·(4πR³/3)=4πR³. Directly on the boundary sphere, F·N=R and the area is 4πR², again giving 4πR³.
The orientation check is immediate: using an inward normal changes the flux sign. The scaling check is also useful: the field magnitude grows with R and the boundary area with R², so a result proportional to R³ is reasonable.
18. Closed and exact forms are not synonyms
A form ω is closed when dω=0. It is exact when ω=dη for a form η of one degree lower. Since d²=0, every exact form is closed.
The converse is a global question about the domain. The Poincaré lemma states that every smooth closed positive-degree form on a star-shaped open subset of Euclidean space is exact. In particular, closed forms are locally exact on sufficiently small coordinate balls.
A star-shaped domain has a point from which every straight segment to another point stays in the domain. That geometry permits an integration-based construction of a primitive. A punctured plane does not satisfy this condition around the missing origin.
Thus checking dω=0 is a local differential test. Establishing a global potential also requires control of the domain. The distinction is developed through exactness and counterexamples in Arapura’s first chapter and the cohomology development in Sadun’s notes.
19. A closed form with a nonzero period
On R² without the origin, define α=(−y dx+xdy)/(x²+y²). Its coefficient functions are smooth on that domain. Write P=−y/(x²+y²) and Q=x/(x²+y²). Differentiation gives Qₓ=Pᵧ=(y²−x²)/(x²+y²)². Therefore dα=0.
Now integrate around a circle of any radius R>0. The numerator pulls back to R²dt and the denominator to R². The resulting integral is ∫₀²πdt=2π.
An exact 1-form would integrate to zero around every closed loop. Hence α is closed but not exact on the punctured plane. The nonzero period has detected a global obstruction that the local derivative test cannot see.
Locally, α=dθ for a suitable angular coordinate. Globally, no single-valued smooth real angle function exists on the whole punctured plane. Going once around the origin changes a continuous lifted angle by 2π, even though the point returns to where it began.
Why Stokes has not failed
A tempting but invalid argument fills the circle with a disc and writes ∫circle α=∫disc dα=0. The form is not defined at the origin inside that disc, so the theorem’s hypothesis is missing.
Use an annulus instead. The form is smooth on a neighbourhood of the annulus, and dα=0 there. Its outer boundary contributes 2π anticlockwise, while its inner boundary contributes −2π because the induced orientation is clockwise. The total is zero, exactly as Stokes predicts.
The problem was not an unreliable theorem. It was an omitted point and an omitted boundary component. Domain and orientation checks are part of the mathematics, not administrative details around it.
20. Constructing a potential step by step
Suppose ω=(2xy+cos x)dx+(x²+3y²)dy on all of R². The cross-partial test gives Pᵧ=2x and Qₓ=2x, so the form is closed. The domain has no missing point or other obstruction, and a potential can be constructed directly.
Integrate the dx coefficient with respect to x: f(x,y)=x²y+sin x+h(y). The integration constant may depend on y because differentiation with respect to x would not detect it.
Differentiate this candidate with respect to y. We obtain fᵧ=x²+h′(y). Matching it to x²+3y² gives h′(y)=3y², so h(y)=y³+C.
The final check differentiates the completed potential in both directions. It recovers both original coefficients. This construction gives stronger evidence than citing closedness alone because it exhibits the globally defined primitive explicitly.
21. De Rham cohomology records what local primitives cannot remove
In degree k, de Rham cohomology is the vector space of closed k-forms modulo exact k-forms. Two closed forms represent the same class when their difference is exact.
This is the differential-form counterpart of the cycles-modulo-boundaries pattern in Homology and Cohomology. Here the coefficients are real. The first de Rham cohomology of a circle is R, whereas its first homology with integer coefficients is Z. Coefficient systems and the kind of invariant must remain explicit.
The angular form supplies a representative of the nontrivial first class of the punctured plane. Adding an exact form changes local expressions but not the integral around a closed loop. Periods therefore provide concrete measurements of cohomology classes.
De Rham’s theorem identifies this cohomology with singular cohomology with real coefficients on smooth manifolds. The theorem connects integration and differential equations to topological information; it does not turn every local formula into a global potential.
22. Which operations need a metric?
Forms, wedge products, exterior differentiation and pullbacks can be defined without choosing a Riemannian metric. Integrating a top-degree form requires orientation and integrability, but not a metric.
By contrast, curve length and unsigned surface area require metric information. Turning df into a gradient vector requires a metric. Identifying a flux form with a vector field in the familiar three-dimensional way uses both metric and orientation.
This separation explains why forms travel so well between coordinate systems. Their basic calculus is organised around multilinearity, alternation and smooth maps. Extra geometric choices can be added when a problem needs distances, angles or metric-dependent operators.
A calculation should therefore name its measurement before choosing a formula: circulation, work, flux, length, area and volume are related but not interchangeable jobs.
23. A dependable integration routine
First identify the degree of the form and the dimension of the integration domain. State the domain where the coefficients are smooth. Then choose an orientation and a parametrisation that respects it.
Pull the form back completely: coefficients and differential factors both need substitution. Expand wedge products before integrating, so repeated factors vanish and sign changes become visible. Check whether the parametrisation covers the region once or with multiplicity.
Before invoking Stokes, inspect every boundary component and confirm that the form is defined across the entire region being used. For a closedness-to-exactness argument, inspect the domain again: a missing point can change the answer even when every local derivative looks harmless.
Finish with an independent check when available. Compare a surface flux with a divergence integral, a Green calculation with an exact-plus-circulating decomposition, or a potential integral with its endpoint difference. Agreement between different routes helps locate sign and domain errors.
24. Independent practice: twelve questions
Use the standard orientations on Euclidean coordinate spaces unless otherwise stated. Anticlockwise means viewed from the positive z-direction for curves in the xy-plane. Show the pullback or derivative before giving the integral.
- For α=3x²y dx+x³dy, compute dα and exhibit a potential.
- Integrate α=xdy along γ(t)=(t,t²), 0≤t≤1. Also compute dα.
- Expand (2dx+dy)∧(dx−3dy). Evaluate it on v=(1,2),w=(3,−1).
- Compute d(xyz dx) in the increasing-coordinate basis dx∧dy,dx∧dz,dy∧dz.
- Compute the pullback of dx∧dy under polar coordinates and under F(u,v)=(v,u). Explain the difference in sign behaviour.
- Integrate −y dx+xdy around a radius-three circle once anticlockwise, then once clockwise.
- Use Green’s theorem to integrate (x²−y)dx+(x+y²)dy around the positively oriented boundary of x²+y²≤4.
- Find the upward flux of F=(x,y,z) through X(u,v)=(u,v,uv), 0≤u,v≤1.
- Find the outward flux of F=(x,y,z) through the sphere of radius two using the divergence theorem.
- Find a potential for (2xy+cos x)dx+(x²+3y²)dy and calculate its integral from (0,0) to (1,1).
- For α=(−y dx+xdy)/(x²+y²), explain why it is closed but not exact on the punctured plane. What is its integral over the entire induced boundary of an annulus centred at the origin?
- On R⁴, let β=dx∧dy+dz∧dw. Compute β∧β. Why does the rule for the square of a 1-form not apply?
25. Worked solutions and checks
1–4. Differentiate and alternate carefully
1. With P=3x²y and Q=x³, we have Qₓ=Pᵧ=3x², so dα=0. The function f=x³y satisfies df=3x²y dx+x³dy. Thus α is exact on all of R², not merely closed. The explicit potential settles the global question.
2. Along the curve, x=t and dy=2t dt, so γ*α=2t²dt and the integral is 2/3. Separately, d(xdy)=dx∧dy. The nonzero derivative shows that xdy is not exact on the plane, although its integral over this particular open path is still perfectly well-defined.
3. The wedge product is −7dx∧dy. On the specified vectors, dx∧dy evaluates to det((1,2),(3,−1))=−7. Therefore the requested value is 49. The coefficient −7 and the determinant −7 play different roles and must both be included.
4. Differentiate the coefficient and wedge with dx: (yz dx+xz dy+xy dz)∧dx. The first term vanishes. Reordering the other two gives −xz dx∧dy−xy dx∧dz. There is no dy∧dz term.
5–7. Preserve orientation through the calculation
5. Polar coordinates give rdr∧dθ. On the chart r>0, the Jacobian determinant is positive. Swapping coordinates gives dv∧du=−du∧dv, with determinant −1. Pullbacks of forms preserve this signed determinant rather than replacing it by an absolute value.
6. The circle pullback is 9dt over an angular interval of length 2π, giving 18π anticlockwise. Reversing orientation gives −18π. The curve’s circumference would be 6π, which is not the quantity asked for.
7. The exterior derivative is 2dx∧dy. The radius-two disc has area 4π, so the integral is 8π. As a separate check, x²dx+y²dy is exact and contributes zero around the loop, while −y dx+xdy contributes 8π.
8–10. Use the matching-dimensional integral
8. Xᵤ×Xᵥ=(−v,−u,1), and F(X)=(u,v,uv). Their dot product is −uv. Integrating over the unit square gives −1/4. The cross product already includes the area scaling, so multiplying by a second surface-area factor would count that scaling twice.
9. The divergence is 3. The radius-two ball has volume 32π/3, so the outward flux is 32π. Directly, F·N=2 on the sphere and its area is 16π, giving the same result.
10. Integrating the dx coefficient gives x²y+sin x+h(y). Matching the dy coefficient requires h′=3y², so f=x²y+sin x+y³+C. The endpoint difference is 2+sin1. The additive constant cancels.
11–12. Do not turn a local identity into a global conclusion
11. The coefficient derivatives Qₓ and Pᵧ agree away from the origin, so dα=0 there. Its integral around any positively oriented circle centred at the origin is 2π, proving nonexactness. On the whole induced annulus boundary, the outer contribution is 2π and the inner contribution is −2π. Their sum is zero, consistent with Stokes because the annulus excludes the singular point.
12. Write A=dx∧dy and B=dz∧dw. Then A∧A=B∧B=0, while A∧B=B∧A because both degrees are two. Thus β∧β=2dx∧dy∧dz∧dw. The square-zero conclusion from graded commutativity applies to odd-degree real forms, not all forms.
26. Questions that reveal the remaining misconceptions
Does dω=0 mean every integral of ω is zero?
No. A closed form can have nonzero integrals over open paths, and a globally nonexact closed form can have nonzero periods over closed loops. Stokes gives zero on a boundary only when the form is defined smoothly over a valid filling region and all orientations are included correctly.
Is a 2-form always a surface-area element?
No. It is an alternating measurement of two tangent vectors. A particular metric and orientation can supply an area form, but an arbitrary 2-form may measure weighted area, projected area or flux and may change sign or vanish.
Why does an integral change sign when orientation is reversed?
Forms retain the ordering of tangent directions through alternation. Reversing the chosen orientation reverses the sign of the corresponding top-dimensional measurement. Unsigned length and area deliberately use different constructions.
What is the practical test of understanding?
Take one integral and evaluate it by two justified routes: direct pullback and a boundary theorem, or direct pullback and a potential. Explain the degree, orientation and domain assumptions in both. Then change one assumption and identify which conclusion no longer follows.
Sources and further study
Donu Arapura, Introduction to Differential Forms, develops concrete calculations, line and surface integrals, exactness and the classical integral theorems. Lorenzo Sadun, Notes on Differential Forms, connects alternating tensors, pullbacks, oriented manifolds, Stokes’ theorem and de Rham cohomology. Brian Conrad, Stokes’ Theorem with Corners, explains the additional care needed for cornered regions.
The four opening differential-geometry guides now connect concrete curves and surfaces to smooth coordinates, tangent and cotangent objects, and oriented integration. They establish a foundation for later work on Riemannian geometry, connections, curvature and bundles; they do not claim that those later topics have already been covered.