Green’s, Stokes’s and divergence theorems

One boundary can encode information about an interior

Around the circle x²+y²=R², the field F=(−y,x) points in the counterclockwise tangent direction. Direct parametrisation gives circulation 2πR². Its planar curl is the constant 2, and integrating that value over the disk gives the same number: 2 times πR². This agreement is a first example of Green’s theorem. It concerns an oriented line integral and a signed measure of local rotation, rather than the field’s magnitude or the circle’s length.

There are three different settings. Green’s theorem relates a planar boundary curve to a two-dimensional region. Stokes’s theorem relates a surface’s oriented boundary to the flux of a curl through that surface. The divergence theorem relates the outward flux through a closed surface to a three-dimensional solid. Naming the objects first prevents a common mistake: a surface with an edge is suitable for a Stokes boundary, while the divergence theorem needs the complete boundary of a solid.

Lessons 131, 133, 134 and 135 supply the derivative, integration, vector-field and orientation language. In particular, a field arrow is a value at an input, curl and divergence are different local derivatives, and scalar area differs from signed flux. Here the new step is a theorem connecting local derivatives to whole-boundary integrals. A correct local calculation does not by itself establish that the required domain or complete boundary has been supplied.

The article uses explicit, convenient sufficient regularity assumptions, rather than the weakest possible versions of these theorems. Coordinates in the polynomial examples are dimensionless unless a physical interpretation is stated. We prove rectangle, graph and rectangular-box cases, explain the cancellation behind compatible finite assemblies, and state the broader classical theorems with their hypotheses. A drawing or a mesh comparison is an illustration; it is not a proof for every admissible curved region.

Orientation belongs to every boundary component

For a planar region in the usual oriented xy-plane, a positive boundary traversal keeps the region on the traveller’s left. An outer boundary therefore runs counterclockwise. The boundary around an excluded hole runs clockwise: the material region is outside that hole. For a region with finitely many holes, the boundary integral is the sum over the outer boundary and all inner boundaries, each with this induced orientation. Omitting the inner curves changes the mathematical boundary.

Let a positively oriented regular boundary piece be r(t)=(x(t),y(t)). Its unit tangent is T=(x′,y′)/‖r′‖. The outward planar unit normal is the right-hand normal n=(y′,−x′)/‖r′‖. Consequently F·n ds=P dy−Q dx for F=(P,Q). On a clockwise inner circle this normal points into the excluded disk, which is outward from the annular material. Outward does not always mean away from the drawing’s centre.

For a surface in three dimensions, first choose a continuous normal direction on every smooth piece, consistently across the surface. The positive boundary direction is the one induced by that orientation. One precise convention is ν×T=n, where ν is the outward-pointing conormal tangent to the surface and perpendicular to its edge, T is the positive unit boundary tangent, and n is the chosen unit normal. For an upward-oriented horizontal disk this gives counterclockwise traversal when viewed from above.

Reversing a surface orientation reverses the induced traversal of its entire boundary. Reversing both sides of Stokes’s equation therefore preserves the equality with both values negated. Reversing only one side generally breaks it. A closed oriented surface has no boundary curve at all; it is not a surface whose boundary is an unspecified large loop. At piecewise-smooth corners the normal or tangent may have one-sided values, and integrals are assembled over their smooth pieces.

Green’s theorem converts planar circulation into accumulated curl

A useful sufficient setting is a bounded planar region D whose closure is compact and whose full boundary consists of finitely many disjoint, piecewise regular C¹ simple closed curves. The curves enclose a usual piecewise-smooth region, possibly with finitely many holes, and are positively oriented relative to D. Let P and Q have continuous first partial derivatives on an open set containing the closure of D. Then Green’s circulation identity is ∮∂D(P dx+Q dy)=∬D(Qₓ−Pᵧ) dA.

The field must be defined and C¹ throughout a neighbourhood of the region, including its boundary. Being smooth only on the outer boundary, or everywhere except at one interior point, is not the stated hypothesis. Conversely, a hole does not automatically forbid Green’s theorem. If the complete multiply connected region and all its boundary components satisfy the assumptions, the theorem applies to that region. Simple connectivity is a common sufficient special setting, not a universal requirement for this identity.

For F=(P,Q,0), the scalar Qₓ−Pᵧ is the z-component of curl F. The sign corresponds to the standard positive orientation of the xy-plane. The area element dA is positive ordinary area; the sign comes from the curl component and the chosen boundary traversal. Reversing every boundary traversal gives the negative of the displayed line integral, so the right side must also carry a minus sign if the same field and positive dA are retained.

For F=(−y,x), Qₓ−Pᵧ=1−(−1)=2. On a disk of radius R, Green’s theorem gives circulation 2πR². A direct circle calculation is a useful implementation and orientation check, but the theorem also handles boundaries that are inconvenient to parametrise. First inspect the field domain, the full region and its orientation; ease of integration is the final consideration, not a replacement for those checks.

A rectangle proof exposes the cancellations

Take D=[a,b]×[c,d] and traverse its four edges counterclockwise. The horizontal-edge contributions to ∮P dx are ∫_a^b P(x,c) dx−∫_a^b P(x,d) dx. The one-variable fundamental theorem gives P(x,d)−P(x,c)=∫_c^d Pᵧ(x,y) dy. Hence the sum of the horizontal contributions is −∫_a^b ∫_c^d Pᵧ(x,y) dy dx. Continuity on the compact rectangle makes these integrals legitimate.

The vertical-edge contributions to ∮Q dy are ∫_c^d Q(b,y) dy−∫_c^d Q(a,y) dy=∫_c^d ∫_a^b Qₓ(x,y) dx dy. Fubini’s theorem permits the order to be matched because the derivatives are continuous on this compact region. Adding the two expressions proves the circulation identity for the rectangle, with precisely the sign Qₓ−Pᵧ. This is an equality obtained from ordinary one-variable theorems, not a limit inferred from a diagram.

If an admissible region is exactly a finite union of such cells with compatible edges, apply the proved identity to each cell. Every shared internal edge is traversed once in each direction, and the same field has opposite line-integral contributions on those two traversals. They cancel. What remains is the complete external boundary, including the boundaries of any omitted holes. The cell integrals add to the region integral when interiors do not overlap.

An arbitrary curved region is not generally an exact finite union of rectangles. Extending this elementary argument to the full stated theorem requires an appropriate approximation or chart argument and control of the resulting integrals. The classical theorem provides that extension under its hypotheses; the finite-cell picture explains its structure without pretending to prove the extension in one sentence. This distinction also matters in computation: a fine grid can support a numerical estimate while leaving a theorem’s assumptions unchecked.

Planar flux uses divergence, and signed area is another choice

Apply the circulation identity to the auxiliary field (−Q,P). Its boundary integrand is −Q dx+P dy, and its scalar curl is Pₓ+Qᵧ. Thus the outward-flux form of Green’s theorem is ∮∂D(P dy−Q dx)=∬D(Pₓ+Qᵧ) dA. The right-normal convention from the earlier section identifies the left side with ∮∂D F·n ds. Both versions need the same region and field regularity checks.

Work and outward flux answer different questions even for the same field and boundary. On the triangle 0≤y≤x≤1, take F=(x²,xy). Its circulation is ∬D y dA=1/6, while its outward flux is ∬_D 3x dA=1. The operators, integrands and values differ. A line integral containing P dx+Q dy is not changed into flux merely by drawing outward arrows beside its curve.

For an area formula, choose P=−y/2 and Q=x/2. The scalar curl is 1, so Area(D)=(1/2)∮∂D(x dy−y dx) with the full positive boundary. Another valid choice, P=0 and Q=x, gives Area(D)=∮∂D x dy. These expressions agree because their difference has zero integral on an admissible closed boundary; they are not formulas for an open path’s geometric area.

For a polygon, summing the first area expression along straight edges yields one half of the cyclic sum xᵢyᵢ₊₁−yᵢxᵢ₊₁. Its sign records traversal orientation. A positively oriented outer polygon contributes positive area, and a negatively oriented hole subtracts its area. Reversing a single simple polygon changes this signed value, while the underlying ordinary geometric area remains positive. Self-intersecting paths require a separate multiplicity interpretation and are outside the simple-boundary version stated here.

Holes and singularities are different kinds of information

On the punctured plane define F=(−y/(x²+y²), x/(x²+y²)). Away from the origin its planar curl is zero. On a circle of radius R centred at the origin, counterclockwise parametrisation gives F·r′=1, so its circulation is 2π, independent of R. A zero curl at every point where the formula is defined therefore does not make every closed-loop integral zero on this domain.

For the annulus 1≤x²+y²≤4, the field is C¹ on an open neighbourhood of the entire annular region. Green’s theorem applies. The outer circle contributes 2π when traversed counterclockwise, and the inner circle contributes −2π when traversed clockwise. Their sum is zero, matching the integral of zero curl. This is a successful use of the theorem on a region with a hole, with both boundary components included.

The disk enclosed by the outer circle is a different proposed region. The same field is undefined at its centre and has no C¹ extension there, so the displayed disk version of Green’s theorem cannot be applied to it. Declaring that one point has zero area does not repair the hypothesis: behaviour near that point can affect boundary circulation. Nor can one silently discard the inner boundary of an annulus while keeping its area integral.

This example connects the local derivatives of lesson 134 with global path dependence. A locally vanishing derivative and a globally admissible potential are distinct assertions. A simply connected domain is a useful sufficient setting for familiar potential tests under suitable regularity, but the annulus calculation above concerns Green’s identity itself. Use the exact domain and exact question rather than substituting a slogan about holes.

Stokes’s theorem uses an oriented surface and its entire edge

Let S be a compact embedded orientable surface, piecewise C², with its full piecewise regular C¹ boundary. Choose a consistent normal orientation n and the induced boundary traversal described earlier. Let F be C¹ on an open neighbourhood of S. A standard classical form of Stokes’s theorem states ∮_∂S F·dr = ∬_S (curl F)·n dS. The boundary can have several components; the line integral includes all of them with their induced orientations.

The surface need not be a closed surface or the boundary of a three-dimensional solid. A disk, a curved patch and an annular surface can all be admissible. The field must be regular near the surface being used; this formulation does not require it to be regular at every point of some additional enclosed volume. That distinction becomes useful when a convenient spanning surface is substituted for another.

Orientability is a real assumption. A Möbius strip has no globally continuous choice of unit normal, so it does not fit this globally oriented surface-flux statement. A surface that crosses itself also needs additional care about parametrisation and multiplicity; it is not automatically covered by the embedded-surface version above. Piecewise smooth edges and corners are handled by compatible smooth pieces, rather than by assigning a unique normal at every corner.

If S is closed and oriented, its boundary is empty, and the total flux of curl F through S is zero under these hypotheses. This conclusion follows directly from Stokes’s theorem on S. It does not require adding a hypothetical solid in which F is defined. If a proposed field has nonzero total flux through such a surface, it cannot be the curl of a C¹ field defined on a neighbourhood of that entire surface.

A graph calculation reduces Stokes’s theorem to Green’s theorem

For a regular C² parametrised patch r(u,v) over an admissible planar domain D, assume the parametrisation covers the patch once and identifies its boundary correctly. Give the surface the orientation rᵤ×rᵥ. Along its boundary the line integral becomes ∮_∂D(A du+B dv), where A=F(r(u,v))·rᵤ and B=F(r(u,v))·rᵥ. For a graph r(u,v)=(u,v,g(u,v)), regularity and single coverage are especially transparent.

The chain rule and equality of the mixed partial derivatives of r give Bᵤ−Aᵥ=(curl F)(r(u,v))·(rᵤ×rᵥ). To see where the terms go, Bᵤ contains (DF(r)rᵤ)·rᵥ+F(r)·rᵥᵤ, while Aᵥ contains (DF(r)rᵥ)·rᵤ+F(r)·rᵤᵥ. The last terms cancel. Expanding the two remaining scalar products gives the stated scalar triple product, with its sign fixed by the order rᵤ×rᵥ.

Apply Green’s theorem in the parameter domain. Its area integral of Bᵤ−Aᵥ is exactly the oriented surface integral of curl F because n dS=(rᵤ×rᵥ) du dv. This proves the identity for this admissible patch. Normalising the cross product and then also multiplying by its full magnitude is one valid equivalent procedure; including that magnitude twice is an error.

For a surface assembled from finitely many compatible such patches, every shared internal edge receives opposite induced traversals. Its two line integrals cancel, leaving the full exterior edge, including any hole boundaries. A proof for every surface in the broader theorem requires an admissible decomposition or chart argument with the relevant regularity. The cancellation explains the mechanism; an arbitrary sketch of tiles does not establish the general theorem by itself.

A spanning surface may be changed only when it remains admissible

Suppose S₁ and S₂ have the same complete oriented boundary C, and each separately satisfies Stokes’s hypotheses for F. Then each curl flux equals ∮_C F·dr, so the two fluxes are equal. This is a useful way to replace an awkward surface by an easier one. Check the field on both surfaces, check each normal orientation and check that the whole boundary really is C.

For F=(−y/2,x/2,0), curl F=(0,0,1). On the unit circle z=0 traversed counterclockwise from above, direct integration gives π. Both the outward-oriented upper unit hemisphere and the upward-oriented unit disk induce that boundary traversal. The disk flux is immediately π. The curved hemisphere therefore has the same curl flux, although its area is 2π and the field’s normal component is not constant on it.

An outward-oriented lower hemisphere induces the opposite equatorial traversal. Its curl flux is −π, not π. Calling two surfaces spanning surfaces does not make their orientations compatible automatically. Similarly, a surface with an extra hole has an extra boundary curve; it cannot replace a disk while that additional curve is ignored.

A forbidden set can defeat a proposed substitution. The field (−y/(x²+y²),x/(x²+y²),0) is smooth off the z-axis, and its integral around a horizontal circle surrounding that axis is 2π. A flat disk spanning the circle crosses the axis and is inadmissible for Stokes’s theorem. A computation of zero curl off the axis does not justify using that disk. Domain checks must precede a surface shortcut.

The divergence theorem requires the closed boundary of a solid

Let V be a bounded solid with compact closure and an admissible piecewise C¹ boundary, which can include finitely many cavity surfaces. Let F be C¹ on an open neighbourhood of the closure of V. With every boundary component oriented outward from V, the divergence theorem states ∬_∂V F·n dS = ∭_V div F dV. The derivative is a scalar, div F=Pₓ+Qᵧ+R_z for F=(P,Q,R).

Outward refers to the material region V. On the exterior surface it points out into the surrounding space. On a cavity wall it points into the empty cavity. The boundary integral includes both kinds of component. Replacing the cavity normal by a radially outward normal from the drawing’s centre can reverse precisely the contribution needed for the theorem.

A hemisphere without its flat closing disk is not a complete closed boundary. To use the divergence theorem for the corresponding half-ball, include the disk with its outward normal, find the total closed flux, and then subtract that disk contribution to obtain the curved-surface flux. The disk term is allowed to be zero in a particular problem, but that must be calculated rather than assumed.

Flux is signed transport through the surface, not surface area and not the integral of the field magnitude. Positive outward flux means a net outward contribution in the chosen units. Zero total flux permits nonzero inward and outward contributions that cancel. The theorem determines an integral balance; it does not say the field vanishes or that every small boundary patch has zero flux.

A box proof explains conservative face cancellation

For V=[a,b]×[c,d]×[e,f], the two faces perpendicular to the x-axis have outward normals −eₓ and +eₓ. Their combined P contribution is ∫_c^d∫_e^f[P(b,y,z)−P(a,y,z)] dz dy. The fundamental theorem replaces the bracket by ∫_a^b Pₓ(x,y,z) dx. Continuity of the derivative and Fubini’s theorem identify the result with ∭_V Pₓ dV.

Apply the same reasoning to the pairs of faces perpendicular to the y- and z-axes. Adding gives ∭_V(Pₓ+Qᵧ+R_z) dV, which is exactly the outward flux through all six faces. This proves the rectangular-box case. The minus signs at the lower-coordinate faces come from their outward normals, rather than from a discretionary convention about whether coordinates are positive.

For an exact finite union of compatible boxes, each internal face has equal field values and opposite normals for the two adjacent cells. Its two flux integrals cancel, while the volume integrals add. A conservative finite-volume method preserves this cancellation by assigning one common face flux with opposite signs to its neighbouring cells. That algebraic conservation does not by itself establish consistency, convergence or accuracy of an approximate flux. Extending the box proof to a general curved solid also needs an actual approximation or geometric argument.

A varying divergence must be integrated. For F=(x³,y³,z³) on [−1,1]³, div F=3(x²+y²+z²). Its value at the centre is zero, yet its volume integral is 24 and its volume average is 3. Multiplying the centre value by the volume would give zero, a failed one-point quadrature rule in this example. The theorem is exact; an independently chosen numerical quadrature rule can still be inaccurate.

Cavities, closing caps and excision keep missing terms visible

Let G(x,y,z)=(x,y,z)/(x²+y²+z²)^(3/2), defined away from the origin. Direct differentiation gives div G=0 on that domain. On a sphere of radius r, its outward normal component is 1/r², so the outward spherical flux is 4π. Radius cancels between that component and the area 4πr². This direct calculation is legitimate without applying the divergence theorem across the origin.

On the shell a≤‖r‖≤b, with 0<a<b, the field is C¹ on a neighbourhood of the shell. The outer boundary contributes 4π, while the inner boundary’s outward-from-material normal is −r/‖r‖ and contributes −4π. Their sum is zero, matching the volume integral of zero divergence. Excision means removing the singular region and keeping its new boundary contribution; it does not mean erasing a term.

On the full ball centred at the origin, G is not defined at an interior point and fails the stated hypotheses. Its nonzero outer flux is therefore no contradiction. A distributional formulation can encode a point source in a more advanced framework, but no undefined derivative at the origin is being integrated as an ordinary classical function here. The shell identity and direct spherical calculation already resolve this example within the classical setting.

For a different closing-surface example, take the constant field F=e_z on an upper hemisphere of radius R. Its divergence is zero throughout the half-ball. The flat disk closing that half-ball has downward normal and flux −πR², so the curved hemisphere has outward flux πR². Its area is 2πR². Confusing the hemisphere’s area with the flux loses both the directional weighting and the closing-disk contribution.

Units distinguish circulation, volume flux and mass balance

If v is a velocity field with units L/T, then ∮v·dr has units L²/T. Curl v has units 1/T, so its surface integral has the same units L²/T. A velocity flux ∬v·n dS has units L³/T, and div v has units 1/T, matching a volume integral with units L³/T. These checks do not prove a formula, but they detect a mistaken exchange of circulation and three-dimensional flux.

For mass density ρ with units M/L³, define mass-flux density j=ρv, with units M/(L²T). Let s be a mass source per unit volume and time, with units M/(L³T). For a fixed spatial control volume V, a smooth conservation law is d/dt ∭_V ρ dV = ∭_V s dV − ∬_∂V j·n dS. Positive outward mass flux reduces the mass inside when there is no source.

If ρ and j have sufficient space-time regularity, the volume is fixed and differentiation may be passed under the integral, the divergence theorem rewrites the balance as ∭_V(∂ρ/∂t+div j−s) dV=0. If this equality holds for every sufficiently small admissible fixed volume and the integrand is continuous, then ∂ρ/∂t+div(ρv)=s pointwise: a nonzero continuous value would keep one sign on a small neighbourhood, giving a nonzero integral. A balance on one volume alone does not imply that local equation.

A moving control surface introduces its own velocity. The relative transport across it depends on v minus the boundary velocity, together with the change of the moving domain. The fixed-volume formula above cannot simply be copied with a moving V(t) while leaving every term unchanged. A material volume moves with the fluid and can retain constant mass while a fixed spatial volume loses mass. The worked example makes this distinction without assuming a transport theorem that has not been derived here.

Read the shared structure without dropping the assumptions

All three identities connect a complete oriented boundary with an accumulated derivative on an interior object. The dimension of that object and the type of integral determine the formula. In the plane, Green’s circulation form uses Qₓ−Pᵧ, and its flux form uses Pₓ+Qᵧ. For a surface in space, Stokes uses curl; for a solid, the divergence theorem uses div. Similar names do not make these operators interchangeable.

A practical checklist is: identify the exact region, surface or solid; state the field’s domain and regularity on a neighbourhood of it; include every boundary component; choose the induced tangents or outward normals; and only then compute. If a convenient replacement surface or closing cap is introduced, repeat the domain and orientation checks for that object. A singularity or omitted component can defeat an otherwise correct symbolic calculation.

The earlier identities curl(grad φ)=0 and div(curl A)=0 are local differential statements under appropriate smoothness. They do not erase global domain information. In particular, the punctured-plane example warns against concluding that a curl-free field has a global potential on every domain. Conversely, Stokes’s theorem shows that a field which is a curl on a neighbourhood of a closed oriented surface must have zero total flux through it.

Differential forms and a generalised Stokes theorem provide a later language for unifying these patterns. This article does not claim to derive that general theorem or all of its manifold hypotheses. Its proved rectangle, graph and box cases, and its stated classical versions, have a narrower explicit scope. Likewise, exact agreement on finitely many numerical examples checks those examples rather than establishing an identity on every admissible domain.

Use the explorer to test a claim before trusting a shortcut

Begin each explorer comparison by naming its theorem, mathematical object, field domain and orientation. Trace all boundary components before comparing a boundary total with an interior total. A hole or cavity has its own visible contribution. A closing disk must be listed even when its eventual contribution is zero. A panel showing a singular field on a filled disk or ball should mark that theorem shortcut as inadmissible.

For the polynomial examples, compare exact fractions or coefficients of π wherever possible. Direct edge or face calculations provide a different route to the same number, while a derivative integral supplies the theorem side. Reverse an allowed orientation and check the associated sign change. Keep ordinary area positive while tracking signed line integrals and flux separately.

In a three-dimensional drawing, projected circles can look elliptical and projected arrows can distort lengths and angles. Treat these diagrams as orientation and domain schematics, not measured geometric proofs. The accompanying formulas and labelled normals carry the quantitative meaning. A diagram of a unit normal and a parametrisation cross product should distinguish their roles so that the surface-area factor is included exactly once.

Finally, inspect the hypotheses when two totals disagree. A wrong sign, omitted inner boundary, missing cap, singularity or inaccurate quadrature has a different remedy. Agreement is also limited evidence: finite model cases and computational checks support the displayed implementation, while the theorem’s reasoning and assumptions explain why its full mathematical conclusion holds.

Worked examples

Try it yourself

1. Which theorem directly connects (a) planar circulation around a region, (b) circulation along a surface’s full edge, and (c) outward flux through a solid’s full boundary to an interior derivative?

   Show the answer and reasoning

   (a) Green’s circulation theorem; (b) Stokes’s theorem; (c) the divergence theorem.

   The interior objects are respectively a planar region, an oriented surface and a solid. The derivatives are scalar planar curl, curl flux and divergence.

2. Viewed from above in the usual xy-plane, how should the two boundary circles of a planar annulus be traversed positively?

   Show the answer and reasoning

   Outer circle counterclockwise; inner circle clockwise.

   The annular material stays on the traveller’s left on both curves. Traversing both counterclockwise does not give the induced full boundary.

3. A positive planar boundary piece has derivative (x′,y′). What is its outward normal times arc-length element, and what is the flux integrand for F=(P,Q)?

   Show the answer and reasoning

   n ds=(y′,−x′) dt, so F·n ds=(P y′−Q x′) dt=P dy−Q dx.

   The outward normal is to the right of a positive tangent. Multiplying its unit form by ds cancels the speed.

4. A field is C¹ near the sphere bounding a ball, but is undefined at its centre. Does the stated divergence theorem automatically apply to the ball?

   Show the answer and reasoning

   No.

   The stated theorem requires C¹ regularity on an open neighbourhood of the closure of the entire solid, including interior points. Boundary regularity alone is insufficient.

5. Why can the origin not simply be ignored when applying Green’s theorem to the vortex field on a filled disk?

   Show the answer and reasoning

   The field is undefined and not C¹ near the origin, violating the theorem’s hypothesis.

   A set of zero area can carry a singularity whose surrounding circulation is nonzero. The ordinary theorem cannot be repaired by discarding its regularity condition.

6. Must a region be simply connected for the full-boundary Green identity used here?

   Show the answer and reasoning

   No; admissible regions with finitely many holes are allowed.

   The field must be regular on a neighbourhood of the whole remaining region, and every inner boundary must be included with its induced orientation.

7. An upward-oriented surface and its induced boundary both give integral 7. What happens if the surface normal and entire boundary traversal are both reversed?

   Show the answer and reasoning

   Both integrals become −7, and Stokes’s equality remains valid.

   An oriented flux and an oriented line integral both change sign on reversal. Reversing only one side would generally break the equality.

8. If A is C¹ on a neighbourhood of a compact closed oriented surface S, what does Stokes’s theorem say about ∬_S curl A·n dS?

   Show the answer and reasoning

   It equals zero.

   The surface’s boundary is empty, so its boundary line integral is zero. This conclusion uses regularity near S and does not require A to be defined inside an additional solid.

9. Which assumption prevents the global oriented-flux form of Stokes’s theorem here from being applied directly to a Möbius strip?

   Show the answer and reasoning

   Orientability: no globally continuous unit normal exists on the Möbius strip.

   Local normal choices cannot be patched consistently around the whole strip. The displayed global flux statement requires such a consistent orientation.

10. For F=(−y/(x²+y²),x/(x²+y²),0), may a circle surrounding the z-axis be filled with a horizontal disk to infer zero circulation from zero curl?

    Show the answer and reasoning

    No; that disk crosses the field’s forbidden z-axis.

    F is not defined on the entire proposed disk, so Stokes’s hypotheses fail there. Direct circle integration gives 2π for counterclockwise traversal.

11. For F=e_z on the upper half-ball of radius R, the curved hemisphere has unknown flux H. Write the complete closed-surface flux equation.

    Show the answer and reasoning

    H−πR²=0, hence H=πR².

    The downward-facing bottom disk contributes −πR². The divergence is zero, so the sum over both boundary pieces is zero.

12. What is the outward-from-solid normal on the inner sphere r=a of the shell a≤r≤b?

    Show the answer and reasoning

    −e_r, pointing into the cavity.

    Leaving the shell across its inner wall means moving toward its empty interior. Using +e_r would give the sphere’s own outward normal, which is opposite to the shell boundary orientation there.

13. For F=(x³,y³,z³) on [−1,1]³, why does div F(0,0,0) times volume fail to give flux? State the correct flux.

    Show the answer and reasoning

    Divergence varies across the cube; its central value is zero while its volume average is 3. The correct flux is 24.

    The integral of 3(x²+y²+z²) is 24. The theorem uses that integral, not an unproved one-point quadrature substitution.

14. A positively traversed simple polygon gives (1/2)∮(x dy−y dx)=12. What are the reversed signed expression and the polygon’s ordinary area?

    Show the answer and reasoning

    The reversed expression is −12; ordinary area is 12.

    Traversal affects the signed integral, not the underlying geometric region. A hole requires its own negative contribution in the complete positive boundary.

15. For r(u,v)=(u,v,u+2v), can the integrand (curl F)·(rᵤ×rᵥ) be multiplied by √6 again to account for surface area?

    Show the answer and reasoning

    No; rᵤ×rᵥ already combines the normal direction and area factor.

    Either use n=(rᵤ×rᵥ)/√6 with dS=√6 du dv, or use the cross product directly with du dv. Combining both procedures double-counts the factor.

16. If density has units M/L³ and velocity L/T, what are the units of mass-flux density ρv, its divergence and its total outward surface flux?

    Show the answer and reasoning

    M/(L²T), M/(L³T) and M/T, respectively.

    A spatial derivative contributes 1/L; surface integration contributes L². The resulting total flux can balance a time derivative of mass.

17. In the smooth radial-expansion example, can a fixed ball lose mass while a ball moving with the fluid keeps constant mass?

    Show the answer and reasoning

    Yes; they are different control volumes.

    Density falls as e^(−3αt), while the material ball’s volume grows as e^(3αt). A fixed ball lacks that volume growth. Applying a fixed-domain formula to the moving domain without its motion term would be incorrect.

18. An explorer matches boundary and interior totals in 50 examples. Does that establish the theorem for every admissible field and region?

    Show the answer and reasoning

    No.

    Finite checks test the displayed examples and implementation. General validity relies on mathematical reasoning and the theorem’s hypotheses; agreement also cannot legitimise a singular or incomplete proposed domain.

What to carry forward

• Identify the planar region, oriented surface or solid before choosing an integral theorem.
• The full boundary includes every hole, cavity and closing surface with its induced orientation.
• Green’s circulation form integrates Qₓ−Pᵧ; its outward-flux form integrates Pₓ+Qᵧ.
• For a positive planar boundary, the outward normal points to the tangent’s right and flux is P dy−Q dx.
• Stokes’s theorem equates boundary circulation with curl flux through an admissible oriented surface.
• A replacement spanning surface must preserve the full oriented boundary and remain inside the field’s regular domain.
• The divergence theorem uses outward flux through a solid’s complete closed boundary and the volume integral of divergence.
• An inner cavity wall’s outward-from-material normal points into the cavity.
• A singular point cannot be ignored merely because it has zero area or volume.
• Rectangle, graph and box proofs reveal exact fundamental-theorem and internal-boundary cancellation mechanisms.
• A nonconstant divergence must be integrated; its value at one convenient point need not equal its average.
• Fixed-volume mass balance distinguishes mass-flux density, outward transport, sources and the time derivative of retained mass.
• Moving material volumes require their motion to be accounted for; their balance is not obtained by blindly copying a fixed-domain formula.
• Exact model cases, diagrams and numerical agreement support specific checks; they do not replace a theorem’s proof or its assumptions.