A simple and more general approach to Stokes' theorem
MMathematical Assoc. of America Mathematics Magazine 88:1 January 29, 2019 1:45 a.m. arxiv.tex page 1
VOL. 88, NO. 1, FEBRUARY 2015 A simple and more general approach toStokes’ theorem
Iosif Pinelis
Michigan Technological [email protected]
Summary
Oftentimes, Stokes’ theorem is derived by using, more or less explicitly, the in-variance of the curl of the vector field with respect to translations and rotations. However, thisinvariance – which is oftentimes described as the curl being a “physical” vector – does not seemquite easy to verify, especially for undergraduate students. An even bigger problem with Stokes’theorem is to rigorously define such notions as “the boundary curve remains to the left of thesurface”. Here an apparently simpler and more general approach is suggested.
In calculus texts (see e.g. [ , § XII.6] or [ , § S be an oriented smooth enough surface in R bounded by asimple closed smooth enough curve C with positive orientation. Then for any smoothenough vector field F (cid:73) C F · d r = (cid:90) (cid:90) S curl F · n dS, (1)where curl F is the curl of the vector field F and n is a continuous field of unit normalvectors on S . The positive orientation of C is “defined” there as the condition for thesurface S to remain on the left (with respect to n ) when C is traced out. In such ele-mentary texts, it is of course impossible to rigorously define such notions as “remainson the left”; cf. e.g. [ , Theorems 5.5 and 5.9], [ , § XXIII.4–XXIII.6], or [ , § ] – possibly with singularities, whichneed to be considered to cover the case of even such simple surfaces as the (convexhulls of) triangles or rectangles; cf. e.g. [ , § XXIII]. Moreover, for the surface integralon the right-hand side of identity (1) to make sense, one has to ensure that the curl andthe unit normal vector field have appropriate invariance properties with respect to thechoice of an atlas for the manifold S .Green’s theorem is the special and comparatively very simple case of Stokes’ the-orem corresponding to the additional condition that the surface S is flat and thus maybe assumed to coincide with a region D ⊂ R . Informally, Green’s theorem may bestated as follows:Let D be a compact subset of R with a smooth enough boundary ∂D , and let P and Q be real continuously differentiable functions on an open neighborhood of D .Then (cid:73) ∂D P du + Q dv = (cid:90) (cid:90) D (cid:16) ∂Q∂u − ∂P∂v (cid:17) du dv, (2)with the line integral (cid:72) ∂D P du + Q dv appropriately defined.One way to make this statement rigorous (and general enough) is to assume that theboundary ∂D of D is a rectifiable Jordan curve Γ oriented so as to make the windingnumber of Γ equal (rather than − ) with respect to any point in the interior of Γ .These conditions on D and ∂D will be assumed in the sequel. a r X i v : . [ m a t h . HO ] J a n athematical Assoc. of America Mathematics Magazine 88:1 January 29, 2019 1:45 a.m. arxiv.tex page 2 MATHEMATICS MAGAZINE
Even though Green’s theorem is only a special case of Stokes’, it is not easy to provethe just mentioned rigorous “Jordan curve” version of it, or even to show that the wind-ing number can be consistently defined; see e.g. [ ] and references there to [
9, 13, 8, 1 ].The best practical approach to teaching Green’s theorem in a calculus course wouldprobably be to restrict the consideration to simple regions, of the form { ( x, y ) : a ≤ x ≤ b, g ( x ) ≤ y ≤ g ( x ) } or { ( x, y ) : a ≤ y ≤ b, g ( y ) ≤ x ≤ g ( y ) } , and possi-bly simple combinations thereof.Anyway, in this note we shall show that, once an appropriate version of Green’stheorem is established, it is then very easy and painless to derive versions of Stokes’theorem, which are even more general, in some aspects, than the conventional one.Indeed, for some open neighborhood U of D and ( u, v ) ∈ U , let ( u, v ) (cid:55)−→ r ( u, v ) ∈ R be a twice continuously differentiable map of U into R . Let C := r ( ∂D ) be theimage of the boundary ∂D of D under the map r . Let G : U → R by a continuously differentiable vector field on U . Since d r = r u du + r v dv , we canwrite/define the line integral (cid:72) C G · d r as follows: (cid:73) C G · d r = (cid:73) ∂D ( G · r u ) du + ( G · r v ) dv. (3)Here, as usual, the subscripts u , v , x , . . . denote the partial differentiation with respectto u, v, x, . . . . One may say that formula (3) reduces the “line integral” (cid:72) C G · d r over the “image-curve” C (which does not have to be a simple curve, without self-intersections) to the well-defined line integral (cid:72) ∂D ( G · r u ) du + ( G · r v ) dv over theproperly oriented rectifiable Jordan curve Γ = ∂D in the domain U of the map r .Letting now P = G · r u and Q = G · r v , we have ∂Q∂u − ∂P∂v = G u · r v + G · r vu − G v · r u − G · r uv = G u · r v − G v · r u , since r vu = r uv . So, Green’s theorem (2) immediately yields the following form ofStokes’ theorem: (cid:73) C G · d r = (cid:90) (cid:90) D (cid:0) G u · r v − G v · r u (cid:1) du dv, (4)which, similarly to (1), reduces a line integral to a double one.This may be compared with the more conventional form of Stokes’ theorem: (cid:73) C F · d r = (cid:90) (cid:90) D ( ∇ × F ) · ( r u × r v ) du dv, (5)where F is a continuously differentiable 3D vector field defined on a neighborhood ofthe “surface” S := r ( D ) , and the line integral may be understood as the one in (4) (or, equivalently, in (3)) with G = F ◦ r . (6) athematical Assoc. of America Mathematics Magazine 88:1 January 29, 2019 1:45 a.m. arxiv.tex page 3 VOL. 88, NO. 1, FEBRUARY 2015 It is not hard to deduce (5) from (4). Indeed, let ( i , j , k ) be any orthonormalbasis in R , in which the cross products ∇ × F and r u × r v can be computed ac-cording to the standard determinant formulas. Let (cid:104) x ( u, v ) , y ( u, v ) , z ( u, v ) (cid:105) and (cid:104) f ( x, y, z ) , g ( x, y, z ) , h ( x, y, z ) (cid:105) be, respectively, the triples of the coordinates of r ( u, v ) and F ( x, y, z ) in the basis ( i , j , k ) . Since both sides of (5) are linear in F ,without loss of generality f = g = 0 , and then the integrands on the right-hand sidesof (4) and (5) become, respectively, I ( u, v ) := ( h x x u + h y y u + h z z u ) z v − ( h x x v + h y y v + h z z v ) z u = ( h x x u + h y y u ) z v − ( h x x v + h y y v ) z u and J ( u, v ) := ( h y i − h x j ) · (cid:0) ( y u z v − z u y v ) i + ( z u x v − x u z v ) j + ( x u y v − y u z v ) k (cid:1) = h y ( y u z v − z u y v ) − h x ( z u x v − x u z v ) . Now it is quite easy to see that I ( u, v ) = J ( u, v ) , which completes the derivation of(5) from (4).The main distinction of (4) and (5) from (1) is that the double integrals in (4) and(5) are taken over the flat preimage-region D ⊂ R – whereas the “image-surface” S = r ( D ) plays no role in (4) and (5), except that the vector field F has to be definedon some neighborhood of S . Therefore, as far as (4) and (5) are concerned, there is noneed to talk about any properties of the image-surface S except for it being a subsetof R . In particular, there is no need to talk about the orientation of S or to use suchhard to define terms as “remains on the left” or to care about the mentioned invarianceproperties of the curl and the unit normal vector field. Moreover, the image-surface S = r ( D ) , as well as the image-curve C = r ( ∂D ) , may be self-intersecting, and S does not have to be a manifold at all. As for the line integrals in (4) and (5), they areover the image-curve C only in form, as they immediately reduce to line integrals overthe pre-image curve ∂D , according to (3). This reduction of integration in the image-space R to that in the flat preimage-space R is natural; it is even unavoidable – forhow else would one actually compute the line and surface integrals in the conventionalform (1) of Stokes’ theorem? (Of course, to do the integration in the preimage-space,we still need a proper orientation of the boundary ∂D of the flat preimage domain D , as was already pointed out in our discussion concerning a general “Jordan curve”version of Green’s theorem and its elementary version for simple regions.)Next, let us consider (4) versus (5). We saw that (4) is very easy to obtain, moduloGreen’s theorem. Arguably, (4) is also easier to remember than (5).An important advantage of (4) is that it is more general than (5). Indeed, on the onehand, (5) follows from (4); on the other hand, in (4) the composition-factorization(6) of the map G is not needed. That is, for (4) one does not need the implica-tion r ( u , v ) = r ( u , v ) = ⇒ G ( u , v ) = G ( u , v ) (which necessarily followsfrom (6)).Plus, one does not need the notion of the curl for (4). On the other hand, (4) by itselfwill not help when proving that a vector field is conservative if its curl is zero. Yet, asshown above, (5) is rather easy to get from (4).We thus have three versions of Stokes’ theorem, corresponding to the three formu-las: • the conventional version (1), which requires comparatively most stringent and evenhard to define conditions, including an appropriate orientability of the image-surfacetogether with the image-curve, and the invariance of the curl and the unit normalvector field; athematical Assoc. of America Mathematics Magazine 88:1 January 29, 2019 1:45 a.m. arxiv.tex page 4 MATHEMATICS MAGAZINE • the “computational”, less conventional version (5), which is not concerned with ori-entability of the images, but needs the composition-factorization condition (6); • (4), which is the most general of the three versions and, at the same time, easiest toobtain – but not useful when, say, the curl is known to be zero.The above discussion is illustrated by Example 1.
Consider the M¨obius strip S δ (of “radius” and half-width δ > ), whichis the image r ( D δ ) of the rectangle D δ := [0 , π ] × [ − δ, δ ] (7)under the map r : D δ → R given by the formula (cf. e.g. [ ]) r ( u, v ) = (cid:0) (1 + v cos u ) cos u, (1 + v cos u ) sin u, v sin u (cid:1) . (8)The image-curve C δ := r ( ∂D δ ) is self-intersecting, a reason being that for all v ∈ [ − δ, δ ] one has r (0 , v ) = r (2 π, − v )[= (1 + v, , – whereas (0 , v ) ∈ ∂D δ , (2 π, − v ) ∈ ∂D δ , and (0 , v ) (cid:54) = (2 π, − v ) . For this reason, the “surface” S δ = r ( D δ ) may be considered self-intersecting as well. If δ > , then S δ is self-intersecting inanother, apparently more interesting manner; cf. [ , Theorem 1.9]. Indeed, assume that δ > and let u = u , u = u + π , v = − u / cos u , and v = − u / cos u ,where u is a small enough positive real number. Then r ( u , v ) = r ( u , v )[=( − , − tan u, − tan u ) ≈ ( − , , – whereas ( u , v ) ∈ D δ , ( u , v ) ∈ D δ , and ( u , v ) (cid:54) = ( u , v ) . Similarly, letting u = 2 π − u , u = π − u , v = 2 cos u / cos u ,and v = 2 sin u / cos u for small enough u > , we have r ( u , v ) = r ( u , v )[=( − , tan u, tan u )] .Letting now, for instance, G ( u, v ) := ( u , , for all ( u, v ) , we will have G ( u , v ) (cid:54) = G ( u , v ) for all pairs of points ( u , v ) and ( u , v ) as above. There-fore, at the self-intersection points r ( u , v ) = r ( u , v ) = ( − , ε tan u, ε tan u ) with ε = ± and small enough u > , one will be unable to define a vector field F so that (6) hold. Thus, formula (5) is not applicable here; of course, formula (1) is notapplicable either, because the M¨obius strip is not orientable.In contrast, (4) applies with no problem in this situation; each side of (4) evaluateshere to − δ/ .The problem of application of Stokes’ theorem to the M¨obius strip was previouslyconsidered in [ ]. The version of the M¨obius strip dealt with in [ ] differs by a com-position of an isometry and a homothety from the apparently more common versiondescribed by (8). So, in the subsequent discussion the considerations in [ ] will betranslated into terms corresponding to (8). Following [ ], let us introduce here thevector field F defined by the formula F ( x, y, z ) = (cid:16) − yx + y , xx + y , (cid:17) (9)for ( x, y, z ) ∈ R such that x + y (cid:54) = 0 . It is noted in [ ] that (i) curl F = wher-ever the vector field F is defined and (ii) the M¨obius strip does not intersect the z -axisif δ < . So, curl F = on the M¨obius strip.It is then concluded in [ ] that the surface integral on the right-hand side of (1)is . However, this conclusion is not quite correct, because the M¨obius strip is notorientable and hence the normal vector field n cannot be appropriately defined on thestrip, so that the surface integral is technically not defined either and thus has no value. athematical Assoc. of America Mathematics Magazine 88:1 January 29, 2019 1:45 a.m. arxiv.tex page 5 VOL. 88, NO. 1, FEBRUARY 2015 Also, it is observed in [ ] that the boundary (say B ) of the M¨obius strip is the imageof the interval [0 , π ] under the map u (cid:55)→ ˜ r ( u ) := r ( u, δ ) , (10)and that the line integral (cid:72) B F · d ˜ r over the so-parameterized curve B is π , whichdiffers from the presumed value of the actually undefined surface integral. This dis-crepancy is ascribed in [ ] to the non-orientability of the M¨obius strip. Now one maywonder as follows:We were told in this note that, as far as (4) and (5) are concerned, there is noneed to talk about the orientation of the images S and C of D and ∂D underthe map r . If so, then the version (5) of Stokes’ theorem must hold even forthe M¨obius strip and the vector field F as in (9). The double integral on theright-hand side of (5) – in contrast to that on the right-hand side of (1) – is welldefined, and it must be , since curl F = on the M¨obius strip. But the lineintegral over the boundary B of the M¨obius strip was found in [ ] to be nonzero.Does this not contradict (5)?In fact, there is no contradiction here. Recall that the line integral in (5) was tobe understood as the one in (3) (cid:0) with G as in (6) (cid:1) and, in turn, the line integral in(3) over the “image-curve” C was defined as the corresponding line integral over thepreimage ∂D of C , where ∂D is the boundary of the preimage-region D . In contrastwith these conditions, the line integral in [ ] was essentially taken over the segment [0 , π ] × { δ } , which is of course not the boundary of D = D δ = [0 , π ] × [ − δ, δ ] .However, the value of the line integral in (5) computed indeed as the one in (3) with G as in (6) is , which is of course the same as the value of the double integral in (5).At this point, one may still wonder:The difficulty with the applicability of Stokes’ theorem to the M¨obius stripwas ascribed in [ ] to the non-orientability. However, the boundary B of the(non-orientable) M¨obius strip can also be the boundary of an orientable surface.Then the line integral (cid:72) B F · d ˜ r as computed in [ ] will have the same value π (cid:54) = 0 , whereas the corresponding surface integral on the right-hand side of(1) will still be , right? So, it seems we have another contradiction here.The answer to this concern is as follows. First here, indeed the boundary B ofthe M¨obius strip can also be the boundary of an orientable surface. To see this, itis more convenient to use the interval [ − π, π ] instead of [0 , π ] . More specifically,recall (8) and note that, as u increases from − π to π , the z -coordinate δ sin u of thevector r ( u, δ ) increases from its minimal value − δ to its maximal value δ ; and as u increases further from π to π , the z -coordinate of r ( u, δ ) decreases from δ backto − δ . Moreover, each value of the z -coordinate of r ( u, δ ) in the interval [ − δ, δ ) istaken at exactly two points u ( z ) and u ( z ) in [ − π, π ] such that − π ≤ u ( z ) <π < u ( z ) = 2 π − u ( z ) ≤ π . The value δ of the z -coordinate of r ( u, δ ) for u ∈ [ − π, π ] is taken only at u = π ; accordingly, assume that u ( δ ) = u ( δ ) = π . Notealso that π − u decreases from π to π as u increases from − π to π . Connectingnow, for each z ∈ [ − δ, δ ] , the points r ( u ( z ) , δ ) and r ( u ( z ) , δ ) by (say) a straightline segment, we do obtain an orientable surface, sat ˆ S , whose boundary is the same asthe boundary B of the M¨obius strip. The surface ˆ S is the image ˆ r ( ˆ D ) of the rectangle ˆ D := [ − π, π ] × [0 , under the map ˆ D (cid:51) ( u, t ) (cid:55)→ ˆ r ( u, t ) := (1 − t ) r ( u, δ ) + t r (2 π − u, δ ) . athematical Assoc. of America Mathematics Magazine 88:1 January 29, 2019 1:45 a.m. arxiv.tex page 6 MATHEMATICS MAGAZINE
To verify that ˆ S is orientable, note that the first two coordinates of the vector c :=ˆ r u ( u, t ) × ˆ r t ( u, t ) are δ cos u sin u and − δ cos u cos u , respectively, whence c (cid:54) = for all ( u, t ) in the interior of ˆ D , and so, one can define the unit normal vectorfield on the image of the interior of ˆ D under the map ˆ r by the formula n = c / | c | .Moreover and more importantly, the image ˆ r ( ∂ ˆ D ) under this map of the boundary ∂ ˆ D of the rectangle ˆ D is the boundary B of the M¨obius strip. So, (5) will hold forany 3D vector field F which is smooth enough, in the sense of being continuouslydifferentiable on a neighborhood of the surface ˆ S = ˆ r ( ˆ D ) . For instance, for the linearvector field defined by the formula F ( x, y, z ) = A ( x y z ) (cid:62) , where A = ( a i,j ) i,j =1 is a constant × real matrix and (cid:62) denotes the transposition, both the left-handside and right-hand side of (5) with D = ˆ D (cid:0) and C = ˆ r ( ∂ ˆ D ) (cid:1) take the same value, π (2 + δ )( a , − a , ) + πδ ( a , − a , ) .Is the particular field F given by (9) continuously differentiable on a neighborhoodof the surface ˆ S ? In other words, is the intersection of the surface ˆ S with the z -axisempty? If that were so, then the right-hand of (5) would be – whereas, as one canrecheck, the value of the line integral in (5) is the same nonzero value, π , as the onefound the other way in [ ]. It follows that the surface ˆ S does intersect the z -axis. Infact, it is not hard to check directly that this intersection contains exactly two points, (0 , , ± δ/ √ .A free bonus of this discussion is the fact that any orientable surface in R whoseboundary coincides with the boundary of the M¨obius strip given by (8) must necessar-ily intersect the z -axis. One may also note here that, for any such orientable surfaceand for F as in (9), the surface integral on the right-hand side of (1) will actually notbe ; rather, it will be undefined.This discussion is partly illustrated in Fig. 1. Figure 1
Left panel: the M ¨obius strip given by (8) (with δ = 3 / ), stretched verticallyby the factor of . Middle panel: the same M ¨obius strip, now stretched vertically by thefactor of . Right panel: the oriented surface ˆ S with the same boundary as the M ¨obiusstrip, stretched vertically by the factor of , for better viewing; shown here are also thetwo points of intersection of the surface ˆ S with the z -axis. REFERENCES
1. T. M. Apostol.
Mathematical analysis: a modern approach to advanced calculus . Addison-Wesley PublishingCompany, Inc., Reading, Mass., 1957.athematical Assoc. of America Mathematics Magazine 88:1 January 29, 2019 1:45 a.m. arxiv.tex page 7
VOL. 88, NO. 1, FEBRUARY 2015
2. H. Federer.
Geometric measure theory . Die Grundlehren der mathematischen Wissenschaften, Band 153.Springer-Verlag New York Inc., New York, 1969.3. J.-P. Gabardo. The M¨obius strip and Stokes theorem, 2006. http://ms.mcmaster.ca/gabardo/moebius.pdf.4. S. Lang.
Calculus of Several Variables . Springer, 1987. Third ed.5. S. Lang.
Real and functional analysis , volume 142 of
Graduate Texts in Mathematics . Springer-Verlag, NewYork, third edition, 1993.6. MathOverflow. Proof of Green’s formula for rectifiable Jordan curves, 2018.https://mathoverflow.net/questions/307713/proof-of-greens-formula-for-rectifiable-jordan-curves.7. I. Pinelis. A simpler R realization of the M¨obius strip. https://arxiv.org/abs/1808.03955 , 2018.8. D. H. Potts. A note on Green’s theorem. J. London Math. Soc. , 26:302–304, 1951.9. J. Ridder. ¨Uber den Greenschen Satz in der Ebene.
Nieuw Arch. Wiskunde (2) , 21:28–32, 1941.10. G. E. Schwarz. The dark side of the Moebius strip.
Amer. Math. Monthly , 97(10):890–897, 1990.11. M. Spivak.
Calculus on manifolds. A modern approach to classical theorems of advanced calculus . W. A.Benjamin, Inc., New York-Amsterdam, 1965.12. J. Stewart.
Calculus . Cengage, 2016. Eighth ed.13. S. Verblunsky. On Green’s formula.
J. London Math. Soc. , 24:146–148, 1949.
IOSIF PINELIS (MR Author ID: 208523) has been with Michigan Technological Universitysince 1992. He has also held visiting positions at the University of Illinois, Urbana–Champaign;(MR Author ID: 208523) has been with Michigan Technological Universitysince 1992. He has also held visiting positions at the University of Illinois, Urbana–Champaign;