A simple and more general approach to Stokes' theorem

TL;DR

This paper proposes a simpler, more general method for deriving Stokes' theorem that avoids complex invariance and boundary orientation concepts, making it more accessible and rigorous.

Contribution

It introduces a novel approach to Stokes' theorem that simplifies its derivation and broadens its applicability compared to traditional methods.

Findings

Provides a more accessible derivation of Stokes' theorem

Eliminates the need for invariance of curl under transformations

Offers a rigorous definition of boundary orientation

Abstract

Oftentimes, Stokes' theorem is derived by using, more or less explicitly, the invariance of the curl of the vector field with respect to translations and rotations. However, this invariance -- which is oftentimes described as the curl being a "physical" vector -- does not seem quite 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 the surface''. Here an apparently simpler and more general approach is suggested.

Figures

Equations

$${\oint }_C\mathbf{F}\cdot d\mathbf{r}={\iint }_S\mathrm{curl}\mathbf{F}\cdot \mathbf{n}dS,$$

$$\oint_{\partial D}P\,du+Q\,dv=\iint_{D}\Big{(}\frac{\partial Q}{\partial u}-\frac{\partial P}{\partial v}\Big{)}du\,dv,$$

$$(u,v)⟼\mathbf{r}(u,v)\in {\mathbb{R}}^3$$

$$\mathbf{G}:U\to {\mathbb{R}}^3$$

$${\oint }_C\mathbf{G}\cdot d\mathbf{r}={\oint }_{\partial D}(\mathbf{G}\cdot {\mathbf{r}}_u)du+(\mathbf{G}\cdot {\mathbf{r}}_v)dv.$$

$$\frac{\partial Q}{\partial u}-\frac{\partial P}{\partial v}={\mathbf{G}}_u\cdot {\mathbf{r}}_v+\mathbf{G}\cdot {\mathbf{r}}_{vu}-{\mathbf{G}}_v\cdot {\mathbf{r}}_u-\mathbf{G}\cdot {\mathbf{r}}_{uv}={\mathbf{G}}_u\cdot {\mathbf{r}}_v-{\mathbf{G}}_v\cdot {\mathbf{r}}_u,$$

$$\oint_{C}\mathbf{G}\cdot d\mathbf{r}=\iint_{D}\big{(}\mathbf{G}_{u}\cdot\mathbf{r}_{v}-\mathbf{G}_{v}\cdot\mathbf{r}_{u}\big{)}du\,dv,$$

$${\oint }_C\mathbf{F}\cdot d\mathbf{r}={\iint }_D(\nabla \times \mathbf{F})\cdot ({\mathbf{r}}_u\times {\mathbf{r}}_v)dudv,$$

$$S:=\mathbf{r}(D),$$

$$\mathbf{G}=\mathbf{F}\circ \mathbf{r}.$$

$$I(u,v)$$

$$=(h_x{x}_u+h_y{y}_u)z_v-(h_x{x}_v+h_y{y}_v)z_u$$

$$J(u,v)$$

$$=h_y(y_u{z}_v-z_u{y}_v)-h_x(z_u{x}_v-x_u{z}_v).$$

$$D_{\delta }:=[0,2\pi ]\times [-\delta ,\delta ]$$

$$\mathbf{r}(u,v)=\big{(}(1+v\cos\tfrac{u}{2})\cos u,(1+v\cos\tfrac{u}{2})\sin u,v\sin\tfrac{u}{2}\big{)}.$$

$$\mathbf{F}(x,y,z)=\Big{(}\frac{-y}{x^{2}+y^{2}},\frac{x}{x^{2}+y^{2}},0\Big{)}$$

$$u\mapsto \tilde{\mathbf{r}}(u):=\mathbf{r}(u,\delta ),$$

$$\hat{D}\ni (u,t)\mapsto \hat{\mathbf{r}}(u,t):=(1-t)\mathbf{r}(u,\delta )+t\mathbf{r}(2\pi -u,\delta ).$$

Full text

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under the map $\mathbf{r}\colon D_{\delta}\to\mathbb{R}^{3}$ given by the formula (cf. e.g. [10])

[TABLE]

The image-curve $C_{\delta}:=\mathbf{r}(\partial D_{\delta})$ is self-intersecting, a reason being that for all $v\in[-\delta,\delta]$ one has $\mathbf{r}(0,v)=\mathbf{r}(2\pi,-v)[=(1+v,0,0)]$ – whereas $(0,v)\in\partial D_{\delta}$, $(2\pi,-v)\in\partial D_{\delta}$, and $(0,v)\neq(2\pi,-v)$. For this reason, the “surface” $S_{\delta}=\mathbf{r}(D_{\delta})$ may be considered self-intersecting as well. If $\delta>2$, then $S_{\delta}$ is self-intersecting in another, apparently more interesting manner; cf. [7, Theorem 1.9]. Indeed, assume that $\delta>2$ and let $u_{1}=u$, $u_{2}=u+\pi$, $v_{1}=-2\cos\tfrac{u}{2}/\cos u$, and $v_{2}=-2\sin\tfrac{u}{2}/\cos u$, where $u$ is a small enough positive real number. Then $\mathbf{r}(u_{1},v_{1})=\mathbf{r}(u_{2},v_{2})[=(-1,-\tan u,-\tan u)\approx(-1,0,0)]$ – whereas $(u_{1},v_{1})\in D_{\delta}$, $(u_{2},v_{2})\in D_{\delta}$, and $(u_{1},v_{1})\neq(u_{2},v_{2})$. Similarly, letting $u_{1}=2\pi-u$, $u_{2}=\pi-u$, $v_{1}=2\cos\tfrac{u}{2}/\cos u$, and $v_{2}=2\sin\tfrac{u}{2}/\cos u$ for small enough $u>0$, we have $\mathbf{r}(u_{1},v_{1})=\mathbf{r}(u_{2},v_{2})[=(-1,\tan u,\tan u)]$.

Letting now, for instance, $\mathbf{G}(u,v):=(u^{2},0,0)$ for all $(u,v)$, we will have $\mathbf{G}(u_{1},v_{1})\neq\mathbf{G}(u_{2},v_{2})$ for all pairs of points $(u_{1},v_{1})$ and $(u_{2},v_{2})$ as above. Therefore, at the self-intersection points $\mathbf{r}(u_{1},v_{1})=\mathbf{r}(u_{2},v_{2})=(-1,\varepsilon\tan u,\varepsilon\tan u)$ with $\varepsilon=\pm 1$ and small enough $u>0$, one will be unable to define a vector field $\mathbf{F}$ so that (6) hold. Thus, formula (5) is not applicable here; of course, formula (1) is not applicable either, because the Möbius strip is not orientable.

In contrast, (4) applies with no problem in this situation; each side of (4) evaluates here to $-160\delta/9$.

The problem of application of Stokes’ theorem to the Möbius strip was previously considered in [3]. The version of the Möbius strip dealt with in [3] differs by a composition of an isometry and a homothety from the apparently more common version described by (8). So, in the subsequent discussion the considerations in [3] will be translated into terms corresponding to (8). Following [3], let us introduce here the vector field $\mathbf{F}$ defined by the formula

[TABLE]

for $(x,y,z)\in\mathbb{R}^{3}$ such that $x^{2}+y^{2}\neq 0$. It is noted in [3] that (i) $\operatorname{curl}\mathbf{F}=\mathbf{0}$ wherever the vector field $\mathbf{F}$ is defined and (ii) the Möbius strip does not intersect the $z$-axis if $\delta<1$. So, $\operatorname{curl}\mathbf{F}=\mathbf{0}$ on the Möbius strip.

It is then concluded in [3] that the surface integral on the right-hand side of (1) is [math]. However, this conclusion is not quite correct, because the Möbius strip is not orientable and hence the normal vector field $\mathbf{n}$ cannot be appropriately defined on the strip, so that the surface integral is technically not defined either and thus has no value.

Also, it is observed in [3] that the boundary (say $B$) of the Möbius strip is the image of the interval $[0,4\pi]$ under the map

[TABLE]

and that the line integral $\oint_{B}\mathbf{F}\cdot d\tilde{\mathbf{r}}$ over the so-parameterized curve $B$ is $4\pi$, which differs from the presumed value [math] of the actually undefined surface integral. This discrepancy is ascribed in [3] to the non-orientability of the Möbius strip. Now one may wonder as follows:

{quoting}

We were told in this note that, as far as (4) and (5) are concerned, there is no need to talk about the orientation of the images $S$ and $C$ of $D$ and $\partial D$ under the map $\mathbf{r}$. If so, then the version (5) of Stokes’ theorem must hold even for the Möbius strip and the vector field $\mathbf{F}$ as in (9). The double integral on the right-hand side of (5) – in contrast to that on the right-hand side of (1) – is well defined, and it must be [math], since $\operatorname{curl}\mathbf{F}=\mathbf{0}$ on the Möbius strip. But the line integral over the boundary $B$ of the Möbius strip was found in [3] to be nonzero. Does this not contradict (5)?

In fact, there is no contradiction here. Recall that the line integral in (5) was to be understood as the one in (3) (with $\mathbf{G}$ as in (6)) and, in turn, the line integral in (3) over the “image-curve” $C$ was defined as the corresponding line integral over the preimage $\partial D$ of $C$, where $\partial D$ is the boundary of the preimage-region $D$. In contrast with these conditions, the line integral in [3] was essentially taken over the segment $[0,4\pi]\times\{\delta\}$, which is of course not the boundary of $D=D_{\delta}=[0,2\pi]\times[-\delta,\delta]$. However, the value of the line integral in (5) computed indeed as the one in (3) with $\mathbf{G}$ as in (6) is [math], which is of course the same as the value of the double integral in (5).

At this point, one may still wonder:

{quoting}

The difficulty with the applicability of Stokes’ theorem to the Möbius strip was ascribed in [3] to the non-orientability. However, the boundary $B$ of the (non-orientable) Möbius strip can also be the boundary of an orientable surface. Then the line integral $\oint_{B}\mathbf{F}\cdot d\tilde{\mathbf{r}}$ as computed in [3] will have the same value $4\pi\neq 0$, whereas the corresponding surface integral on the right-hand side of (1) will still be [math], right? So, it seems we have another contradiction here.

The answer to this concern is as follows. First here, indeed the boundary $B$ of the Möbius strip can also be the boundary of an orientable surface. To see this, it is more convenient to use the interval $[-\pi,3\pi]$ instead of $[0,4\pi]$. More specifically, recall (8) and note that, as $u$ increases from $-\pi$ to $\pi$, the $z$-coordinate $\delta\sin\tfrac{u}{2}$ of the vector $\mathbf{r}(u,\delta)$ increases from its minimal value $-\delta$ to its maximal value $\delta$; and as $u$ increases further from $\pi$ to $3\pi$, the $z$-coordinate of $\mathbf{r}(u,\delta)$ decreases from $\delta$ back to $-\delta$. Moreover, each value of the $z$-coordinate of $\mathbf{r}(u,\delta)$ in the interval $[-\delta,\delta)$ is taken at exactly two points $u_{1}(z)$ and $u_{2}(z)$ in $[-\pi,3\pi]$ such that $-\pi\leq u_{1}(z)<\pi<u_{2}(z)=2\pi-u_{1}(z)\leq 3\pi$. The value $\delta$ of the $z$-coordinate of $\mathbf{r}(u,\delta)$ for $u\in[-\pi,3\pi]$ is taken only at $u=\pi$; accordingly, assume that $u_{1}(\delta)=u_{2}(\delta)=\pi$. Note also that $2\pi-u$ decreases from $3\pi$ to $\pi$ as $u$ increases from $-\pi$ to $\pi$. Connecting now, for each $z\in[-\delta,\delta]$, the points $\mathbf{r}(u_{1}(z),\delta)$ and $\mathbf{r}(u_{2}(z),\delta)$ by (say) a straight line segment, we do obtain an orientable surface, sat $\hat{S}$, whose boundary is the same as the boundary $B$ of the Möbius strip. The surface $\hat{S}$ is the image $\hat{\mathbf{r}}(\hat{D})$ of the rectangle $\hat{D}:=[-\pi,\pi]\times[0,1]$ under the map

[TABLE]

To verify that $\hat{S}$ is orientable, note that the first two coordinates of the vector $\mathbf{c}:=\hat{\mathbf{r}}_{u}(u,t)\times\hat{\mathbf{r}}_{t}(u,t)$ are $\delta\cos\frac{u}{2}\,\sin u$ and $-\delta^{2}\cos^{2}\frac{u}{2}\,\cos u$, respectively, whence $\mathbf{c}\neq\mathbf{0}$ for all $(u,t)$ in the interior of $\hat{D}$, and so, one can define the unit normal vector field on the image of the interior of $\hat{D}$ under the map $\hat{\mathbf{r}}$ by the formula $\mathbf{n}=\mathbf{c}/|\mathbf{c}|$.

Moreover and more importantly, the image $\hat{\mathbf{r}}(\partial\hat{D})$ under this map of the boundary $\partial\hat{D}$ of the rectangle $\hat{D}$ is the boundary $B$ of the Möbius strip. So, (5) will hold for any 3D vector field $\mathbf{F}$ which is smooth enough, in the sense of being continuously differentiable on a neighborhood of the surface $\hat{S}=\hat{\mathbf{r}}(\hat{D})$. For instance, for the linear vector field defined by the formula $\mathbf{F}(x,y,z)=A(x\ y\ z)^{\top}$, where $A=(a_{i,j})_{i,j=1}^{3}$ is a constant $3\times 3$ real matrix and ⊤ denotes the transposition, both the left-hand side and right-hand side of (5) with $D=\hat{D}$ (and $C=\hat{\mathbf{r}}(\partial\hat{D})$) take the same value, $2\pi(2+\delta^{2})(a_{1,2}-a_{2,1})+\pi\delta^{2}(a_{1,3}-a_{3,1})$.

Is the particular field $\mathbf{F}$ given by (9) continuously differentiable on a neighborhood of the surface $\hat{S}$? In other words, is the intersection of the surface $\hat{S}$ with the $z$-axis empty? If that were so, then the right-hand of (5) would be [math] – whereas, as one can recheck, the value of the line integral in (5) is the same nonzero value, $4\pi$, as the one found the other way in [3]. It follows that the surface $\hat{S}$ does intersect the $z$-axis. In fact, it is not hard to check directly that this intersection contains exactly two points, $(0,0,\pm\delta/\sqrt{2})$.

A free bonus of this discussion is the fact that any orientable surface in $\mathbb{R}^{3}$ whose boundary coincides with the boundary of the Möbius strip given by (8) must necessarily intersect the $z$-axis. One may also note here that, for any such orientable surface and for $\mathbf{F}$ as in (9), the surface integral on the right-hand side of (1) will actually not be [math]; rather, it will be undefined.

This discussion is partly illustrated in Fig. 1.