El teorema de Stokes

El teorema de Stokes relaciona integrals de superfıcie amb integrals sobre corbes a $\ensuremath{\mathbbm{ R}^3}$. En una versió més moderna [Spi], el teorema de Stokes conté com a casos particulars ``tots" els teoremes integrals, desde la regla de Barrow, passant pel teorema de Green, el teorema de Stokes clàssic i el teorema de Gauss, fins a teoremes corresponents en espais de dimensió superior. Nosaltres, però, ens limitarem al teorema de Stokes clàssic.

Sigui S una superfıcie amb parametrització $\Phi:D\longrightarrow\ensuremath{\mathbbm{ R}^3}$, de classe C², injectiva a quasi tot arreu, i tal que D té per vora una corba tancada $\partial$D de classe C¹ a trosos i que no es talla a ella mateixa. Sigui $\vec{F}$ un camp vectorial de classe C¹. Aleshores

$$\int_{{S^+}}^{}$$($$\vec{\nabla}$$×$$\vec{F}$$) ^. d$$\vec{s}$$ = $$\oint_{{C^+}}^{}$$$$\vec{F}$$ ^. d$$\vec{l}$$,

on C = $\Phi$($\partial$D) és la vora de S, C⁺ és el sentit induıt al recórrer $\partial$D en sentit antihorari amb els paràmetres (u, v) i S⁺ està orientada segons $\vec{n}$ = $\vec{T}_{u}^{}$×$\vec{T}_{v}^{}$. $\Box$

Els elements gràfics del teorema es troben a la figura 28.

Figura 28: Per il-l ^. lustrar el teorema de Stokes

La demostració és relativament senzilla, utilitza el teorema de Green, i l'única dificultat està en les orientacions. Sigui $\vec{F}$ = (P, Q, R). Tenim que

$$\int_{{S^+}}^{} \vec{\nabla} \vec{F} \vec{s} \int \int_{D}^{} \left[\vphantom{ \left(\frac{\partial R}{\partial y}-\frac{\parti... ... v} -\frac{\partial y}{\partial v}\frac{\partial z}{\partial u}\right) }\right. \left(\vphantom{\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}}\right. {\frac{{\partial R}}{{\partial y}}} {\frac{{\partial Q}}{{\partial z}}} \left.\vphantom{\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}}\right) \left(\vphantom{\frac{\partial y}{\partial u}\frac{\partial z}{\partial v} -\frac{\partial y}{\partial v}\frac{\partial z}{\partial u}}\right. {\frac{{\partial y}}{{\partial u}}} {\frac{{\partial z}}{{\partial v}}} {\frac{{\partial y}}{{\partial v}}} {\frac{{\partial z}}{{\partial u}}} \left.\vphantom{\frac{\partial y}{\partial u}\frac{\partial z}{\partial v} -\frac{\partial y}{\partial v}\frac{\partial z}{\partial u}}\right)$$ =

$$\left(\vphantom{\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}}\right. {\frac{{\partial P}}{{\partial z}}} {\frac{{\partial R}}{{\partial x}}} \left.\vphantom{\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}}\right) \left(\vphantom{\frac{\partial z}{\partial u}\frac{\partial x}{\partial v} -\frac{\partial z}{\partial v}\frac{\partial x}{\partial u}}\right. {\frac{{\partial z}}{{\partial u}}} {\frac{{\partial x}}{{\partial v}}} {\frac{{\partial z}}{{\partial v}}} {\frac{{\partial x}}{{\partial u}}} \left.\vphantom{\frac{\partial z}{\partial u}\frac{\partial x}{\partial v} -\frac{\partial z}{\partial v}\frac{\partial x}{\partial u}}\right)$$ +

$$\left.\vphantom{ \left(\frac{\partial Q}{\partial x}-\frac{\parti... ... v} -\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}\right) }\right. \left(\vphantom{\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}}\right. {\frac{{\partial Q}}{{\partial x}}} {\frac{{\partial P}}{{\partial y}}} \left.\vphantom{\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}}\right) \left(\vphantom{\frac{\partial x}{\partial u}\frac{\partial y}{\partial v} -\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}}\right. {\frac{{\partial x}}{{\partial u}}} {\frac{{\partial y}}{{\partial v}}} {\frac{{\partial x}}{{\partial v}}} {\frac{{\partial y}}{{\partial u}}} \left.\vphantom{\frac{\partial x}{\partial u}\frac{\partial y}{\partial v} -\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}}\right) \left.\vphantom{ \left(\frac{\partial Q}{\partial x}-\frac{\parti... ... v} -\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}\right) }\right]$$ +

Si aconseguim veure que

$$\oint_{{C^+}}^{} \int \int_{D}^{} \left[\vphantom{ \frac{\partial P}{\partial z} \left(\frac{\par... ...} -\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}\right) }\right. {\frac{{\partial P}}{{\partial z}}} \left(\vphantom{\frac{\partial z}{\partial u}\frac{\partial x}{\partial v} -\frac{\partial z}{\partial v}\frac{\partial x}{\partial u}}\right. {\frac{{\partial z}}{{\partial u}}} {\frac{{\partial x}}{{\partial v}}} {\frac{{\partial z}}{{\partial v}}} {\frac{{\partial x}}{{\partial u}}} \left.\vphantom{\frac{\partial z}{\partial u}\frac{\partial x}{\partial v} -\frac{\partial z}{\partial v}\frac{\partial x}{\partial u}}\right) {\frac{{\partial P}}{{\partial y}}} \left(\vphantom{\frac{\partial x}{\partial u}\frac{\partial y}{\partial v} -\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}}\right. {\frac{{\partial x}}{{\partial u}}} {\frac{{\partial y}}{{\partial v}}} {\frac{{\partial x}}{{\partial v}}} {\frac{{\partial y}}{{\partial u}}} \left.\vphantom{\frac{\partial x}{\partial u}\frac{\partial y}{\partial v} -\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}}\right) \left.\vphantom{ \frac{\partial P}{\partial z} \left(\frac{\par... ...} -\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}\right) }\right]$$ (1)

i anàlogament pels termes proporcionals a Q i R, haurem demostrat el teorema de Stokes. Sigui

g(u, v) = P(x(u, v), y(u, v), z(u, v)).

Llavors

$${\frac{{\partial}}{{\partial u}}} \left(\vphantom{g\frac{\partial x}{\partial v}}\right. {\frac{{\partial x}}{{\partial v}}} \left.\vphantom{g\frac{\partial x}{\partial v}}\right) {\frac{{\partial}}{{\partial v}}} \left(\vphantom{g\frac{\partial x}{\partial u}}\right. {\frac{{\partial x}}{{\partial u}}} \left.\vphantom{g\frac{\partial x}{\partial u}}\right)$$

$${\frac{{\partial g}}{{\partial u}}} {\frac{{\partial x}}{{\partial v}}} {\frac{{\partial^2 x}}{{\partial u\partial v}}} {\frac{{\partial g}}{{\partial v}}} {\frac{{\partial x}}{{\partial u}}} {\frac{{\partial^2 x}}{{\partial v\partial u}}}$$ =

Si ara emprem que $\Phi$ de classe C² i, en particular, té derivades segones creuades iguals, queda

$${\frac{{\partial}}{{\partial u}}} \left(\vphantom{g\frac{\partial x}{\partial v}}\right. {\frac{{\partial x}}{{\partial v}}} \left.\vphantom{g\frac{\partial x}{\partial v}}\right) {\frac{{\partial}}{{\partial v}}} \left(\vphantom{g\frac{\partial x}{\partial u}}\right. {\frac{{\partial x}}{{\partial u}}} \left.\vphantom{g\frac{\partial x}{\partial u}}\right)$$

$${\frac{{\partial g}}{{\partial u}}} {\frac{{\partial x}}{{\partial v}}} {\frac{{\partial g}}{{\partial v}}} {\frac{{\partial x}}{{\partial u}}}$$ =

$$\left(\vphantom{ \frac{\partial P}{\partial x}\frac{\partial x}{\... ...partial u}+ \frac{\partial P}{\partial z}\frac{\partial z}{\partial u} }\right. {\frac{{\partial P}}{{\partial x}}} {\frac{{\partial x}}{{\partial u}}} {\frac{{\partial P}}{{\partial y}}} {\frac{{\partial y}}{{\partial u}}} {\frac{{\partial P}}{{\partial z}}} {\frac{{\partial z}}{{\partial u}}} \left.\vphantom{ \frac{\partial P}{\partial x}\frac{\partial x}{\... ...partial u}+ \frac{\partial P}{\partial z}\frac{\partial z}{\partial u} }\right) {\frac{{\partial x}}{{\partial v}}}$$ =

$$\left(\vphantom{ \frac{\partial P}{\partial x}\frac{\partial x}{\... ...partial v}+ \frac{\partial P}{\partial z}\frac{\partial z}{\partial v} }\right. {\frac{{\partial P}}{{\partial x}}} {\frac{{\partial x}}{{\partial v}}} {\frac{{\partial P}}{{\partial y}}} {\frac{{\partial y}}{{\partial v}}} {\frac{{\partial P}}{{\partial z}}} {\frac{{\partial z}}{{\partial v}}} \left.\vphantom{ \frac{\partial P}{\partial x}\frac{\partial x}{\... ...partial v}+ \frac{\partial P}{\partial z}\frac{\partial z}{\partial v} }\right) {\frac{{\partial x}}{{\partial u}}}$$ -

$${\frac{{\partial P}}{{\partial y}}} \left(\vphantom{\frac{\partial x}{\partial u}\frac{\partial y}{\partial v} -\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}}\right. {\frac{{\partial x}}{{\partial u}}} {\frac{{\partial y}}{{\partial v}}} {\frac{{\partial x}}{{\partial v}}} {\frac{{\partial y}}{{\partial u}}} \left.\vphantom{\frac{\partial x}{\partial u}\frac{\partial y}{\partial v} -\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}}\right) {\frac{{\partial P}}{{\partial z}}} \left(\vphantom{\frac{\partial z}{\partial u}\frac{\partial x}{\partial v} -\frac{\partial z}{\partial v}\frac{\partial x}{\partial u}}\right. {\frac{{\partial z}}{{\partial u}}} {\frac{{\partial x}}{{\partial v}}} {\frac{{\partial z}}{{\partial v}}} {\frac{{\partial x}}{{\partial u}}} \left.\vphantom{\frac{\partial z}{\partial u}\frac{\partial x}{\partial v} -\frac{\partial z}{\partial v}\frac{\partial x}{\partial u}}\right)$$ =

Per tant, el membre dret de (1) és

$$\int$$$$\int_{D}^{}$$$$\left[\vphantom{ \frac{\partial}{\partial u}\left(g\frac{\partia... ...\frac{\partial}{\partial v}\left(g\frac{\partial x}{\partial u}\right)}\right.$$$${\frac{{\partial}}{{\partial u}}}$$$$\left(\vphantom{g\frac{\partial x}{\partial v}}\right.$$g$${\frac{{\partial x}}{{\partial v}}}$$$$\left.\vphantom{g\frac{\partial x}{\partial v}}\right)$$ - $${\frac{{\partial}}{{\partial v}}}$$$$\left(\vphantom{g\frac{\partial x}{\partial u}}\right.$$g$${\frac{{\partial x}}{{\partial u}}}$$$$\left.\vphantom{g\frac{\partial x}{\partial u}}\right)$$$$\left.\vphantom{ \frac{\partial}{\partial u}\left(g\frac{\partia... ...\frac{\partial}{\partial v}\left(g\frac{\partial x}{\partial u}\right)}\right]$$ dudv,

i, aplicant el teorema de Green al camp vectorial (g${\frac{{\partial x}}{{\partial u}}}$, g${\frac{{\partial x}}{{\partial v}}}$), cosa que podem fer ja que aquest camp és C¹ a D, serà

$$\oint_{{\partial D}}^{} {\frac{{\partial x}}{{\partial u}}} {\frac{{\partial x}}{{\partial v}}}$$ (2)

amb $\partial$D en sentit antihorari.

Anem ara a transformar el membre esquerra de (2), treient profit del fet que C⁺ és la imatge de $\partial$D per $\Phi$. Sigui $\alpha:[a,b]\longrightarrow\ensuremath{\mathbbm{ R}^2}$ una parametrització de $\partial$D. Llavors $\Phi$o$\alpha$ serà una parametrització de C⁺, $\Phi\circ\alpha:[a,b]\longrightarrow\ensuremath{\mathbbm{ R}^3}$ amb components, tenint en compte que $\alpha$(t) = (u(t), v(t)),

($$\Phi$$o$$\alpha$$)(t) = (x(u(t), v(t)), y(u(t), v(t)), z(u(t), v(t))),

i amb vector velocitat

($$\Phi$$o$$\alpha$$)'(t) = $$\left(\vphantom{ \frac{\partial x}{\partial u}\frac{\text{d}u}{... ...l x}{\partial v}\frac{\text{d}v}{\text{d}t} , \mbox{altres components}}\right.$$$${\frac{{\partial x}}{{\partial u}}}$$$${\frac{{\text{d}u}}{{\text{d}t}}}$$ + $${\frac{{\partial x}}{{\partial v}}}$$$${\frac{{\text{d}v}}{{\text{d}t}}}$$,altres components$$\left.\vphantom{ \frac{\partial x}{\partial u}\frac{\text{d}u}{... ...l x}{\partial v}\frac{\text{d}v}{\text{d}t} , \mbox{altres components}}\right)$$.

Llavors

$$\oint_{{C^+}}^{}$$P dx = $$\int_{a}^{b}$$P(x(u(t), v(t)), y(u(t), v(t)), z(u(t), v(t)))$$\left(\vphantom{ \frac{\partial x}{\partial u}\frac{\text{d}u}{\text{d}t} + \frac{\partial x}{\partial v}\frac{\text{d}v}{\text{d}t} }\right.$$$${\frac{{\partial x}}{{\partial u}}}$$$${\frac{{\text{d}u}}{{\text{d}t}}}$$ + $${\frac{{\partial x}}{{\partial v}}}$$$${\frac{{\text{d}v}}{{\text{d}t}}}$$$$\left.\vphantom{ \frac{\partial x}{\partial u}\frac{\text{d}u}{\text{d}t} + \frac{\partial x}{\partial v}\frac{\text{d}v}{\text{d}t} }\right)$$ dt.

Però això és precissament l'expressió paramètrica de la integral sobre $\partial$D (en el pla) del camp vectorial

$$\left(\vphantom{ P(x(u,v),y(u,v),z(u,v))\frac{\partial x}{\partial u}, P(x(u,v),y(u,v),z(u,v))\frac{\partial x}{\partial v} }\right. {\frac{{\partial x}}{{\partial u}}} {\frac{{\partial x}}{{\partial v}}} \left.\vphantom{ P(x(u,v),y(u,v),z(u,v))\frac{\partial x}{\partial u}, P(x(u,v),y(u,v),z(u,v))\frac{\partial x}{\partial v} }\right)$$

$$\left(\vphantom{ g\frac{\partial x}{\partial u},g\frac{\partial x}{\partial v} }\right. {\frac{{\partial x}}{{\partial u}}} {\frac{{\partial x}}{{\partial v}}} \left.\vphantom{ g\frac{\partial x}{\partial u},g\frac{\partial x}{\partial v} }\right)$$ =

que és el que tenim a (2). Queda per tant demostrat el que voliem.

$\Diamond$ A l'enunciat del teorema de Stokes està implıcit que la superfıcie S ha de ser orientable.

$\Diamond$ El teorema de Stokes també és vàlid si D, i, per tant, S, tenen forats. L'orientació és llavors segons s'havia discutit per al teorema de Green amb forats.

$\Diamond$ Si S és una superfıcie tancada no té vora i, aleshores

$$\oint_{S}^{}$$($$\vec{\nabla}$$×$$\vec{F}$$) ^. d$$\vec{s}$$ = 0.

$\Diamond$ En alguns casos la imatge de $\partial$D per $\Phi$ és quelcom més que la vora de S. En concret, $\Phi$($\partial$D) pot recòrrer la vora, moure's sobre l'interior de S i tornar a la vora (vegeu l'exemple 20). En aquest cas, el teorema de Stokes segueix éssent vàlid, ja que el que es fa de més es fa dues vegades, una en cada sentit, i en total s'anul-l ^. la.

$\Diamond$ Si no tenim tercera dimensió, de manera que S = D, C⁺ = $\partial$D, x = u, y = v, el teorema de Stokes es redueix al teorema de Green.

Exemple 20   Sigui el cilindre sense tapadores

$$\left\{\vphantom{\begin{array}{l} x=\cos\phi\\ y=\sin\phi\\ z=z \end{array} }\right.$$$$\begin{array}{l} x=\cos\phi\\ y=\sin\phi\\ z=z \end{array}$$$$\begin{array}{l} \phi\in[0,2\pi)\\ z\in[0,1] \end{array}$$

els elements gràfics del qual es troben a la figura 29.

Figura 29: Un cilindre obert per dalt i per baix

El tros de cilindre té una vora formada per dues circumferències. La imatge de $\partial$D conté, però, a més, dues rectes, que es mostren a la figura 30.

Figura 30: Imatge de la vora de D

Les dues rectes són, però, la mateixa, recorreguda en els dos sentits, i qualsevol integral d'un camp vectorial s'anul-l ^. la quan es consideren les dues rectes. Cal notar que les dues circumferències tenen orientacions ben definides, fixades per l'orientació antihorària de $\partial$D. El vector normal que fixa l'orientació de S és, amb aquesta elecció dels paràmetres,

$$\vec{n}$$ = $$\vec{T}_{\phi}^{}$$×$$\vec{T}_{z}^{}$$ = (- sin$$\phi$$, cos$$\phi$$, 0)×(0, 0, 1) = (cos$$\phi$$, sin$$\phi$$, 0),

que correspon a una normal ``cap a fora", tal com es pot veure, per exemple, dibuixant-lo per a $\phi$ = 0.