Kenneth Kuttler

290 CHAPTER 12. THEOREMS INVOLVING LINE INTEGRALS

Theorem 12.2.4 (Stoke’s Theorem) Let U be any region in R2 for which the con-clusion of Green’s theorem holds and let R∈C2

(U ,R3

)be a one to one function satisfying

|(Ru×Rv)(u,v)| ̸= 0 for all (u,v) ∈U and let S denote the surface

S≡ {R(u,v) : (u,v) ∈U} , ∂S≡ {R(u,v) : (u,v) ∈ ∂U}

where the orientation on ∂S is consistent with the counter clockwise orientation on ∂U (Uis on the left as you walk around ∂U). Then for F a C1 vector field defined near S,∫

∂SF ·dR=

∫S

curl(F ) ·ndσ

where n is the normal to S defined by n≡ Ru×Rv|Ru×Rv| .

Proof: Letting C be an oriented part of ∂U having parametrization, r (t)≡ (u(t) ,v(t))for t ∈ [α,β ] and letting R(C) denote the oriented part of ∂S corresponding to C,

∫R(C)F ·

dR=

=∫

F (R(u(t) ,v(t))) ·(Ruu′ (t)+Rvv′ (t)

)dt

=∫

F (R(u(t) ,v(t)))Ru (u(t) ,v(t))u′ (t)dt

+∫

F (R(u(t) ,v(t)))Rv (u(t) ,v(t))v′ (t)dt

=∫

C((F ◦R) ·Ru,(F ◦R) ·Rv) ·dr.

Since this holds for each such piece of ∂U , it follows∫∂SF ·d R=

∫∂U

((F ◦R) ·Ru,(F ◦R) ·Rv) ·dr.

By the assumption that the conclusion of Green’s theorem holds for U , this equals∫U[((F ◦R) ·Rv)u− ((F ◦R) ·Ru)v]dm2

=∫

U[(F ◦R)u ·Rv +(F ◦R) ·Rvu− (F ◦R) ·Ruv− (F ◦R)v ·Ru]dm2

=∫

U[(F ◦R)u ·Rv− (F ◦R)v ·Ru]dm2

the last step holding by equality of mixed partial derivatives, a result of the assumption thatR is C2. Now by Lemma 12.2.2, this equals∫

U(Ru×Rv) · (∇×F )dm2 =

∫U

∇×F ·(Ru×Rv)dm2 =∫

S∇×F ·ndσ

because dσ = |(Ru×Rv)|dm2 and n= (Ru×Rv)|(Ru×Rv)| . Thus

(Ru×Rv)dm2 =(Ru×Rv)

|(Ru×Rv)||(Ru×Rv)|dm2 = ndσ .

This proves Stoke’s theorem. ■Note that there is no mention made in the final result that R is C2. Therefore, it is not

surprising that versions of this theorem are valid in which this assumption is not present. Itis possible to obtain extremely general versions of Stoke’s theorem.

290 CHAPTER 12. THEOREMS INVOLVING LINE INTEGRALSTheorem 12.2.4 (Stoke’s Theorem) Let U be any region in R? for which the con-clusion of Green’s theorem holds and let R.€ C? (U, R3) be aone to one function satisfying|(R, x R,) (u,v)| 40 for all (u,v) € U and let S denote the surfaceS={R(u,v): (u,v) €U}, ODS={R(u,v) : (u,v) € dU}where the orientation on OS is consistent with the counter clockwise orientation on OU (Uis on the left as you walk around OU). Then for F aC! vector field defined near S,F-dR= [cust (F)-ndoJOS JS— R,xRy—~ |RyxR,y|*where n is the normal to S defined by nProof: Letting C be an oriented part of JU having parametrization, r (t) = (u(t) ,v(t))for t € [@, B] and letting Re (C) denote the oriented part of 0S corresponding to C, fc) F’-dR=BI F(R(u(t),v(t)))- (Ryu! (t) + Ryv' (0) dtB[. FRUW) 90) Buu) vO) HatB+f F(R(u(t),v(t))) Ry (u(t) ,v(t)) (de_ [Pe R): Ry (Fo R)-Ry)-dr.Since this holds for each such piece of dU, it follows[ FaR= |) (FoR) Ry.(FeR):Ry)-dr.By the assumption that the conclusion of Green’s theorem holds for U, this equals[Pe R)- Ry), — (Fe R)- Ry),[Po By B+ (FoR): Ruy — (FoR): Ruy — (FR), Raldmy= | (oR), -R,-(FoR),-Ryldmthe last step holding by equality of mixed partial derivatives, a result of the assumption thatRis C’. Now by Lemma 12.2.2, this equalsJ (Bux By) (Vx P)dm = | Vx F(R, x Ry)dm = [Vx F-ndoU U Ss_ — (Rix Ry)because do = |(R, x R,)|dmz andn = I(R.xR,|° Thus(R, x R,)R, x R,) dm = ————TR Ry)|(R, x R,)|dmz = ndo.This proves Stoke’s theorem. MHNote that there is no mention made in the final result that R is C2. Therefore, it is notsurprising that versions of this theorem are valid in which this assumption is not present. Itis possible to obtain extremely general versions of Stoke’s theorem.