Kenneth Kuttler

600 CHAPTER 30. STOKES AND GREEN’S THEOREMS

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]dA

=∫

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

=∫

U[(F ◦R)u ·Rv − (F ◦R)v ·Ru]dA

the last step holding by equality of mixed partial derivatives, a result of the assumption thatR is C2. Now by Lemma 30.3.2, this equals∫

U(Ru ×Rv) · (∇×F )dA

=∫

U∇×F ·(Ru ×Rv)dA =

∫S

∇×F ·ndS

because dS = |(Ru ×Rv)|dA and n= (Ru×Rv)|(Ru×Rv)| . Thus

(Ru ×Rv)dA =(Ru ×Rv)

|(Ru ×Rv)||(Ru ×Rv)|dA = ndS.

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 if you use the Lebesgueintegral.

30.3.1 The Normal and the OrientationStoke’s theorem as just presented needs no apology. However, it is helpful in applicationsto have some additional geometric insight.

600 CHAPTER 30. STOKES AND GREEN’S THEOREMSProof: Letting C be an oriented part of JU having parametrization, r (t) = (u(t) ,v(t))for ¢ € [a, B] and letting R(C) denote the oriented part of 0S corresponding to C, [ ROE:dR=“B/ F(R(u(t),v(t))) (Ryu! (t) + Ryv' (0) dtBF(R(u(t),v(t))) Ru (u(t), v())u! ode" B+f F(R(u(t),v(1))) Ry (u(t) ,v())v' (de[\(PeR)-Ry,(FoR): Ry)-dr.Since this holds for each such piece of OU, it follows| F.dR= [ ((FoR)-R,,(FoR)-R,)-dr.as auBy the assumption that the conclusion of Green’s theorem holds for U, this equals[\(PeR): Ry), (FeR)-Ry),ldA= I [((FoR),:-R,+(FoR)-Ry—(FoR)-Rw—(FoR),:R,|dA= | (FoR),-Ry— (FoR), RyjdAUthe last step holding by equality of mixed partial derivatives, a result of the assumption thatR is C?. Now by Lemma 30.3.2, this equalsI (R, x R,)-(V x F)dA- [Y xP (Rix R))dA = [Vx F nasU SsRuxRybecause dS = |(R, x R,)|dA and n = ao. Thus(R, x R,)R, x R,) dA = ————( A= TR xR.)|(R, x R,)|\dA = nds.This proves Stoke’s theorem. IfNote 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 if you use the Lebesgueintegral.30.3.1 The Normal and the OrientationStoke’s theorem as just presented needs no apology. However, it is helpful in applicationsto have some additional geometric insight.