488 CHAPTER 18. DIFFERENTIAL FORMS

and so

0 =∫(â,b̂)

fn(r−1 ◦r̂ (t)

)(∣∣det(D(r−1◦r̂

)(t))∣∣−det

(D(r−1◦r̂

)(t)))

dt.

Using the monotone convergence theorem, it follows that

0 =∫(â,b̂)

(∣∣det(D(r−1◦r̂

)(t))∣∣−det

(D(r−1◦r̂

)(t)))

dt.

and so det(D(r−1◦r̂

)(t))≥ 0 a.e. This is the condition given in the above proposition.

Thus this condition which applies to Lipschitz functions follows from a statement thatd(r−1◦r̂,Ω,u

)= 1. Conversely, if the condition det

(D(r−1◦r̂

)(t))≥ 0 a.e. it will follow

from the above formula that d(r−1◦r̂,Ω,u

)= 1 since it cannot equal −1.

18.8 Examples of Stoke’s TheoremHere the attempt is made to tie this formalism to the usual things studied in calculus.

18.8.1 Fundamental Theorem of CalculusFirst let k = p = 1. What does Stoke’s theorem say? In this case, dω is a 1 form and so ω

should be a 0 form which is just a function. ω = a(x) , dω = a′ (x)dx. Then if r : [a,b]→Ris C1, ∫

rdω ≡

∫r

a′ (x)dx≡∫ b

aa′ (r (u))

drdu

du = a(r (b))−a(r (a))∫∂r

ω =∫

∂ra(x) = (−1)2 a(r (b))− (−1)2 a(r (a))

which is the same thing. It is just the fundamental theorem of calculus essentially.

18.8.2 Line Integrals for Conservative FieldsWhat if p = 3,k = 1? This is similar. You need to have ω a zero form. Thus ω =a(x1, ...,xm) . Then dω = ∑i axidxi. Letting r : [a,b]→ Rm,∫

rdω =

∫ b

a∇a ·r′du = a(r (b))−a(r (a))

∫∂r

ω = (−1)(1+1) a(r (b))− (−1)(1+1) a(r (a))

which is the same thing. This is the case of a conservative vector field, the potential functionbeing a.

18.8.3 Green’s TheoremNext ω be a 1 form and let p = k = 2 so dω is a 2 form and ω is a 1 form. Say

ω = P(x,y)dx+Q(x,y)dy

dω = Py (x,y)dy∧dx+Qxdx∧dy