Page 109 - Vector Analysis

§4.1 The Line Integrals 105

Line integrals of vector fields

We recall that a vector field is a vector-valued function whose domain and co-domain are
subsets of identical Euclidean space Rn.

Let C be a simple regular curve parameterized by γ : I Ñ Rn, and F : C Ñ Rn be a
vector field. The line integral of F along C in the direction of γ (or the oriented

line integral of F along C) is defined as the line integral of the scalar function F ¨ T
along C, where T is the unit tangent of C given by

T = γ1 ˝ γ´1 on C. (4.3)
}γ 1}Rn

Given another parametrization ϕ : Ir Ñ Rn of C such that (ϕ1 ˝ ϕ´1) ¨ (γ 1 ˝ γ´1) ą 0 (that is,
the orientation of C given by ϕ and γ are the same), using the chain rule we obtain that

γ1 = d ˝ ϕ´1 ˝ γ)(t) = (ϕ 1 ˝ ϕ´1 ˝ γ)(t)(ϕ´1 ˝ γ)1(t) . (4.4)
(ϕ

dt

Since ϕ´1 ˝ γ : I Ñ Ir, (ϕ´1 ˝ γ)1 is a scalar function; thus (4.4) and the fact that (ϕ1 ˝ ϕ´1) ¨

(γ 1 ˝ γ´1) ą 0 imply that γ 1 ˝ γ´1 = c(ϕ1 ˝ ϕ´1) for some positive scalar function c : C Ñ R.

Therefore,

ϕ1 ˝ ϕ´1 = γ1 ˝ γ´1 on C. (4.5)
}ϕ 1}Rn }γ 1}Rn

In other words, the tangent vector T is well-defined on C; thus the line integral of F along

C in the direction of the parametrization γ is a well-defined quantity.

Suppose that I = [a, b]. Using (4.1), we find that

ż F ¨ T ds = żb ˝ γ)(t) ¨ γ 1(t) }γ 1(t)}Rn dt = żb ˝ γ)(t) ¨ γ 1(t) dt .
(F }γ 1(t)}Rn (F
C
a a

Let r : Ir Ñ Rn be an arc-length parametrization of C such that (r1 ˝ r´1) ¨ (γ 1 ˝ γ´1) ą 0
on C. Then (4.5) implies that T = dr . In terms of notation, we also write T ds as dr; thus

ds

żż żb

F ¨ dr = F ¨ T ds = (F ˝ γ)(t) ¨ γ 1(t) dt .

CC a

Remark 4.26 (The interpretation of line integrals of vector fields). Consider the work done
by moving an object along a smooth curve C parameterized by γ : I Ñ Rn with a continuous
variable force F : C Ñ Rn from γ(a) to γ(b) (that is, in the direction of the parametrization