Multivariate Calculus - Bruce E. Shapiro

LECTURE 22. LINE INTEGRALS 185
where the last step follows from the chain rule. By the fundamental theorem of
calculus,
∫
∫ b
∇f(r) · r = D v f(r(t))dt (22.27)
C
=
a
∫ b
a
df(r(t))
dt (22.28)
dt
= f(r(b)) − f(r(a)) (22.29)
= f(B) − f(A) (22.30)
Theorem 22.2 Path Independence Theorem. Let D be an open connected set,
and suppose that F : D ↦→ R 2 3 is continous on D. Then F is a gradient field, i.e,
there exists some scalar function f such that F = ∇f, if and only if ∫ C
F(r) · dr is
independent of path, i.e.,
∫
F(r) · dr = f(B) − f(A)
for all smooth paths C in D,
with A = r(a) and B = r(b).
C
C = {r(t), a ≤ t ≤ b} ⊂ S
Proof. Suppse that F = ∇f, i.e., that F is a gradient field. Then
∫
∫
F(r) · dr = ∇f(r) · dr
C
C
= f(B) − f(A)
by the fundamental theorem of calculus for line integrals. Hence the line integral
depends only its path.
Now suppose that the line integral depends only on its path. To complete the
proof we need to show that there exists some function f such that F = ∇f. Let
A = (x a , y a , z a ) and let B = (x, y, z) denote the end points of C. Then
∫
f(x, y, z) − f(A) = f(B) − f(A) =
Define the components of F as
C
F · dr =
∫ (x,y,z)
(x a,y a,z a)
F · dr
F(x, y, z) = (M(x, y, z), N(x, y, z), P (x, y, z)) (22.31)
Since the domain is connected we can pick a point (x 1 , y, z) (close to (x, y, z)) so
that
f(x, y, z) − f(A) =
∫ (x1 ,y,z)
(x a,y a,z a)
F · dr +
∫ (x,y,z)
(x 1 ,y,z)
F · dr
Math 250, Fall 2006 Revised December 6, 2006.