Multivariate Calculus - Bruce E. Shapiro

186 LECTURE 22. LINE INTEGRALS
The first integral does not depend on x; in fact the only place that x appears on the
right hand side of the equation is in the upper limit of the second integral. But in
the second integral the values of y and z are constant, hence F · dr = M(x, y, z)dx.
∂f(x, y, z)
∂x
= ∂
∂x
∫ x
x 1
M(u, y, z)du = M(x, y, z) (22.32)
where the last equality follows from the fundamental theorem of calculus. Similarly
we can pick a point (x, y 1 , z) such that
f(x, y, z) − f(A) =
∫ (x,y1 ,z)
(x a,y a,z a)
F · dr +
∫ (x,y,z)
(x,y 1 ,z)
F · dr
Now the first integral does not depend on y, and the second integral has x and z
fixed, so that F · dr = N(x, y, z)dy in the second integrand and
∂f(x, y, z)
∂y
= ∂ ∂y
∫ y
y 1
N(x, u, z)du = N(x, y, z) (22.33)
Finally, we can pick a point (x, y, z 1 ) such that
f(x, y, z) − f(A) =
∫ (x,y,z1 )
(x a,y a,z a)
F · dr +
∫ (x,y,z)
(x,y,z 1 )
F · dr
Now the first integral does not depend on z, and the second integral has x and y
fixed, so that F · dr = P (x, y, z)zy in the second integrand and
∂f(x, y, z)
∂z
= ∂ ∂z
∫ z
z 1
P (x, y, u)du = P (x, y, z) (22.34)
Combining equations 22.31, 22.32, 22.33, and 22.34, we have
( )
∂f
F = (M, N, P ) =
∂x , ∂f
∂y , ∂f
= ∇f (22.35)
∂z
Thus F = ∇f, i.e, F is a gradient field.
Theorem 22.3 A vector field F is a gradient field if and only if
∮
F · dr = 0
for every closed curved C.
C
Proof. F is a gradient field if and only if it is path independent. Pick any two points
on the closed curve C, and call them A and B. Let C 1 be one path from A to B
along C, and let C 2 be the path from A to B along the other half of C. Then
∮
F · dr =
C
∫
F · dr −
C 1
∫
F · dr
C 2
Revised December 6, 2006. Math 250, Fall 2006