Section 14.3: The Fundamental Theorem for Line Integrals
Theorem: (Fundamental Theorem for Line Integrals)
~
Let C be a smooth curve defined by the vector function R(t)
for a ≤ t ≤ b. Let F~ be a
continuous, conservative vector field with scalar potential f . Then
Z
~ = f (R(b))
~
~
F~ · dR
− f (R(a)).
C
Example: Show that the vector field F~ (x, y) = h2xyZ3 , 3x2 y 2 i is conservative and find a scalar
~ where C is any smooth path
F~ · dR,
potential for F~ . Then evaluate the line integral
C
connecting (0, 0) to (1, 1).
The partial derivatives of the component functions are
∂P
∂Q
= 6xy 2 =
.
∂y
∂x
Thus, F~ is a conservative vector field. Now let
Z
Z
f (x, y) = fx (x, y)dx = 2xy 3 dx = x2 y 3 + K(y).
Then let
fy (x, y) = 3x2 y 2
3x2 y 2 + K 0 (y) = 3x2 y 2
K(y) = C.
Thus, a scalar potential for F~ is f (x, y) = x2 y 3 . By the Fundamental Theorem for Line
Integrals,
Z
~ = f (1, 1) − f (0, 0) = 1.
F~ · dR
C
Z
Definition: The line integral
~ is independent of path in a region D if
F~ · dR
C
Z
C1
~ =
F~ · dR
Z
~
F~ · dR
C2
for any two paths C1 and C2 in D that have the same initial and terminal points.
Note: The Fundamental Theorem for Line Integrals states that line integrals ofI conservative
~ =0
vector fields are independent of path. Moreover, if F~ is a conservative, then
F~ · dR
C
for every closed path C in D.
Example: Show that the vector field F~ (x, y, z) = h2xz + sin y,Zx cos y, x2 i is conservative and
~ where C is the helix
F~ · dR,
find a scalar potential for F~ . Then evaluate the line integral
C
~
R(t)
= hcos t, sin t, ti for 0 ≤ t ≤ 2π.
The curl of F~ is
~
~
~
i
j
k
∂
∂
∂ ~
~
curlF = = h0, 2x − 2x, cos y − cos yi = 0.
∂x
∂y
∂z
2xz + sin y x cos y x2 Thus, F~ is conservative. If f is a scalar potential for F~ , then
fx = 2xz + sin y
fy = x cos y
fz = x2 .
Then,
Z
f (x, y, z) =
2xz + sin ydx = x2 z + x sin y + K1 (y, z).
Now let fy (x, y, z) = x cos y. That is
∂K1
= x cos y
∂y
∂K1
= 0
∂y
K1 (y, z) = K2 (z).
x cos y +
Now let fz (x, y, z) = x2 . That is
x2 + K20 (z) = x2
K20 (z) = 0
K2 (z) = C.
Thus, a scalar potential for F~ is
f (x, y, z) = x2 z + x sin y.
By the Fundamental Theorem for Line Integrals,
Z
~ = f (R(2π))
~
~
F~ · dR
− f (R(0))
= f (1, 0, 2π) − f (1, 0, 0) = 2π.
C
Example: Find the work done by the force field
√
F~ (x, y) = h6y 3/2 + 9x yi
in moving an object from (1, 1) to (2, 4).
The partial derivatives of the component functions satisfy
∂P
∂Q
√
=9 y=
.
∂y
∂x
Thus, F~ is conservative. Now let
Z
Z
f (x, y) = fx (x, y)dx =
dx = 6xy 3/2 + K(y).
6y 3/2
Then let
√
fy (x, y) = 9x y
√
√
9x y + K 0 (y) = 9x y
K(y) = C.
A scalar potential for F~ is f (x, y) = 6xy 3/2 . By the Fundamental Theorem for Line Integrals,
the work done is
Z
~ = f (2, 4) − f (1, 1) = 96 − 6 = 90.
F~ · dR
W =
C