Section 14.2: Line Integrals
~
Consider a smooth plane curve C defined by R(t)
= hx(t), y(t)i, where a ≤ t ≤ b. A
partition of the interval [a, b]
a = t0 < t1 < t2 < · · · < tn = b
determines a partition P of the curve C by points (xi , yi ), where xi = x(ti ) and yi = y(ti ),
into n subarcs with lengths ∆si . Choose a representative point (x∗i , yi∗ ) on each subarc and
form the Riemann sum
n
X
f (x∗i , yi∗ )∆si .
i=1
Let ||P || denote the norm of the partition P which is the length of the longest subarc.
Definition: If f is defined on the smooth plane curve C, then the line integral of f along
C is
Z
n
X
f (x, y)ds = lim
f (x∗i , yi∗ )∆si
C
||P ||→0
i=1
provided that
I the limit exists. If the curve C is a closed curve, then the line integral is
denoted by
f (x, y)ds.
C
Theorem: If f is a continuous function on the plane curve C, then
s 2
Z
Z b
2
dx
dy
f (x, y)ds =
f (x(t), y(t))
+
dt.
dt
dt
C
a
Z
x2 yds, where C is the upper half circle x2 + y 2 = 4,
Example: Evaluate the line integral
C
y ≥ 0.
The curve C can be parameterized by x = 2 cos t, y = 2 sin t, for 0 ≤ t ≤ π. Then
Z π
Z
p
2
(4 cos2 t)(2 sin t) 4 cos2 t + 4 sin2 tdt
x yds =
0
C
Z π
= 16
cos2 t sin tdt
0
π
16
= − cos3 t
3
0
32
=
.
3
Note: The definition of a line integral can be extended to piecewise-smooth curves
C = C1 ∪ C2 ∪ · · · ∪ Cn .
In particular,
Z
Z
f (x, y)ds =
Z
f (x, y)dx + · · · +
f (x, y)ds +
C
C1
Z
C2
f (x, y)ds.
Cn
Z
xyds, where C consists of the line segments from
Example: Evaluate the line integral
C
(0, 0) to (1, 0) and from (1, 0) to (2, 1).
The curve C can be expressed as C = C1 ∪ C2 , where C1 is the segment from (0, 0) to (1, 0)
and C2 is the segment from (1, 0) to (2, 1). The segments can be parameterized by
C1 : x = t, y = 0, 0 ≤ t ≤ 1
C2 : x = t + 1, y = t, 0 ≤ t ≤ 1.
Then
Z
Z
xyds =
C
Z
xyds +
xyds
C2
Z 1
√
=
0dt +
(t2 + t) 2dt
0
0
√ 1 3 1 2 1
=
2
t + t 3
2
0
√
5 2
=
.
6
C
Z 11
Theorem: If f is a continuous function on the space curve C defined by the vector function
~
R(t)
= hx(t), y(t), z(t)i, then
s 2 2
Z b
Z
2
dy
dz
dx
f (x(t), y(t), z(t))
f (x, y, z)ds =
+
+
dt.
dt
dt
dt
a
C
Z
Example: Evaluate the line integral
xyzds, where C is the part of the helix defined by
π
x = 2t, y = 3 sin t, z = 3 cos t, for 0 ≤ t ≤ .
2
C
The curve C has already been parameterized. Then
Z
√
18t sin t cos t 13dt
0
√ Z π/2
t sin t cos tdt
= 18 13
Z
π/2
xyzds =
C
0
π/2
√
1
1
1
2
= 18 13
t sin t − t + sin 2t 2
4
8
0
√
9 13π
=
.
4
Theorem: (Center of Mass of a Wire)
Suppose that ρ(x, y) is the density of a thin wire is shaped like a curve C at the point (x, y).
Then the mass of the wire is
Z
m=
ρ(x, y)ds
C
and the center of mass of the wire is the point (x̄, ȳ), where
Z
Z
1
1
x̄ =
xρ(x, y)ds,
and
ȳ =
yρ(x, y)ds.
m C
m C
Example: A thin wire is bent into the shape of a semicircle x2 + y 2 = 9, y ≥ 0. If the density
is ρ(x, y) = y, find the mass and center of mass of the wire.
The wire can be parameterized by x = 3 cos t, y = 3 sin t, for 0 ≤ t ≤ π. The mass of the
wire is
Z
m =
yds
C
Z π
= 9
sin tdt
0
= −9 cos t|π0
= 18.
By symmetry, the center of mass of the wire is the point (0, ȳ), where
Z
1
ȳ =
y 2 ds
18 C
Z π
√
1
9 sin2 t 9dt
=
18 0
Z
3 π 2
sin tdt
=
2 0
Z
3 π
=
1 − cos(2t)dt
4 0
π
3
1
=
t − sin(2t) 4
2
0
3π
=
.
4
Definition: There are other line integrals called the line integrals of f along C with
respect to x, y, and z which are given by
Z
Z b
f (x, y, z)dx =
f (x(t), y(t), z(t))x0 (t)dt
C
a
Z
Z b
f (x, y, z)dy =
f (x(t), y(t), z(t))y 0 (t)dt
C
a
Z
Z b
f (x, y, z)dz =
f (x(t), y(t), z(t))z 0 (t)dt.
C
a
For distinction, the original line integral
respect to arc length.
R
C
f (x, y, z)ds is called the line integral with
Z
Example: Evaluate the line integral
0 ≤ x ≤ π2 .
xydx + ydy, where C is the sine curve y = sin x for
C
The curve C can be parameterized by x = t, y = sin t, for 0 ≤ t ≤ π2 . Then
Z π/2
Z
t sin t + sin t cos tdt
xydx + ydy =
0
C
π/2
1 2 = −t cos t + sin t + sin t
2
0
3
=
.
2
A given parameterization x = x(t), y = y(t), a ≤ t ≤ b determintes the orientation of
a curve C, with the positive direction corresponding to increasing values of t. The curve C
with the opposite orientation is denoted by −C and
Z
Z
f (x, y, z)dx = − f (x, y, z)dx,
ZC
Z−C
f (x, y, z)dy = − f (x, y, z)dy,
ZC
Z−C
f (x, y, z)dz = − f (x, y, z)dz,
−C
Z
Z C
f (x, y, z)ds =
f (x, y, z)ds.
−C
C
Definition: Let F~ be a continuous vector field defined on a smooth curve C given by a vector
~
function R(t)
for a ≤ t ≤ b. Then the line integral of F~ along C is
Z b
Z
~
~ 0 (t)dt.
~
~
F~ (R(t))
·R
F · dR =
a
C
Z
Example: Evaluate the line integral
~ where F~ (x, y, z) = hx2 , xy, z 2 i and C is defined
F~ · dR,
C
~
by R(t)
= hsin t, cos t, t2 i for 0 ≤ t ≤ 1.
~ 0 (t) = hcos t, − sin t, 2ti. Then F~ (t) · R
~ 0 (t) = 2t5 and
First, F~ (t) = hsin2 t, sin t cos t, t4 i and R
1
Z
Z 1
1 6 1
5
~
~
F · dR =
2t dt = t = .
3 0 3
C
0
Theorem: (Work as a Line Integral)
Suppose that F~ is a continuous force field on R3 . The work done by the force field F~ in
~
moving a particle along a smooth curve C defined by R(t)
for a ≤ t ≤ b is given by
Z
~
W =
F~ · dR.
C
Example: Find the work done by the force field F~ (x, y, z) = hxz, xy, yzi on a particle that
~
moves along the curve defined by R(t)
= ht2 , −t3 , t4 i for 0 ≤ t ≤ 1.
~ 0 (t) = h2t, −3t2 , 4t3 i. Then F~ (t) · R
~ 0 (t) = 5t7 − 4t10 and the
First, F~ (t) = ht6 , −t5 , −t7 i and R
work done is
Z
~
W =
F~ · dR
C
Z 1
=
(5t7 − 4t10 )dt
0
1
4 11 5 8
t − t =
8
11
0
23
=
.
88