14.2 Line Integrals
Line integrals on plane
Let C be a plane curve with parametric equations:
x = x(t),
a≤t≤b
y = y(t),
or, equivalently, by the vector equation
a≤t≤b
r(t) = x(t)i + y(t)j,
A partition of the parameter interval [a, b] by points ti with
a = t0 < t1 < ... < tn−1 < tn = b
determine a partition P of the curve by points Pi (xi , yi ), where xi = x(ti ) and yi =
y(ti ). These points divide C into n subarcs with length ∆s1 , ∆s2 , . . . , ∆sn . The norm
kP k of the partition is the longest of these lengths. We choose any point Pi∗ (x∗i , yi∗ ) in
the i-th subarc.
Denition. If f (x, y) is dened on a smooth curve C given by
x = x(t),
a≤t≤b
y = y(t),
then the line integral of f along C (with respect to arc length) is
ˆ
f (x, y) ds := lim
C
if this limit exists.
kP k→0
n
X
i=1
f (x∗i , yi∗ )∆si
Recall that the length of C is
ˆ
b
L=
ds
a
where
s
ds =
Then
ˆ
ˆ
f (x, y) ds =
C
dx
dt
2
+
dy
dt
2
s
b
f (x(t), y(t))
a
dt
dx
dt
2
+
dy
dt
2
dt
If we use the vector form of the parametrization we can simplify the notation up
noticing that
r0 (t) = x0 (t)i + y 0 (t)j and ds = |r0 (t)|dt
The line integral becomes
ˆ
ˆ
f (r(t))|r0 (t)| dt
f (x, y) ds =
C
b
a
Remark. The value of the line integral does not depend on the parametrization of
the curve, provided that the curve is traversed exactly once as t increases from a to b.
ˆ
Example 1. Evaluate the line integral
y ds, where C : x = t, y = t3 , 0 ≤ t ≤ 1.
C
Suppose now that C is a piecewise-smooth curve; that is, C is a union of a nite
number of smooth curves C1 , C2 , . . . , Cn .
Then
ˆ
ˆ
f (x, y) ds =
ˆ
f (x, y) ds + · · · +
f (x, y) ds +
C
C1
ˆ
C2
f (x, y) ds
Cn
ˆ
Example 2. Evaluate
(x + y) ds if C consists of line segments: C1 from (1, 0) to
C
(0, 1), C2 from (0, 1) to (0, 0) and C3 from (0, 0) to (1, 0).
Physical interpretation of a line integral
Suppose that ρ(x, y) represents the linear density at a point (x, y) of a thin wire shaped
like a curve C . Then the mass of wire is
ˆ
m=
ρ(x, y) ds
C
The center of mass of the wire with density function ρ(x, y) is at the point (x̄, ȳ),
where
ˆ
ˆ
x̄ =
1
m
xρ(x, y) ds
C
ȳ =
1
m
yρ(x, y) ds
C
Example 3. A thin wire is bent into the shape of a semicircle x2 + y 2 = 4, x ≥ 0 If the
linear density is a constant k, nd the mass and center of mass of the wire.
Line integrals of f along C with respect to x and y
ˆ
n
X
f (x, y) dx := lim
kP k→0
C
∆xi = xi − xi−1
f (x∗i , yi∗ )∆yi ,
∆yi = yi − yi−1
i=1
ˆ
n
X
f (x, y) dy := lim
kP k→0
C
f (x∗i , yi∗ )∆xi ,
i=1
Let x = x(t), y = y(t), a ≤ t ≤ b be parametric equations of C . Then
ˆ
ˆ
b
f (x(t), y(t))x0 (t)dt
f (x, y) dx =
C
a
ˆ
ˆ
b
f (x(t), y(t))y 0 (t)dt
f (x, y) dy =
C
a
In general, these two integral often appear together by the following notation:
ˆ
ˆ
P (x, y)dx + Q(x, y)dy =
C
Example 4. Evaluate
ˆ
P (x, y) dx +
C
ˆ
C
Q(x, y) dy
C
x
−y
dx + 2
dy
2
+y
x + y2
x2
where C is the circle x2 + y 2 = 1 oriented in the counterclockwise direction.
Line integrals in space
Suppose that C is a smooth space curve given by the parametric equations
x = x(t),
y = y(t),
a≤t≤b
z = z(t),
or by a vector equation r(t) = x(t)i + y(t)j + z(t)k. We dene the line integral of
along C (with respect to arc length) in a manner similar to that for plane curves
ˆ
f (x, y, z) ds := lim
kP k→0
C
n
X
f (x∗i , yi∗ , zi∗ )∆si
i=1
We evaluate it as
ˆ
ˆ
f (x(t), y(t), z(t))
f (x, y, z) ds =
C
s
b
ˆ
a
dx
dt
2
+
dy
dt
2
+
dz
dt
2
dt
b
f (r(t))|r0 (t)| dt
=
a
In particular, if f (x, y, z) = 1, we have the length of the curve C is
ˆ
L=
ˆ
|r0 (t)| dt =
ds =
C
ˆ
C
a
b
s
dx
dt
2
+
dy
dt
2
+
dz
dt
2
dt
Line integrals along with respect to x, y and z can also be dened. For example,
ˆ
f (x, y, z) dz := lim
C
kP k→0
n
X
ˆ
f (x∗i , yi∗ , zi∗ )∆zi
i=1
b
f (x(t), y(t), z(t))z 0 (t)dt
=
a
In general, we will evaluate integrals of the form
ˆ
P (x, y, z)dx + Q(x, y, z)dy + R(x, y, z)dz
ˆCb
[P (x(t), y(t), z(t))x0 (t) + Q(x(t), y(t), z(t))y 0 (t) + R(x(t), y(t), z(t))z 0 (t)] dt
=
a
ˆ
Example 5. Evaluate
0 ≤ t ≤ π/4.
x2 z ds if C is given by x = cos(2t), y = 3t, z = sin(2t) for
C
ˆ
Example 6. Evaluate
ydx + zdy + xdz where C consists of the line segment C1
C
from (1, 0, 0) to (2, 3, 4), followed by the vertical line segment C2 from (2, 3, 4) to
(2, 3, 0).
Line integrals of vector elds
Problem. Given a continuous force eld F(x, y, z) such as a gravitational eld. Find
the work done by the force F in moving a particle along a curve C given by
a≤t≤b
r(t) = x(t)i + y(t)j + z(t)k,
We divide C into subarcs with lengths ∆si by means of a partition of the parameter
interval [a, b]. Choose a point Pi∗ (x∗i , yi∗ , zi∗ ) on the i-th subarc corresponding to the
parameter value ti .
Let T(x, y, z) be the unit tangent vector at the point (x, y, z). Then the work done
by the force eld F is the limit of Riemann sums
W = lim
kP k→0
n
X
ˆ
[F(x∗i , yi∗ , zi∗ )
·
T(x∗i , yi∗ , zi∗ )] ∆si
F(x, y, z) · T(x, y, z) ds
=
C
i=1
r0 (t)
and ds = |r0 (t)|dt. We have
Note that T(t) = 0
|r (t)|
ˆ
ˆ b
ˆ b
ˆ
r0 (t) 0
0
F · T ds =
F(r(t)) · 0
|r (t)| dt =
F(r(t)) · r (t) dt =
F · dr
|r (t)|
C
a
a
C
Denition. Let F be a continuous vector eld dened on a smooth curve C given by
a vector function r(t), a ≤ t ≤ b. Then the line integral of F along C is
ˆ
ˆ
0
F · dr =
C
ˆ
b
F(r(t)) · r (t) dt =
a
F · T ds
C
Suppose the vector eld F is given in component form by
F(x, y, z) = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k
Then
ˆ
ˆ
F · dr =
(P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k) · (dxi + dyj + dzk)
ˆC
C
=
P (x, y, z)dx + Q(x, y, z)dy + R(x, y, z)dz
C
Example 7. Find the work done by the force eld F(x, y, z) = xyi + yzj + xzk in
moving a particle along the curve C : r(t) = ti + t2 j + t3 k, 0 ≤ t ≤ 1.