16.2 Line Integrals
Lukas Geyer
Montana State University
M273, Fall 2011
Lukas Geyer (MSU)
16.2 Line Integrals
M273, Fall 2011
1 / 21
Scalar Line Integrals
Definition
Z
f (x) ds =
C
lim
{∆si }→0
N
X
f (Pi )∆si
i=1
Riemann sum illustration.
Subdivision into arcs by red
crosses, sample points Pi as
green stars, the ∆si are the
length of the red line
segments.
Lukas Geyer (MSU)
16.2 Line Integrals
M273, Fall 2011
2 / 21
Applications of Scalar Line Integrals
Length
Z
length(C) =
1 ds
C
Mass of a wire of linear density ρ(x)
Z
mass = ρ(x)ds
C
Surface area of (one side of) a fence of height h(x)
Z
area = h(x)ds
C
Electric Potential of a charged wire with linear charge density ρ(x)
Z
ρ(x)
V (y) = k
ds, where k is Coulomb’s constant.
C kx − yk
Lukas Geyer (MSU)
16.2 Line Integrals
M273, Fall 2011
3 / 21
Calculation of Scalar Line Integrals
Theorem
Let c(t) be a parametrization of a curve C for a ≤ t ≤ b. If f (x) and c0 (t)
are continuous, then
Z
Z b
f (x) ds =
f (c(t))kc0 (t)kdt.
C
a
Differential version
ds = kc0 (t)kdt.
Lukas Geyer (MSU)
16.2 Line Integrals
M273, Fall 2011
4 / 21
Scalar Line Integral Example I
Example
R
Calculate C xy + z ds over the helix C parametrized by
c(t) = hcos 3t, sin 3t, 4ti for 0 ≤ t ≤ π.
Compute ds
c0 (t) = h−3 sin 3t, 3 cos 3t, 4i
p
√
kc0 (t)k =
9 sin2 3t + 9 cos2 3t + 16 = 25 = 5
ds = kc0 (t)kdt = 5dt.
Lukas Geyer (MSU)
16.2 Line Integrals
M273, Fall 2011
5 / 21
Scalar Line Integral Example II
Example
R
Calculate C xy + z ds over the helix C parametrized by
c(t) = hcos 3t, sin 3t, 4ti for 0 ≤ t ≤ π.
Compute ds
ds = kc0 (t)kdt = 5dt.
Write out integrand and evaluate
Z
Z
π
xy + z ds =
C
Lukas Geyer (MSU)
(cos 3t sin 3t + 4t)5 dt
Z π
Z π
= 5
cos 3t sin 3t dt + 20
t dt
0
π
π0
1 2
1
= 5
sin 3t + 20 t 2 = 10π 2
6
2
0
0
0
16.2 Line Integrals
M273, Fall 2011
6 / 21
Scalar Line Integral Applied Example I
Example
Find the mass of a wire in the shape of a circle of radius 2 centered at
(3,4) with linear mass density ρ(x, y ) = y 2 .
Parametrize C
c(t) = h3, 4i + 2hcos t, sin ti
= h3 + 2 cos t, 4 + 2 sin ti,
where 0 ≤ t ≤ 2π.
Lukas Geyer (MSU)
16.2 Line Integrals
M273, Fall 2011
7 / 21
Scalar Line Integral Applied Example II
Example
Find the mass of a wire in the shape of a circle of radius 2 centered at
(3,4) with linear mass density ρ(x, y ) = y 2 .
Parametrize C
c(t) = h3 + 2 cos t, 4 + 2 sin ti,
0 ≤ t ≤ 2π.
Compute ds
c0 (t) = h−2 sin t, 2 cos ti
p
√
kc0 (t)k =
4 sin2 t + 4 cos2 t = 4 = 2
ds = kc0 (t)kdt = 2dt.
Lukas Geyer (MSU)
16.2 Line Integrals
M273, Fall 2011
8 / 21
Scalar Line Integral Applied Example III
Example
Find the mass of a wire in the shape of a circle of radius 2 centered at
(3,4) with linear mass density ρ(x, y ) = y 2 .
Parametrize C, compute ds
c(t) = h3 + 2 cos t, 4 + 2 sin ti,
0 ≤ t ≤ 2π,
ds = 2dt.
Write out integral and evaluate
Z
m =
y 2 ds =
C
Z
=
Z
2π
(4 + 2 sin t)2 · 2 dt
0
2π
32 + 32 sin t + 8 sin2 t dt
0
= [32t − 32 cos t + 4(t − sin t cos t)]2π
0
= 64π + 8π = 72π
Lukas Geyer (MSU)
16.2 Line Integrals
M273, Fall 2011
9 / 21
Vector Line Integrals
Definition
The line integral of a vector field F along an oriented curve C is the scalar
line integral of the tangential component of F.
Z
Z
F · ds = (F · T)ds.
C
C
Vector field F(x, y ) = h0, −1i
with an oriented curve C (in
green) and one unit tangent
vector (in red).
Question
R
Is C F · ds positive, negative,
or zero? Positive
Lukas Geyer (MSU)
16.2 Line Integrals
M273, Fall 2011
10 / 21
Applications of Vector Line Integrals
Work done by a force field F on a particle moving along C
Z
W = F · ds.
C
Work done against a force field F, moving along C
Z
W = − F · ds.
C
Lukas Geyer (MSU)
16.2 Line Integrals
M273, Fall 2011
11 / 21
Computing a Vector Line Integral
Theorem
If c(t) is a regular parametrization of an oriented curve C for a ≤ t ≤ b,
then
Z
Z b
F · ds =
F(c(t)) · c0 (t) dt.
C
a
Differential version
ds = c0 (t)dt
c(t) is regular if it is smooth and c0 (t) 6= 0.
Lukas Geyer (MSU)
16.2 Line Integrals
M273, Fall 2011
12 / 21
Scalar vs. Vector Line Integrals
Orientation matters for vector line integrals: If −C is the curve C in
negative orientation, then
Z
Z
F · ds = − F · ds
−C
C
Orientation does not matter for scalar line integrals:
Z
Z
F ds = Fds
−C
C
Z
Z
F · ds ≤ kFkds
C
C
Why? The tangential component F · T always satisfies |F · T| ≤ kFk.
Lukas Geyer (MSU)
16.2 Line Integrals
M273, Fall 2011
13 / 21
Vector Line Integral Example I
Example
R
Evaluate C F · ds, where F(x, y ) = h0, −1i, and C is the part of the graph
of y = 21 x 3 − x from (2, 2) to (−2, −2).
Parametrize C
1
c(t) = ht, t 3 − ti,
2
−2 ≤ t ≤ 2.
Problem
The orientation is wrong. What to
do about this? The easiest solution
is to calculate with the wrong
orientation and reverse the sign of
the result.
Lukas Geyer (MSU)
16.2 Line Integrals
M273, Fall 2011
14 / 21
Vector Line Integral Example II
Example
R
Evaluate C F · ds, where F(x, y ) = h0, −1i, and C is the part of the graph
of y = 21 x 3 − x from (2, 2) to (−2, −2).
Parametrize −C
1
c(t) = ht, t 3 − ti,
2
−2 ≤ t ≤ 2.
Write out integral and evaluate
Z
Z
Z
2
3
h0, −1i · h1, t 2 − 1i dt
2
−2
−2
2
Z 2
3
1
=
− t 2 + 1 dt = − t 3 + t
2
2
−2
−2
= −4 + 2 − 4 + 2 = −4
F · ds =
−C
2
Lukas Geyer (MSU)
0
F(c(t)) · c (t) dt =
16.2 Line Integrals
M273, Fall 2011
15 / 21
Vector Line Integral Example III
Example
R
Evaluate C F · ds, where F(x, y ) = h0, −1i, and C is the part of the graph
of y = 21 x 3 − x from (2, 2) to (−2, −2).
Write out integral and evaluate
Z
F · ds = −4
−C
Fix the orientation
Z
Z
F · ds = −
C
Lukas Geyer (MSU)
F · ds = −(−4) = 4.
−C
16.2 Line Integrals
M273, Fall 2011
16 / 21
Alternative Notation for Vector Line Integrals
2-D Version
For F(x, y ) = hF1 (x, y ), F2 (x, y )i we write ds = hdx, dy i and
Z
Z
F · ds = F1 dx + F2 dy
C
C
3-D Version
For F(x, y , z) = hF1 (x, y , z), F2 (x, y , z), F3 (x, y , z)i we write
ds = hdx, dy , dzi and
Z
Z
F · ds = F1 dx + F2 dy + F3 dz
C
Lukas Geyer (MSU)
C
16.2 Line Integrals
M273, Fall 2011
17 / 21
Alternative Notation II
3-D Parametrized Version
For F(x, y , z) = hF1 (x, y , z), F2 (x, y , z), F3 (x, y , z)i and a regular
parametrization c(t) = hx(t), y (t), z(t)i, a ≤ t ≤ b of C,
Z
Z
F · ds =
F1 dx + F2 dy + F3 dz
C
C
Z b
=
F1 (c(t))x 0 (t) + F2 (c(t))y 0 (t) + F3 (c(t))z 0 (t) dt
a
Differential version
dx = x 0 (t) dt,
Lukas Geyer (MSU)
dy = y 0 (t) dt,
16.2 Line Integrals
dz = z 0 (t) dt
M273, Fall 2011
18 / 21
3-D Line Integral Example I
Example
Calculate
(3, 2, 0).
R
C
yz dx + xz dy + xy dz over the line segment from (1, 1, 1) to
Parametrize C
c(t) = h1, 1, 1i + t(h3, 2, 0i − h1, 1, 1i) = h1 + 2t, 1 + t, 1 − ti,
0 ≤ t ≤ 1.
Write out differentials
Lukas Geyer (MSU)
dx
= x 0 (t) dt = 2 dt
dy
= y 0 (t) dt = dt
dz
= z 0 (t) dt = −dt
16.2 Line Integrals
M273, Fall 2011
19 / 21
3-D Line Integral Example II
Example
Calculate
(3, 2, 0).
R
C
yz dx + xz dy + xy dz over the line segment from (1, 1, 1) to
Parametrize C, write out differentials
c(t) = h1 + 2t, 1 + t, 1 − ti,
0 ≤ t ≤ 1,
dx = 2 dt, dy = dt, dz = −dt
Write out integral and evaluate
Z
C
yz dx + xz dy + xy dz
Z 1
=
(1 + t)(1 − t) · 2 + (1 + 2t)(1 − t) + (1 + 2t)(1 + t)(−1) dt
0
Lukas Geyer (MSU)
16.2 Line Integrals
M273, Fall 2011
20 / 21
3-D Line Integral Example III
Example
Calculate
(3, 2, 0).
R
C
yz dx + xz dy + xy dz over the line segment from (1, 1, 1) to
Write out integral and evaluate
Z
C
yz dx + xz dy + xy dz
Z 1
=
(1 + t)(1 − t) · 2 + (1 + 2t)(1 − t) + (1 + 2t)(1 + t)(−1) dt
0
Z 1
Z 1
2
2
2
=
2 − 2t + 1 + t − 2t − 1 − 3t − 2t dt =
2 − 2t − 6t 2 dt
0
0
1
= 2t − t 2 − 2t 3 0 = 2 − 1 − 2 = −1.
Lukas Geyer (MSU)
16.2 Line Integrals
M273, Fall 2011
21 / 21