Line integrals - University of Alberta

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
MATH 209—Calculus, III
Volker Runde
University of Alberta
Edmonton, Fall 2011

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Line integrals, I
The setting
Let C be a smooth curve in R 2 given by the parametric
equations
or by the vector equation
x = x(t), y = y(t), t ∈ [a, b]
r(t) = x(t)i + y(t)j.
We want to integrate a function f along C and define the line
integral �
C f (x, y) ds.
Geometric interpretation
If f ≥ 0, then �
C f (x, y) ds is the area of the curtain with base
C and whose height above (x, y) is f (x, y).

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Line integrals, II
The construction
Divide [a, b] into n subintervals [tj−1, tj].
Set xj = x(tj) and yj = y(tj). The corresponding points
Pj(xj, yj) divide C into n subcurves with lengths ∆s1, . . . , ∆sn.
) on each subcurve.
Pick a support point Pj(x ∗ j
Definition
, y ∗
j
The line integral of f along C is defined as
�
C
f (x, y) ds = lim
n→∞
n�
j=1
f (x ∗ j , y ∗
j )∆sj.

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Line integrals, III
Theorem
For continuous f :
�
�
� b
�dx
f (x, y) ds = f (x(t), y(t))
dt
C
Important
a
� 2
+
� �2 dy
dt.
dt
The value of the integral does not depend on the
parametrization of C as long as C is traversed exactly once as
t increases from a to b.

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Line integrals, IV
To remember. . .
Let s(t) be the length of the curve from r(a) to r(t). Then
ds
dt = s′ �
�dx �2 � �2 dy
(t) = + ,
dt dt
so that
ds =
� �dx
dt
� 2
+
� �2 dy
dt.
dt

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Examples, I
Example
Let C be the right half of the circle x 2 + y 2 = 16.
What is �
C xy 4 ds?
Set
�
x = 4 cos t, y = 4 sin t, t ∈
Then:
�
C
xy 4 � π
2
ds = 1024
− π
2
π
�
, .
2
cos t sin 4 �
t 16 sin2 t + 16 cos2 t dt
− π
2
� π
2
= 4096
− π
cos t sin
2
4 � 1
t dt = 4096
−1
�u=1
= 4046
5 u5
�
�
�
u 4 du
u=−1
= 8192
5 .

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Examples, II
Example
Let C be the line segment from (0, 0) to (1, 1).
Evaluate �
C xy ds.
Set
x = t, y = t, t ∈ [0, 1].
Then:
�
C
xy ds =
� 1
0
t 2√ 1 + 1 dt = √ 2
� 1
0
t 2 dt =
√ 2
3 .

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Examples, III
Example
Let C be the line segment from (a, 0) to (b, 0).
Evaluate �
C f (x, y) ds for arbitrary continuous f .
Set
x = t, y = 0, t ∈ [a, b].
Then: �
C
� b
f (x, y) ds = f (t, 0) dt.
a

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Examples, IV
Example
Let C consist of the paraboloa y = x 2 from (0, 0) to (1, 1)
followed by the vertical line segment from (1, 1) to (1, 2).
Find �
C 2x ds.
The curve C is not smooth, but piecewise smooth, i.e., of the
form C = C1 ⊕ C2 with C1 and C2 smooth.
C1: the parabola y = x 2 from (0, 0) to (1, 1):
x = t, y = t 2 , t ∈ [0, 1].
C2: the line segment from (1, 1) to (1, 2):
r(t) = 〈1, 1〉 + t(〈1, 2〉 − 〈1, 1〉) = 〈1, 1 + t〉
for t ∈ [0, 1], i.e., x = 1, y = 1 + t, and t ∈ [0, 1].

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Examples, V
Example (continued)
Then:
� �
2x ds =
C
= 1
4
� 5
1
C1
=
�
2x ds +
� 1
0
√ u du+2 = u 3
2
C2
2x ds
2t � 1 + 4t 2 dt +
6
�
�
�
�
�
u=5
u=1
� 1
0
+2 = 5√ 5 − 1
6
2dt
+2 = 5√5 − 11
.
6

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Types of line integrals, I
More line integrals
We call �
C f (x, y) ds the line integral with respect to arclength.
Suppose that C is given by the parametric equations
Then
so that
x = x(t), y = y(t), t ∈ [a, b].
dx
dt = x ′ (t) and
dy
dt = y ′ (t),
dx = x ′ (t) dt and dy = y ′ (t) dt.

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Types of line integrals, II
Definition
The line integrals of f with respect to x and with respect to y,
respectively, are defined as
�
C
Shorthand
�
C
� b
f (x, y) dx := f (x(t), y(t))x ′ (t) dt;
and
�
P(x, y) dx + Q(x, y) dy =
�
a
C
� b
f (x, y) dy := f (x(t), y(t))y ′ (t) dt.
C
a
�
P(x, y) dx + Q(x, y) dy.
C

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Examples, VI
Example
Evaluate �
(−5, −3) to (0, 2).
Parametrize the curve as
C y 2 dx + x dy where C is the line segment from
r(t) = 〈−5, −3〉 + t(〈0, 2〉 − 〈−5, −3〉) = 〈−5 + 5t, −3 + 5t〉,
so that
x = −5 + 5t, y = −3 + 5t, t ∈ [0, 1].

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Examples, VII
Example (continued)
Thus:
�
C
y 2 dx + x dy =
= 5
= 5
� 1
0
� 1
0
= − 5
6 .
(5t − 3) 2 5 dt +
�
25t 3
� 1
0
25t 2 − 25t + 4 dt
3
− 25t2
2
�
�
+ 4t�
�
(5t − 5)5 dt
t=1
t=0
�

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Examples, VIII
Example
Evaluate �
x = 4 − y 2 from (−5, −3) to (0, 2).
The parametric equations are
C y 2 dx + x dy where C is the arc of the parabola
x = 4 − t 2 , y = t, t ∈ [−3, 2].

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Examples, IX
Example (continued)
Then:
�
C
y 2 dx + x dy =
=
� 2
Important
−3
� 2
−3
−2t 3 − t 2 + 4 dt = − t4
2
t 2 � 2
(−2t) dt + 4 − t
−3
2 dt
− t3
3
�
�
+ 4t�
�
t=2
t=−3
= 40 5
6 .
Both currves in the last two examples have the same starting
point and endpoint. Still, the two line integrals are different.

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Examples, X
Example
Evaluate �
from (0, 2) to (−5, −3).
Parametrize the curve as
so that
C y 2 dx + x dy where C is the line segment from
r(t) = 〈0, 2〉 + t(〈−5, −3〉 − 〈0, 2〉) = 〈−5t, 2 − 5t〉,
x = −5t, y = −5t + 2, t ∈ [0, 1].

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Examples, XI
Example (continued)
We obtain:
�
y 2 dx + x dy =
C
� 1
0
= −5
= 5
6 .
(−5t + 2) 2 (−5) dt +
� 1
Note
This is precisely the negative of �
line segment from (−5, −3) to (0, 2).
0
25t 2 − 25t + 4 dt
� 1
0
25t dt
C y 2 dx + x dy where C is the

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Properties of line integrals
Definition
If C is any curve in R 2 , we write −C for the curve with
reversed orientation.
Properties
We have �
and �
−C
−C
but �
−C
�
f (x, y) dx = − f (x, y) dx
C
�
f (x, y) dy = − f (x, y) dy,
C
�
f (x, y) ds =
C
f (x, y) ds.

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Line integrals in R 3 , I
As in R 2 . . .
�
f (x, y, z) ds
C
�
� b
�dx
= f (x(t), y(t), z(t))
dt
and. . .
a
� 2
+
� �2 dy
+
dt
� �2 dz
dt
dt

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Line integrals in R 3 , II
As in R 2 . . . (continued)
and. . .
�
P(x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz
C
� b
= P(x(t), y(t), z(t))x ′ (t) dt
a
� b
+ Q(x(t), y(t), z(t))y ′ (t) dt
a
� b
+ R(x(t), y(t), z(t))z
a
′ (t) dt.

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Examples, XII
Example
Let C be the helix given by
Then:
�
C
x = cos t, y = sin t, z = t, t ∈ [0, 2π].
y sin z ds =
= √ 2
� 2π
0
� 2π
0
sin 2 �
t sin2 t + cos2 t + 1 dt
� 2π
sin 2 t dt = 1
√ 1 − cos 2t dt
2 0
= 1
� � �
sin 2t �t=2π
√ t − �
2 2 � = √ 2π.
t=0

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Examples, XIII
Example
Let C = C1 ⊕ C2 with:
C1 = line segment from (2, 0, 0) to (3, 4, 5),
C2 = line segment from (3, 4, 5) to (3, 4, 0).
Evaluate
�
y dx + z dy + x dz
C �
�
= y dx + z dy + x dz +
C1
C2
y dx + z dy + x dz.

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Examples, XIV
Example (continued)
Parametrize C1:
i.e.,
Thus:
�
r(t) = 〈2, 0, 0〉 + t(〈3, 4, 5〉 − 〈2, 0, 0〉) = 〈2 + t, 4t, 5t〉,
C1
x = 2 + t, y = 4t, z = 5t, t ∈ [0, 1].
y dx +z dy +x dz =
=
� 1
0
� 1
0
4t dt +4
� 1
0
5t dt +5
10 + 29t dt = 10t + 29t2
2
� 1
0
t=1
�
�
�
�
t=0
2+t dt
= 49
2 .

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Examples, XV
Example (continued)
Parametrize C2:
Thus:
All in all:
x = 3, y = 4, z = 5 − 5t, t ∈ [0, 1].
�
C2
�
y dx + z dy + x dz = −15
C
y dx + z dy + x dz = 49
2
� 1
0
dt = −15.
− 15 = 19
2 .

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Line integrals of vector fields, I
Problem
Let F = Pi + Qj + Rk be a continuous vector field on R 3 which
moves a particle along a smooth curve C. What is the work W
done?
Easy case
If F is constant and moves the particle along a line segment
from P to Q,
W = F · D,
where D = −→
PQ is the displacement vector.

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Line integrals of vector fields, II
The general case
Divide C into n subarcs Pj−1, Pj with lengths ∆sj by dividing
the parameter interval [a, b] into n subintervals of equal length.
, y ∗ ) on the j-th subarc, and let
Choose a point P∗ j (x ∗ j j , z∗ j
t∗ j ∈ [tj−1, tj] be the corresponding parameter.
If ∆sj is small: as the particle moves from Pj−1 to Pj, it
), the unit
proceeds approximately in the direction of T(t∗ j
tangent vector to C at P∗ j (x ∗ j , y ∗
j , z∗ j ).

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Line integrals of vector fields, III
The general case (continued)
If Wj is the work to move the particle from Pj−1 to Pj, then
Hence,
and thus
W = lim
n→∞
Wj ≈ F(x ∗ j , y ∗
j , z ∗ j ) · ∆sjT(t ∗ j ).
W ≈
n�
j=1
n�
j=1
F(x ∗ j , y ∗
j , z ∗ j ) · ∆sjT(t ∗ j ),
F(x ∗ j , y ∗
j , z ∗ j ) · ∆sjT(t ∗ j )
�
=
C
F(x, y, z) · T(x, y, z) ds.

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Line integrals of vector fields, IV
The general case (continued)
Let C be given by the vector equation
Then:
Thus:
�
C
r(t) = x(t)i + y(t)j + z(t)k, t ∈ [a, b].
T(t) = r′ (t)
|r ′ (t)| .
� b
F · T ds = F(r(t)) ·
a
r′ (t)
|r ′ (t)| |r′ (t)| dt
� b
= F(r(t)) · r ′ (t) dt = · · ·
a

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Line integrals of vector fields, V
The general case (continued)
� b
· · · = F(r(t)) · r ′ (t) dt
a
� b
= P(x(t), y(t), z(t))x ′ (t) dt
a
� b
+ Q(x(t), y(t), z(t))y ′ (t) dt
a
� b
+ R(x(t), y(t), z(t))z
a
′ (t) dt
�
= P(x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz.
C

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Line integrals of vector fields, VI
Definition
Let F = Pi + Qj + Rk be a continuous vector field on R 3 , and
let C be a smooth curve given by the vector function r(t) for
t ∈ [a, b]. The line integral of F along C is
�
C
� b
F · dr = F(r(t)) · r
a
′ (t) dt
�
= P(x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz.
C

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Examples, XVI
Example
A force field
F(x, y) = x sin yi + yj
moves a particle from (−1, 1) to (2, 4) along the parabola
y = x 2 . Compute the total work W .
Parametrize C as
Then:
and
r(t) = ti + t 2 j, t ∈ [−1, 2].
r ′ (t) = i + 2tj
F(r(t)) = t sin(t 2 )i + t 2 j.

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Examples, XVII
Example (continued)
Thus:
�
W =
=
C
� 2
−1
F · dr =
� 2
−1
(t sin(t 2 )i + t 2 j) · (i + 2tj) dt
t sin(t 2 ) + 2t 3 dt = 1
2
cos u
�
�
= − �
2
u=4
+
u=1
15
2
� 4
1
sin u du + t4
2
�
�
�
�
t=2
t=−1
1
= (15 − cos 4 + cos 1).
2

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Examples, XVIII
Example
Let
and let C be given by
i.e., by
Thus:
and
F(x, y, z) = xyi + yzj + zxk,
x = t, y = t 2 , z = t 3 , t ∈ [0, 1].
r(t) = ti + t 2 j + t 3 k, t ∈ [0, 1].
r ′ (t) = i + 2tj + 3t 2 k
F(r(t)) = t 3 i + t 5 j + t 4 k.

MATH 209—
Calculus,
III
Volker Runde
Line integrals
in R 2
Types of line
integrals
Line integrals
in R 3
Line integrals
of vector fields
Examples, XIX
Example (continued)
Therefore:
�
C
F · dr =
=
� 1
0
� 1
0
= t4
4
= 27
28 .
t 3 + 2t 6 + 3t 6 dt
t 3 + 5t 6 dt
+ 5t7
7
�
�
�
�
t=1
t=0