28 work and line integrals

Work
The work W that is done by a constant force F
moving an object of a distance d is defined to be
W = F * d

Work
W = F * d
For example, a book that weights 10 lb is moved
vertically by a distance of 5 ft. The work that's done is
W = 5*10 = 50 (ft-lb)

Work
W = F * d
W = 5*10 = 50 (ft-lb)We are to calculate the work done on an object under
a non–constant force.

Work
W = F * d
a non–constant force. Some examples of these are:
* tension of a stretched spring
* work done on an object’s movement in a force field
such as a current flow, or a gravitational force field.

Work
W = F * d
a non–constant force. Some examples of these are:
* tension of a stretched spring
* work done on an object’s movement in a force field
such as a current flow, or a gravitational force field.
Let’s clarity the notion of a force field.

Vector Fields, Work and Line Integrals

A 2D vector field is a function that assigns a vector
at each point (x, y).

The vector assigned at (x, y) is denoted as
F(x, y) = <f(x, y), g(x, y)> = f(x, y)i + g(x, y)j

Example: Draw the vector field F(x, y) = <x, y>

Example: Draw the vector field F(x, y) = <x, y>
Make a table:
(x, y) <x, y>
(1, 1) <1, 1>
(1, -1) <1, -1>
(-1, 1) <-1, 1>
(-1, -1) <-1, -1>
(3, 3) <3, 3>

Example: Draw the vector field F(x, y) = <x, 1>

Make a table:
(x, y) <x, y>

Make a table:
(x, y) <x, y>
(0, y) <0, 1>

Make a table:
(x, y) <x, y>
(0, y) <0, 1>
(1, y) <1, 1>

Make a table:
(x, y) <x, y>
(0, y) <0, 1>
(1, y) <1, 1>
(2, y) <2, 1>

Make a table:
(x, y) <x, y>
(0, y) <0, 1>
(1, y) <1, 1>
(2, y) <2, 1>
(-1, y) <-1, 1>
(-2, y) <-2, 1>

Vector fields are used to represent force fields such
as gravitational, magnetic fields or wind speed,
pressure charts.
Make a table:
(x, y) <x, y>
(0, y) <0, 1>
(1, y) <1, 1>
(2, y) <2, 1>
(-1, y) <-1, 1>
(-2, y) <-2, y>

The work W of a force F
vector acting on a particle
moving it from P to Q along
a straight line is given by the
dot product
F
P
Q
θ

dot product
W = F • PQ = |PQ|(|F|cos(θ)).
F
P
Q
θ
|F|cos(θ) = force projected
in the direction PQ

dot product
W = F • PQ = |PQ|(|F|cos(θ)).
F
P
Q
θ
in the direction PQ
Note that the work W may be negative meaning
the angle θ is more than 90o
.

dot product
W = F • PQ = |PQ|(|F|cos(θ)).
F
P
Q
θ
in the direction PQ
We want to generalize this
computation for work done
moving a particle along a
curve from points P to Q in a
vector field. P
Q
Note that the work W may be negative meaning
the angle θ is more than 90o
.

Let C(t) = (x(t), y(t)) be a paramterization of the
curve C where a < t < b with C(a) = P and C(b) = Q.

Let F(x, y) = <f(x, y), g(x, y)> = f(x, y)i + g(x, y)j be
a vector-valued function defines over C that gives
the force at each point on C. (F: R2
R2
)
C(a)=P
C(b)=Q
C(t) = (x, y)
F(x, y)=force
at (x, y)

R2
)
C(a)=P
C(b)=Q
C(t) = (x, y)
F(x, y)=force
at (x, y)
f(x, y), g(x, y) the coordinate functions of the forces
are functions from R2
R .

R2
)
C(a)=P
C(b)=Q
C(t) = (x, y)
F(x, y)=force
at (x, y)
F(x, y) can be viewed as
F(x(t), y(t)) as a function from
R  R2
with domain a < t < b.
f(x, y), g(x, y) the coordinate functions of the forces
are functions from R2
R .

C(a)=P
C(b)=Q
To find the total work done from
P to Q, we partition the interval
[a, b] by Δt and write the partition
as a=t0, t1, t2,…tn=b.

C(a)=P
C(b)=Q
This gives a partition C.

C(a)=P
C(b)=Q
C(ti)
Let C(ti) and C(ti +Δt) be a two
points in the partition of C.
C(ti+Δt)

C(a)=P
C(b)=Q
C(ti)
F(x(ti), y(ti))
=force at C(ti)
C(ti+Δt)
Then the work required to move the particle from
C(ti) to C(ti+Δt) is approximately the dot product
F(x(ti), y(ti)) • C(ti+Δt) – C(ti)
C(ti+Δt) – C(ti)

C(a)=P
C(b)=Q
C(ti)
F(x(ti), y(ti))
=force at C(ti)
C(ti+Δt)
Then the work required to move the particle from
C(ti) to C(ti+Δt) is approximately the dot product
F(x(ti), y(ti)) • C(ti+Δt) – C(ti)
C(ti+Δt) – C(ti)
Hence the total work
W = lim Σ F(x(ti), y(ti)) • C(ti+Δt) – C(ti)Δt0

W = lim Σ F(x(ti), y(ti) • C(ti+Δt) – C(ti)Δt0

= lim Σ F(x(ti),y(ti)) •
Δt0
C(ti+Δt) – C(ti)
Δt
Δt

Δt0
C(ti+Δt) – C(ti)
Δt
Δt
As limit Δt  0, the Σ tunes into integral,

Δt0
C(ti+Δt) – C(ti)
Δt
Δt
As limit Δt  0, the Σ tunes into integral, and
C(ti+Δt) – C(ti)
Δt
 C'(t).

C(ti+Δt) – C(ti)
Δt
 C'(t).
Hence W = ∫ F • C' dt.
t = a
b
Δt0
C(ti+Δt) – C(ti)
Δt
Δt

C(ti+Δt) – C(ti)
Δt
 C'(t).
t = a
b
C'(t) dt = dC
dt
*dt = dC
Δt0
C(ti+Δt) – C(ti)
Δt
Δt

C(ti+Δt) – C(ti)
Δt
 C'(t).
t = a
b
C'(t) dt = dC
dt
*dt = dC
So we write W = ∫ F • dC where dC = C'(t) dt
t = a
b
Δt0
C(ti+Δt) – C(ti)
Δt
Δt

C(ti+Δt) – C(ti)
Δt
 C'(t).
t = a
b
C'(t) dt = dC
dt
*dt = dC
t = a
b
Δt0
C(ti+Δt) – C(ti)
Δt
Δt
Its also written as W = ∫C
F • dC

C(ti+Δt) – C(ti)
Δt
 C'(t).
t = a
b
C'(t) dt = dC
dt
*dt = dC
t = a
b
Δt0
C(ti+Δt) – C(ti)
Δt
Δt
Its also written as W = ∫C
F • dC
and it is called the line integral of F over C.

Theorem: The line integral is independent
of the parameterizations of the curve C.
∫C
F • dC

Example: Find the total work done from (-2, 4) to
(1, 1) along the parabola y= x2
subjected to the
force function F(x, y) = xyi + (x – y)j
∫C
F • dC

∫C
F • dC
We may parameterize y = x2
from (-2, 4) to (1, 1)
as x = t, y = t2
for -2 < t < 1.
subjected to the

∫C
F • dC
from (-2, 4) to (1, 1)
as x = t, y = t2
for -2 < t < 1. Hence C(t) = <t, t2
>
subjected to the

∫C
F • dC
from (-2, 4) to (1, 1)
as x = t, y = t2
for -2 < t < 1. Hence C(t) = <t, t2
>
F(x, y) = <xy, x – y>
subjected to the

∫C
F • dC
from (-2, 4) to (1, 1)
as x = t, y = t2
for -2 < t < 1. Hence C(t) = <t, t2
>
F(x, y) = <xy, x – y> = <t3
, t – t2
>.
subjected to the

∫C
F • dC
from (-2, 4) to (1, 1)
as x = t, y = t2
for -2 < t < 1. Hence C(t) = <t, t2
>
F(x, y) = <xy, x – y> = <t3
, t – t2
>.
The integrand F • C'(t) = <t3
, t – t2
>•<1, 2t>
subjected to the

∫C
F • dC
from (-2, 4) to (1, 1)
as x = t, y = t2
for -2 < t < 1. Hence C(t) = <t, t2
>
F(x, y) = <xy, x – y> = <t3
, t – t2
>.
, t – t2
>•<1, 2t> = t2
– t3
subjected to the

∫C
F • dC
from (-2, 4) to (1, 1)
as x = t, y = t2
for -2 < t < 1. Hence C(t) = <t, t2
>
F(x, y) = <xy, x – y> = <t3
, t – t2
>.
, t – t2
>•<1, 2t> = t2
– t3
Hence W = ∫ F • C' dt =
b
t=a
∫ t2
– t3
dt =
1
t=-2
subjected to the

∫C
F • dC
from (-2, 4) to (1, 1)
as x = t, y = t2
for -2 < t < 1. Hence C(t) = <t, t2
>
F(x, y) = <xy, x – y> = <t3
, t – t2
>.
, t – t2
>•<1, 2t> = t2
– t3
Hence W = ∫ F • C' dt =
b
t=a
∫ t2
– t3
dt =
1
t=-2
27
4
subjected to the

Theorem: If D is a parameterization of C in the
opposite directiont, then = – .∫C
F • dC∫D
F • dD

F • dC∫D
F • dD
The line integrals may be written as the sum of two integrals:

F • dC∫D
F • dD
C(t) = <x(t), y(t)>  C'(t) = <x'(t), y'(t)>, with a < t < b.

F • dC∫D
F • dD
F(x, y) = <f(x, y), g(x, y)> = <f(x(t), y(t)), g(x(t), y(t))>

F • dC∫D
F • dD
∫C
F • dC = ∫t=a
F • C' dt =
b

F • dC∫D
F • dD
∫C
F • dC = ∫t=a
F • C' dt =
b
<f(x(t), y(t)), g(x(t), y(t))>•<x'(t), y'(t)> dt∫t=a
b

F • dC∫D
F • dD
∫C
F • dC = ∫t=a
F • C' dt =
b
b
= f(x(t), y(t)) x'(t)dt + g(x(t), y(t)) y'(t)dt∫t=a
b
∫t=a
b

F • dC∫D
F • dD
∫C
F • dC = ∫t=a
F • C' dt =
b
b
= f(x(t), y(t)) x'(t)dt + g(x(t), y(t)) y'(t)dt∫t=a
b
∫t=a
b
or simply as f(x, y)dx + g(x, y)dy∫C
∫C
where dx = x'(t)dt, dy = y'(t)dt.

Example: Let C be the curve from (0, 1) to (1, e)
along the y = ex
and h(x, y) = xy. Find ∫C
h(x, y) dy.

along the y = ex
and h(x, y) = xy. Find
We may parameterize y = ex
from (0, 1) to (1, e)
as x = t, y = et
with 0 < t < 1.
∫C
h(x, y) dy.

along the y = ex
from (0, 1) to (1, e)
as x = t, y = et
with 0 < t < 1.
Therefore h(x, y) = tet
and dy = y'(t)dt = et
dt.
∫C
h(x, y) dy.

along the y = ex
from (0, 1) to (1, e)
as x = t, y = et
with 0 < t < 1.
dt.
∫C
h(x, y) dy.
∫C
h(x, y) dy =So ∫t=0
h(x, y) dy
1
∫t=0
tet
* et
dt =
1
=

along the y = ex
from (0, 1) to (1, e)
as x = t, y = et
with 0 < t < 1.
dt.
∫C
h(x, y) dy.
∫C
h(x, y) dy
1
∫t=0
tet
* et
dt =
1
= ∫t=0
te2t
dt =
1

along the y = ex
from (0, 1) to (1, e)
as x = t, y = et
with 0 < t < 1.
dt.
∫C
h(x, y) dy.
∫C
h(x, y) dy
1
∫t=0
tet
* et
dt =
1
= ∫t=0
te2t
dt = ½ (te2t
– ½ e2t
)
1
integration by parts

along the y = ex
from (0, 1) to (1, e)
as x = t, y = et
with 0 < t < 1.
dt.
∫C
h(x, y) dy.
∫C
h(x, y) dy
1
∫t=0
tet
* et
dt =
1
= ∫t=0
te2t
dt = ½ (te2t
– ½ e2t
) | =
1
t=0
1
e2
+ 1
4
integration by parts

We define the line integral in 3D space similarly .

C(t) = <x(t), y(t), z(t)> be a parameterized space curve with
a < t < b and f(x, y, z), g(x, y, z), and h(x, y, z) be functions on
C.

C. Then
∫C
f(x, y, z) dx = ∫t=a
f(x, y, z) x'(t) dt
b

C. Then
∫C
f(x, y, z) x'(t) dt
b
∫C
g(x, y, z) dy = ∫t=a
g(x, y, z) y'(t) dt
b

C. Then
∫C
f(x, y, z) x'(t) dt
b
∫C
g(x, y, z) y'(t) dt
b
∫C
h(x, y, z) dz = ∫t=a
h(x, y, z) z'(t) dt
b

C. Then
These are the lines integrals for calculating the work of
moving a particle in 3D space.
∫C
f(x, y, z) x'(t) dt
b
∫C
g(x, y, z) y'(t) dt
b
∫C
h(x, y, z) dz = ∫t=a
h(x, y, z) z'(t) dt
b

Let F(x, y, z) = <f(x, y, z), g(x, y, z), h(x, y, z)>
be a 3D vector field.

C(t) = <x(t), y(t), z(t)> is a parameterized space curve with
a < t < b.

a < t < b. Let P = C(a) and Q = C(b).

Then the work done for moving the particle from P to Q is

∫C
F • dC = ∫C
F • C' dt =

∫C
F • dC = ∫C
F • C' dt = <f, g, h> • <x', y', z'> dt∫C

∫t=a
f(x, y, z) x'(t) dt +
b
∫t=a
g(x, y, z) y'(t) dt +
b
= ∫t=a
h(x, y, z) z'(t) dt
b
∫C
F • dC = ∫C
F • C' dt = <f, g, h> • <x', y', z'> dt∫C

A piece-wise curve C is the joint of
finitely many curves
C1, C2, .. Cn end to end.
W write C = C1 + C2 + … + Cn

W write C = C1 + C2 + … + Cn
C1
C2
C3
C1 + C2 + C3

W write C = C1 + C2 + … + Cn
Then the work done for moving the particle over C is
∫
C1
F • dC1 +∫F • C' dt =
C1
C2
C3
C1 + C2 + C3
C1+C2+…+Cn
∫
C2
F • dC2 + .. ∫
Cn
F • dCn∫
C
F • dC =

W write C = C1 + C2 + … + Cn
Then the work done for moving the particle over C is
∫
C1
F • dC1 +∫F • C' dt =
C1
C2
C3
C1 + C2 + C3
C1+C2+…+Cn
∫
C2
F • dC2 + .. ∫
Cn
F • dCn∫
C
F • dC =
Example: Evaluate
Write C = C1 + C2 + … + Cn
∫C
f dy where f(x, y) = xy – y
with C as shown.
1
1
C

C = C1 + C2 + C3
1
1
C
C1
C2
C3

C = C1 + C2 + C3 where
1
1
C
C1 = <x(t) = 0, y(t) = t> and 0 < t < 1
C1
C2
C3

1
1
C
C1 = <x(t) = 0, y(t) = t> and 0 < t < 1
C2 = <x(t) = t, y(t) =1 – t> and 0 < t < 1 C1
C2
C3

1
1
C
C1 = <x(t) = 0, y(t) = t> and 0 < t < 1
C2 = <x(t) = t, y(t) =1 – t> and 0 < t < 1
C3 = <x(t) = t, y(t) = 0> and t from 1 to 0
C1
C2
C3

1
1
C
C1 = <x(t) = 0, y(t) = t> and 0 < t < 1
C2 = <x(t) = t, y(t) =1 – t> and 0 < t < 1
For C1: f(x, y) = xy – y = -t, and dy = y'dt = dt.
C1
C2
C3

1
1
C
C1 = <x(t) = 0, y(t) = t> and 0 < t < 1
C2 = <x(t) = t, y(t) =1 – t> and 0 < t < 1
So ∫C1
f dy = ∫t=0
-t dt
1
C1
C2
C3

1
1
C
C1 = <x(t) = 0, y(t) = t> and 0 < t < 1
C2 = <x(t) = t, y(t) =1 – t> and 0 < t < 1
So ∫C1
f dy = ∫t=0
-t dt = -t2
/2 | = -1/2
1
t=0
1
C1
C2
C3

1
1
C
C1 = <x(t) = 0, y(t) = t> and 0 < t < 1
C2 = <x(t) = t, y(t) =1 – t> and 0 < t < 1
So ∫C1
f dy = ∫t=0
-t dt = -t2
/2 | = -1/2
1
t=0
1
For C2: f(x, y) = -(1 – t)2
, and dy = y'dt = -dt.
C1
C2
C3

1
1
C
C1 = <x(t) = 0, y(t) = t> and 0 < t < 1
C2 = <x(t) = t, y(t) =1 – t> and 0 < t < 1
So ∫C1
f dy = ∫t=0
-t dt = -t2
/2 | = -1/2
1
t=0
1
So ∫C2
f dy = ∫t=0
(1 – t)2
dt =
1
C1
C2
C3
For C2: f(x, y) = -(1 – t)2

1
1
C
C1 = <x(t) = 0, y(t) = t> and 0 < t < 1
C2 = <x(t) = t, y(t) =1 – t> and 0 < t < 1
So ∫C1
f dy = ∫t=0
-t dt = -t2
/2 | = -1/2
1
t=0
1
So ∫C2
f dy = ∫t=0
(1 – t)2
dt = -(1 – t)3
/3 | = 1/3
1
t=0
1
C1
C2
C3
For C2: f(x, y) = -(1 – t)2

1
1
C
C1 = <x(t) = 0, y(t) = t> and 0 < t < 1
C2 = <x(t) = t, y(t) =1 – t> and 0 < t < 1
So ∫C1
f dy = ∫t=0
-t dt = -t2
/2 | = -1/2
1
t=0
1
So ∫C2
f dy = ∫t=0
(1 – t)2
dt = -(1 – t)3
/3 | = 1/3
1
t=0
1
For C3: f(x, y) = 0, so ∫C3
f dy = 0.
C1
C2
C3
For C2: f(x, y) = -(1 – t)2

1
1
C
C1 = <x(t) = 0, y(t) = t> and 0 < t < 1
C2 = <x(t) = t, y(t) =1 – t> and 0 < t < 1
So ∫C1
f dy = ∫t=0
-t dt = -t2
/2 | = -1/2
1
t=0
1
So ∫C2
f dy = ∫t=0
(1 – t)2
dt = -(1 – t)3
/3 | = 1/3
1
t=0
1
For C3: f(x, y) = 0, so ∫C3
f dy = 0.
Therefore ∫C
f dy = -1/2 +1/3 = -1/6
C1
C2
C3
For C2: f(x, y) = -(1 – t)2

28 work and line integrals

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