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Chapter 9. Line Integrals
9.1 Introduction
9.1.1 Work Done I
(i) Let F be a constant force acting on a particle
in the displacement direction as shown in figure
(i) above. Suppose the distance moved by the
particle iss. The work done is given by
W = F s.
(ii) Let F be a constant force acting on a particle
in the direction which form an angle against
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the displacement direction (see figure (ii) above).
Suppose the distance moved by the particle is s.
The work done is given by
W = F cos s= (F T) s=F sTwhere T is the unit vector in the displacement
direction.
9.1.2 Work Done II
LetF(x,y,z) be a variable force acting on a particle
which moves along the curveCwith vector equation
r(t) = x(t)i+y(t)j+z(t)k as shown in the figure
below. Suppose the particle moves from point P to
pointQ. What is the work done?
Solution: To solve this problem, we divide the
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curve C into n segments. If a segment i is small
enough, it can be treated as a straight line segment
and the force within which can be assumed to be con-
stantFi. Then the work done for such a segment is
approximately given by
Wi Fi ri
where ri = sTi and Ti is the unit tangent vector
along this segment.
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9.2 Vector Fields
9.2.1 Vector field (two variables)
Let R be a region in xy-plane. A vector field on
R is a vector function F that assigns to each point
(x, y) inRa two-dimensional vectorF(x, y).
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We may write F(x, y) in terms of its component func-
tions. That is
F(x, y) =P(x, y)i +Q(x, y)j
or simplyF=Pi +Qj.
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9.2.2 Vector field (three variables)
Let Dbe a solid region in xyz-space. A vector field
onDis a vector functionFthat assigns to each point
(x,y,z) in D a three-dimensional vector F(x,y,z).
That is, F(x,y,z) =P(x,y,z)i +Q(x,y,z)j +R(x,y,z)k.
9.2.3 Example
A vector field in xy-plane is defined by F(x, y) =
(y)i+xj. Show thatF(x, y) is always perpendic-
ular to the position vector of the point (x, y).
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F(x, y) = (y)i+ xj
The diagram above shows the vector fieldF.
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9.2.5 Example
The gradient field off(x, y) =xy2 +x3 is
f(x, y) = (y2 + 3x2)i + (2xy)j.
9.2.6 Conservative fields
A vector field F is called a conservative vector field
if it is the gradient of some (scalar) function. In other
words, there is a function f such that F =f. In
this case,f is called apotentialfunction forF.
9.2.7 Example
By Example 9.2.5,F(x, y) = (y2 + 3x2)i + (2xy)jis
conservative since it has a potential function f(x, y) =
xy2
+x3
.
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9.2.8 Example
Let F(x, y) = (3 + 2xy)i + (x2 3y2)j. Find a
potential functionf forF.
Solution: Asf=F, we have fx(x, y) = 3+2xy.Integrating with respect tox, we getf(x, y) = 3x +
x2y +g(y), whereg(y) is an integration constant, but
it could be a function ofy.
Thus fy(x, y) = x2 +g(y) so that x2 + g(y) =
x2 3y2. That is,g(y) = 3y2.
Integratingg (y) with respect toy, we obtaing(y) =
y3 +K, whereKis a constant.
Consequently,f(x, y) = 3x+x2y y3 +K.
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9.2.9 Example
The gravitational field given by
G=
m1m2Kx(x2 +y2 +z2)
32
i
+
m1m2Ky(x2 +y2 +z2)
32
j+
m1m2Kz(x2 +y2 +z2)
32
k
is conservative because it is the gradient of the grav-
itational potential function
g(x,y,z) = m1m2K
x2 +y2 +z2,
where K is the gravitational constant, m1 and m2
are the masses of two objects. Think of the mass
m1 at the origin that creates the field and g is the
potential energy attained by the massm2situated at
(x,y,z).
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9.2.10 Criteria of conservative fields
Throughout this chapter, we will assume the compo-
nent functions of any vector field to have continuous
partial derivatives, unless otherwise stated.
(a) Let F(x, y) = P(x, y)i+Q(x, y)j be a vector
field on thexy-plane.
If P
y =
Q
x, thenFis conservative.
(b) Let F(x,y,z) =P(x,y,z)i+Q(x,y,z)j+R(x,y,z)k
be a vector field on the xyz-space.
If P
y =
Q
x,
P
z =
R
x,
Q
z =
R
y,
thenFis conservative.
The converse of (a) and (b) also hold.
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9.2.11 Example
Consider the vector field
F(x, y) = (3 + 2xy)i + (x2 3y2)j.
As
(x2
3y2)
x = 2x=
(3 + 2xy)
y ,
Fis conservative.
9.2.12 Example
Show thatF(x,y,z) =xzi + xyzj y2kis not con-
servative.
Solution: For example,(xyz)
x =yzwhich is not
equal to(xz)
y = 0.
SoFis not conservative.
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9.2.13 Exercise
Show that the vector field F(x,y,z) =x2i+y2j+z2k
is conservative. Find a functionfsuch that f=F.
9.3 Line Integrals
We have mentioned in section 9.1 that a line integral
refers to an integration along a curve C. There are
two types of line integrals. One is for vector fields
and the other is for scalar functions.
9.3.1 Line integrals of scalar functions (Two
variables)
Suppose we want to find the area of the following
surface with the base, a plane curve C on the xy-
plane and the top is described by a functionf(x, y).
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Let C be described by the vector function r(t) =
x(t)i+y(t)j for a t b. We assume C is asmooth curve (meaning that r(t)= 0) and r(t) is
continuous for allt.
To find the surface area along C, we subdivide the
curve from r(a) to r(b) into n small arcs of length
si,i= 1, n. Pick an arbitrary point (xi , yj ) in-
side the ith small arc and form the sumn
i=1 f(xi , yj )si.14
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The surface area is given by
limn
ni=1
f(xi , y
j )si.
The above limit of sum is called the line integral
of the scalar function f along the plane curve
C, and is denoted by
C f(x, y) dsHere,sdenotes the arc length ofC.
Recall from Chapter 6 that, ifr(t) is the vector func-
tion of a curve Cwith a
t
b, the arc length of
Cis given by
s=
ba
r(t)dt.
If we replace the endpoint bby a variable t, then we
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obtain the arc lengths(t) as a function oft:
s(t) =
ta
r(u)du.
Then, by Fundamental Theorem of Calculus, ds
dt =
r(t).Therefore, we can rewrite the line integral in terms
oft:
C
f(x, y) ds= b
af(x(t), y(t))r(t)dt
=
ba
f(x(t), y(t))
dx
dt
2+
dy
dt
2dt
where r(t) = x(t)i+y(t)j is the vector equation of
the plane curveC.
In other words, the formula above allows us to com-
pute the line integral in terms of ordinary integration
of single variable function (int).
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9.3.2 Example
Evaluate
C
(2y+x2y)ds, whereCis the upper half
of the unit circle centered at the origin.
Solution: The vector function of C is given by
r(t) = cos ti + sin tjwith 0 t .
Thus r(t) =
sin2 t+ cos2 tand
C
(2y+x2y)ds=
0
(2sin t+ cos2 t sin t)
sin2 t+ cos2 tdt
=
0
(2sin t+ cos2 t sin t) dt
=
2cos t 1
3cos3 t
0
=14
3
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9.3.3 Line integrals of scalar functions (Three
variables)
For line integral of a functionf(x,y,z) along a space
curveC, we have the similar definitions:
C
f(x,y,z) ds=
ba
f(x(t), y(t), z(t))
dx
dt
2+
dy
dt
2+
dz
dt
2dt
9.3.4 Example
Evaluate
C
xy sin z ds, whereCis the circular helix
r(t) = cos ti + sin tj +tk,t
[0, /2].
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Solution:
C
xy sin z ds
=
/20
(cos t)(sin t)(sin t)
sin2 t+ cos2 t+ 1 dt
=2 /20
cos t sin2 tdt
=
2
3
sin3 t
/20
=
2
3
9.3.5 Piecewise smooth curves
We denote the union of a finite number of (smooth)
curvesC1, C2, , Cnby
C=C1+C2+ +Cn.
We sayC is apiecewise-smoothcurve.
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Then the line integralf alongC is defined to be
C
f(x, y) ds=
C1
f(x, y) ds + +
Cn
f(x, y) ds.
9.3.6 Example
Evaluate
C
9y ds, whereCconsists of the arcC1of
the cubicy=x3 from (0, 0) to (1, 1) followed by the
vertical line segmentC2 from (1, 1) to (1, 5).
Solution: We first obtain the vector function forC1
andC2:
ForC1, the Cartesian equation isy =x3. So we may
let x = t and get y = t3 with 0 t 1. Hence
r1(t) =ti +t3jand r1(t) =
1 + (3t2)2
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9.3.7 Line integrals of vector fields
Let F be a continuous (2 or 3 variable) vector field
defined on a domain containing a smooth curve C
given by a vector function r(t), t[a, b]. Thelineintegral of the vector field Falong the curveC
is
C F dr=
ba F(r(t)) r(t) dt.
Ifr(t) =x(t)i +y(t)j +z(t)k, then
F(r(t)) =F(x(t), y(t), z(t))
refers to the vector field along the curve.
Geometrically, the line integral ofF over C is sum-
ming up the tangential component ofF with respect
to the arc length ofC.
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9.3.8 Example
Evaluate
C
F dr, where
F(x,y,z) =xi +xyj +xyzk
andCis the curver(t) =ti +t2j +t3k,t [0, 2].
Solution: Firstr(t) =i + 2tj + 3t2k. Thus
F(r(t)) r(t) = (ti +t t2
j +t t2
t3
k) (i + 2tj + 3t2
k)
=t+ 2t4 + 3t8.
Therefore,
C
F dr=
2
0
F(r(t)) r(t) dt
=
2
0
(t+ 2t4 + 3t8) dt= 2782/15.
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9.3.9 Orientation of curves
The vector equation of a curveCdetermines anori-
entation(direction) ofC. The same curve with the
opposite orientation ofCis denoted by C.
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BC
A
BC
We have C
F dr=
C
F dr
asr(t) changes sign in C.
On the other hand, for line integral of scalar func-
tions,
Cf(x,y,z) ds=
Cf(x,y,z) ds
since the arc length is always positive.
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9.3.10 Line integrals in component form
Suppose
F(x, y) =P(x, y)i +Q(x, y)j
andC :r(t) =x(t)i +y(t)j, t [a, b].
Then we may write the line integral as
C F dr= C P dx+Qdy.Indeed
C
F dr
= ba
F(r(t))
r(t) dt
=
ba
[P(r(t))i +Q(r(t))j]
dx
dti +
dy
dtj
dt
=
ba
P(r(t))
dx
dt+Q(r(t))
dy
dt
dt
=
CP dx+Qdy.
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Similarly, for three variable vector field
F=Pi +Qj +Rk,
we can write the line integral as
C F dr= C P dx+Qdy+Rdz.
9.3.11 Example
Evaluate the line integral C y2dx+xdy, where
(a) C = C1 is the line segment from (5,3) to
(0, 2),
(b) C = C2 is the arc of the parabola x = 4 y2
from (5,3) to (0, 2).
Solution:
(a) C1 is a line passing through the point (5,3)
and parallel to the vector (2j) (5i3j) = 5i + 5j.26
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So the vector function ofC1 is given by
r(t) = (5i 3j) + t(5i + 5j) = (5t 5)i + (5t 3)j
with 0 t 1. Thus,
C1 y
2
dx+xdy = 10 (5t 3)
2dx
dt dt+ 10 (5t 5)
dy
dt dt
=
1
0
(5t 3)25dt+
1
0
(5t 5)5dt
=5/6.
(b) By settingy=t, we have the vector function of
C2 given by
r(t) = (4 t2)i +tj with 3 t 2. Thus
C2
y2dx+xdy =
2
3t2
dx
dtdt+
2
3(4 t2)dy
dtdt
=
2
3t2(2t)dt+
2
3(4 t2)dt
= 245/6.
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9.3.12 The fundamental theorem for line in-
tegrals
Recall the fundamental theorem for Calculus:
b
a
F(x) dx=F(b)
F(a).
It has the following generalization in terms of line
integrals:
Let Cbe a smooth curve with vector function r(t),
t [a, b].
Iff is a function of 2 or 3 variables whose gradient
f is continuous. ThenC
f dr=f(r(b)) f(r(a)).
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9.3.13 Example
Find the work done by the (earth) gravitational field
(see Example 9.2.9) in moving a particle of mass m
from the point (3, 4, 12) to the point (1, 0, 0) along a
curveC.
Solution:
W C
G dr= C
g dr=g(1, 0, 0)g(3, 4, 12).
Since the potential function g(x,y,z) = mM K
x2 +y2 +z2
whereM is the mass of the earth and Kthe gravi-
tational constant, we have W = 12mMK/13.
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9.3.14 Consequences of conservative fields
(I) IfFis a conservative vector field, then
C
F dr
isindependent of path,
i.e.
C1F dr = C2 F dr for any 2 paths C1 and
C2 that have the same initial and terminal points.
(II) IfF is a conservative vector field, then
F dr= 0
for any closed curve (i.e. a curve with terminal
point coincides with its initial point).
Notation: If a curve is closed, we write the line
integral as
F dr.
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9.3.15 Example
Let F(x, y) = (y2+3x2)i+(2xy)j. Show that the line
integral
C
F dr is independent of path and evaluate
this integral over the curveCwhereC is
(i) given byr(t) = cos ti +et sin tj, t [0, ];
(ii) the unit circle.
Solution: We have seen in Example 9.2.5 that
f = F where f(x, y) = xy2 +x3 is the potential
function ofF. So F is conservative. By section 9.3.14
(I), the line integral
CF dris independent of path.
(i) The initial point ofCis givenr(0) =iwhich cor-
responds to the coordinates (1, 0); and the terminal
point is given r() =i which corresponds to the31
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coordinates (
1, 0). SinceF=
f, by Fundamental
Theorem, we have
C
F dr=f(1, 0) f(1, 0) = 2.
(ii) Since the unit circle is a closed path and F is
conservative, by section 9.3.14 (II),
C F dr= 0.9.4 Greens Theorem
LetD be a bounded region in the xy-plane andD
the boundary of D. Suppose P(x, y) and Q(x, y)
has continuous partial derivatives onD. Then
D
P dx+Qdy =
D
Q
x P
y
dA.
The orientation ofD is such that, as one traverses
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along the boundary in this direction, the region D
is always on the left hand side. We call this the
positive orientationof the boundary.
9.4.1 Example
Evaluate
C
2xy dx+xy2dy, where C is the trian-
gular curve consisting of the line segments from (0, 0)
to (2, 0), from (2, 0) to (0, 2) and from (0, 2) to (0, 0).
Solution: The functions
P(x, y) = 2xy andQ(x, y) =xy2
have continuous partial derivatives on the xy-plane.
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.......
(0, 2)
(0, 0) (2, 0)x
y= 2 x
C
D
33
-
8/10/2019 NUS - MA1505 (2012) - Chapter 9
34/38
34 MA1505 Chapter 9. Line Integrals
The regionDis given by: 0
y
2
x, 0
x
2.
By Greens Theorem,
C
2xy dx+xy2dy =
D
(xy2)
x 2xy
y
dA
=
D(y2
2x) dydx=
2
0
2x
0
(y2 2x) dydx
=43
.
9.4.2 Example
Evaluate
C
(4y ex2)dx+ (9x+ sin(y2 1))dy, where
Cis the circlex2 +y2 = 4.
Solution:Cbounds the circular diskDof radius 2
and is given the positive orientation.
34
-
8/10/2019 NUS - MA1505 (2012) - Chapter 9
35/38
35 MA1505 Chapter 9. Line Integrals
By Greens Theorem,
C
(4y ex2)dx+ (9x+ sin y2 1)dy
=
D
(9x+ sin y2 1)
x (4y e
x2)
y
dA
=
D
5 dA= 5
D
dA
= 5 (area ofD) = 5(22) = 20.
9.4.3 Exercise
Evaluate by Greens Theorem
C
ex sin y dx+ex cos y dy
whereCis the rectangle with vertices at (0, 0), (, 0),
(,/2), (0, /2).
[Answer: 2(e 1)]
35
-
8/10/2019 NUS - MA1505 (2012) - Chapter 9
36/38
36 MA1505 Chapter 9. Line Integrals
9.4.4 Example
Let F(x, y) = yi+yj and D a region in xy-plane
bounded by the two circles centered at the origin with
radius 1 and 2.
......
......
......
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..................
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.....
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......
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..............
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...................
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.....................
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...............
..............
............................................................................................
......
......
......
......
......
......
.....
......
......
......
......
......
......
......
.....
......
......
......
......
......
......
......
.....
......
................
......
......
......
.....
......
......
......
......
......
......
......
.....
......
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......
......
......
......
......
......
......
......
......
......
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.....
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...................
................
y
................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... ................ x
...........................................
.....
...................
................
C2
C1
D
1 2
Verify Greens Theorem.
Solution:
(i) Compute
D
F drdirectly:
The boundary of D is made up of two disjoint
curvesC1 andC2.
36
-
8/10/2019 NUS - MA1505 (2012) - Chapter 9
37/38
37 MA1505 Chapter 9. Line Integrals
NowC1: r1= cos ti + sin tj and C2: r2=
2cos ti + 2 sin tj with t [0, 2]. Note that
the equations give counterclockwise orientation to
both curves.
However, to get positive orientation for the bound-
ary of D, the outer boundary should traverse
counterclockwise while the inner boundary should
traverse clockwise.
HenceD=C2 C1.
C1
F
dr =
2
0
(sin ti + sin tj)
(
sin ti + cos tj) dt
=
2
0
( sin2 t+ sin t cos t) dt
=
2
0
1
2(cos2t 1 + sin 2t) dt
= 1
2sin2t
2 t cos2t
220
= 37
-
8/10/2019 NUS - MA1505 (2012) - Chapter 9
38/38
38 MA1505 Chapter 9. Line Integrals
Similarly,
C2F dr= 4.
So
D
F dr=
C2
F dr
C1
F dr= 3.
(ii) Using Greens Theorem, we have
D
Fdr=
D
yx
yy
dA=
D
(1) dA.
In polar coordinates,D is given by
1 r 2, 0 2.
So we have
D
(1) dA= 2
0 2
1
r dr d= 3.