CALCULUS II

Chapter 3: LINE AND SURFACE INTEGRALS

LE THAI THANH

HCMC UNIVERSITY OF TECHNOLOGY

Unit 5: LINE INTEGRALS

LINE INTEGRALS
Suppose a plane curve C is given by the parametric equations

x  x pt q , y  y pt q a¤t
¤b (1)

or, equivalently, Ýr pt q  x pt qÑ
by the vector equation Ñ
Ýi
Ñ
Ý
y pt q j , and we assume that C is a smooth curve. If we divide
the parameter interval ra, b s into n sub-intervals rti
1 , ti s of
equal width and we let xi  x pti q and yi  y pti q, then the
corresponding points Pi pxi , yi q divide C into n sub-arcs with
lengths ∆s1 , ∆s2 , . . . , ∆sn (See Figure). We choose any point
Pi pxi , yi q in the i-th sub-arc. (This corresponds to a point
ti in rti
1 , ti s.) Now if f is any function of two variables
whose domain includes the curve C , we evaluate f at the
point pxi , yi q, multiply by the length ∆si of the sub-arc,
and form the sum
ņ
f pxi , yi q ∆si

i 1
Then we take the limit of these sums and make the following
definition by analogy with a single integral.
Definition 1: If f is defined on a smooth curve C given by
Equations (1), then the line integral of f along C is
» ņ
f px , y q ds  nlim f pxi , yi q ∆si
Ñ8 i 1 
C

if this limit exists.

We can show that if f is a continuous function, then the limit

in the definition always exists and the following formula can
be used to evaluate the line integral:

» »b d 
2 2
f px , y q ds  f px pt q, y pt qq
dx dy
dt (2)
dt dt
C a
NOTES
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.
If s pt q is the length of C between Ñ
Ýr paq and Ñ
Ýr pt q, then
d 
2 2
ds
dt
 dx
dt
dy
dt
d 
2 2
ñ ds  dx
dt
dy
dt
dt
NOTES
In the special case where C is the line segment that
joins pa, 0q to pb, 0q, using x as the parameter, we can
write the parametric equations of C as follows: x  x,
y  0, a ¤ x ¤ b. Formula (2) then becomes
» »b
f px , y q ds  f px , 0qdx
C a

and so the line integral reduces to an ordinary single

integral in this case.
 »
Example 1: Evaluate I  p2 x 2 y qds , where C is the
C
upper half of the unit circle x 2 y2  1.

Solution : The upper half of the unit circle can be

parametrized by means of the equations x  cos t, y  sin t,
0¤t ¤ π. Therefore, Formula (2) gives
»π a
I  p2 cos2 t sin t q sin2 t cos2 tdt
0
»π  t π
 p2 cos t sin t qdt
2
 2t

cos3 t   2π 2
3 t 0 3
0
Suppose now that C is a piecewise-smooth curve; that is, C
is a union of a finite number of smooth curves C1 , C2 ,. . . ,
Cn , where the initial point of Ci 1 is the terminal point of
Ci . Then we define the integral of f along C as the sum of
the integrals of f along each of the smooth pieces of C :
» » » »
f px , y qds  f px , y qds f px , y qds  f px , y qds
C C1 C2 Cn

»
Example 2: Evaluate I  2xds , where C consists of the
C
arc C1 of the parabola y  x 2 from p0, 0q to p1, 1q followed
by the vertical line segment C2 from p1, 1q to p1, 2q.
Solution : For C1 we can choose x as the parameter and
the equations for C1 become: x  x , y  x 2, 0 ¤ x ¤ 1.
Therefore
»1 a »1 a
I1  2x p1q2 p2x q2dx  2x 1 4x 2 dx
0
1 ?0
 14  23 p1 4x 2q3{2  5 56
1
0

On C2 we choose y as the parameter, so the equations of C2

³ ³2
are: x  1, y  y , 1 ¤ y ¤ 2 and I2  2xds  2dy  2.
C 1
?
2
Thus
5 5
1
I  I1 I2  2
6
Suppose that ρpx , y q represents the linear density at a point
px , y q of a thin wire shaped like a curve C . Then the mass
m of the wire is: »
m ρpx , y q ds
C

and the center of mass of the wire is located at the point

px , y q, where » »
x x ρpx , y q ds,  y ρpx , y q ds
1 1
y
m m
C C

Example 3: A wire takes the shape of the semicircle x 2 y 2 

1, y ¥ 0, and is thicker near its base than near the top. Find
the center of mass of the wire if the linear density at any
point is proportional to its distance from the line y  1.
Solution : We use the parametrization x  cos t, y  sin t,
0 ¤ t ¤ π. The linear density is ρpx , y q  k p1
y q, where k
is a constant, and so the mass of the wire is
» »π π
m k p1
y qds  k p1
sin t qdt  k pt cos t q  k pπ
2 q
0
C 0

By symmetry we see that x  0 and

» »
y  m
1
y ρpx , y qds  π
2 1
y p1
y qds
C C
»π
 π
2
1
psin t
sin2 t qdt  2p4π

π2q  0.38
0

The center of mass is p0; 0.38q.

Two other line integrals are obtained by replacing ∆si by
either ∆xi  xi
xi
1 or ∆yi  yi
yi
1 in the definition of
the line integral. They are called the line integrals of f along
C with respect to x and y :
» ņ
f px , y q dx  nlim f pxi , yi q ∆xi (3)
Ñ8 
i 1
C
» ņ
f px , y q dy  nlim f pxi , yi q ∆yi (4)
Ñ8 
i 1
C
When we want to distinguish the original line integral
»
f px , y q ds from those in Equations (3) and (4), we call it
C
the line integral with respect to arc length.
The following formulas say that line integrals with respect to
x and y can also be evaluated by expressing everything in
terms of t: x  x pt q, y  y pt q, dx  x 1pt qdt, dy  y 1pt qdt
» »b
f px , y q dx  f px pt q, y pt qqx 1 pt qdt (5)
C a
» »b
f px , y q dy  f px pt q, y pt qqy 1 pt qdt (6)
C a

It frequently happens that line integrals with respect to x

and y occur together. When this happens, it is customary to

»
abbreviate »
by writing »
P px , y qdx Q px , y qdy  P px , y qdx Q px , y qdy
C C C
 »
Example 4: Evaluate I  px 2 y 2 qdx xydy , where (a)

 C1 is the line segment from p0, 0q to p1, 1q and (b)

C
C
C  C2 is the arc of the parabola y  x 2 from p0, 0q to
p1, 1q.

Solution : (a) A parametric representation for the line seg-

ment is x  x , y  x , 0 ¤ x ¤ 1. Therefore dy  dx and
»1 »1 1
I  px 2
x qdx
2
x dx2
 3x 2
dx  x   1
3
0
0 0
(b) A parametric representation for the parabola is x  x,
y  x 2, 0¤x ¤ 1. Therefore dy  2xdx and
»1 »1
I  px 2
px q qdx
2 2
x px q2xdx
2
 px 2 3x 4 qdx

0 1 0

 x3 3x 5   14
3 5 0 15

NOTES
In general, the value of a line integral depends not just
on the endpoints of the curve but also on the path.
The value of a line integral depends on the direction, or
orientation, of the curve.
If
C denotes the curve consisting of the same points as C
but with the opposite orientation, then we have
» »
Pdx Qdy 
Pdx Qdy

C C

But if we integrate with respect to arc length, the value of the

line integral does not change when we reverse the orientation
of the curve.
» »
f px , y qds  f px , y qds

C C

This is because ∆si is always positive, whereas ∆xi and ∆yi

change sign when we reverse the orientation of C .
LINE INTEGRALS
IN SPACE
We now suppose that C is a smooth space curve given by the
parametric equations

x  x pt q,  y pt q, z  z pt q,
y a¤t ¤b
Ýr pt q  x pt qÑ
or by a vector equation Ñ
Ýi Ñ
Ý z pt qÑ
y pt q j
Ýk .
If f is a function of three variables that is continuous on
some region containing C , then we define the line integral of
f along C (with respect to arc length) in a manner similar
to that for plane curves and we evaluate it using a formula
similar to Formula (2):
» »b
f px , y , z q ds  f px pt q, y pt q, z pt qq 
C
d
a
  
2 2 2
 dx
dt
dy
dt
dz
dt
dt (7)

Observe that the integrals in both Formulas (2) and (7) can
be written in the more compact vector notation

» »b  
f px , y , z q ds  f pÑ
Ýr pt qq Ñ
Ýr 1pt q dt
C a
For the special case f px , y , z q  1, we get

» »b 
ds  Ñ Ýr 1pt q dt  L
C a
where L is the length of the curve C . Line integrals along
C with respect to x , y , and z can also be defined similarly.
Therefore, as with line integrals in the plane, we evaluate
integrals of the form
»
P px , y , z qdx Q px , y , z qdy R px , y , z qdz
C

by expressing everything px , y , z, dx , dy , dz q in terms of the

parameter t.
 »
Example 5: Evaluate I  y sin zds , where C is the cir-

 cos t, y  sin t, z  t,
C
cular helix given by the equation x
0¤t ¤ 2π.

Solution : Formula (7) gives

»2π a ? »2π
I  psin t qpsin t q sin2 t cos2 t 1dt  2 sin2 tdt
0 0
? »2π ?  2π ?
 2 p1
cos 2t qdt  2 t
2 sin 2t  π 2
2 2 1
0
0
Now suppose that F
Ñ
Ý  PÑ
Ýi Ñ
Ý
Q j
Ñ
Ý
R k is a continuous
vector field on R3 , such as the force field, the gravitational
field, the electric force field, etc., defined on a smooth curve
C given by a vector function Ñ
Ýr pt q, a ¤ t ¤ b. Then the line
Ñ
Ý
integral of F along C is

» » »
Ñ
Ý Ñ
Ý Ýr pt qq  Ñ
b

Pdx Qdy Rdz  F  d r  F pÑ

Ñ
Ý Ýr 1pt q dt
C C a

Ñ
Ý
In particular, the work done by the force field F in moving a
particle along a smooth curve C is
»
W  Ñ
ÝF  d Ñ
Ýr
C
Example 6: Find the work done by the force field
Ñ
ÝF px , y , z q  xy Ñ
Ýi Ñ
Ý
yz j
Ñ
Ý
zx k in moving a particle along
the twisted cubic given by x  t, y  t 2, z  t 3, 0 ¤ t ¤ 1.

Solution : We have Ñ  tÑ
Ýr pt q Ýi t 2Ñ Ýj t 3Ñ
Ýk , Ñ
Ýr 1pt q 
Ñ
Ýi 2t Ñ
Ýj 3t 2Ñ
Ýk , and Ñ
ÝF pÑ
Ýr pt qq  t 3Ñ
Ýi t 5Ñ
Ýj t 4Ñ Ýk .
Thus
» » »
Ñ
Ý Ñ
Ý
1 1
Ñ
Ý Ñ
Ý Ñ
Ý 1
W  F d r  F p r pt qq r pt q dt  pt 3 5t 6 qdt  28
27

C 0 0
INDEPENDENCE OF PATH
Theorem 1: Let C be a smooth curve given by the vector
function Ñ
Ýr pt q, a ¤ t ¤ b. Let f be a differentiable function
of two or three variables whose gradient vector ∇f is contin-
uous on C . Then
»
∇f  d Ñ
Ýr  f pÑ
Ýr pbqq
f pÑ
Ýr paqq
C

NOTE
Theorem says that we can evaluate the line integral of a con-
servative vector field (the gradient vector field of the potential
function f ) simply by knowing the value of f at the endpoints
of C . If f is a function of two variables and C is a plane curve
with initial point Apx1 , y1 q and terminal point B px2 , y2 q, as
in Figure (a), then Theorem becomes
 »
∇f  d Ñ
Ýr  f px2, y2q
f px1, y1q
C

If f is a function of three variables and C is a space curve

joining the point Apx1 , y1 , z1 q to the point B px2 , y2 , z2 q, then
»
we have ∇f d Ñ
Ýr  f px2, y2, z2q
f px1, y1, z1q
C
Example 7: Find the work done by the gravitational field

Ñ
ÝF pÑ
Ýr q 
mMG3 ÑÝ
|Ñ
Ýr | r
in moving a particle with mass m from the point p3, 4, 12q to
the point p2, 2, 0q along a piecewise-smooth curve C .

Ñ
Ý
Solution : We know that F is a conservative vector field
Ñ
Ý  ∇f , where
and, in fact, F

f pr q  ô f px , y , z q  a
mMG mMG
r x2 y2 z2
Solution : Therefore, the work done is
» »
W  Ñ
ÝF  d Ñ
Ýr  ∇f  d Ñ
Ýr  f p2, 2, 0q
f p3, 4, 12q
C C

 ? 2 mMG2 2
? 2 mMG
2
2 0 3 42 122
 mMG ?1
131
2 2
Suppose C1 and C2 are two piecewise-smooth curves (which
are called paths) that have the same initial point A and ter-
minal point B. In general,
» »
Ñ
ÝF  d Ñ
Ýr  ÑÝF  d Ñ
Ýr
C1 C2

But one implication of the above theorem is that

» »
∇f  d Ñ
Ýr  ∇f  d Ñ
Ýr
C1 C2

whenever ∇f is continuous. In other words, the line integral

of a conservative vector field depends only on the initial point
and terminal point of a curve.
 Ñ
Ý
» vector field
Definition 2: In general, if F is a continuous
with domain D, we say that the line integral
Ñ
ÝF  d Ñ
Ýr is
C
independent of path if
» »
Ñ
ÝF  d Ñ
Ýr  ÑÝF  d Ñ
Ýr
C1 C2

for any two paths C1 and C2 in D that have the same initial
and terminal points.
With this terminology we can say that line integrals of con-
servative vector fields are independent of path.

Definition 3: A curve is called closed if its terminal point

coincides with its initial point, that is, Ñ
Ýr paq  Ñ
Ýr pbq.
We have the following theorems.
»
Theorem 2:
Ñ
ÝF d Ñ
Ýr is independent of path in D if and
»
Ñ
ÝF  d Ñ
C

only if Ýr  0 for every closed path C in D.

C

Ñ
Ý
» field that is continuous on
Theorem 3: Suppose F is a vector
an open connected region D. If
Ñ
ÝF  d Ñ
Ýr is independent
Ñ
Ý C
of path in D, then F is a conservative vector field on D; that
is, there exists a function f such that ∇f Ñ
ÝF .
 Ñ
Ý Ñ
Ý
Theorem 4: If F px , y q  P px , y q i
Ñ
Ý
Q px , y q j is a con-
servative vector field, where P and Q have continuous first-
order partial derivatives on a domain D, then throughout D
we have
BP  BQ
By Bx

Theorem 5: Let F
Ñ
Ý  PÑ
Ýi Ñ
Ý
Q j be a vector field on an
open simply-connected region D. Suppose that P and Q have
continuous first-order derivatives and
BP  BQ
By Bx throughout D

Ñ
Ý
Then F is conservative.
Example 8:
Ñ
Ý
(a) Given F px , y q  p3
Ý px 2
3y 2qÑ
Ñ
2xy q i
Ýj .
Ñ
Ý
Determine whether or not the vector field F is
conservative.
Ñ
Ý  ∇f .
(b) Find a function f such that F
»
(c) Evaluate the line integral F  dr , where C is the
curve given by Ñ
Ýr pt q  et sin t i
C
Ñ
Ý Ñ
Ý
et cos t j ,
0¤t ¤π

Solution :
(a) We have P px , y q  3 2xy and Q px , y q  x 2
3y 2 ,
BP  2x  BQ .
and
By Bx
 Ñ
Ý
Also, the domain of F is the entire plane D  R2, which is
open and simply-connected.
Ñ
Ý
Therefore we conclude that F
is conservative.
Ñ
Ý
(b) F is conservative and so there exists a function f with
Ñ
Ý
∇f  F , that is,
B f  P px , y q, B f  Q px , y q .
Bx By
Integrating the first equation with respect to x , we
obtain
f px , y q  3x x 2y g py q

Notice that the constant of integration is a constant

with respect to x , that is, a function of y , which we
have called g py q.
Next we differentiate both sides of this equation with respect
to y :

f y px , y q  x 2 g 1 py q  Q px , y q  x 2
3y 2

We obtain g 1 py q 
3y 2 ñ g py q 
y 3 C , where C is a

constant. Finally

f px , y q  3x x 2y
y3 C

(c) We have to calculate the initial and terminal points of

C: Ñ
Ýr p0q  p0; 1q and Ñ
Ýr pπq  p0;
eπ q. Therefore
» »
Ñ
ÝF  d Ñ
Ýr  ∇f  d Ñ Ýr  f p0,
eπ q
f p0, 1q
C C
 r
p
eπ q3 C s
r
p1q C s  e3π 1
 Ñ
Ý Ñ
Ý p2xy
Example 9: If F px , y , z q  y 2 i
Ñ
Ý
e3z q j
Ñ
Ý
3y e3z k ,
find a function f
Ñ
Ý
such that F  ∇f .

Solution : If there is such a function f , then

Bf  y 2, Bf  2xy e3z ,
Bf  3y e3z
Bx By Bz
Integrating the first equation with respect to x , we get

f px , y , z q  xy 2 g py , z q

Thus
Bf  2xy gy py , z q  2xy e3z ñ gy py , z q  e3z
By
So

g py , z q  y e3z hpz q ñ f px , y , z q  xy 2 y e3z h pz q

Therefore
Bf  3y e3z h1 pz q  3y e3z ñ h 1 pz q  0 ñ h pz q  C
Bz
Finally,
f px , y , z q  xy 2 y e3z C
It is easily verified that F
Ñ
Ý  ∇f .
GREEN’S THEOREM
Green’s Theorem gives the relationship between a line integral
around a simple closed curve C and a double integral over the
plane region D bounded by C (C is a boundary of D). We
assume that D consists of all points inside C as well as all
points on C .
In stating Green’s Theorem we
use the convention that the posi-
tive orientation of a simple closed
curve C refers to a single coun-
terclockwise traversal of C . Thus
if C is given by the vector func-
tion Ñ
Ýr pt q, a ¤ t ¤ b, then the
region D is always on the left as
the point Ñ Ýr pt q traverses C .
Theorem 6 (Green’s Theorem): Let C be a positively ori-
ented, piecewise-smooth, simple closed curve in the plane and
let D be the region bounded by C . If P and Q have contin-
uous partial derivatives on an open region that contains D,
then ¾ ¼  BQ BP
Pdx Qdy 
Bx
By dA
C D

NOTE

¾
The notation Pdx Qdy is used to indicate that the
C
line integral is calculated using the positive orientation of the
closed curve C .
 ¾
Example 10: Evaluate I  x 4 dx xydy , where C is the

triangular curve consisting of the line segments from p0, 0q to

C

p1, 0q, from p1, 0q to p0, 1q, and from p0, 1q to p0, 0q.

Solution : Although the given line integral could be evaluated

directly as usual, that would involve setting up three separate
integrals along the three sides of the triangle, so let’s use
Green’s Theorem instead. Notice that the region D enclosed
by C is simple and C has positive orientation.
If we let P px , y q  x 4 and Q px , y q  xy , then we have
BQ  y , BP  0
Bx By
and
¾ ¼ » 1 » 1
x
I  4
x dx xydy  py
0qdA  ydydx
0 0
» 1  y 2 1
x
C D
»1 1 
 2
dx  1
2
p1
x q dx 
16 p1
x q3
2
 16
0 0 0 0
Example 11: Evaluate
¾ a
I  p3y
esin x qdx p7x y4 1qdy
C

where C is the circle x 2 y2  9.

Solution : Let
a
P px , y q  3y

esin x , Q px , y q  7x y4 1
BQ  7, BP  3 and
we have
Bx By
¼
I  p7
3qdA  4ApD q  4  32π  36π
D
 ¼ Theorem
One application of the reverse direction of Green’s

is in computing areas. Since the area of D is 1 dA ,

BQ
BP  1.
D

we wish to choose P and Q so that

Bx By There
are several possibilities:

P  0, Q  x or P 
y , Q  0 or P 
y2 , Q  x2
Then Green’s Theorem gives the following formulas for the
area of D:
¾ ¾ ¾
A pD q  

y dx
1
x dy y dx x dy
2
C C C
 x2 y2
Example 12: Find the area enclosed by the ellipse
a2 b2

1.

Solution : The ellipse has parametric equations x  a cos t

and y  b sin t, where 0 ¤ t ¤ 2π. We have
¾
A  1
2
x dy
y dx
1 2π
»
C

 2 0
rpa cos t qpb cos t q
pb sin t qp
a sin t qsdt
»
ab 2π
 2 0
dt  πab
 Ñ
Ý
Example 13: If F px , y q  x 2
yy 2 Ñ
Ýi x Ñ
Ýj , show
» x2 y2
that
Ñ
ÝF  d Ñ
Ýr  2π for every positively oriented simple
C
closed path that encloses the origin.

Solution :
Since C is an arbitrary closed path
that encloses the origin, it’s dif-
ficult to compute the given in-
tegral directly. So let’s consider
a counterclockwise-oriented circle
C 1 with center the origin and ra-
dius a, where a is chosen to be
small enough that C 1 lies inside
C . (See Figure)
Let D be the region bounded by C and C 1 . Then its positively
oriented boundary is C Yp
C 1 q and so Green’s Theorem gives
» »
I  Pdx Qdy Pdx Qdy
¼  BQ
C
C 1


BP dA
Bx By
¼  y2
x2
D
y2
x2

 px 2 y 2q2
px 2 y 2q2 dA  0
D

Therefore
» » » »
Pdx Qdy  Pdx Qdy ô Ñ
ÝF  d Ñ
Ýr  ÑÝF  d Ñ
Ýr
C C1 C C1
We now easily compute this last integral using the
parametrization given by Ñ
Ýr pt q  a cos t
Ñ
Ýi a sin t
Ñ
Ýj ,
0¤t ¤ 2π. Thus
» »
Ñ
ÝF  d Ñ
Ýr  Ñ
ÝF  d Ñ
Ýr
»C2π
1
C

 Ñ
ÝF pÑ
Ýr pt qq  Ñ
Ýr 1pt q dt
»02π p
asint qp
a sin t q pa cos t qpa cos t q
 a2 cos2 t a2 sin2 t
dt
» 2π
0

 dt  2π
0
That’s all. Thanks a lot

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