nus - ma1505 (2012) - chapter 9

8/10/2019 NUS - MA1505 (2012) - Chapter 9

1/38

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

8/10/2019 NUS - MA1505 (2012) - Chapter 9

2/38

2 MA1505 Chapter 9. Line Integrals

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

2

8/10/2019 NUS - MA1505 (2012) - Chapter 9

3/38


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.

3

8/10/2019 NUS - MA1505 (2012) - Chapter 9

4/38

8/10/2019 NUS - MA1505 (2012) - Chapter 9

5/38


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).

......

.....

......

......

......

......

......

......

.....

......

......

......

......

......

......

.....

......

......

......

......

......

......

.....

......

......

......

......

......

......

.....

......

......

......

......

......

......

.............

......

......

......

......

......

.....

......

......

......

......

......

......

.....

......

......

......

......

......

......

.....

......

......

......

......

......

......

.....

......

......

......

......

.................

................

y

....................................................................................................................................................................................................................................................................................................................................................................................................................................... ................ x

...........................................................

...........................................................

............................................................................

.........................

....

................

...........................................................................................

................

.....................

......................................................

.......... .......................................

...........................

...

................

......................

......................

......................

.......................................

R

F(x, y)..................

............

.............

...............

...................

.........................

.........................................................................................................................................................................................................................................

..................................................................................................................................................................................................................................................................................................................................

..................

..............

...............

.....................................................................................................

.......................

............

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.

5

8/10/2019 NUS - MA1505 (2012) - Chapter 9

6/38


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).

......

......

.....

......

......

......

......

......

......

.....

......

......

......

......

......

......

.....

......

......

......

......

......

......

......

.....

......

......

......

......

......

......

......

......

......

......

......

......

......

......

.....

......

......

......

......

......

......

.....

.............

......

......

......

......

......

.....

..................

................

y

...................................................................................................................................................................................................................................................................................................................................................... ................ x........

......

......

......

......

.....

......

......

......................

................

.................

................

.................................

................

..................................................................

..............

..

..................................................................

................

..............

..............

..............

........................

................

.............................................................

..........

......

..........

.......................................................................................

....................................................................... ................

.....

......

......

......

.....................

................

........................................................................................................

......

..........

............................................ ................

F(x, y) = (y)i+ xj

The diagram above shows the vector fieldF.

6

8/10/2019 NUS - MA1505 (2012) - Chapter 9

7/38

8/10/2019 NUS - MA1505 (2012) - Chapter 9

8/38


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

.

8

8/10/2019 NUS - MA1505 (2012) - Chapter 9

9/38


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.

9

8/10/2019 NUS - MA1505 (2012) - Chapter 9

10/38


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).

10

8/10/2019 NUS - MA1505 (2012) - Chapter 9

11/38


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.

11

8/10/2019 NUS - MA1505 (2012) - Chapter 9

12/38


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.

12

8/10/2019 NUS - MA1505 (2012) - Chapter 9

13/38


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).

13

8/10/2019 NUS - MA1505 (2012) - Chapter 9

14/38


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

8/10/2019 NUS - MA1505 (2012) - Chapter 9

15/38


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

15

8/10/2019 NUS - MA1505 (2012) - Chapter 9

16/38


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).

16

8/10/2019 NUS - MA1505 (2012) - Chapter 9

17/38


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

17

8/10/2019 NUS - MA1505 (2012) - Chapter 9

18/38


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].

18

8/10/2019 NUS - MA1505 (2012) - Chapter 9

19/38


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.

................

.................

...................

.........................

.....................................................................................................

................ ..................................................................

..............

................

........................

..............

.......................................

........................................................................................................................

.......

...................... ...............

.

C1

C2

C3

19

8/10/2019 NUS - MA1505 (2012) - Chapter 9

20/38


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

20

8/10/2019 NUS - MA1505 (2012) - Chapter 9

21/38

8/10/2019 NUS - MA1505 (2012) - Chapter 9

22/38


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.

22

8/10/2019 NUS - MA1505 (2012) - Chapter 9

23/38


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.

23

8/10/2019 NUS - MA1505 (2012) - Chapter 9

24/38


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.

......

.......................................................................................

.................

.......................

............................................................

.................

..............

..................

...................... ................

.............................................................................................

.................

.......................

............................................................

.................

..............

..................

......................................

A

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.

24

8/10/2019 NUS - MA1505 (2012) - Chapter 9

25/38


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.

25

8/10/2019 NUS - MA1505 (2012) - Chapter 9

26/38


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

8/10/2019 NUS - MA1505 (2012) - Chapter 9

27/38


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.

27

8/10/2019 NUS - MA1505 (2012) - Chapter 9

28/38


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)).

28

8/10/2019 NUS - MA1505 (2012) - Chapter 9

29/38


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.

29

8/10/2019 NUS - MA1505 (2012) - Chapter 9

30/38


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.

30

8/10/2019 NUS - MA1505 (2012) - Chapter 9

31/38


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

8/10/2019 NUS - MA1505 (2012) - Chapter 9

32/38


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

32

8/10/2019 NUS - MA1505 (2012) - Chapter 9

33/38


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.

......

.....

......

......

......

......

......

......

......

......

......

......

......

..............

......

.....

......

......

......

......

......

......

......

......

......

......

......

......

......

......

.....

......

......

......

......

......

......

......

.....

......

......

......

......

......

......

......

.....

......

......

......

.....................

................

y

.................................................................................................................................................................................................................................................................................................................................... ................ x........

......

......

......

.....

......

......

...........

......

.....

......

......

......

......

......

......

......

......

......

......

.....

......

......

......

......

.....

......

......

......

......................................................................................................................................................................................................................................................................................................................................................................................................................................................... ................................... .......

.........

...............

...........................................

.........................

..........

......

..........

.......................................................

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

..

.

.

.

.

.

.

.

..

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.......

(0, 2)

(0, 0) (2, 0)x

y= 2 x

C

D

33

8/10/2019 NUS - MA1505 (2012) - Chapter 9

34/38


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


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


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.

......

......

......

............

.............

...............

..................

................................................................................................................................................................................................................................................................................................................

..................

..............

........................................... ......

.....

......

......

......

......

........................

............

............

..............

..............

.................

...................

........................

.............................................

.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

.............................

.....................

.................

...............

..............

............................................................................................

......

......

......

......

......

......

.....

......

......

......

......

......

......

......

.....

......

......

......

......

......

......

......

.....

......

................

......

......

......

.....

......

......

......

......

......

......

......

.....

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

.....

......

......

......

......

......

......

......

.....

......

......

......

......

......

......

......

......

.....

......

......

......

......

...................

................

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


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


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.

nus - ma1505 (2012) - chapter 9

Documents