CHAPTER 5 VECTOR ANALYSIS

5.2 Line Integrals

Definition

The line integral of the scalar function ( , )f x y

along a smooth and orientable curve C is given by

* *

1

( , ) lim ( , )n

k k kn

kC

f x y ds f x y s

Evaluation of Line Integrals

Theorem

Let C be a smooth curve with parametric

representation

( ), ( ) ( )x x t y y t a t b

and ( , )f x y is a continuous function over C. The

line integral is given as:

2 2( , ) ( ( ), ( )) ( ( )) ( ( ))b

C a

f x y ds f x t y t x t y t dt

If C is a curve in 3-space with parametric

representation

( ), ( ), ( )( )x x t y y t z z t a t b

and ( , , )f x y z is continuous on C, then

( , , ) ( ( ), ( ), ( ))b

C a

dsf x y z ds f x t y t z t dtdt

where

2 2 2( ( )) ( ( )) ( ( ))ds x t y t z tdt

Example

Evaluate x

C

ye ds , where C is the segment

joining (1, 2) and (4, 7).

Example

Evaluate 2 2( )

C

x y ds , along the helix

( ) (cos 4 ) (sin 4 )r t t i t j tk , 0 2t

Line Integrals in a Vector Field

Definition

Let F be a vector field and let C be the curve with

parametric representation

( ) ( ), ( ), ( )r t x t y t z t for a t b

Then the line integral of F over C is

C

F T ds ( )b

a

drF t dtdt

C

F dr

where T is the tangent vector of ( )r t .

Connection between the line integrals of vector

fields and line integrals of scalar fields:

Let

( , , ) ( , , ) ( , , )F P x y z i Q x y z j R x y z k

t b

C t a

F dr P dx Q dy Rdz

[ ( ( ), ( ), ( ))

( ( ), ( ), ( ))

( ( ), ( ), ( )) ]

t b

t a

b

a

b

a

dxP x t y t z t dt

dyQ x t y t z t dt

dzR x t y t z t dtdt

Example

Let ( ) ( )F x y i y x j . Compute

C

F dr where C is the path along the parabola

2y x from (1, 1) to (4, 2).

Example

Compute

C

F dr where F yi x j and C

is the closed path shown in the figure.

(0, 0)

(0, 1)

(1, 0)

Work as a Line Integral

Definition

The work done by a force F Pi Qj Rk

over a smooth curve ( )r t from t = a to t = b is

C

W F dr

What this say:

If F is a force field and C(t) represents the position

of a mass in the plane at time t, then the dot

product gives the the amount of force exerted on

the mass in its direction of travel, how much of a

push the force field will give the mass at the point.

Example

If 22 ( ) (3 2 4 )F x y z i x y z j x y z k ,

find the work done by F in moving a particle once

around a circle C in the xy-plane, if the circle has

radius 3 and centre at the origin.

Example

Find the work done by the force

2 2 3F x z i yx j xz k ,

in moving a particle along a straight line segments

from (1,1,0) to (1,1,1) and then to (0,0,0).

5.2.1 Green’s Theorem

Theorem

Let R be a plane region with a positively oriented

piecewise smooth, simple closed curve boundary C.

If the vector field

( , ) ( , ) ( , )F x y P x y i Q x y j

is continuously differentiable on R, then

C C R

Q PF dr P dx Qdy dAx y

R is a 2D space

enclosed by a simple

closed curve C

C

R

Note

R is required to be simply connected with a positively oriented boundary C.

The notation

C

P dx Q dy or

C

P dx Q dy

is used to indicate that the line integral is

calculated using the positive (counter-

clockwise) orientation of the close curve C.

Example

Use Green’s Theorem to evaluate the line integral

C

F dr

along the positively oriented triangular path shown

and that 2( , )F x y x yi x j .

Example

Find the work done by the force field

1 2 2( , ) tan ln( )yF x y i x y jx

when an object moves counterclockwise around

the circular path 2 2 4x y .

(1, 2)

(1, 0) (0, 0)

5.2.2 Conservative Fields

Definition

A vector field F is said to be conservative if there

exists a differentiable function such that the

gradient of is F . That is

F

The function is called the scalar potential

function for F .

Not all vector fields are conservative.

Terminology

i. A path C is called closed if its initial and terminal points are the same point.

ii. A path C is simple if it doesn’t cross itself.

iii. A region D is open if it doesn’t contain any of its boundary points.

iv. A region D is connected if any two points in D can be joined by a path that lies entirely within D.

v. D is simply connected if every closed curve in D encloses only points in D.

simply connected region is one with “no holes”

A simply connected region A region that is not

simply connected

Theorem: Testing for conservative field

Let ( , ) ( , )F P x y i Q x y j be a vector field

where P and Q have continuous first partials in the

open connected region D. Then F is conservative

in D if and only if

Q Px y

throughout D.

By similar reasoning, it can be shown that

( , , ) ( , , ) ( , , )F P x y z i Q x y z j R x y z k

is conservative if and only if

QRy z

, R Px z

, Q Px y

This is equivalent to

conservative curl 0F F .

Example

Determine whether the vector field

y yF e i xe j

is conservative. If it is, find a potential.

Example

Show that the vector field

3 2 22 3F xy z i x j xz k

is conservative and then find a scalar potential

function for F .

Theorem: Fundamental theorem for line

integrals

Let F be a conservative vector field continuous on

an open connected region D. That is, there exists a

function such that F . Then, if C is any

smooth curve from A to B lying entirely in D, we

have

( ) ( )B

A

F dr B A

What this say:

Recall FTC,

( ) ( ) ( )b

a

F x dx F b F a

A version of this for line integrals over certain kinds

of vector fields,

( ( )) ( ( ))C

dr r b r a

Example

Let 2 22 ( 2 )F xyzi x z j x y z k be a

vector field describing a force.

(a) Show that F is conservative.

(b) Find the work done by F in moving an object along the line segment beginning at (1, 1, 1) and ending at (2, 2, 4).

Theorem: Independence of path

Let F be a vector field continuous on an open

connected region D.

C

F dr is independent of

path in D, if and only if F is a conservative vector

field.

What this says:

CF dr is independent of path if

1 2C CF dr F dr for any two paths 1C

and 1C in D with the same endpoints.

C

dr is independent of path.

If F is conservative then CF dr is

independent of path.

Theorem: Closed-loop property

C

F dr is independent of path in D if and only if

0C

F dr for every closed path C in D.

This is equivalent to,

conservative 0C

F F dr

The symbol indicate that the curve C must

be closed.

Example

Show that

C

F dr is independent of path.

Hence find a potential function and evaluate the

line integral along C.

(a) 10 7 7 2F x y i x y j and C is

the curve 21y x from (0,1) to (1,0).

(b) F yz i xz j xy k and C is the

the line segment from (0,0,0) to (1,1,1), and

then to (0,0,1).

5.3 Surface Integrals

Surface Integrals of Scalar Fields

Definition

Suppose f is defined and continuous on a surface .

The surface integral of f over is denoted by

( , , )f x y z dS

where S is the area of the surface.

When the surface projects onto the region Rxy in

the xy-plane and has the representation

( , )z f x y then

22

1z zdS dAx y

where dA is either dx dy or dy dx (or rdrd )

The surface integral is

22

( , , )

( , , ( , )) 1R

f x y z dS

z zf x y g x y dAx y

can obtain similar formulas for surfaces given by ( , )y h x z (with R in the xz-plane) and

( , )x k y z (with R in the yz-plane).

Formula for Surface Integrals

1. ( , )z g x y , R in the xy-plane

22

( , , )

( , , ( , )) 1R

f x y z dS

z zf x y g x y dAx y

2. ( , )y h x z , R in the xz-plane

2 2

( , , )

( , ( , ), ) 1R

f x y z dS

y yf x h x z z dAx z

3. ( , )x k y z , R in the yz-plane

2 2

( , , )

( ( , ), , ) 1R

f x y z dS

x xf k y z y z dAy z

Example

Evaluate the surface integral

2 2y z dS

where is part of the cone 2 2z x y that

lies between the planes 1z and 2z .

Example

Evaluate

2x z dS

where is the portion of the cylinder 2 2 1x y between 0z and 1z .

Surface Integrals of Vector Fields

For a general surface in space, each element of

surface dS has a vector area dS such that

dS n dS .

If F is a vector field, the surface integral

F dS F n dS

where n is the outward unit normal to the surface

.

Definition

If F Pi Qj Rk is a continuous vector

field on an oriented surface with outward unit

normal vector n , then the surface integral of F

over is

F dS F n dS

22

1R

z zF n dAx y

where ( , )z g x y

Note

This integral is also called the flux of F across .

Example:

Calculating the flux of a vector field outward

through the surface S.

Other forms

1. F dS F n dS

2 2

1R

y yF n dAx z

where ( , )y h x z .

2. F dS F n dS

2 2

1R

x xF n dAy z

where ( , )x k y z

Note

To work with surface integrals of vector fields we need to be able to write down a formula for the unit normal vector corresponding to the chosen orientation.

depends on how the surface is given

Let’s suppose that the surface is given by

( , )z g x y . We define a new function,

( , , ) ( , )x y z z g x y

Thus, in term of the new function, the surface is

a level surface for ( , , ) 0x y z .

Recall: will be normal to at ( , )x y . This

means that we have a normal vector to the surface.

We obtain a unit normal vector:

n

To compute the gradient vector:

( , , ) ( , )x y z z g x y

g g z zi j k i j kx y x y

2 2 1

x y

x y

z i z j kn

z z

Notice that the component of the normal vector

in the z-direction (identified by the k in the normal vector) is always positive and so this normal vector will generally point upwards.

Multiplying by -1 produces the negative orientation.

Likewise for surfaces in the form

( , )y h x z , so ( , , ) ( , )x y z y h x z

y yh hi j k i j kx z x z

Thus, 2 21x z

y yi j kx zny y

For surfaces in the form

( , )x k y z , so ( , , ) ( , )x y z x k y z

2 21 y z

x xi j ky znx x

Surfaces positive

orientation

negative

orientation

( , )z f x y 2 2 1

x y

x y

z z

z z

i j kn

2 2 1

x y

x y

z z

z z

i j kn

( , )y g x z 2 21x z

x z

y y

y y

i j kn

2 21x z

x z

y y

y y

i j kn

( , )x h y z 2 21

y z

y z

x x

x x

i j kn

2 21

y z

y z

x x

x x

i j kn

Theorem

Let be a smooth surface of the form

( , )z g x y , ( , )y h x z or ( , )x k y z and

let R be the projection of on the xy-plane, xz-

plane and yz-plane respectively. Suppose that the

surfaces can be rewritten as ( , , ) 0x y z . If is

continuous on R, then

i.

R

F n dS F dA

if has a positive orientation.

ii.

R

F n dS F dA

if has a negative orientation.

Example

Let be the part of the surface of the cone

2 2z x y from 0z to 1z and let

( , , )F x y z i j k . Evaluate

F n dS

Example

Let is the closed surface of the tetrahedron with

vertices (1,0,0), (0,3,0), (0,0,2) and (0,0,0). Evaluate

the surface integral

F n dS

where 2F x i xy j xzk .

5.3.1 Stokes’ Theorem

Let C be any closed curve in 3D space, and let be

an oriented surface bounded by C with unit normal

vector n . If F is a vector field that is continuously

differentiable on , then

C

F dr F ndS

can be any surface bounded by the curve.

Stokes Theorem reduces to Green's Theorem when the curve is a 2D curve.

The integral gives the circulation of the vector

field F around C.

Example

Evaluate

C

F dr where

2 3 24F x i xy j xy k and C is the

close curve on the plane z y with vertices (1,3,3),

(0,3,3) and (0,0,0) with the orientation

counterclockwise when viewed from the positive z-

axis.

Example

Evaluate

C

F dr where F zi x j yk and C

is the intersection of the xy-plane with the paraboloid 2 29y x z with the orientation

counterclockwise direction when viewed from the

positive y-axis.

5.3.2 Gauss’ Theorem

Gauss’ Theorem (Divergence Theorem)

Let be a smooth orientable surface that encloses

a solid region G in 3

. If F Pi Qj Rk

is a vector field whose components P, Q, and R

have continuous partials derivatives in G, then

divG

F n dS F dV

where n is an outward unit normal vector.

Note

Under appropriate conditions, the flux of a vector field across a closed surface with outward orientation is equal to the triple integral of the divergence of the field over the solid region enclosed by the surface.

Flux integral Flow out of small cubes

divG

F ndS F dV

Flux integral (left) - measures total fluid flow across

the surface per unit time.

Right integral – measures the fluid flow leaving the

volume dV

For divergence free vector field F , the flux through a closed surface area is zero. Such fields are also called incompressible or source free.

Example

Use the Divergence Theorem to calculate the

surface integral F n dS

where 3 2 2( , , ) 2 3F x y z x i xz j y zk and

is the surface of the solid bounded by the

paraboloid 2 24z x y and the xy-plane.

Example

Let be the surface of solid enclosed by

2 2 2z a x y and 0z oriented

outward. If 3 3 3F x i y j z k , evaluate

.F n dS