Analysis –Basics of Chapter 1’ Calculus Vector...

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Chapter 1’

Vector Calculushttp://ctaps.yu.edu.jo/physics/Courses/Phys201/Chapter1P

Analysis – Basics of

Calculus

© Dr. Nidal M. Ershaidat Phys. 201 Chapter 1': Vector Analysis

3

Derivative

The infinitesimal rate of change in a

function with respect to one of its

parameters.

Derivative

!" #$%& ! !" #$%& ! !" #$%& ! !" #$%& ! . . . .

The derivative is one of the key

concepts in calculus.


4

Integral:

A mathematical object that can be

interpreted as an area or a

generalization of area. Integrals and

derivatives are the fundamental objects

of calculus.

Integral

2


5

Limits

A limit is the value a function

approaches as the variable approaches

some point.

If the function is not continuous, the

limit could be different from the value

of the function at that point.

Differentiation of

Vectors


7

The derivative of a (real) scalar function φφφφ is

defined as follows:

This definition means that this operation is a

local one, i.e. it’s only defined at a given point

in space (x here).

Derivative of a Real Function

( ) ( )x

xxx

x x ∆φ−∆+φ

=φ

→∆ 0lim

d

d

Actually the derivative at some point x is

nothing else but the slope of the tangent to the

function at that particular point


8

Geometrical Representation of a Derivative

The derivative at x is the slope of the tangent

to the function at x. (And we know how to

compute the slope of a straight line!)

(x,φφφφ(x))

x

φφφφ

x

3


9

The definition of the derivative of a vector

(function) is defined in a similar way. If

vector is function of the variable x then we

have:

(1st)Derivative of a Vector

→→→→A

( ) ( )x

xxx

x x ∆−∆+

=

→→

→∆

→AA

limd

Ad

0

It defines the infinitesimal rate of change in

this vector with respect to x.


10

The derivative of the sum of vectors follow

also the same rules as for scalar functions.

Sum of Vectors

xxx d

Bd

d

Ad

d

BAd )(→→→→

+=+

If vectors and are function of the variable x

and and exist then: xd

Bd→

→B

xd

Ad→

→A


11

Product of VectorsThe derivative of the product of vectors

follow also the same rules as for scalar

functions, provided some attention due to

the definition of products in the vectors

space.

Thus we should write:

xxx d

BdAB

d

Ad

d

)BA(d→

→→→→→

•+•=•

and

xxx d

BdAB

d

Ad

d

)BA(d→

→→→→→

×+×=×


12

The derivative of a vector is also the sum of

the derivatives of its components, i.e.

We can also use the definition of the unit

vector in the direction of a vector to

calculate the (1st) derivative

-Derivative using Unit Vectors

xxxx d

dAaA

d

ad

d

A)ad(

d

Ad →→→→

+==

,( )kjitt

zyxÂÂÂ

d

d

d

Ad++=

→

kt

jt

itt

zyx ˆd

dAˆ

d

dAˆ

d

dA

d

Ad++=

→

4

13

The 2nd, 3rd , … and higher order derivatives of

a vector follow the same rules as for scalars.

This is based on the fact the 1st derivative of a

vector is, itself, a vector.

-Higher Orders Derivatives

Hence:

kt

jt

itt

zyx ˆd

Adˆ

d

Adˆ

d

Ad

d

Ad2

2

2

2

2

2

2

2

++=→

( )kjitttt

zyxÂÂÂ

d

d

d

Ad

d

d

d

Ad2

2

2

2

++=

=

→→

And more generally:

kt

jt

itt n

zn

n

yn

nx

n

n

n

ˆd

Adˆ

d

Adˆ

d

Ad

d

Ad++=

→

Application in

Physics


15

(((( )))) (((( ))))dt

tvdttv

dt

→→→→→→→→

→→→→

−−−−++++====

0lim

In mechanics, the instanteous velocity of a

mechanical system at some instant t is simply

the derivative of the position vector of the

system at that same instant t.

Velocity and Acceleration

dt

rdv

→→→→→→→→

====(((( )))) (((( ))))

dt

trdttr

dt

→→→→→→→→

→→→→

−−−−++++====

0lim

dt

vda

→→→→→→→→

====2

2

dt

rd→→→→

====

→→→→v

The instanteous acceleration of the system at

some instant t is simply the derivative of the

velocity at that same instant t, i.e.

→→→→a

→→→→v


16

The 1st derivative of φφφφ is sometimes denoted φφφφ’.

The 2nd derivative is denoted φφφφ’’, etc ... In

mechanics we sometimes use the symbol •••• to

indicate the (1st) derivative with respect to

time, the symbol •••••••• To indicate the 2nd

derivative, etc …

The “dot” notation

Thus we write:

••••→→→→→→→→

==== rv

Which we read “ r dot ” and:••••→→→→→→→→

==== va

••••••••→→→→

==== r

See Examples 1.14 and 1.15 - Important

5

17

0====••••====••••====→→→→→→→→→→→→

→→→→

vdvmvddt

rdmdW

We can rewrite the previous expression as follows:

A central force , i.e. it is perpendicular to the

velocity , is exerted on an object of mass m. The

trajectory of the body is circular in a plane

containing both vectors.

Example: Circular Motion

Proof:

→v

0====••••====••••====→→→→

→→→→→→→→→→→→

rddt

vdmrdFdW

→→→→F

Since is perpendicular to the displacement the

work done by the force on the body is zero.

→→→→F

→→→→rd


18

0=•⇒→→vdv

Circular Motion

02)(2

====••••====••••++++••••====••••====→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→vdvvdvvvdvvdvd

This means that the magnitude of the velocity

is constant. The change caused by the force

concerns only the direction of the vector .

The locus of where it keeps its magnitude

constant and changes only its direction is a

circle of radius R given by:

→v

→v

R

vmF

2

====

→→→→Gradient ∇∇∇∇∇∇∇∇


20

The Gradient Operator→→→→∇∇∇∇

Thus dφφφφ appears as “a scalar product” of

two vectors: by (dx1,dx2,dx3).

∂

φ∂∂

φ∂∂

φ∂

321,,xxx

kx

jx

ix

ˆˆˆ

321 ∂∂∂∂∂∂∂∂++++

∂∂∂∂∂∂∂∂++++

∂∂∂∂∂∂∂∂====∇∇∇∇

→→→→

φ∇=∂

φ∂+

∂φ∂

+∂

φ∂ →k

xj

xi

xˆˆˆ

321

by the scalar function φφφφ, i.e.

In the same logic, the vector can

be considered as the multiplication of a

vector denoted

∂

φ∂∂

φ∂∂

φ∂

321,,xxx

→∇

6


21

The vector, symbol nabla or del

The Gradient (del) Operator→→→→∇∇∇∇

kx

jx

ix

ˆˆˆ

321 ∂∂∂∂∂∂∂∂++++

∂∂∂∂∂∂∂∂++++

∂∂∂∂∂∂∂∂====∇∇∇∇

→→→→

ix∂∂ is a differential change in the direction xi.

is called “gradient”.

kx

jx

ix

ˆˆˆ

321 ∂∂∂∂φφφφ∂∂∂∂++++

∂∂∂∂φφφφ∂∂∂∂++++

∂∂∂∂φφφφ∂∂∂∂====φφφφ∇∇∇∇

→→→→

For a scalar function φφφφ we write (read del φφφφ)φ∇→

G


22

φφφφ=CThe position vector

of P is thus:

A surface in a space is determined by an

(implicit) equation of the following type:

- Properties of ∇φ∇φ∇φ∇φ

φφφφ(x, y, z) = C

Where C is a constant.

Consider P(x, y, z) a point on this surface. x,

y and z verify the surface equation I.

I

r

O

P

kzjyixr ˆˆˆ ++++++++====→→→→

→→→→


23

Consider a differential displacement ( )

from P to the point Q which belongs to the

surface φφφφ=C

- Properties of ∇φ∇φ∇φ∇φ→→→→

O

r

P

Q

φφφφ=C

Since we did not leave the surface

0dd =⋅φ∇=φ→→r

→rd

dφφφφ= 0Then we have:

Which means that the 2

vectors and are

perpendicular

→rdφ∇

→


24

The displacement vector has no privileged

direction. We just want it to lie in a plane

tangent to the surface at P.

- Properties of ∇φ∇φ∇φ∇φ→→→→

O

r

P

Q

φφφφ=C

→rd

Conclusion: is perpendicular to the surface.φ∇→

7


25

- Maximum of ∇φ∇φ∇φ∇φ→→→→

This leads us to define the direction of the

gradient of φφφφ as the direction of the maximum

space rate of change of φφφφ

For an arbitrary differential displacement→→→→rd

θθθθφφφφ∇∇∇∇====⋅⋅⋅⋅φφφφ∇∇∇∇→→→→→→→→→→→→→→→→

cos|||| rdrd

θθθθ being the angle between the 2 vectors.

dφφφφ has its maximum value for angle θθθθ = 0°°°°, i.e.for a displacement which is perpendicular to

the surface. In other words should be in the

same direction as φ∇→

→rd

→→→→→→→→Gradient Gradient ∇∇∇∇∇∇∇∇ in Physicsin Physics


27

In Physics - Mechanics

UF→→→→→→→→∇∇∇∇−−−−====

This means that the magnitude of the force

equals the maximum space rate of change

of U and its direction is opposite to the

direction of .→∇

In an isolated mechanical system the

conservative force equals minus the

gradient of the potential energy of the

system.

→∇


28

In Physics - EM→∇

φφφφ∇∇∇∇−−−−====→→→→→→→→

E

This means that the magnitude of the electric

field equals the maximum space rate of change

of φφφφ and its direction is opposite to the

direction of .→∇

In an isolated system the electric field

equals minus the gradient of the electric

potential of the system.

8


29

kzjyixr ˆˆˆ ++=→

Find a)

Example 1 – (Problem 1.28a)

r→∇

kzyx

zj

zyx

yi

zyx

x ˆˆˆ

222222 222 +++

+++

++=∇

→r

kzyxz

jzyxy

izyxx

ˆˆˆ 222222 222 ++∂∂

+++∂∂

+++∂∂

=∇→r

( ) rr

rkzjyix

rr ˆˆˆˆ ==++=∇

→→ 1

222zyxr ++++++++====⇒⇒⇒⇒where


30

Find b) where

Example 1 – (Problem 1.28b)

r

1→∇

(((( )))) 21222

222

11 −−−−++++++++====

++++++++==== zyx

zyxr

(((( )))) (((( )))) (((( ))))k

zyx

zj

zyx

yi

zyx

x ˆˆˆ232222322223222 ++++++++

−−−−++++++++

−−−−++++++++

−−−−====

kzyxz

jzyxy

izyxxr

ˆ1ˆ1ˆ11222222222 ++++++++∂∂∂∂

∂∂∂∂++++++++++++∂∂∂∂

∂∂∂∂++++++++++++∂∂∂∂

∂∂∂∂====∇∇∇∇→→→→

(((( ))))233

ˆˆˆˆ11

r

r

r

rkzjyix

rr−−−−====−−−−====++++++++−−−−====∇∇∇∇

→→→→→→→→


31

We know from our studies in electricity that:

This equation gives the corresponding electric

field at P. We find exactly the same result

using Coulomb’s law.

Electric field of a point charge

r

qrq K)( =φ →

For a positive point charge (or for a spherically

symmetric distribution of charges), q > 0 the

electric potential at a given point in space P

defined by the vector , is given by:→→→→r

φφφφ∇∇∇∇−−−−====→→→→→→→→

E

φφφφ∇∇∇∇−−−−====∴∴∴∴→→→→→→→→

E rr

qK

rqK ˆ

12

====∇∇∇∇−−−−====→→→→

r

qq K)P( =φwhich we also write as


32

→E

Example – Lines of Force

VE→→→→→→→→∇∇∇∇−−−−====

For a static distribution of charge, The lines

of field (force) are oriented from the regions in

space where the electric potential is high to

regions where it is lower.

The figure shows the

electric field inside a

capacitor.

V1 > V2 > V3 > V4

-- -- -- -- -- -- -- -- -- -- -- -- --

+ + + ++ + + + + + ++ + + ++ + + + + + +

9


33

Equipotential Surfaces

VVVV1111

VVVV2222

VVVV3333

VVVV4444

!( )$ *+,

+ + + + + + + + + + + + + + + + + + + + + + + + + + + +

-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --

→E

→∇

V1 > V2 > V3 > V4


34

Find b) where

Example 1 – (Problem 1.28c)nr

→∇

(((( )))) (((( )))) 2222222 nnn

zyxzyxr ++++++++====++++++++====

(((( )))) (((( )))) (((( )))) kzyxzn

jzyxyn

izyxxn

nnn

ˆ22

ˆ22

ˆ22

122221

222212222 −−−−−−−−−−−−

++++++++++++++++++++++++++++++++====

(((( )))) (((( )))) (((( )))) kzyxz

jzyxy

izyxx

rnnnn ˆˆˆ 222222222222 ++++++++

∂∂∂∂∂∂∂∂++++++++++++

∂∂∂∂∂∂∂∂++++++++++++

∂∂∂∂∂∂∂∂====∇∇∇∇

→→→→

(((( )))) rrnrrnkzjyixrnr nnnn ˆˆˆˆ 122 −−−−→→→→−−−−−−−−→→→→++++========++++++++====∇∇∇∇

Exercise: Using the definition G check that:

rrr ˆ22 ====∇∇∇∇→→→→

rrr ˆ3 23 ====∇∇∇∇→→→→

, rrr ˆ4 34 ====∇∇∇∇→→→→

and

Vector Integration Generalities

10


37

Three vector integrals:

Vector Integrals

The line integrals

The surface integrals

The volume integrals Line Integrals


39

Line integrals can take one of the following

three forms, depending on the form of the

integrand:

Line Integrals

1. Integration of a scalar function over a vector

∫∫∫∫→→→→φφφφ

C

rd

2. The integration of a dot product

∫∫∫∫→→→→→→→→

⋅⋅⋅⋅C

rdF

3. The integration of a vector product

∫∫∫∫→→→→→→→→

××××C

rdF


40

The line can be one of two kinds:

Closed and Open Line

Open Line: it is a line which has a beginning

and an end. It is not possible to return back

to the starting point if the curve is traversed

in a certain sense.

Closed Line: it is a line which begins at any

point of a curve and returns back to the

starting point regardless the sense in which

the curve is traversed.

The symbol is used to indicate an integral

taken over a closed line.∫

11


41

∫∫∫∫ φφφφ drThe integral can be written as∫∫∫∫

→→→→φφφφC

rd

The result is obtained by evaluating the 3

integrals over the path C which should be

defined.

Generally the result depends on the path

taken.

→→→→

(((( ))))∫∫∫∫∫∫∫∫ ++++++++φφφφ====φφφφ →→→→

CC

zdkydjxdird ˆˆˆ

∫∫∫ φ+φ+φ=CCC

zkyjxi dˆdˆdˆ


42

Example ∫∫∫∫ φφφφdr – Problem (1.35)

Evaluate the line integral ( )∫ ++C

zyxxy dd2d2

Where C connects the two points (0,0,0) and (1,1,1):

a) Along a straight line from (0,0,0) to (1,0,0) to (1,0,1)

to (1,1,1)

→→→→

( )∫ ++1

dd2d2

Czyxxy ( )∫ ++=

)0,0,1(

)0,0,0(

2 dd2d zyxxy

( )∫ +++)1,0,1(

)0,0,1(

2 dd2d zyxxy

( )∫ +++)1,1,1(

)1,0,1(

2 dd2d zyxxy

Solution: We shall call this path C1


43

We actually have 9 integrals to be evaluated:

Problem (1.35) – Solution1

( )∫ ++=),,(

),,(A dzdyxdxyI

001

000

22

I1 = 0 because y = 0

I2 = 0 because we are integrating over y and y = 0

I3 = 0 because we are integrating over z and z = 0

2

001

000

2

I

z,y,x

z,y,xdyx∫

===

===+

3

001

000

I

z,y,x

z,y,xdz∫

===

===+

1

001

000

2

I

z,y,x

z,y,xdxy∫

===

====


44


( )

654

101

001

101

001

101

001

2

101

001

2

2

2

I

,zy,x

,zy,x

I

z,y,x

z,y,x

I

z,y,x

z,y,x

),,(

),,(B

dzdyxdxy

dzdyxdxyI

∫∫∫

∫===

===

===

===

===

===++=

++=

I4 = 0 because y = 0 and we are integrating over x

and x = 1,

I5 = 0 because we are integrating over y and y = 0 ,

11

0

101

0016

=== ∫===

===zdzI

,zy,x

,zy,x

12


45


( )

987

111

101

111

101

111

101

2

111

101

2

2

2

I

,zy,x

,zy,x

I

z,y,x

z,y,x

I

z,y,x

z,y,x

),,(

),,(C

dzdyxdxy

dzdyxdxyI

∫∫∫

∫===

===

===

===

===

===++=

++=

I7 = 0 – same argument as for I4.

I9 = 0 because we are integrating over z and z = 1

111

0

111

1018

=×== ∫===

===ydyxI

,zy,x

,zy,x

Result: IA + IB + IC = 3


46

Example ∫∫∫∫ φφφφdr – Problem (1.35b)

Evaluate the line integral ( )∫ ++2

22

Cdzdyxdxy

Where C2 connects the two points (0,0,0) and (1,1,1):

b) Along a straight line from (0,0,0) to (1,1,1)

→→→→

( )∫ ++2

22

Cdzdyxdxy ( )∫ ++=

),,(

),,(dzdyxdxy

111

000

22

Solution

In order to evaluate the previous integral we

need the equation of the line.

The equation of the line is x = y = z

Which means that we can write dx = dy = dz


47

Problem (1.35b) – Solution

1

0

1

0

21

0

3

22

3z

xy++=

∫∫∫ ++=222

22

CCCdzdxxdyy

3

711

3

1=++=

( )∫ ++2

22

Cdzdyxdxy

The integral can be written as:

Note that ( )∫1C⋯ ( )∫≠

2C⋯


48

The integral is a scalar integral.

If the dependence of on x, y and z is known,

the integral is a classical scalar integral.

∫∫∫∫ F . dr - Work

→→→→F

→→→→→→→→

∫∫∫∫→→→→→→→→

⋅⋅⋅⋅C

rdF

The work done by a force on a system causing

a displacement of the system is given by:→rd

→→→→F

∫∫∫∫→→→→→→→→

⋅⋅⋅⋅====C

rdFW

Generally the result depends on the path

taken. If W is independent of the path then we

conclude that is a conservative force.→→→→F

13

Surface Integrals


50

A surface is a two-dimensional geometric

figure (a collection of points) in three-

dimensional space.

Surfaces and Areas

The simplest example is a plane - a flat surface.

Some other common surfaces are spheres,

cylinders, and cones, the names of which are

also used to describe the three-dimensional

geometric figures that are enclosed (or partially

enclosed) by those surfaces. In a similar way,

cubes, parallelepipeds are surfaces.

Any bounded plane region has a measure

called the area.


51

Types of Surfaces – Closed

Closed surface

Examples:

The surface of a

sphere, a cylinder,

the outer surface of

a parallelepiped.

A closed surface is a surface which divides the

whole (physical) space into two parts: one

completely inside the surface and the other

one outside it.


52

Types of Surfaces - Open

Open surface

Examples:

Page of a book.

An open surface is a surface which is

bounded by a curve.

14


53

θθθθ

In many cases in physics we need to deal with

surfaces. And we often need to determine the

normal to a surface.

We have seen, when we studied the vector

products, that the area is considered as a

vector whose magnitude is the area and its

direction is normal to the area.

Normal to a Surface

→AFor a parallelogram the area is

represented by the vector →A

→b

→c

is nothing else but the cross

product : (see figure)→→

× cb

→A


54

The direction of the normal to a surface

depends on the type of surface.

Direction of the Normal to a Surface

For a closed surface the (positive) normal at a

certain element on the surface is taken

conventionally to point outward.


56

When you curl the fingers of your right hand in

the direction of traversing the perimeter of the

element then the thumb ( ام) will point in the

positive direction of .n

Direction of the Normal to a SurfaceThe figure shows a surface element of an open

surface. The vector is a unit vector normal to the

surface at the location of the area element da and

its direction is defined using a right-hand rule.

n

n


57

The flux of a vector is a surface integral.

Surface Integral in Physics

The most famous example is that of Gauss’s law

in electromagnetism.

∫ ⋅=Φ→

danA

Gauss’s law states that the flux of the electric

field from a closed surface is proportional to the

charge Q enclosed inside the surface, i.e.

ε

QAEdanE enc

AA====⋅⋅⋅⋅====⋅⋅⋅⋅ ∫∫∫∫∫∫∫∫

→→→→→→→→→→→→ˆ

Where εεεε is the permittivity.

The flux of vector is given by:→A

15

Solid Angle


59

Surface integrals become simpler when

using solid angles.

Solid Angle - Definition

The notion of solid angle is a “kind of”

generalization of plane (1D) angle in a 2D

space.

Solid angle can be interpreted as the angle

with which a surface is seen by a point.

This geometrical concept is very common

and important in physics.


60

Consider the element of area da. Let C be

the center of coordinates and the vector

representing the position of da.

Solid Angle – Geometrical Definition

The (element of) solid angle

is defined as:

2

cosd

r

da θ=Ω

Where θθθθ is the angle between

the direction of and that of

the normal to the element of

area .→n

→r

→r

→r


61

In a vector notation, dΩΩΩΩ can be written as:

Solid Angle –Definition using vectors

20dr

danr→⋅

=Ω

Where is the unit vector in

the direction of→r

0r →r

3r

danr→→ ⋅

=

•Solid angle is dimensionless

•dΩΩΩΩ can be positive or negative

(following the sign of )→→ ⋅ nr

•The unit of solid angles is

called the steradian (symbol sr)

16


62

Consider a sphere, center C and radius R

The meaning of Solid Angle

∫∫→⋅

=Ω=Ω2

0dr

danr

The whole solid angle is given by adding all

the elements of solid angle, i.e.

ππππ====ππππ××××========ΩΩΩΩ⇒⇒⇒⇒ ∫∫∫∫ 4411 2

22R

RRda

10 =⋅ →nr

All elements of area are “seen” by C as being at distance r = R. And the normal to element of area is parallel to the vector .

Thus we have:

→r

∫→→ ⋅

=3r

danr


63

The previous result is generalized to any

closed surface and we always have:

ΩΩΩΩ = 4ππππ

ππππ====⋅⋅⋅⋅

====ΩΩΩΩ====ΩΩΩΩ ∫∫∫∫∫∫∫∫→→→→

4ˆ

20

r

danrd

The solid angle subtended by any simple

closed surface at any interior point is 4ππππ sr.

If C lies outside the closed surface then

ΩΩΩΩ = 0.


64

2220

1

0cosˆ

r

da

r

da

r

danrdΩ ========

⋅⋅⋅⋅====

→→→→

As we can see from the figure, the element

of solid angle dΩΩΩΩ1 = dΩΩΩΩ2 in magnitude.

ΩΩΩΩ = 0 if C lies outside the surface

But while dΩΩΩΩ1 is positive, dΩΩΩΩ2 is negative.

is anti-parallel to →n 01r

2222

022

cosˆd

r

da

r

da

r

danr−=

π=

⋅=Ω

→

Integrating over all the dΩΩΩΩ’s gives zero.

Volume Integrals

17


67

Since the volume element dττττ is a scalar quantity,

the volume integral involving vectors are not

different from scalar volume as we integrate

using components, i.e.

Volume Integral

∫∫∫∫ ++=→

Vz

Vy

Vx

VdτAkdτAjdτAidτA ˆˆˆ

68

Divergence

Divergence in Physics


70

Flux = The rate of flow of fluid, particles, or energy per unit volume per time unit.

Flux

[ ][ ]

13TL

Time

Volume[Flux]

−×==

By dimensional analysis it is easy to see

that for a fluid the flux can be written as the

product of its velocity by the surface it

traverses, i.e.

[ ] 21 LTL[Area]Velocity[Flux] ××=×= −

18


71

If there is no source or no

sink then the total flux

through the surface is zero,

since the “lines of flux”

entering the surface will leave

it.

If the flux lines flow out through the surface, the surface is said to be a source of vectors.

- Sources and Sinks

If the flux lines flow into the surface, the

surface is said to be a sink of vectors.


73

If there is no source or no

sink then the total flux

through the surface is zero,

since the “lines of flux”

entering the surface will leave

it.

If the flux lines flow out through the surface,

the surface is said to be a source of vectors.

- Sources and Sinks

If the flux lines flow into the surface, the

surface is said to be a sink of vectors.


74

y

x

z

Consider a compressible fluid of velocity

flowing through a certain volume.

div

The velocity components vx, vy and vz are

functions of the space coordinates x, y and z.

At the center the velocity is

given by:

kvjvivv zyxˆˆˆ ++=

→

→v

In the following we only

consider the 1st order

variations of the velocity.


75

Consider an element of area in the (front wall)

direction x+.

Flux through the front wall

This element is defined by the vector:

The velocity at the center of the front wall is:

izdyd ˆ

y

x

z

kvjvixd

x

vvv fzfy

xxf

ˆˆˆ2

++++++++

∂∂∂∂∂∂∂∂

++++====→→→→

izdydda ˆ====→→→→

33

22

11

dddd xx

xx

xx ∂

φ∂+

∂φ∂

+∂

φ∂=φ

19


76

The flux through the front wall is:

izdydkvjvidx

x

vvF fzfy

xxf

ˆˆˆˆ2

⋅⋅⋅⋅

++++++++

∂∂∂∂∂∂∂∂

++++====

Flux through the front wall

y

x

zzdyddx

x

vv xx

∂∂∂∂∂∂∂∂

++++====2


77

Flux through the back wallThe element of area in the back wall is defined

by the vector:

(((( ))))izdydkvjvidx

x

vvF bzby

xxb

ˆˆˆˆ2

−−−−⋅⋅⋅⋅

++++++++

∂∂∂∂

∂∂∂∂−−−−====

The velocity at the center of the back wall is:

idzyd ˆ−−−−

kvjvidx

x

vvv bzby

xxb

ˆˆˆ2

++++++++

∂∂∂∂∂∂∂∂

−−−−====→→→→

The flux through the back

wall is:

zdyddx

x

vv xx

∂∂∂∂∂∂∂∂

++++−−−−====2


78

zdydxdz

vF ztbnet ∂∂∂∂

∂∂∂∂====

zdydxdx

vx

∂∂∂∂∂∂∂∂

====

The net flux is:

Net Flux

In a similar manner one can prove that the net

flux through the top and bottom sides and the

net flux through the left and right sides are,

respectively, given by:

zdydxdy

vF

yrlnet ∂∂∂∂

∂∂∂∂====

zdyddx

x

vvzdyd

dx

x

vvF x

xx

xfbnet

∂∂∂∂∂∂∂∂

++++−−−−++++

∂∂∂∂∂∂∂∂

++++====22

and


79

z

v

y

v

x

v zyx

∂∂∂∂

∂∂∂∂++++

∂∂∂∂

∂∂∂∂++++

∂∂∂∂∂∂∂∂

The net flux through the small volume dxdydz

is:

Net Flux through the small volume

which also represents the outgoing flux per

unit volume.

z

v

y

v

x

v zyx

∂∂∂∂

∂∂∂∂++++

∂∂∂∂

∂∂∂∂++++

∂∂∂∂∂∂∂∂ can be written in a vector form as:

z

v

y

v

x

vv

zyx

∂∂∂∂

∂∂∂∂++++

∂∂∂∂

∂∂∂∂++++

∂∂∂∂∂∂∂∂

====⋅⋅⋅⋅∇∇∇∇→→→→→→→→

Which we call the divergence of the vector .→v

We sometimes also write div→v

20


80

The divergence of a vector appears as a

measurement of its flux through a certain

surface.

The divergence of a vector is a direct

measurement of the extent to which there

are local sources or sinks

Divergence of a vector

<⋅∇

→→0v

>⋅∇

→→0v


81

In electrostatics (Maxwell’s equations)

The divergence of the electric field is

proportional to the distribution of charge ρρρρ

(proportionality constant = 1/εεεε0).

Divergence of a vector - Examples

0εεεερρρρ

====⋅⋅⋅⋅∇∇∇∇→→→→→→→→E

If ρρρρ is positive then we say that we have a

source of electric field (lines) and if it is

negative then we say that we have a sink of

electric filed lines.


82

In Magnetostatics we always have:

The divergence of the magnetic field is

always zero. In source-sink terms we do not

have neither a source nor a sink.

Divergence – Magnetic Monopoles

0====⋅⋅⋅⋅∇∇∇∇→→→→→→→→B

This equation implies the non existence of

magnetic monopoles (source of field lines)

The Continuity Equation

21


84

The net flux of velocity through the volume is:

If ρρρρ(t) is the density of the fluid at instant t,

then the net flux of mass per unit time through

the volume is:

The Continuity Equation

ττττ⋅⋅⋅⋅∇∇∇∇→→→→→→→→dv

ττττρρρρ⋅⋅⋅⋅∇∇∇∇→→→→→→→→dv )(

equals strictly the decrease of mass

per unit time, i.e. Which gives:

ττττρρρρ⋅⋅⋅⋅∇∇∇∇→→→→→→→→dv )(

ττττ∂∂∂∂ρρρρ∂∂∂∂−−−− dt

)(→→→→→→→→

ρρρρ⋅⋅⋅⋅∇∇∇∇ v 0====∂∂∂∂ρρρρ∂∂∂∂++++t

This equation is called the continuity equation.


85

Prove that

a)

Example – Problem 1.38

→→→→→→⋅∇φ+⋅φ∇=φ⋅∇ AAA

( )kjikz

jy

ix

zyxˆˆˆˆˆˆ AAAA φ+φ+φ⋅

∂∂

+∂∂

+∂∂

=φ⋅∇→→

( ) ( ) ( )zyxzyxAAA φ

∂∂

+φ∂∂

+φ∂∂

=

∂φ∂

+∂φ∂

+∂φ∂

= zyxzyxAAA

∂

∂φ+

∂

∂φ+

∂∂

φzyx

zyx AAA

→→⋅φ∇ A

→→⋅∇φ A

+

+=

Solution:


86

Example – Problem 1.38Prove that

b)→→→→→→→→→→→→→→→→→→→→→→→→→→→→

⋅⋅⋅⋅∇∇∇∇++++⋅⋅⋅⋅∇∇∇∇====

++++⋅⋅⋅⋅∇∇∇∇ BABA

We leave the proof as an exercise.

See also exercises 1.18 and 1.19

87

z

v

y

v

x

v zyx

∂

∂+

∂

∂+

∂∂

We saw that the outgoing flux per volume unit

is:

Net Flux and divergence

And the flux through a surface Si limiting an

element of volume dττττi is given by:

For a finite volume V limited by a finite surface

S, we have:

∫→→

⋅iS

iav d

→→⋅∇= v

iv τ⋅∇=→→

d

In other words we have:

∫→→

⋅iS

iav d

∫∫∫∫∫∫∫∫ ττττ⋅⋅⋅⋅∇∇∇∇====⋅⋅⋅⋅→→→→→→→→→→→→→→→→

Vdvadv

S

22


88

The previous relation is called the Gauss’s

divergence theorem

Gauss’s divergence theorem

This theorem is valid even if the volume is

multiply connected i.e., even it has holes in it.

In such cases the surface integral is taken over

the sum of the surface bounding the volume.

∫∫ τ⋅∇=⋅→→→→

V

dd vavS

The integral at the lhs must be taken only on

the outer surface limiting the volume V. Simply

because the integration on the elements of the

interior of V cancel in pairs.

89

Evaluate Over the unit cube defined by the

point (0,0,0) and the point intercepts on the positive

x, y and z axes.

y(0,1,0)

x

(1,0,0)

(0,0,1)

z

Example 3 (see Example 1.20)

∫∫∫∫→→→→→→→→

⋅⋅⋅⋅S

adr

(0,0,0)


90

Example 3 –Solution Solution:

∫∫∫∫→→→→→→→→

⋅⋅⋅⋅ adr ∫∫∫∫∫∫∫∫→→→→→→→→→→→→→→→→

⋅⋅⋅⋅++++⋅⋅⋅⋅====wallbackwallfront

adradr

∫∫∫∫∫∫∫∫→→→→→→→→→→→→→→→→

⋅⋅⋅⋅++++⋅⋅⋅⋅++++wallbottomwalltop

adradr

∫∫∫∫∫∫∫∫→→→→→→→→→→→→→→→→

⋅⋅⋅⋅++++⋅⋅⋅⋅++++wallleftwallright

adradr


91

Front and Back Walls

(((( )))) idzdykzjyixdrwallfront

ˆˆˆˆa

1

0

1

0

⋅⋅⋅⋅++++++++====⋅⋅⋅⋅ ∫∫∫∫ ∫∫∫∫∫∫∫∫→→→→→→→→

(((( )))) (((( )))) 0ˆˆˆˆ0

1

0

1

0

====−−−−++++++++====⋅⋅⋅⋅ ∫∫∫∫ ∫∫∫∫∫∫∫∫→→→→→→→→

idzdykzjyiadrwallback

i

11

1

0

1

0

1

0

1

0

====××××======== ∫∫∫∫∫∫∫∫∫∫∫∫ ∫∫∫∫ dzdydzdyx

23


92

Other Walls

1====⋅⋅⋅⋅∫∫∫∫→→→→→→→→

walltopadr

0====⋅⋅⋅⋅∫∫∫∫→→→→→→→→

wallbottomadr

1====⋅⋅⋅⋅∫∫∫∫→→→→→→→→

wallrightadr

0====⋅⋅⋅⋅∫∫∫∫→→→→→→→→

wallleftadr

Similar relations are obtained for the other walls

0====⋅⋅⋅⋅∫∫∫∫→→→→→→→→

wallbackadr1====⋅⋅⋅⋅∫∫∫∫

→→→→→→→→

wallfrontadr

3adreurfacswhole

====⋅⋅⋅⋅⇒⇒⇒⇒ ∫∫∫∫→→→→→→→→


93

Using Gauss’s Divergence Theorem

∫∫∫∫∫∫∫∫ ττττ⋅⋅⋅⋅∇∇∇∇====⋅⋅⋅⋅→→→→→→→→→→→→→→→→

V

dradrS

We need to calculate (See Example 1.19)→→

⋅∇ r

(((( ))))kzjyixkz

jy

ix

r ˆˆˆˆˆˆ ++++++++⋅⋅⋅⋅

∂∂∂∂∂∂∂∂++++

∂∂∂∂∂∂∂∂++++

∂∂∂∂∂∂∂∂====⋅⋅⋅⋅∇∇∇∇

→→→→→→→→

3====∂∂∂∂∂∂∂∂

++++∂∂∂∂∂∂∂∂

++++∂∂∂∂∂∂∂∂====

z

z

y

y

x

x

According to Gauss’s Divergence Theorem

unitsddradrS

311133 ====××××××××××××====ττττ====ττττ⋅⋅⋅⋅∇∇∇∇====⋅⋅⋅⋅ ∫∫∫∫∫∫∫∫∫∫∫∫→→→→→→→→→→→→→→→→

VV


94

Find the flux of the vector out

of the unit sphere (center (0,0,0) and radius unity).

Example

Solution:

22223333 rzxy

z

F

y

F

x

FFdiv zyx ====++++++++====

∂∂∂∂

∂∂∂∂++++

∂∂∂∂

∂∂∂∂++++

∂∂∂∂∂∂∂∂

====

kzjyxiyxF ˆˆ3ˆ3322 ++++++++====

5

1222

5

3

cos5

3sin3

2

00

1

0

51

0 0

2

0

22

ππππ====ππππ××××××××====

φφφφ××××θθθθ××××====φφφφθθθθθθθθ====ττττ⋅⋅⋅⋅∇∇∇∇====⋅⋅⋅⋅ππππππππ

ππππ ππππ

∫∫∫∫ ∫∫∫∫ ∫∫∫∫∫∫∫∫∫∫∫∫ rdddrrrdFadFV

S

Curl (rot)

24


101

is a vector:

Curl (rot) is a vector

zyx FFF

zyx

kji

F∂∂∂∂∂∂∂∂

∂∂∂∂∂∂∂∂

∂∂∂∂∂∂∂∂

====××××∇∇∇∇→→→→→→→→

ˆˆˆ

→→→→→→→→××××∇∇∇∇ F

Note: if then the vector is said to be irrotational.→→→→F0====××××∇∇∇∇

→→→→→→→→F

The curl is related to the measurement of the

“strength” of rotation of a vector at a point in

space.

and we call it curl or rot →→→→F

→→→→F


102

We can prove (in fluid mechanics) that

Flux of Curl (rot)

Important:

See examples 1.22 and 1.23 in the textbook

The line integral of vector around a closed

path is equal to the flux of its curl.

→→→→F

→→→→→→→→→→→→→→→→→→→→⋅⋅⋅⋅××××∇∇∇∇====⋅⋅⋅⋅∫∫∫∫ adFrdF

Stokes’ Theorem


104

Stokes’ theorem states that:

Stokes’ Theorem

∫∫∫∫∫∫∫∫→→→→→→→→→→→→→→→→→→→→

⋅⋅⋅⋅====⋅⋅⋅⋅××××∇∇∇∇CS

dFadF ℓ

Where S is a surface bounded (limited) by a

curve C

Proof:

Let’s consider the surface in the

figure. Let’s divide the area into

differential (infinitesimal) area

elements

25


105

Stokes’ Theorem - ProofFor the ith element of area the equation we

obtained in the previous paragraph can be written

here as: →→→→→→→→→→→→→→→→→→→→⋅⋅⋅⋅××××∇∇∇∇====⋅⋅⋅⋅∫∫∫∫ ii adFdF ℓ

∑∑∑∑∑∑∑∑∫∫∫∫→→→→→→→→→→→→→→→→→→→→

⋅⋅⋅⋅××××∇∇∇∇====⋅⋅⋅⋅i

i

i

i adFdF ℓ

∫∫∫∫∫∫∫∫→→→→→→→→→→→→→→→→→→→→

⋅⋅⋅⋅××××∇∇∇∇====⋅⋅⋅⋅SC

adFdF ℓ

Summation over all the area elements gives:

Which tends toward the limit when the number of

elements become large enough and the summation

becomes simply an integral, i.e.


106

Stokes’ Theorem - Note

Note: When making the sum of the linear integral of the

vector the terms involving common boundaries between

the area elements will vanish because they cancel in

pairs.

∑∑∑∑∑∑∑∑∫∫∫∫→→→→→→→→→→→→→→→→→→→→

⋅⋅⋅⋅××××∇∇∇∇====⋅⋅⋅⋅i

i

i

i adFdF ℓ

∫∫∫∫∑∑∑∑∫∫∫∫→→→→→→→→→→→→→→→→

⋅⋅⋅⋅====⋅⋅⋅⋅C

i

i dFdF ℓℓ


107

Stokes’ Theorem

Irish mathematician and physicist. Stokes

studied fluorescence and the luminiferous ether, questioning Young and Fresnel's ether drag theory. Stokes systematically studied fluid mechanics, elastic solids, waves in elastic media, and diffraction. In

fact, the basic equations of fluid mechanics are called the Navier-Stokes equations in his honor. He also formulated the three-dimensional analog of Green's theorem known as Stokes' theorem, and was a

pioneer in the use of divergent series.

Stokes, George (1819-1903)

The Laplacian

26


109

In physics and especially in Electromagnetism we

need to apply the operator more than once.→∇

Successive Application of ∇∇∇∇

φ∇=φ∇⋅∇→→→ 2

→→→→

AA-- The Gradient: Two possibilitiesThe Gradient: Two possibilities

1- The Laplacian

2

2

2

2

2

2

zyx ∂φ∂

+∂

φ∂+

∂φ∂

=

∂φ∂

+∂φ∂

+∂φ∂

⋅

∂∂

+∂∂

+∂∂

= kz

jy

ix

kz

jy

ix

ˆˆˆˆˆˆ

φ

∂∂

+∂∂

+∂∂

=2

2

2

2

2

2

zyx

∆=∇=∇=∇⋅∇→→→

22

2∇ © Dr. Nidal M. Ershaidat Phys. 201 Chapter 1': Vector Analysis

110

The Laplacian is an important operator. It is defined

by:

The Laplacian

∂∂

+∂∂

+∂∂

=∇⋅∇=∇→→

2

2

2

2

2

22

zyx

Famous Example:

Schrödinger Wave Equation:

Ψ

∂∂

−=

Ψ

+

Ψ∇−

→→→→

t,rt

t,rt,rVt,r2

22

ℏℏ

im


111

Laplace proved the stability of the solar system.

In analysis, Laplace introduced the potential

function and Laplace coefficients. He also put the

theory of mathematical probability on a sound

footing.

Laplace, Pierre Simon 1749 - 1827


112

Gradient of a scalar is irrotational

0====

2-

∂∂∂∂φφφφ∂∂∂∂++++

∂∂∂∂φφφφ∂∂∂∂++++

∂∂∂∂φφφφ∂∂∂∂××××

∂∂∂∂∂∂∂∂++++

∂∂∂∂∂∂∂∂++++

∂∂∂∂∂∂∂∂====φφφφ∇∇∇∇××××∇∇∇∇

→→→→→→→→kz

jy

ix

kz

jy

ix

ˆˆˆˆˆˆ

0====φφφφ∇∇∇∇××××∇∇∇∇→→→→→→→→

We leave the easy calculations as an exercise.

27


113

B- The divergence: One possibility

Case of the divergence

⋅∇∇

→→→A

iAAA ˆ

∂∂∂

+∂∂

∂+

∂∂

=zxyxx

zyx22

2

2

∂

∂+

∂

∂+

∂∂

∂∂

+∂∂

+∂∂

=zyx

kz

jy

ix

zyx AAAˆˆˆ

jzyyxy

zyx ˆ2

2

22

∂∂

∂+

∂

∂+

∂∂∂

+AAA

kzyzxz

zyx ˆ2

222

∂

∂+

∂∂

∂+

∂∂∂

+AAA


114

C- The curl: Two possibilities

Case of the Curl

∂∂

∂∂

∂∂

⋅∇=

×∇⋅∇

→→→→

zyx

zyx

kji

AAA

A

ˆˆˆ

xzxzzyxyzxyx

xyxzyz

∂∂∂

−∂∂

∂−

∂∂∂

+∂∂

∂−

∂∂

∂−

∂∂

∂=

AAAAAA 222222

0=

Assuming that the 1st derivative of the components

are continuous, which allows us to write:

α∂β∂∂

=β∂α∂

∂ 22

αααα , ββββ ∈∈∈∈ x,y,z


115

Case of the Curl –

∇⋅∇−

⋅∇∇=

×∇×∇

→→→→→→→→→AAA

Using the BAC-CAB rule.

)(→→→

×∇×∇ A

×∇×∇−

⋅∇∇=∇

→→→→→→→AAA2

The previous equation can be rewritten as:

This result (identity) is used extensively in

electromagnetism in deriving the wave equation –

see Phys. 207 and Phys. 332.

Green’s TheoremPotential Theorem

28


117

Green’s theorem is a useful corollary of

Gauss’s divergence theorem.

Consider two scalar functions φφφφ1 and φφφφ2

defined over a region of space of volume V

surrounded by a surface S.

We shall assume that the 2 functions and

their first derivatives are finite and

continuous over V.

Green’s Theorem


118

We start from the identity:

Green’s Theorem 1

22

12121 φφφφ∇∇∇∇φφφφ++++φφφφ∇∇∇∇⋅⋅⋅⋅φφφφ∇∇∇∇====

φφφφ∇∇∇∇φφφφ⋅⋅⋅⋅∇∇∇∇

→→→→→→→→→→→→→→→→

Now Let’s integrate this equation over the

volume V:

∫∫∫∫∫∫∫∫∫∫∫∫ ττττφφφφ∇∇∇∇φφφφ++++ττττφφφφ∇∇∇∇⋅⋅⋅⋅φφφφ∇∇∇∇====ττττ

φφφφ∇∇∇∇φφφφ⋅⋅⋅⋅∇∇∇∇

→→→→→→→→→→→→→→→→

VVVddd 2

212121

The integral on the lhs of the previous equation can

be transformed to a surface integral using Gauss’s

divergence theorem and we can, thus, write:

∫∫∫∫∫∫∫∫∫∫∫∫ ττττφφφφ∇∇∇∇φφφφ++++ττττφφφφ∇∇∇∇⋅⋅⋅⋅φφφφ∇∇∇∇====

φφφφ∇∇∇∇φφφφ

→→→→→→→→→→→→

VVS

ddad 22

12121


119

And this is the Green’s Theorem in the first form:

Green’s Theorem 2



→→→→→→→→→→→→

VVS

ddad 22

12121

By exchanging φφφφ1 and φφφφ2 we obtain the following

equation G2



→→→→→→→→→→→→

VVS

ddad 12

21212

G1

G2

Subtracting equation G2 from equation G1 we get:

(((( ))))∫∫∫∫∫∫∫∫ ττττφφφφ∇∇∇∇φφφφ−−−−φφφφ∇∇∇∇φφφφ====⋅⋅⋅⋅

φφφφ∇∇∇∇φφφφ−−−−φφφφ∇∇∇∇φφφφ

→→→→→→→→

VS

dad 12

222

11221 G3

G3 is known as the second form of of Green’s

Theorem.

Potential Theorem

29


121

Consider a force which satisfies the equation:

Which means as we have said before that is

irrotational.

This suggests the existence of a scalar function

φφφφ such that:

In terms of gradient, the magnitude of the force

equals the magnitude of the gradient of the scalar

function φφφφ and its direction is the opposite

direction of this gradient. (See lecture on Gradient

and the negative sign)

Scalar Potential

0====××××∇∇∇∇→→→→→→→→F

We saw that , φφφφ being a scalar function,0====φφφφ∇∇∇∇××××∇∇∇∇→→→→→→→→

→→→→F

→→→→F

φφφφ∇∇∇∇−−−−====→→→→→→→→

F


122

And the scalar function φφφφ that satisfies the

equation:

We have seen that such a force is said to be

conservative. According to the previous

discussion a force is conservative if it

satisfies the irrotational relation:

Alternatively we can associate to any force

satisfying the irrotational relation a scalar

function (which depends on space coordinates)

is called a scalar potential.

Scalar Potential

0====××××∇∇∇∇→→→→→→→→F

→→→→F

φφφφ∇∇∇∇−−−−====→→→→→→→→

F


123

Using Stokes’ theorem

Conservative forces and Stokes’ Theorem

∫∫∫∫∫∫∫∫→→→→→→→→→→→→→→→→→→→→

⋅⋅⋅⋅××××∇∇∇∇====⋅⋅⋅⋅SC

adFrdF

0====⋅⋅⋅⋅∫∫∫∫→→→→→→→→

CrdF

Which means that the work done by a

conservative force over a closed path is zero.

We also equivalently say that the work done by

a conservative force does not depend on the

nature of the path but only on its initial and

final points.

gives in the case of a conservative force )( 0====××××∇∇∇∇→→→→→→→→F


124

Tow kinds of potential appear in physics:

scalar potential and vector potential

In general the term potential includes the

effects of any external force (action) on a

given system.

We often express this influence in terms of

“potential energy”.

This energy is “stored” in the system and

when the external agent “leaves” the

system, the system uses the “potential

energy” to perform actions itself.

Scalar Potential

30


125

We can create a vector potential in a similar

manner to the case of the scalar potential

Vector Potential

For a vector satisfying the relation:→→→→B

0====

××××∇∇∇∇⋅⋅⋅⋅∇∇∇∇

→→→→→→→→→→→→A

Consider a vector which satisfies the equation:→→→→A

0====⋅⋅⋅⋅∇∇∇∇→→→→→→→→B

We can define a vector such that:→A

→→→→→→→→→→→→××××∇∇∇∇==== AB

Theoretically there are an infinite number of

potential vectors satisfying the equation VP.

VP

is called the vector potential.→→→→A

Poisson’s Equation and

Laplace’s Equation


127

From Gauss’s law we have

If ρρρρ=0 then the resulting equation is called

Laplace’s equation

Poisson’s Equation

We can write:

And from the potential theory we have:0εεεερρρρ====⋅⋅⋅⋅∇∇∇∇

→→→→→→→→E

φφφφ∇∇∇∇−−−−====→→→→→→→→

E

P

This is Poisson’s equation.

0εεεερρρρ

−−−−====φφφφ∇∇∇∇⋅⋅⋅⋅∇∇∇∇→→→→→→→→

0

2

εεεερρρρ

−−−−====φφφφ∇∇∇∇

L02 ====φφφφ∇∇∇∇


12

9

Problems

Vector Calculus

31


130

Evaluate

Problem 1.46→→

×∇ r

Solution:

zyx

zyx

kji

r∂∂

∂∂

∂∂

=×∇→→

ˆˆˆ

ky

x

x

yj

x

z

z

xi

z

y

y

z ˆˆˆ

∂∂

−∂∂

+

∂∂

−∂∂

+

∂∂

−∂∂

=

2r

r,

→→

×∇


131

222222222

2

zyx

z

zyx

y

zyx

x

zyx

kji

r

r

++++++

∂∂

∂∂

∂∂

=×∇→

→

ˆˆˆ

Problem 1.46

( )( ) ( )( )i

z

zyxy

y

zyxzˆ

∂++∂

−∂

++∂=

222222

( )( ) ( )( )j

x

zyxz

z

zyxx ˆ

∂++∂

−∂

++∂+

222222

( )( ) ( )( )k

y

zyxx

x

zyxyˆ

∂++∂

−∂

++∂+

222222


133

Given

kx

yyx

zyx

kji

2

2 0

−−−−====

−−−−∂∂∂∂∂∂∂∂

∂∂∂∂∂∂∂∂

∂∂∂∂∂∂∂∂

====××××∇∇∇∇→→→→→→→→F

Problem 1.50

a) Find

Solution:

ji ˆyˆyx −−−−====→→→→

2F

→→→→→→→→××××∇∇∇∇ F


134

b) Find over a rectangle bounded by the

lines x = 0 , x = b, y = 0 and y = c

Stokes’ Theorem

IIII

IIIIII

P(x,y)

II

IVIV

∫∫∫∫→→→→→→→→

⋅⋅⋅⋅S

adF

0,0 b,0

b,c0,c

0====⋅⋅⋅⋅∫∫∫∫→→→→→→→→

SadF

kdydxd ˆ====→→→→a

0====⋅⋅⋅⋅⇒⇒⇒⇒→→→→→→→→daF

x

y

O

32


135

Stokes’ Theorem

IIII

IIIIII

P(x,y)

II

x

y

OIVIV

→F

∫∫∫∫→→→→→→→→

⋅⋅⋅⋅C

drF ∫ →→⋅+

IIIdrF∫ →→

⋅+II

drF∫ →→⋅=

IdrF ∫ →→

⋅+V

drI

F

0,0 b,0

b,c0,c

136

(((( )))) 0ˆ

0

00

2

0

00

0

00

========⋅⋅⋅⋅====⋅⋅⋅⋅====⋅⋅⋅⋅ ∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫========

========

========

========

→→→→========

========

→→→→→→→→→→→→→→→→bx

y

yx

bxy

yx

bxy

yx

IdxyxidxFdlFdrF

Stokes’ Theorem

IIII

IIIIII

P(x,y)

II

x

y

0

IVIV

→F

0,0 b,0

b,c0,c

(((( ))))

2

ˆ

2

0

0

c

dyy

jdyFdrF

bxcy

ybx

bxcy

ybx

II

−−−−====

−−−−====

⋅⋅⋅⋅====⋅⋅⋅⋅

∫∫∫∫

∫∫∫∫∫∫∫∫

========

========

========

========

→→→→→→→→→→→→


137

Stokes’ Theorem

(((( ))))(((( ))))3

ˆ3

2

00bcdxyxidxFdrF

xcy

cybx

xcy

cybx

III====−−−−====−−−−⋅⋅⋅⋅====⋅⋅⋅⋅ ∫∫∫∫∫∫∫∫∫∫∫∫

========

========

========

========

→→→→→→→→→→→→

(((( ))))(((( )))) (((( ))))2

ˆ20

0

0

00

0

cdyyjdyFdrF

xy

cyx

xy

cyx

IV====−−−−−−−−====−−−−⋅⋅⋅⋅====⋅⋅⋅⋅ ∫∫∫∫∫∫∫∫∫∫∫∫

========

========

========

========

→→→→→→→→→→→→

32320

3232 bc

cbc

cdr

C

====++++++++−−−−====⋅⋅⋅⋅∫∫∫∫→→→→→→→→

F


138

Stokes’ Theorem

kx

yyx

yx

kji

ˆ

ˆˆˆ

2

2 0

F −=

−∂∂

∂∂

∂∂

=×∇→→

z

( )∫∫ ⋅−=⋅×∇ →→→

SSa kdydxkxdF ˆˆ2

∫ ∫=

=

=

=−=

cy

y

bx

xdydxx

0 0

2 [[[[ ]]]]33

3

0

0

3bcy

x cy

y

bx

x

−−−−====

−−−−==== ====

====

====

====

33

Potential Theorem


141

Given the force

Problem 1. 61

( ) ( ) ( )kzyxjxyzxiyzyxF 2322223 ++−+−=

→

Solution: This force is conservative if and

only if

z

z

FFF

yx

kji

yx

∂∂

∂∂

∂∂

=×∇→→

ˆˆˆ

F

0FFFFFF

=

∂∂

−∂

∂+

∂

∂−

∂∂

+

∂

∂−

∂∂

= kyx

jx

iy

xyxyz ˆˆˆ z

zz

a) Show that this force is conservative


142

Problem 1. 61

( ) 222 xzyx

yy

Fz =+∂∂

=∂∂ ( ) 222

3 xxyzxzz

Fy =−∂∂

=∂

∂,

( ) yxyzyxzz

Fx22

3 =−∂∂

=∂∂ ( ) yxzyx

xx

Fz22

2 =+∂∂

=∂∂

,

( ) 222323 yzxxyzx

xx

Fy −=−∂∂

=∂

∂

( ) 23322 yzxyzyx

yy

Fx −=−∂∂

=∂∂

0=×∇⇒→→F


143

⇒⇒⇒⇒

b) Find a potential φφφφ(x,y,z) such that

Problem 1. 61 (b)

φ∇−=→→

F

Solution:

φ−=

∂φ∂

+∂

φ∂+

∂φ∂

−=→

→

rd

dkj

yix

ˆˆˆ

zF

32 yxyx +−=φ z 32 yxyx +− z 2zz −− yx2

23223 zyxzyx −+−=φ

∫→→

⋅−=φ rdF ( )dyxyzx∫ −− 223

( )dzzyx∫ +− 22

( )dxyzyx∫ −−= 32

Analysis –Basics of Chapter 1’ Calculus Vector...

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