EMT Electromagnetic Theory MODULE I

MODULE I

VECTOR ALGEBRA, VECTOR CALCULUS, COORDINATE SYSTEMS, VECTOR FIELDS

Compiled by: MKP for CEC S5 EC - July 2008

Syllabus – Module IVector analysis:

Vector algebra, Coordinate systems and transformations-Cartesian, cylindrical and spherical coordinates. Constant coordinate surfaces. Vector calculus-Differential length, area and volume. Line, surface and volume integrals. Del operator. Gradient of a scalar, Divergence of a vector, Divergence theorem, Curl of a vector. Stock’s theorem, Laplacian of a scalar. Classification of vector fields.


ReferencesText Books:

1. Mathew N.O. Sadiku, Elements of Electromagnetics, Oxford University Press

2. Jordan and Balmain, Electromagnetic waves and radiating systems,Pearson Education PHI Ltd.References:

1. Kraus Fleisch, Electromagnetics with applications, McGraw Hill2. William.H.Hayt, Engineering Electromagnetics, Tata McGraw Hill3. N.Narayana Rao, Elements of Engineering Electromagnetics, Pearson

Education PHI Ltd. 4. D.Ganesh Rao, Engineering Electromagnetics, Sanguine Technical

Publishers.5. Joseph.A.Edminister, Electromagnetics, Schaum series-McGraw Hill


Scalars and vectorsA scalar is a quantity that has only magnitude.

TimeDistanceTemperatureSpeed

A vector is a quantity that has both magnitude and direction.

ForceDisplacementVelocity


Unit vectorA vector has both magnitude and direction.

A unit vector along is defined as a vector whose magnitude is unity and whose direction is along vector . It is denoted by

Vector is completely specified in terms of its magnitude A and direction

A

Magnitude of A A A= =

AA ˆAa

ˆ AA Aa

AA= =

ˆ AA A a=

ˆAaA


Vectors represented in rectangular coordinate systems

Any vector in space can be uniquely expressed in terms of x, y and z coordinates using a rectangular coordinate system.

Y

X

Z

xA

yA

zA

ˆxaˆ ya

ˆza A

ˆ ˆ ˆx x y y z zA A a A a A a= + +


Vectors represented in rectangular coordinate systems

ˆ ˆ ˆx x y y z zA A a A a A a= + +, , , ,x y zA A A Components of A in the direction of x y z⇒

ˆ ˆ ˆ, ,

, , x y za a a Unit vectors specifying the direction

of x y z axes

⇒


Position vector of a point in spaceA point P in Cartesian coordinate system may be expressed as itsx,y,z coordinates. The position vector of a point P is the directed distance from the origin O to the point P.A point P (3,4,5) has the position vector

Y

X

Z

xA

yA

zA

ˆxaˆ ya

ˆza

ˆ ˆ ˆp x x y y z zr OP A a A a A a= = + +P

ˆ ˆ ˆ3 4 5p x y zr a a a= + +


Vector addition and subtraction

ˆ ˆ ˆ

ˆ ˆ ˆ x x y y z z

x x y y z z

If A A a A a A a and

B B a B a B a

= + +

= + +

( ) ( ) ( )ˆ ˆ ˆx x x y y y z z zC A B A B a A B a A B a= + = + + + + +

( ) ( ) ( )ˆ ˆ ˆx x x y y y z z zD A B A B a A B a A B a= − = − + − + −


Distance vectorDistance vector is the displacement from one point to another.If two points A (Ax,Ay,Az) and B (Bx,By,Bz) are given, the distance vector from A to B is given by

( ) ( ) ( )ˆ ˆ ˆAB x x x y y y z z zr B A a B A a B A a= − + − + −


Unit vector in the direction of given vector

Let be a vector in space given by

A unit vector in the direction of is given by

ˆ ˆ ˆx x y y z zA A a A a A a= + +A

A

2 2 2

ˆ ˆ ˆˆ x x y y z z

A

x y z

A a A a A aa

A A A

+ +=

+ +


Example 1ˆ ˆ ˆ 10 4 6 x y zIf A a a a and= − + ˆ ˆ2 x yB a a find= +

ˆ( )

( ) 3

( ) 2

yi Component of A along a

ii Magnitude of A B

iii A unit vector along A B

−

+

:Answer( 4) -i

( ) ( )ˆ ˆ ˆ ˆ ˆ( ) 3 30 12 18 2x y z x yii A B a a a a a− = − + − +ˆ ˆ ˆ28 13 18x y za a a= − +

2 2 23 28 13 18 35.74A B− = =+ +


Example 1ˆ ˆ ˆ( ) 2 14 2 6x y ziii A B a a a+ = − +

ˆ ˆ ˆ ˆ0.9113 0.1302 0.3906x y zc a a a= − +

2ˆ 22

A BA unit vector c along A BA B+

+ =+

2 2 2

ˆ ˆ ˆ14 2 6

14 2 6x y za a a− +

=+ +


Vector multiplication-dot productScalar product or dot product: It is defined as the product of magnitudes of the two vectors and the cosine of the angle between them.

Properties:

cos ABA B AB θ⋅ =

AB is the smaller angle between themθ

( ) : i Commutative Prope ty A B B Ar ⋅ = ⋅

( ) ii When two vectors are perpendicular the angle =90 cos90 0between them is θ =

cos90 0A B AB⋅ = = , If the dot product of two vectors are zero they are

. perpendicular


Vector multiplication-dot productˆ ˆ ˆ( ) , , x y ziii Since a a a are mutually perpendicular

ˆ ˆ ˆ ˆ ˆ ˆ = 0x y y z z xa a a a a a⋅ ⋅ = ⋅ =

( ) iv When two vectors are parallel the angle between them 0 180is either or

cos0 A B AB AB or⋅ = =cos180A B AB AB⋅ = = −

( ) .v The square of a vector is the square of its magnitude2cos0 A A AA A⋅ = =

2 2A A=


Vector multiplication-dot product

ˆ ˆ ˆ


x x y y z z

If A A a A a A a and

B B a B a B a

= + +

= + +

( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ. x x y y z z x x y y z zA B A a A a A a B a B a B a= + + ⋅ + + x x y y z zA B A B A B= + +

( ) vi Scalar product is equal to the sum of products of their.corresponding components


Vector Product or cross productVector Product or cross product: Vector product of two vectors and is denoted as and is defined as

Where is a unit vector perpendicular to and such that forms a right handed system.

Geometrically the cross product can be defined as a vector whose magnitude is equal to the area of the parallelogram formed by

and whose direction is in the direction of advance of a right handed screw as is turned in to through the smaller angle.

AB A B×

sin ˆAB nA B A B aθ× =A Bˆna

ˆ, nA B and a

A B A and B


Vector Product or cross product

A

B

θ

A B×

ˆnaAB sinθ

AB sinθ


Vector Product or cross productProperties:

( ) : i Anti commutative A B B A× = − ×

( ) ( ) ( )( ) : ii Distributive A B C A B A C× + = × + ×

( ) ( )( ) : iii Not Associative A B C A B C× × ≠ × ×

( ) V .iv ector product of two parallel vectors is zero

ˆ ˆsin0sin 0AB n nA B A B a A B aθ× = = =

ˆ( ) sin0 0nv A A A A a× = =



ˆ ˆ ˆ ˆ ˆ ˆ( ) 0x x y y z zvi a a a a a a× = × = × =

ˆ ˆ ˆ ˆ( ) 1.1.sin 90 x y z zvii a a a a and× = =

ˆ ˆ ˆ y z xa a a× =

ˆ ˆ ˆ z x ya a a× = ˆ ˆ ˆy x za a a× = −

ˆ ˆ ˆ z y xa a a× = −

ˆ ˆ ˆx z ya a a× = −



ˆxa

ˆ yaˆza

ˆxa

ˆ ya ˆzaˆxa

ˆ yaˆza

ˆxa−

ˆ ya− ˆza−


Vector Product or cross product

ˆ ˆ ˆ( )


x x y y z z

viii If A A a A a A a and

B B a B a B a

= + +

= + +

ˆ ˆ ˆx y z

x y z

x y z

a a aA B A A A

B B B× =


Projection of a vector on another vectorScalar component of along is called projection of on and is given by

A B A B

cosB ABA A θ=

ˆ cosB ABA a θ=

ˆBA a= ⋅

A

B

ABθ

cos ABA θˆBa


Projection of a vector on another vectorVector component of along is the scalar component multiplied by a unit vector along

A BB

( )ˆ ˆ ˆB B B B BA A a A a a= = ⋅


Scalar triple productGiven three vectors the scalar triple product is defined as

and is represented as

Geometrically the scalar triple product is equal to the volume of a parallelepiped having as sides Properties:

( ) ( ) ( )A B C B C A C A B⋅ × = ⋅ × = ⋅ ×

, A B and C

A B C⎡ ⎤⎣ ⎦

, A B and C

( ) i A B C B C A C A B⎡ ⎤ ⎡ ⎤ ⎡ ⎤= =⎣ ⎦ ⎣ ⎦ ⎣ ⎦

( ) ( ) ( ). . A B C Bi e C A C A B⋅ × = ⋅ × = ⋅ ×


Scalar triple product(ii) A change in the cyclic order of vectors changes the sign of scalar

triple product.

A B C B A C⎡ ⎤ ⎡ ⎤= −⎣ ⎦ ⎣ ⎦

ˆ ˆ ˆ( )

ˆ ˆ ˆ

ˆ ˆ ˆ

x x y y z z

x x y y z z

x x y y z z

iii If A A a A a A a and

B B a B a B a

C C a C a C a

= + +

= + +

= + +

x y z

x y z

x y z

A A AA B C B B B

C C C⎡ ⎤ =⎣ ⎦


Vector triple product

, ,For any three vectors A B C

( ) ( ) ( )A B C B A C C A B× × = ⋅ − ⋅

bac-cab rule


Cylindrical Coordinate SystemsAny point in space is considered to be at the intersection of three mutually perpendicular surfaces:

A circular cylinder (ρ=constant)A vertical plane (Φ=constant)A horizontal plane (z=constant)

Any point in space is represented by three coordinates P(ρ,Φ,z)ρ denotes the radius of an imaginary cylinder passing through P, or the radial distance from z axis to the point P.Φ denotes azimuthal angle, measured from x axis to a vertical intersecting plane passing through P.z denotes distance from xy-plane to a horizontal intersecting plane passing through P. It is the same as in rectangular coordinate system.


Cylindrical Coordinate Systems

P(ρ,Φ,z)

z=constant

ρ=constant

Φ =constant

z

ρ

Φ

X

Y

Z

: 0 0 2

Ranges

z

ρπ

≤ < ∞≤ Φ <−∞ < < ∞



P(3,45,8)

z=8

ρ=3

Φ=45

X

Y

ZP(3,45,8)



z

ρ

Φ

X

Y

Z

aρ

aΦˆza

aρ

aΦˆza


Cylindrical Coordinate SystemsA vector in cylindrical coordinate system may be specified using three mutually perpendicular unit vectors

form a right handed system because an RH screw when rotated from

These unit vectors specify directions along ρ,Φ and z axes.Using these unit vectors any vector A may be expressed as

The magnitude of the vector is given by

ˆ ˆ ˆ, , za a aρ Φ

ˆ ˆ ˆz zA A a A a A aρ ρ Φ Φ= + +

2 2 2zA A A Aρ Φ= + +

ˆ ˆ ˆ za to a aρ Φ moves towards

ˆ ˆ ˆ, , za a aρ Φ



ˆ ˆ ˆ ˆ ˆ ˆ 1z za a a a a aρ ρ Φ Φ⋅ = ⋅ = ⋅ =

ˆ ˆ ˆ ˆ ˆ ˆ 0z za a a a a aρ ρΦ Φ⋅ = ⋅ = ⋅ =

ˆ ˆ ˆza a aρ Φ× =ˆ ˆ ˆza a aρΦ × =

ˆ ˆ ˆza a aρ Φ× =


Relationship between cylindrical and rectangular coordinate systems

X

Y

Z

x

y

z

Φ ρ

siny ρ= Φ

cosx ρ= Φ

P(ρ,Φ,z) or (x,y,z)

z

2 2x yρ = +1tan y

x− ⎛ ⎞Φ = ⎜ ⎟⎝ ⎠

z z=

cosx ρ= Φsiny ρ= Φ

z z=

aρ

aΦˆza



X

Y

Φ

aρ

aΦ

aΦ− aρ

ˆxaΦ

ˆ sinaΦ− Φ

ˆ cosaρ Φ

ˆ ˆ ˆcos sinxa a aρ Φ= Φ − Φ



X

Y

Φ

aρaΦ

aρˆ ya

Φ

ˆ sinaρ Φ

ˆ cosaΦ Φ

ˆ ˆ ˆcos sinya a aρΦ= Φ + Φ



X

Y

Φ

aρˆ ya

Φ

ˆ ˆ ˆcos sinx ya a aρ = Φ + Φ

ˆxa



X

Y

Φ

ˆ ya

Φ

ˆ ˆ ˆcos siny xa a aΦ = Φ − Φ

ˆxaˆxa−

aΦ Φaρ



ˆ ˆ ˆcos sinxa a aρ Φ= Φ − Φ

ˆ ˆ ˆcos sinya a aρΦ= Φ + Φ

ˆ ˆ ˆcos sinx ya a aρ = Φ + Φˆ ˆ ˆcos siny xa a aΦ = Φ − Φ

ˆ ˆz za a=

ˆ ˆz za a=

⎫⎪⎬⎪⎭

⎫⎪⎬⎪⎭

(2)equations

(3)equations


Transformation of vectors between cylindrical and rectangular coordinate systems

Substituting the equations (2) in the general equation for a vector in rectangular coordinates,


( ) ( )ˆ ˆ ˆ ˆ ˆcos sin cos sinx y z zA A a a A a a A aρ ρΦ Φ= Φ − Φ + Φ + Φ +

( ) ( )ˆ ˆ ˆcos sin sin cosx y x y z zA A A a A A a A aρ Φ= Φ + Φ + − Φ + Φ +

cos sinx yA A Aρ = Φ + Φ

sin cosx yA A AΦ = − Φ + Φ

z zA A=

⎫⎪⎬⎪⎭

(4)equations



Substituting the equations (3) in the general equation for a vector in cylindrical coordinates,

ˆ ˆ ˆx y z zA A a A a A aρ Φ= + +

( ) ( )ˆ ˆ ˆ ˆ ˆcos sin sin cosx y x y z zA A a a A a a A aρ Φ= Φ + Φ + − Φ + Φ +

( ) ( )ˆ ˆ ˆcos sin sin cosx y z zA A A a A A a A aρ ρΦ Φ= Φ − Φ + Φ + Φ +

cos sinxA A Aρ Φ= Φ − Φ

sin cosyA A Aρ Φ= Φ + Φ

z zA A=

⎫⎪⎬⎪⎭

(5)equations



Transformation of a vector expressed in rectangular coordinates (Ax.Ay,Az) to cylindrical coordinates (Aρ.AΦ,Az) can be achieved using equations (4). The set of equations (4) can be expressed in matrix form as

cos sin 0sin cos 00 0 1

=

z

x

y

z

AAA

AAA

ρ

Φ

Φ Φ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥−

⎡ ⎤⎢ ⎥⎢ Φ Φ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦⎥⎣ ⎦ ⎦⎥

⎢ ⎣



Transformation of a vector expressed in cylindrical coordinates (Aρ.AΦ,Az) to rectangular coordinates (Ax.Ay,Az) can be achieved using equations (5). The set of equations (5) can be expressed in matrix form as

cos sin 0sin cos 0

0 0 1 =

z

x

y

z

AA

AAA A

ρ

Φ

Φ − Φ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥Φ

⎡ ⎤⎢ ⎥⎢ Φ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣⎥⎦ ⎦⎢ ⎥⎣


Example-2

Convert the following points expressed in cylindrical coordinates to rectangular coordinate system.

(i) P(2,5π/6,3)(ii) Q(4,4π/3,-1) cosx ρ= Φ

siny ρ= Φz z=

52co s 1.7( s 36

o) c 2xi πρ = −= =Φ52sis nin 16

y πρ= Φ = =

( 1.732,1,35(2, ,3)6

)P Pπ−⇒

4( ) (4, , 1) ( 2,3

3.464, 1)i P Pi π− − −− ⇒


Example-1

Convert the following point expressed in rectangular coordinate system to cylindrical coordinates and sketch the location of the point. P(x=3,y=4,z=5)

2 2x yρ = +1tan y

x− ⎛ ⎞Φ = ⎜ ⎟⎝ ⎠

z z=

22 22 3 4 5x yρ = ++ ==

1 1 4tat n 53.1an3

yx

− − ⎛ ⎞= =⎜ ⎟⎝ ⎠

⎛ ⎞Φ = ⎜ ⎟⎝ ⎠

(5( ,3, 534,5) .1 ,5)PP ⇒

P(5,53.1,5)

ρ=5

Φ=53.1

Y

Z

z=5

X


Example-3

cos sin 0 = sin cos 0

0 0 1

x

y

z z

A AA AA A

ρ

Φ

Φ Φ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥− Φ Φ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

cos sin 0 4= sin cos 0 2

0 0 1 4

Φ Φ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥− Φ Φ −⎢ ⎥ ⎢ ⎥

−⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

4zA = −4cos 2sinAρ = Φ − Φ 4sin 2cosAφ = − Φ − Φ

P(2,3,5)At the point 1 1 3tat n 56.3an2

yx

− − ⎛ ⎞= =⎜ ⎟⎝ ⎠

⎛ ⎞Φ = ⎜ ⎟⎝ ⎠

( ) ( )4cos56.3 2sin56.3 4sin56.3 2cˆ ˆ ˆos56.3 4 zF a a aρ Φ− − − −= +

ˆ ˆ ˆ0.556 4.44 4 zF a a aρ Φ= − −

4cos56.3 2sin56.3Aρ = − 4sin56.3 2cos56.3Aφ = − − 4zA = −

Convert the vector located at A(2,3,5) in to cylindrical coordinates.

ˆ ˆ ˆ4 2 4x y zF a a a= − −


Example-4

Express the vector in cylindrical coordinates.

2 2 2ˆ ˆ ˆx y zxy za x yza xyz a+ +

siny ρ= Φcosx ρ= Φ


=

z

x

y

z

AAA

AAA

ρ

Φ

Φ Φ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥−

⎡ ⎤⎢ ⎥⎢ Φ Φ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦⎥⎣ ⎦ ⎦⎥

⎢ ⎣2

2

2

cos sin 0sin cos 00

=0

1

z

xy zx yzxy

A

zAA

ρ

Φ

⎡ ⎤Φ Φ⎡ ⎤⎢ ⎥⎢ ⎥− Φ Φ

⎡ ⎤⎢ ⎥⎢ ⎥⎢

⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦⎥ ⎣⎦


Example-4 (Contd…)

2zA xyz=

2 2cos sinA xy z x yzρ Φ += Φ2 2sin cosxy z x yzAΦ Φ += − Φ

siny ρ= Φ cosPut x ρ= Φ

3 2 2 3 2 2cos sin sin cosz zAρ ρ ρΦ Φ + Φ= Φ3 2 2cos sin2 zρ= Φ Φ3 3 3 3cos sin cos sinz zAφ ρ ρΦ Φ + Φ= − Φ

2 2 sin coszA zρ Φ= Φ


Spherical coordinate systemAny point in space is represented as the intersection of three surfaces:

A sphere of radius r from the origin (r=constant)A cone centered around the z axis (θ=constant)A vertical plane (Φ=constant)

Any point in spherical coordinate system is considered to be at the intersection of the above three planes.

θ

θ=constant

X Y

Z


Spherical coordinate system

X

Y

Z

r

X

Y

Z

Φ

r=constant

Φ=constant



X

Y

Z

θ

Φ

r

P(r, θ, Φ)

: 0 0 0 2

Ranges rθ πφ π

≤ < ∞≤ ≤< <



X

Y

Z

θ

Φ

r

ˆra

aθ

aΦ

ˆra

aθ

aΦ

: 0 0 0 2

Ranges rθ πφ π

≤ < ∞≤ ≤< <


Spherical coordinate systemThree unit vectors of the spherical coordinate system are shown in the figure.Unit vector lies along the radially outward direction to the spherical surface. It lies on the cone θ=constant and the plane Φ=constantThe unit vector is normal to the conical surface and lies in Φ=constant plane and is tangential to the spherical surface.Unit vector is the same as in cylindrical coordinate system. It is normal to Φ=constant plane and is tangential to both the cone and the sphere.The unit vectors are mutually perpendicular and forms a right handed set. An RH screw when rotated from will move it towards direction.

ˆra

aθ

aΦ

ˆ ˆ ra to aθaΦ


Spherical coordinate systemA vector in spherical coordinate system may be expressed as

Magnitude of the vector is given by

The unit vectors are mutually orthogonal. Thus

Aˆ ˆ ˆr rA A a A a A aθ θ φ φ= + +

ˆ ˆ ˆ, , , , ra a a are unit vectors along r directionsθ φ θ φ

2 2 2rA A A Aθ φ= + +

ˆ ˆ ˆ, ,ra a aθ φ

ˆ ˆ ˆ ˆ ˆ ˆ 1r ra a a a a aθ θ φ φ⋅ = ⋅ = ⋅ =ˆ ˆ ˆ ˆ ˆ ˆ 0r ra a a a a aθ θ φ φ⋅ = ⋅ = ⋅ =

ˆ ˆ ˆra a aθ φ× =ˆ ˆ ˆra a aθ φ× =ˆ ˆ ˆra a aφ θ× =


Transformation of variables

X

Y

Z

x

y

Φ ρ

siny ρ= Φ

cosx ρ= Φ

P(x,y,z) or (ρ,Φ,z) or (r,θ,Φ)

z

θr

sinrρ θ=

cosz r θ=

, ,, , in ter rms fx oy z θ φ ,, ,, in termr s f xo y zθ φ2 2 2r x y z= + +

2 21tan x y

zθ − +=

1tan yx

φ −=

sin cosx r θ φ=sin siny r θ φ=

cosz r θ=


Transformation of variables

2 2 2r x y z= + +

2 21tan x y

zθ − +=

1tan yx

φ −=

sin cosx r θ φ=

sin siny r θ φ=

cosz r θ=


Transformation of vectors

X

Y

Z

x

y

Φ ρ

siny ρ= Φ

cosx ρ= Φ

z

θr

sinrρ θ=

cosz r θ=

ˆra

aθ

aΦ

90 θ−

θ

θ

ˆ ˆ ˆ ˆsin cos cos cos sinx ra a a aθ φθ φ θ φ φ= + −

ˆ ˆ ˆ ˆsin sin cos sin cosy ra a a aθ φθ φ θ φ φ= + +ˆ ˆ ˆcos sinz ra a aθθ θ= −

ˆxa ˆ ya

ˆza

90 θ−



X

Y

Z

x

yΦ

ρ

siny ρ= Φ

cosx ρ= Φ

z

θr

sinrρ θ=

cosz r θ=

ˆra

aθ

aΦ

90 θ−

θ

θ


ˆ ˆ ˆ ˆsin sin cos sin cosy ra a a aθ φθ φ θ φ φ= + +ˆ ˆ ˆcos sinz ra a aθθ θ= −

ˆxa ˆ ya

ˆza

aΦ−

aΦ−aΦ

Φ


Transformation of vectorsThe unit vectors are to be expressed in terms of unit vectors in spherical coordinates

consists of the projections of on the x axis.In order to find this projection, first find the projection on the xy plane and then on to the required unit vectors.

ˆ ˆ ˆ, , ra a aθ φ

ˆ ˆ ˆ, , x y za a a

ˆxa ˆ ˆ ˆ, , ra a aθ φ


ˆ ˆ ˆ ˆsin sin cos sin cosy ra a a aθ φθ φ θ φ φ= + +

ˆ ˆ ˆcos sinz ra a aθθ θ= −

⎫⎪⎪⎬⎪⎪⎭

(1)equation



X

Y

Z

x

y

Φ ρ

siny ρ= Φ

cosx ρ= Φ

z

θr

sinrρ θ=

cosz r θ=

ˆra

aθ

aΦ

90 θ−

θ

θ

ˆ ˆ ˆ ˆcos sin sin sin cosr x y za a a aφ θ φ θ θ= + +ˆ ˆ ˆ ˆcos cos sin cos sinx y za a a aθ φ θ φ θ θ= + −

ˆ ˆ ˆsin cosx ya a aφ φ φ= − +

ˆxa ˆ ya

ˆza

90 θ−



X

Y

Z

x

yΦ

ρ

siny ρ= Φ

cosx ρ= Φ

z

θr

sinrρ θ=

cosz r θ=

ˆra

aθ

aΦ

90 θ−

θ

θ

ˆ ˆ ˆ ˆcos sin sin sin cosr x y za a a aφ θ φ θ θ= + +

ˆ ˆ ˆ ˆcos cos sin cos sinx y za a a aθ φ θ φ θ θ= + −ˆ ˆ ˆsin cosx ya a aφ φ φ= − +

ˆxa ˆ ya

ˆza

aΦ−

aΦ

Φ

ˆxa


Transformation of vectorsThe unit vectors are to be expressed in terms of unit vectors in rectangular coordinates

consists of the projections of on the r axis.In order to find this projection, first find the projection on the Φ=constant plane and then on to the required unit vectors.

ˆ ˆ ˆ, , ra a aθ φ ˆ ˆ ˆ, , x y za a aˆra

ˆ ˆ ˆ ˆsin cos sin sin cosr x y za a a aθ φ θ φ θ= + +

ˆ ˆ ˆ ˆcos cos cos sin sinx y za a a aθ θ φ θ φ θ= + −

ˆ ˆ ˆsin cosx ya a aφ φ φ= − +

ˆ ˆ ˆ, , x y za a a

⎫⎪⎪⎬⎪⎪⎭

(2)equation


Transformation of vectorsSubstituting eq(1) in the general equation of a vector in rectangular coordinates,


( )ˆ ˆ ˆsin cos cos cos sinx ra aA aA θ φθ φ θ φ φ= + −

( )ˆ ˆcos sinrz a aA θθ θ−+

( )ˆ ˆ ˆsin sin cos sin cosy r aA a aθ φθ φ θ φ φ+ ++

( ) ˆsin cos sin sin cosx y z rA A A A aθ φ θ φ θ= + +

( ) ˆcos cos cos sin sinx y zA A A aθθ φ θ φ θ+ + −

( ) ˆsin cosx yA A aφφ φ+ − +



This can be represented in matrix form as

sin cos sin sin cosr x y zA A A Aθ φ θ φ θ= + +

cos cos cos sin sinx y zA A A Aθ θ φ θ φ θ= + −sin cosx yA A Aφ φ φ= − +

sin cos sin sin coscos cos cos sin sin

sin cos 0

x

y

z

rAAA

AAA

θ

θ φ θ φ θθ φ θ φ θ

φ φφ

⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥− ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎦⎣ ⎦−⎣


Transformation of vectorsSubstituting eq(2) in the general equation of a vector in spherical coordinates,

ˆ ˆ ˆr rA A a A a A aθ θ φ φ= + +

( )ˆ ˆ ˆsin cos sin sin cosr x y za aA A aθ φ θ φ θ= + +

( )ˆ ˆsin cosx yaA aφ φ φ++ −

( )ˆ ˆ ˆcos cos cos sin sinx y zA a a aθ θ φ θ φ θ−+ +

( ) ˆsin cos cos cos sinr xA A A A aθ φθ φ θ φ φ= + −

( ) ˆsin sin cos sin cosr yA A A aθ φθ φ θ φ φ+ + +

( ) ˆcos sinr zA A aθθ θ+ −



This can be represented in matrix form as

sin cos cos cos sinx rA A A Aθ φθ φ θ φ φ= + −

sin cos cos cos sinsin sin cos sin cos

cos sin 0

x

y

rAAA Az

AAθφ

θ φ θ φ φθ φ θ φ φθ θ

−

−

⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥= ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

sin sin cos sin cosy rA A A Aθ φθ φ θ φ φ= + +cos sinz rA A Aθθ θ= −


Example 1Given the point P(-2,6,3)and vector express P and in Cartesian, cylindrical and spherical coordinates.

ˆ ˆ( )x yA ya x z a= + +A

:Solution ( -2, 6, 3)At point P x y z= = =

2 2 4 36 6.32x yρ + = + ==

11 6tan 108. 3t 42

an yx

−− ⎛ ⎞= =⎜ ⎟−⎝ ⎠⎛ ⎞Φ = ⎜ ⎟⎝ ⎠3z z= =

2 2 2 4 36 9 7r x y z= + + = + + =2 2

1 1 40tta an 64. 23

n 6x yz

θ − − =+

= =


Example 1 (Contd…)( 2,6,3) (6.32,108.43,3) (7,64.62 ,108.43 )P P P− = =

cos sin 0sin cos 0

0 0 1 =

z

x

y

z

AAA

AAA

ρ

Φ

Φ Φ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥−

⎡ ⎤⎢ ⎥⎢ Φ Φ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦⎥⎣ ⎦ ⎦⎥

⎢ ⎣


0

=

z

AAA

yx z

ρ

Φ

Φ Φ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥− Φ Φ +⎢ ⎥

⎡ ⎤

⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣

⎢ ⎥⎢⎢ ⎥ ⎦

⎥⎣ ⎦

cos , sin . ,But x y Substitutingρ φ ρ φ= =cos sin 0 sinsin cos 0 cos0 0 1 0

=

z

zAAA

ρ ρ φρ φΦ

Φ Φ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥− Φ Φ +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦⎦

{ } ˆcos sin ( cos ) sinA z aρρ φ φ ρ φ φ= + +

{ }2 ˆsin ( cos ) cosz aφρ φ ρ φ φ+ − + +


Example 1 (Contd…)

, =6.32, =108.43 , z=3At P ρ φ

cos 0.316 sin 0.9487 φ φ= − =

,Substituting

ˆ ˆ0.9487 6.008A a aρ φ= − −

,In the spherical system

sin cos sin sin coscos cos cos sin sin

sin cos 0

x

y

z

r AA

A

A AAθ

θ φ θ φ θθ φ θ φ θφ φφ

− −−

⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦


Example 1 (Contd…)sin cos sin sin coscos cos cos sin sin

sin cos 0 0

r yx z

AAAθ

θ φ θ φ θθ φ θ φ θφ φφ

⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎣

+

⎦

− −−

sin cos , sin sin , cos . ,But x r y r z r Substitutingθ φ θ φ θ= = =

sin cos sin sin cos sin sincos cos cos sin sin sin cos cos

sin cos 0 0

r rr r

AAAθ

θ φ θ φ θ θ φθ φ θ φ θ θ φ θφ φφ

⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎣

+

⎦

− −−

{ }2sin cos sin (sin cos cos )si ˆn sin rA ar rθ φ φ θ φ θ θ φ= + +

{ }sin cos sin cos (sin cos co ˆs )cos sinr r aθθ θ φ φ θ φ θ θ φ+ + +

{ }2sin sin ( ˆsin cos cos )cos ar r φθ φ θ φ θ φ+ − + +



, =7, =108.43 , =64.62At P r φ θcos 0.316 sin 0.9487 φ φ= − =cos 0.4286 sin 0.903 θ θ= =

,Substitutingˆ ˆ ˆ0.8571 0.4066 6.008rA a a aθ φ= − − −


Example 2Express the vector in

Cartesian coordinates. Find

1 0 ˆ ˆ ˆc o srB a r a ar θ φθ= + +

( 3,4,0) B −

sin cos cos cos sinsin sin cos sin cos

cos sin 0

x

y

r

z

AAA

AAAθ

φ

θ φ θ φ φθ φ θ φ φθ θ

−

−

⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥= ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

:Solution

sin cos cos cos sin 10 /sin sin cos sin cos cos

cos sin 0 1

x

y

z

BB

rr

B

θ φ θ φ φθ φ θ φ φ θθ θ

⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎣

−

−⎦ ⎣ ⎦ ⎦



210sin cos sin cos sinxB rr

θ φ θ φ φ= + −

210sin sin cos sin cosyB rr

θ φ θ φ φ= + +

10cos cos sinzB rr

θ θ θ= −

2 2 2r x y z= + +2 2

1tan x yz

θ − += 1tan y

xφ −=

sin cos cos cos sin 10 /sin sin cos sin cos cos

cos sin 0 1

x

y

z

BB

rr

B

θ φ θ φ φθ φ θ φ φ θθ θ

⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎣

−

−⎦ ⎣ ⎦ ⎦


Example 2 (Contd…)2 2

2 2 2sin x y

r x y zρθ

+= =

+ + 2 2 2cos z z

r x y zθ = =

+ +

2 2sin y y

x yφ

ρ= =

+ 2 2cos x x

x yφ

ρ= =

+

2 2 2 2 2 2

2 2 2 2 2 22 2 2 2 2 2

10x

x y x y zx z x yx y z x y zx y x y x

By

+ + += ⋅ + ⋅ −

+ + + ++ + +

( )( )2

2 2 2 2 22 2 2 2 2

10x xz yx y z x yx y z x y

= + −+ + ++ + +

2 2 2 2 2 2

2 2 2 2 2 22 2 2 2 2 2

10y

x y x y zy z y yx y z x y zx y x y x

By

+ + += ⋅ + ⋅ −

+ + + ++ + +

( )( )2

2 2 2 2 22 2 2 2 2

10y yz xx y z x yx y z x y

= + ++ + ++ + +


Example 2 (Contd…)2 2

2 2 2 2 2 2

10z

z x yzx y z z

Bx y

+= −

+ + + +

ˆ ˆ ˆ x yx y z zand B a B aB aB= + +

( 3,4,0) 3, 4, 0At x y z− = − = =

30 402

25 5xB = − + = −−

40 3025

1 5yB = + − =

0 yB =

ˆ ˆ, 2 x ySubstituting B a a= − + ˆ ˆ2 x yB a a= − +


Constant coordinate surfacesIf we keep one of the coordinate variables constant and allow the other two to vary, constant coordinate surfaces are generated inrectangular, cylindrical and spherical coordinate systems.In the Cartesian system, if we keep x constant and allow y and z to vary, an infinite plane x=constant is generated.Thus we can have infinite planes

X=constantY=constantZ=constant

These surfaces are perpendicular to x, y and z axes respectively.Intersection of two planes is a line. x=constant, y=constant is the line RPQ parallel to z axis.Intersection of three planes is a point. x=constant, y=constant,z=constant is the point P(x,y,z). Any point P may be defined as the intersection of three orthogonal planes.


Constant coordinate surfaces

x=constant

z=constant

y=constant

X

Z

Y



x=constant

z=constant

y=constantX

Z

Y

P

Q

R

x=constant

z=constant

y=constantX

Z

Y

P

Q

R


Constant coordinate surfacesOrthogonal surfaces in cylindrical coordinate system can be generated as

ρ=constantΦ=constantz=constant

ρ=constant is a circular cylinder, Φ=constant is a semi infinite plane with its edge along z axis, z=constant is an infinite plane as in the rectangular system.The intersection of two surfaces z=constant, ρ=constant is the circle QPR of radius ρThe intersection of surfaces z=constant, Φ=constant is a semi infinite line. The intersection of three surfaces produces a point. ρ=constant, Φ=constant, z=constant is the point P(ρ,Φ,z)



z=constant

ρ=constant

Φ =constantX

Y

Z

p qr



z=constantρ=constant

Φ =constantX

Y

Z

pq

r


Constant coordinate surfacesOrthogonal surfaces in spherical coordinate system can be generated as

r=constantθ=constantΦ=constant

r=constant is a sphere with its centre at the origin, θ =constant is a circular cone with z axis as its axis and origin at the vertex, Φ =constant is a semi infinite plane as in the cylindrical system.The intersection of two surfaces r=constant, Φ =constant is a semi circle passing through Q an P The intersection of three surfaces produces a point. r=constant,θ =constant, Φ=constant is the point P(r,θ, Φ)Any point P may be defined as the intersection of these orthogonal planes



r=constant

Φ=constant

θ=constantZ

p

q


Differential elements in rectangular coordinate systems

X

Y

Z

x

y

z

z

dx

dz

dyA B

CD

P Q

RS

ˆxa ˆ yaˆza

figure(1)


Differential length, area and volume in Cartesian coordinates

Differential displacement is given by

Differential normal area is given by

Differential volume is given by

are vectors where as dv is a scalar.

ˆ ˆ ˆx y zdl dxa dya dza= + +

ˆ xdS dydza= ˆ ydxdza= ˆ zdzdya=

dv dxdydz=

dl and dS


Differential length, area and volume in Cartesian coordinates

If we move from P to Q, If we move from Q to S,If we move from D to Q,

In general the differential surface area is defined as where dS is the area of the surface element and is a unit vector normal to the surface dS.The different surfaces in figure(1) is described as

ˆ ydl dya=ˆ ˆy zdl dya dza= +ˆ ˆ ˆx y zdl dxa dya dza= + +

ˆndS dsa=ˆna

ˆxABCD dS dydza⇒ =

ˆxPQRS dS dydza⇒ = −

ˆ yBCRQ dS dydza⇒ =

ˆ yADSP dS dydza⇒ = −

ˆzABQP dS dxdya⇒ =

ˆ zDCRS dS dxdya⇒ = −


Differential normal areas in rectangular coordinate systems

X

Y

Zz

dy

dz

dx

dzdy dx

ˆxa

ˆ ya ˆza


Differential elements in cylindrical coordinate systems

X

Y


Differential elements in cylindrical coordinate systems

AB

C

DP

QR

Sdρ

dz

dρ φ

Φ

aρ

aφ

ˆzaX

Y

Z


Differential normal areas in cylindrical coordinate systems




ˆ ˆ ˆ zdl d a d a dzaρ φρ ρ φ= + +

ˆdS d dzaρρ φ= ˆd dzaφρ= ˆ zd d aρ φ ρ=

dv d d dzρ ρ φ=


Differential normal areas in cylindrical coordinate systems

X

Y

Zz

aρ

aΦˆza

dρ Φ

dzdz

dρ dρ Φ

dρ


Differential elements in spherical coordinate systems

X

Y

Z

Φ

θ

dφ

dθ

drrdθ

sinr dθ φ


Differential elements in spherical coordinate systems

X

Y

Z

Φ

θ

dφ

dθ

drrdθ sinr dθ φ


Differential normal areas in spherical coordinate systems

X

Y

Z

rdθ

sinr dθ φ ˆra

aθ

sinr dθ φ

dr

aΦ

drrdθ


Differential normal areas in spherical coordinate systems




ˆ ˆ ˆsinrdl dra rd a r d aθ φθ θ φ= + +

2 ˆsin rdS r d d aθ θ φ= ˆsinr drd aφθ φ= ˆrdrd aφθ=

2 sindv r drd dθ θ φ=


Example 1For the object shown below calculate:(i) The distance BC(ii) The distance CD(iii) The surface area ABCD(iv) The surface area ABO(v) The surface area AOFD(vi) The volume ABDCFO

A(5,0,0)

B(0,5,0)

C(0,5,10)

D(5,0,10)

O(0,0,0)

F(0,0,10)


Example 1Solution:The object has cylindrical symmetry hence it is convenient to solve the problem using cylindrical coordinates. For this first we have to convert all the points to cylindrical coordinates.

(5,0,0) (5,0 ,0)A A⇒

(0,5,0) 5, ,02

B B π⎛ ⎞⇒ ⎜ ⎟⎝ ⎠

(0,5,10) 5, ,102

C C π⎛ ⎞⇒ ⎜ ⎟⎝ ⎠

( )(5,0,10) 5,0 ,10D D⇒

(5,0 ,0)A

5, ,02

B π⎛ ⎞⎜ ⎟⎝ ⎠

5, ,102

C π⎛ ⎞⎜ ⎟⎝ ⎠( )5,0 ,10D

F(0,0,10)

O(0,0,0)


Example 1(Contd….)

( ) , ; i Along BC dl dz hence=10

010BC dl dz= = =∫ ∫

( ) , , 5. ii Along CD dl d Henceρ φ ρ= =/2 /2

005 2.5CD d

π π πρ φ φ= = =∫( ) , ; 5. iii For ABCD dS d dz Henceρ φ ρ= =

/2 10

0 0

zArea ABCD dS d dz

π

φρ φ

= == =∫ ∫ ∫

/2 10

0 05 25d dz

ππφ= =∫ ∫

( ) , ; 0. iv For ABO dS d d z Henceρ φ ρ= =/2 5

0 0 Area ABO dS d d

π

φ ρρ φ ρ

= == =∫ ∫ ∫ /2 5

0 06.25d d

πφ ρ ρ π= =∫ ∫


Example 1(Contd….)( ) , ; 0. v For AOFD dS d dz Henceρ φ= =

5 10

0 050

zArea AOFD dS d dz

ρρ

= == = =∫ ∫ ∫

( ) , . vi For volume ABDCFO dv d dzd Henceρ φ ρ=5 /2 10

0 0 0

zVolume ABDCFO dv dzd d

π

ρ φρ φ ρ

= = == =∫ ∫ ∫ ∫

5 /2 10

0 0 062.5

zd d dz

πρ ρ πφ

== =∫ ∫ ∫


Example 2The object given below is part of a spherical shell described as

Calculate(i) The distance DH(ii) The distance FG(iii) The surface area AEHD(iv) The surface area ABDC(v) The volume of the object

BA

C

D

E

F

G

H

3 560 9045 60

rθ

φ

≤ ≤

≤ ≤

≤ ≤


Example 2:Solution

( ) sin ; 3, 90i Along DH dl r d rθ φ θ= = =60

453 sin 90D H dl d

φφ

== =∫ ∫

60

453 sin 90 dφ= ∫

312

0.785π= × =

( ) ; 5, 60 , 60 90ii Along GF dl rd r toθ φ θ= = = ⇒90

605G F dl d

θθ

== =∫ ∫ 65 . 1

62 7π

= × =



2( ) sin ; 3iii On AEHD dS r d d rθ θ φ= =

60 90 , 45 60to toθ φ⇒ ⇒60 90

45 60 = 9sinArea AEHD dS d d

φ θθ θ φ

= ==∫ ∫ ∫

60 90

45 609 sind d

φ θφ θ θ

= == ∫ ∫

[ ]90

609 cos

12π θ= × × − 1.178=



( ) ; 45 iv On ABCD dS rd drθ φ= =

60 90 , 3 5to r toθ ⇒ ⇒5 90

3 60 =

rArea ABCD dS rd dr

θθ

= ==∫ ∫ ∫

90 5

60 3d rdrθ= ∫ ∫

52

36 2rπ ⎡ ⎤

= × ⎢ ⎥⎣ ⎦

4 183

64 .π= =


Example 2 (Contd…)2( ) Volume sin v On ABCDEFGH dv r drd dθ θ φ=

3 5, 60 90 , 45 60r to to toθ φ⇒ ⇒ ⇒5 90 60 2

3 60 45 = sin

rVolume dv r drd d

θ φθ θ φ

= = ==∫ ∫ ∫ ∫

4936

4.276π= =

5 90 602

3 60 45sinr dr d dθ θ φ= ∫ ∫ ∫

[ ] [ ]53

90 60

60 453

cos3r θ φ⎡ ⎤

= −⎢ ⎥⎣ ⎦


Line integralsLine integral is defined as any integral that is to be evaluated along a line. A line indicates a path along a curve in space.

represents a line integral where each element of length on the curve is multiplied according to scalar dot product rule by the local value of and then these products are summed to get the value of the integral.Let be a vector field in space and ab a curve from point a to point b. Let the curve ab is subdivided in to infinitesimally small vector elementsLet the dot products are taken where are the value of the vector field at the junction points of the vector elementsThen the sum of these products along the entire length of the curve is known as the line integral of along the curve ab.

b

aA d l⋅∫ d l

A

1 2 3, , ,........, rdl dl dl dl

A

1 2 31 2 3. , . , . , ........, . rrA d l A d l A d l A d l

1 2 3, , , ........, rA A A A1 2 3, , , ........, rdl dl dl dl

b

rra

A d l⋅∑A


Line integrals

Y

X

Z

a

b

1dl

2dl3dl

4dl

1A

rdl

2A

3A

rA

bbrra

a

A d l A dl⋅ = ⋅∑∫

A


Line integrals

a

b

rdl rAθ

A


Line integralsAs an example if represents the force acting on a moving particle along a curve ab, then the line integral of over the path described by the particle represents the work done by the force in moving the particle from a to b.The line integral around a closed curve is called closed line integral

FF

a

bA dl⋅∫


Surface integralsConsider a vector field continuous in a region of space containing a smooth surface S.The surface integral of through S can be defined as

A

A sA dSψ = ⋅∫

ˆnad S

ASurface S

θ


Surface integrals

Consider a small incremental surface area on the surface S denoted by dS. Let be a unit normal to the surface dS.The magnitude of flux crossing the unit surface normally is given by

Where denote the vector area having magnitude equal to dSand whose direction is in the direction of the unit normal.

The total flux crossing the surface is obtained by integrating over the surface of interest.

ˆna

dS

ˆnS Sad d=

cosA dSθ ˆ nA a dS= ⋅ ˆ nA dSa= ⋅ A dS= ⋅

A dS⋅

sA dSψ = ⋅∫


Surface integralsFor a closed surface defining a volume the surface integral becomes closed surface integral and is denoted by

It represents the net outward flow of flux from surface SS

A dSψ = ⋅∫


Volume integrals

Let V be a volume bounded by the surface S. Let be a

function of position defined over V. If the volume V is subdivided in

to n elements of volumes

In each part let us choose an arbitrary point

Then the limit of the sum as is

called the volume integral of over V and is denoted by

vd vϕ∫

( , , )x y zϕ

1 2 3, , ,........, ndV dV dV dV

( , , )i i ix y zϕ

( , , )i i i ix y z dVϕ∑ 0in and dV→∞ →( , , )x y zϕ


Scalar and vector fields

Let R be a region of space at each point of which a scalar

Φ = Φ (x,y,z) is given, then Φ is called a scalar point function and R is called a scalar field.Examples:

Temperature distribution in a mediumDistribution of atmospheric pressure in space.

Let R be a region of space at each point of which a vector is

given, then is called a vector point function and R is called a scalar field.Examples:

The velocity of a moving fluid at any instantGravitational force in a region.

( , , )v v x y z=

v


DEL OperatorThe del operator is the vector differential operator and is denoted by . In Cartesian coordinates

The vector differential operator is not a vector in itself, but when it operates on a scalar function the result is a vector.This operation is useful in defining

The gradient of a scalarThe divergence of a vectorThe curl of a vectorThe Laplacian of a scalar

ˆ ˆ ˆx y za a ax y z∂ ∂ ∂

∇ = + +∂ ∂ ∂

∇

V∇.A∇

A∇×2V∇


DEL Operator in cylindrical coordinatesThe expression for del operator in other coordinate systems can be obtained using the transformation equations derived earlier.

2 2x yρ = + tan yx

φ =

cosx ρ φ= siny ρ φ=sincos

xφφ

ρ ρ φ∂ ∂ ∂= −

∂ ∂ ∂cossin

yφφ

ρ ρ φ∂ ∂ ∂= +

∂ ∂ ∂

⎫⎪⎬⎪⎭

(1)equation

ˆ ˆ ˆ x y zSubstituting in a a ax y z∂ ∂ ∂

∇ = + +∂ ∂ ∂

sin cosˆ ˆ ˆ cos sin (2)x y za a az

φ φφ φρ ρ φ ρ ρ φ

⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂∇ = − + + + − − − −⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠


DEL Operator in cylindrical coordinates

( ) sin cosˆ ˆ ˆ ˆ ˆ cos sin (3)x y x y za a a a az

φ φφ φρ φ ρ ρ

⎛ ⎞∂ ∂ ∂∇ = + + − + + −−−−⎜ ⎟∂ ∂ ∂⎝ ⎠

aρ aφ

( ) ( )ˆ ˆ ˆ ˆcos sin s1 ˆi s (4co )nx y x y za a az

a aρ ρ

φ φ φ φφ

∂ ∂ ∂∇ = + + −−−− +

∂+ −

∂ ∂

1ˆ ˆ ˆ (5)za a azρ φρ ρ φ

∂ ∂ ∂∇ = + + − − − −

∂ ∂ ∂1ˆ ˆ ˆ za a a

zρ φρ ρ φ∂ ∂ ∂

∇ = + +∂ ∂ ∂


DEL Operator in spherical coordinatesThe expression for del operator in other coordinate systems can be obtained using the transformation equations derived earlier.

cos cos sinsin cosx r r

θ φ φθ φθ ρ φ

∂ ∂ ∂ ∂= + −

∂ ∂ ∂ ∂ ⎫⎪⎬⎪⎭

(1)equation

ˆ ˆ ˆ x y zSubstituting in a a ax y z∂ ∂ ∂

∇ = + +∂ ∂ ∂

2 2 2r x y z= + +

2 21tan x y

zθ − +=

1tan yx

φ −= sin cosx r θ φ=sin siny r θ φ=cosz r θ=

cos sin cossin siny r r

θ φ φθ φθ ρ φ

∂ ∂ ∂ ∂= + +

∂ ∂ ∂ ∂sincos

z r rθθ

θ∂ ∂ ∂= −

∂ ∂ ∂


DEL Operator in spherical coordinatescos cos sin ˆ sin cos xa

r rθ φ φθ φ

θ ρ φ⎛ ⎞∂ ∂ ∂

∇= + −⎜ ⎟∂ ∂ ∂⎝ ⎠

cos sin cos sin ˆsin sin cos zar r r r

θ φ φ θθ φ θθ ρ φ θ

⎛ ⎞∂ ∂ ∂ ∂ ∂⎛ ⎞+ + + + −⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠

( )ˆ ˆ ˆ sin cos sin sin cosx y za a ar

θ φ θ φ θ∂∇ = + +

∂cos cos cos sin sin sin cosˆ ˆ ˆx z ya a a

r r rθ φ θ φ θ φ φ

θ φ ρ ρ⎛ ⎞∂ ∂⎛ ⎞+ + − + − +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠


DEL Operator in spherical coordinates

( )ˆ ˆ ˆsin cos sin sin co sx y zra a aθ φ θ φ θ+

∂∇ =

∂+

( ) ( )ˆ ˆ ˆ ˆcos cos cos sin sin sin co1 s1x z x ya a a a

r θθ φ

φφ φ

ρθ θ φ∂ ∂

− +∂

++ +∂

−

ˆra

aθ aφ

1 1ˆ ˆ ˆ (5)ra a ar r θ φθ ρ φ∂ ∂ ∂

∇ = + + − − − −∂ ∂ ∂

1 1ˆ ˆ ˆ (6)sinra a a

r r rθ φθ θ φ∂ ∂ ∂

∇ = + + − − − −∂ ∂ ∂

1 1ˆ ˆ ˆ sinra a a

r r rθ φθ θ φ∂ ∂ ∂

∇ = + +∂ ∂ ∂


Gradient of a scalar field

Y

X

Z

1V

2 1V V V= + Δ

3V

VΔ

VΔ

1P

1P

dl θG

A Level Surface


Gradient of a scalar fieldThe gradient of a scalar field V is a vector that represents themagnitude and direction of the maximum space rate of increase of V.Let V be a scalar field and let V1,V2 and V3 be contours on which V is constant.Consider the difference in the field dV between points P1 and P2

V V VdV dx dy dzx y z

∂ ∂ ∂= + +∂ ∂ ∂

( )ˆ ˆ ˆ ˆ ˆ ˆx y z x y zV V Va a a dxa dya dzax y z

⎛ ⎞∂ ∂ ∂= + + ⋅ + +⎜ ⎟∂ ∂ ∂⎝ ⎠

ˆ ˆ ˆ x y zFor c V V Va a aonvenienc GLex y

ez

t ⎛ ⎞∂ ∂ ∂+ + =⎜ ⎟∂ ∂ ∂⎝ ⎠

cosThen dV G dl G dlθ= ⋅ =



cosdVThen Gdl

θ=

0 . ., dl dV is maximum when i e when is in the direction of Gdl

θ =

MAX

dV Gdl

=

Magnitude of G is equal to the maximum space rate of change of V

Direction of G is along the maximum space rate of change of V

G is defined as the gradient of V and is denoted by grad V

ˆ ˆ ˆ V = V= x y zV V Vgrad a a ax y z

∂ ∂ ∂∇ + +

∂ ∂ ∂



,For cylindrical coordinates1ˆ ˆ ˆ V = V= z

V V Vgrad a a azρ φρ ρ φ

∂ ∂ ∂∇ + +

∂ ∂ ∂

,For spherical coordinates

1 1ˆ ˆ ˆ V = V=sinr

V V Vgrad a a ar r rθ φθ θ φ

∂ ∂ ∂∇ + +

∂ ∂ ∂


Gradient of a scalar field-Important Relations

( )V U V U∇ + = ∇ + ∇∗ ( )VU V U U V∇ = ∇ + ∇∗

2- V U V V U

U U∇ ∇⎛ ⎞∇ =⎜ ⎟

⎝ ⎠∗

1 ( )n nV nV V−∇ = ∇∗


Gradient of a scalar field-Important Points

1. Magnitude of V is equal to the maximum space rate of change of V∇

2. Direction of V is along the maximum space rate of change of V∇

at any point 3. V is perpendicular to the constant V surface that passes ∇

through that point. , 4. If A V V is called the scalar potential of A= ∇

ˆ ˆ 5. The projection of V in the direction of a given unit vector a is V a∇ ∇ ⋅

ˆ It is called the directional derivative of V along aˆ It indicates the rate of change of V in the direction of a


Example 12 3 2 3Find grad when x y y zφ φ = −

:Solutionˆ ˆ ˆ = = x y zgrad a a a

x y zφ φ φφ φ ∂ ∂ ∂

∇ + +∂ ∂ ∂

2 3 2 2 3 2 2 3 2(3 ) (3 ) (3 )ˆ ˆ ˆ = = x y zx y y z x y y z x y y zgrad a a a

x y zφ φ ∂ − ∂ − ∂ −

∇ + +∂ ∂ ∂

( ) ( ) ( )2 2 2 3ˆ ˆ ˆ 6 3 3 2x y zgrad xy a x y z a y z aφ φ=∇ = + − + −

Find the directionalˆ ˆ ˆ 3 4 12 (2,-1,0)x y zderivative in the direction of a a a at+ +

( ) ( )ˆ ˆ (2,-1,0) 12 12x yAt a aφ∇ = − +

( ) ( )ˆ ˆ ˆ3 4 12 ˆ ˆ ˆ 12 12

9 16 144x y z

x y

a a aDirectional derivative a a aφ

+ +∇ ⋅ = − +

+ +( )( )ˆ ˆ ˆ ˆ ˆ12 12 0.23 0.31 0.92 x y x y za a a a a= − + + +

0.96=


Example 22 V V 10 sin cosFind grad when r θ φ=

:Solution

1 1ˆ ˆ ˆ V = V=sinr

V V Vgrad a a ar r rθ φθ θ φ

∂ ∂ ∂∇ + +

∂ ∂ ∂

2 ˆ ˆ ˆ10sin cos 10sin 2 cos 10sin sinrV a a aθ φθ φ θ φ θ φ∇ = + −


Divergence of a vectorThe divergence of a vector quantity at a given point P is the outward flux per unit volume over a closed incremental surface as the volume shrinks about P.

A

0lim Sv

A dSdivA A

vδ →

⋅= ∇ ⋅ =

Δ∫

S

A dS is the net outflow of flux of a vector field A from⋅∫ a closed surface S


Divergence of a vector in Cartesian coordinates

dxdy

dz

X

Y

Z

0 0 0( , , )P x y z1F 2F



0 0 0 ( , , ) To evaluate the divergence of a vector field A at point P x y z♣ first construct a differential volume around point P

The closed surface integral of A is obtained as♣

( )S FRONT BACK LEFT RIGHT TOP BOTTOMA dS⋅ = + + + + +∫ ∫ ∫ ∫ ∫ ∫ ∫

xA three dimensional Taylors series expansion of A about P is♣

0 0 0 0 0 0( , , ) ( , , ) ( ) ( ) ( )x x xx x

P PP

A A AA x y z A x y z x x y y z zx y z

∂ ∂ ∂= + − + − + −

∂ ∂ ∂

higher order terms+



0 ˆ ˆ , ,2 x x x

dxFor the front side x x A A a dS dydza= + = =

0 0 0( , , ) 2

xR NT

PF xO

A dx AA x y z dydz higher ordd er termsSx

⎛ ⎞∂+ +⎜ ⎟∂⎝

⋅⎠

=∫

( ) ( )0 ˆ ˆ , ,2 x x xdxFor the back side x x A A a dS dydz a= − = − = −

0 0 0( , , ) 2

xCK

PBA xA dx AA x y z dydz higher ord der ms

xS ter

⎛ ⎞∂− − +⎜ ⎟∂⎝

⋅⎠

=∫

FRONT BACK

x

P

Adxdydz higher order terA dS A d msx

S ∂+

∂⋅ + ⋅ =∫ ∫



Similarly

LEFT RIGHT

y

P

Adxdydz higher order terms

yA dS A dS

∂+

∂⋅ + ⋅ =∫ ∫

TOP BOTTOM

z

P

Adxdydz higher order teA dS A d r sz

S m∂+

∂⋅ + ⋅ =∫ ∫

yx z

P PPS

AA Adxdydz dxdydz dxdydz higher orA d der termsx y z

S∂∂ ∂

+ + +∂ ∂

⋅∂

=∫

yx zS

P

AA A v higher order termsx y

dz

A S∂∂ ∂

+ + Δ +∂

⋅∂

=∂∫

0 lim S

v

A dSSubstituting in

vδ →

⋅

Δ∫



0 0 lim lim

yx z

PSv v

A dS

v

AA Ax y z

vvδ δ→ →

∂∂ ∂+ +∂ ∂ ∂⋅

= ΔΔΔ

∫ yx z

P

AA Ax y z

∂∂ ∂= + +∂ ∂ ∂

yx zAA AAx y z

∂⎛ ⎞∂ ∂∇ ⋅ = + +⎜ ⎟∂ ∂ ∂⎝ ⎠

0Since higher order terms vanish as vΔ →

0 0 0 ( , , ) Divergence of A at P x y z in Cartesian coordinates is


Divergence of a vector in other coordinates

In cylindrical coordinates

( )1 1 zA AA Az

φρρ

ρ ρ ρ φ∂⎛ ⎞∂ ∂

∇ ⋅ = + +⎜ ⎟∂ ∂ ∂⎝ ⎠

In spherical coordinates

( ) ( )22

1 1 1sinsin sinr

AA r A A

r r r rφ

θ θθ θ θ φ

∂⎛ ⎞∂ ∂∇ ⋅ = + +⎜ ⎟∂ ∂ ∂⎝ ⎠


Divergence of a vector - Physical significance

xδyδ

zδ

X

Y

Z

yv y yv δ+

( , , )P x y z1F 2F



Consider the motion of a fluid having velocity at a point P(x,y,z).Consider a small parallelepiped with edges parallel to the axes enclosing the point P.The mass of fluid entering through the face F1 per unit time is given by velocity x areaThe mass of fluid flowing out through face F2 isBy Taylor’s theorem (neglecting higher order terms)

ˆ ˆ ˆx x y y z zV v a v a v a= + +

, y zx andδ δ δ

y x zv δ δy y x zv δ δ δ+

yy y y

vv v y

yδ δ+

∂= +

∂

yy y y

vv x z v y x z

yδ δ δ δ δ δ+

∂⎛ ⎞= +⎜ ⎟∂⎝ ⎠


Divergence of a vector - Physical significanceThe net decrease in the mass of fluid flowing across these two faces is

Similarly by considering the other two faces we get the total decrease in the mass of fluid inside the parallelepiped per unit time

Rate of loss of fluid per unit volume is

The above quantity is defined as the divergence of fluid velocity at the point P and is denoted by div V or

yy y

vv y x z v x z

yδ δ δ δ δ

∂⎛ ⎞+ −⎜ ⎟∂⎝ ⎠

yvx y z

yδ δ δ

∂=∂

yx zvv v x y zx y z

δ δ δ∂⎛ ⎞∂ ∂

+ +⎜ ⎟∂ ∂ ∂⎝ ⎠yx zvv v

x y z∂⎛ ⎞∂ ∂

+ +⎜ ⎟∂ ∂ ∂⎝ ⎠

V∇⋅



EXAMPLES

Divergence of the velocity of water in a container after the outlet has opened is zero because water is an incompressible fluid. Volume of water entering and leaving different regions of the closed surface is equal.When the valve on a steam boiler is opened there is a net outward flow of steam for each elemental volume. In this case the divergence has a positive value. It indicates a source of vectorquantity at that point.When an evacuated glass bulb is broken there is a sudden inrush of air and there is a net inward flow of air for each elemental volume. In this case the divergence has a negative value. It indicates a sink of that vector quantity.



ZERO DIVERGENCE



POSITIVE DIVERGENCE

SOURCE

NEGATIVE DIVERGENCE

SINK


ExamplesDetermine the divergence of the vector fields

2 ˆ ˆ( ) x zi P x yza xza= +

2ˆ ˆ ˆ( ) sin cos zii Q a za z aρ φρ φ ρ φ= + +

yx zPP PPx y z

∂⎛ ⎞∂ ∂∇ ⋅ = + +⎜ ⎟∂ ∂ ∂⎝ ⎠

( ) ( ) ( )2 0P x yz xzx y z∂ ∂ ∂

∇ ⋅ = + +∂ ∂ ∂

2P xyz x∇⋅ = +

( )1 1 zQ QQ Qz

φρρ

ρ ρ ρ φ∂⎛ ⎞∂ ∂

∇⋅ = + +⎜ ⎟∂ ∂ ∂⎝ ⎠

( )2 21 1sin ( ) ( cos )Q z zz

ρ φ ρ φρ ρ ρ φ

⎛ ⎞∂ ∂ ∂∇ ⋅ = + +⎜ ⎟∂ ∂ ∂⎝ ⎠

2sin cosQ φ φ∇ ⋅ = +


Examples

21 ˆ ˆ ˆ( ) T cos sin cos cosriii a r a ar θ φθ θ φ θ= + +

( ) ( )22

1 1 1sinsin sinr

TT r T T

r r r rφ

θ θθ θ θ φ

∂⎛ ⎞∂ ∂∇ ⋅ = + +⎜ ⎟∂ ∂ ∂⎝ ⎠

( ) ( ) ( )22

1 1 1cos sin cos cossin sin

T rr r r r

θ θ φ θθ θ θ φ

⎛ ⎞∂ ∂ ∂∇ ⋅ = + +⎜ ⎟∂ ∂ ∂⎝ ⎠

1 2 sin cos cossin

rr

θ θ φθ

=

2cos cosT θ φ∇ ⋅ =


Gauss’s Divergence theoremThe integral of the normal component of any vector field over a closed surface is equal to the integral of the divergence of this vector field throughout the volume enclosed by the closed surface.

The total outward flux of a vector field through the closed surface S is the same as the volume integral of the divergence of

Volume integrals are easier to evaluate than surface integrals. Using divergence theorem we can convert surface integral to a volume integral and then easily evaluate it.

AA

S VA dS AdV⋅ = ∇ ⋅∫ ∫


Divergence theorem

yx zV V V V

AA AA dV dxdydz dxdydz dxdydzx y z

∂∂ ∂∇ ⋅ = + +

∂ ∂ ∂∫ ∫ ∫ ∫

yx zV

AA AA dV dxdydz dxdydz dxdydzx y z

∂∂ ∂∇ ⋅ = + +

∂ ∂ ∂∫ ∫∫∫ ∫∫∫ ∫∫∫

Proof

yx zAA AAx y z

∂∂ ∂∇ ⋅ = + +

∂ ∂ ∂

yx zV V

AA AA dV dxdydzx y z

∂∂ ∂∇ ⋅ = + +

∂ ∂ ∂∫ ∫


Divergence theorem

x x y y z zVA dV A dS A dS A dS∇⋅ = + +∫ ∫∫ ∫∫ ∫∫ A dS= ⋅∫∫

S VA dS A dV⋅ = ∇ ⋅∫ ∫

S VA dS A dV⋅ = ∇ ⋅∫ ∫

xx

A dx Ax

∂=

∂∫y

y

Ady A

y∂

=∂∫

zz

A dz Az

∂=

∂∫

x y zVA dV A dydz A dxdz A dxdy∇⋅ = + +∫ ∫∫ ∫∫ ∫∫

xdydz dS= ydxdy dS= zdxdz dS=


Divergence theorem-Explanation

X

Y

Z Closed surface S

V


Divergence theorem-ExplanationLet the volume V bounded by the surface S is subdivided in to a number of elementary volumes The flux diverging from each such cell enters or converges on the adjacent cells unless the cell contains a portion of the outer surface.As a result the divergence of the flux density throughout the volume leads to the same result as determining the net flux crossing the enclosing surface.

VΔ


Example 1

X

Y

ZTψ

Bψ

Sψ

ˆ ˆ2 2ρDetermine the flux of D = ρ cos a + zsin aφφ φ

over the surface of the cylinder 0 1, 4z ρ≤ ≤ = .Verify divergence theorem


Example 1If is the flux through the given surfaceψ

+ T B Sψ ψ ψ ψ= +ˆ , 1,T zFor z dS d d aψ ρ ρ φ= =

( )4 2

0 0ˆ ˆ ˆ 02 2

T ρ zD dS ρ cos a + zsin a d d aπ

φρ φψ φ φ ρ ρ φ

= == ⋅ = ⋅ =∫ ∫ ∫

( ) ( )4 2

0 0ˆ ˆ ˆ 02 2

B ρ zD dS ρ cos a + zsin a d d aπ

φρ φψ φ φ ρ ρ φ

= == ⋅ = ⋅ − =∫ ∫ ∫

( )ˆ , 0,B zFor z dS d d aψ ρ ρ φ= = −

ˆ , 4,SFor dS d dzaρψ ρ ρ φ= =

( )1 2

0 0ˆ ˆ ˆ2 2

S ρzD dS ρ cos a + zsin a d dza

π

φ ρφψ φ φ ρ φ

= == ⋅ = ⋅∫ ∫ ∫


Example 11 2

0 064 2

S zcos d dz

π

φψ φ φ

= == ∫ ∫

2

064 2cos d

π

φφ φ

== ∫

2

0

642

1+ cos2 dπ

φφ φ

== ∫

2

032 641d

π

φφ π

== =∫

Applying divergence theoremS V

D dS D dV⋅ = ∇ ⋅∫ ∫( )1 1

zD D D Dzρ φρ

ρ ρ ρ φ∂ ∂ ∂

∇ ⋅ = + +∂ ∂ ∂

( )3 21 1cos sinzρ φ φρ ρ ρ φ

∂ ∂= +

∂ ∂


Example 12 13 cos cosD zρ φ φ

ρ∇ ⋅ = +

4 2 1 2

0 0 0

1 3 cos cosV z

D dV z d d dzπ

ρ φρ φ φ ρ ρ φ

ρ= = =

⎛ ⎞∇ ⋅ = +⎜ ⎟

⎝ ⎠∫ ∫ ∫ ∫

432 1 2

0 00

3 cos cos3z

z d dzπ

φ

ρ φ ρ φ φ= =

⎛ ⎞= +⎜ ⎟

⎝ ⎠∫ ∫

( )2 1 2

0 064cos 4 cos

zz d dz

π

φφ φ φ

= == +∫ ∫

122 2

00

464 cos cos2zz d

π

φφ φ φ

=

⎛ ⎞= +⎜ ⎟

⎝ ⎠∫

( )2 2

064cos 2cos d

π

φφ φ φ

== +∫


Example 1

2 22

0 064cos 2cosd d

π π

φ φφ φ φ φ

= == +∫ ∫

( )2

0

64 1 cos22

dπ

φφ φ

== +∫

2 2

0 032 cos2d d

π π

φ φφ φ φ

= == +∫ ∫

2

032 64d

π

φφ π

== =∫


Example 2Verify divergence theorem for the flux of for the volume of cube with 1unit for each side. The cube is situated in the first octant of the coordinate system with one corner on the origin.

2 ˆ ˆ ˆx y zD x a xya yza= + +

X

Y

Z

A B

CD

E F

GH1 1

1


Example 22 ˆ ˆ ˆx y zD x a xya yza= + +

S VD dS D dV⋅ = ∇ ⋅∫ ∫

( )S FRONT BACK LEFT RIGHT TOP BOTTOMD dS⋅ = + + + + +∫ ∫ ∫ ∫ ∫ ∫ ∫

( )1 1 2

0 0ˆ ˆ ˆ ˆ 1x y z xFRONT y z

D dS x a xya yza dydza and x= =

⋅ = + + ⋅ =∫ ∫ ∫1 1

0 01 dydz= =∫ ∫

( ) ( )1 1 2

0 0ˆ ˆ ˆ ˆ 0x y z xBACK y z

D dS x a xya yza dydz a and x= =

⋅ = + + ⋅ − =∫ ∫ ∫0 =


Example 2

( ) ( )1 1 2

0 0ˆ ˆ ˆ ˆ 0x y z yLEFT z x

D dS x a xya yza dxdz a and y= =

⋅ = + + ⋅ − =∫ ∫ ∫0 =

( )1 1 2

0 0ˆ ˆ ˆ ˆ 1x y z yRIGHT z x

D dS x a xya yza dxdza and y= =

⋅ = + + ⋅ =∫ ∫ ∫1 1

0 0

1 2

xdxdz= =∫ ∫

( )1 1 2

0 0ˆ ˆ ˆ ˆ 1x y z zTOP y x

D dS x a xya yza dxdya and z= =

⋅ = + + ⋅ =∫ ∫ ∫1 1

0 0

1 2

ydxdy= =∫ ∫


Example 2( ) ( )

1 1 2

0 0ˆ ˆ ˆ ˆ 0x y z zBOTTOM y x

D dS x a xya yza dxdy a and z= =

⋅ = + + ⋅ − =∫ ∫ ∫0 =

1 112

2 2S

D dS⋅ = + + =∫yx zDD DD

x y z∂∂ ∂

∇ ⋅ = + +∂ ∂ ∂ ( ) ( ) ( )2 3x xy yz

x y zx y∂ ∂ ∂

= + + =∂ ∂ ∂

+

1 1 1

0 0 0(3 )

VDdV x y dxdydz∇⋅ = +∫ ∫ ∫ ∫ 2 =

S VD dS D dV⋅ = ∇ ⋅∫ ∫

.Thus divergence theorem is verified


Curl of a vectorThe curl of a vector is an axial or rotational vector whose magnitude is the maximum circulation (closed line integral) of per unit area as the area tends to zero and whose direction is the normal direction of the area when the area is oriented so as to make the circulation maximum.The circulation of a vector field around a closed path L is the integral

AA

AL

A dL⋅∫

0ˆ lim L

nS

MAX

A dlCurl A A a

SΔ →

⎛ ⎞⋅⎜ ⎟= ∇ × =

Δ⎜ ⎟⎝ ⎠

∫

Where the area S is bounded by the curve LΔˆ is the unit vector normal to the surface S and is determined by right na Δ

hand rule.


Curl of a vectorConsider the differential area in the yz plane.

a b

cd

X

Y

dz

dy


Curl of a vectorClosed line integral of vector around abcd is obtained as below: A

( ) -------(1)l ab bc cd daA dl A dl⋅ = + + + ⋅∫ ∫ ∫ ∫ ∫

0ˆ 2ydzAlong ab dl dya and z z= = −

0 0 0( , , ) -------(a)2

yyab

P

AdzA dl A x y z dyz

∂⎡ ⎤⋅ = −⎢ ⎥∂⎣ ⎦

∫

0 0 0 0 0 0( , , ) ( , , ) ( ) ( ) ( )y y yy y

P P P

A A AA x y z A x y z x x y y z z

x y z∂ ∂ ∂

= + − + − + −∂ ∂ ∂

0 0 0 0 0 0( , , ) ( , , ) ( ) ( ) ( )z z zz z

P PP

A A AA x y z A x y z x x y y z zx y z

∂ ∂ ∂= + − + − + −

∂ ∂ ∂


Curl of a vector

0 0 0( , , ) -------(b)2

zzbc

P

dy AA dl A x y z dzy

⎡ ⎤∂⋅ = +⎢ ⎥∂⎣ ⎦

∫

0ˆ 2ydzAlong cd dl dya and z z= = +

0 0 0( , , ) -------(c)2

yycd

P

AdzA dl A x y z dyz

∂⎡ ⎤⋅ = − +⎢ ⎥∂⎣ ⎦

∫

0ˆ da y2zdyAlong dl dza and y= = −

0 0 0( , , ) -----(d)2

zzda

P

dy AA dl A x y z dzy

⎡ ⎤∂⋅ = − −⎢ ⎥∂⎣ ⎦

∫

0ˆ y2zdyAlong bc dl dza and y= = +


Curl of a vector , , (1)Substituting equations a b c and d in

yzl

AAA dl dydz dydzy z

∂∂⋅ = −

∂ ∂∫

yzlA dl AAS y z

⋅ ∂⎛ ⎞∂= −⎜ ⎟Δ ∂ ∂⎝ ⎠

∫

yz AA Sy z

∂⎛ ⎞∂= − Δ⎜ ⎟∂ ∂⎝ ⎠

0lim yzlS

A dl AAS y zΔ →

⋅ ∂⎛ ⎞∂= −⎜ ⎟Δ ∂ ∂⎝ ⎠

∫

By definition the above equation represents curl of vector about x axis

( ) ( ) yzx x

AAcurl A Ay z

∂⎛ ⎞∂= ∇× = −⎜ ⎟∂ ∂⎝ ⎠


Curl of a vector .y and z components of the curl of A can similarly obtained

( ) ( ) x zy y

A Acurl A Az x

∂ ∂⎛ ⎞= ∇× = −⎜ ⎟∂ ∂⎝ ⎠

( ) ( ) y xz z

A Acurl A Ax y

∂⎛ ⎞∂= ∇× = −⎜ ⎟∂ ∂⎝ ⎠

The resultant curl will be the sum of component curls about x, y,z axes

ˆ ˆ ˆy yz x z xx y z

A AA A A AA a a ay z z x x y

∂ ∂⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂⎛ ⎞∇× = − + − + −⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠ˆ ˆ ˆx y z

x y z

a a a

yA A

Ax z

A

∂×

∂ ∂∂ ∂ ∂

∇ =


Curl of a vector in other coordinates The expression for curl in cylindrical and spherical coordinates are

ˆ ˆ ˆ z

z

a a a

zA A A

A

ρ φ

ρ φ

ρ

ρ φρ

∂×

∂ ∂∂ ∂ ∂

∇ =

ˆ ˆ ˆsin

sin

r

r

a ra r a

rA rA r A

A

θ φ

θ φ

θ

θ φθ

∂×

∂ ∂∂ ∂ ∂

∇ =


Curl of a vector- Some properties The curl of a vector field is another vector field♣

The curl of a scalar field does not exist♣

( ) A B A B♣ ∇× + = ∇× +∇×

( ) ( ) ( ) A B A B B A∇× × = ∇⋅♣ − ∇⋅

( ) VA V A V A∇× = ∇× +∇ ×♣

The divergence of the curl of a vector field vanishes♣

( ) 0A∇⋅ ∇× =

.The curl of the gradient of a scalar field vanishes♣0V∇×∇ =


Curl of a vector

Y

Z

a b

cd


Curl of a vector

Y

Z


Curl of a vector

Y

Z

a b

cd


Curl of a vector

Y

Z


Curl of a vector-Physical interpretation

Y

Z

X



Z

Y

XRotation of the

paddle wheel along x,y and z axes



Curl Direction


Example 1Determine the curl of the following vector fields

2 ˆ ˆ( ) x zi P x yza xza= +2ˆ ˆ ˆ( ) sin cos zii Q a za z aρ φρ φ ρ φ= + +


:Solution2 ˆ ˆ( ) x zi P x yza xza= +

ˆ ˆ ˆx y z

x y z

a a a

yP P

Px z

P

∂×

∂ ∂∂ ∂ ∂

∇ = ˆ ˆ ˆy yz x z xx y z

P PP P P Pa a ay z z x x y

∂ ∂⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂⎛ ⎞= − + − + −⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠


Example 1ˆ ˆ ˆy yz x z x

x y z

P PP P P PP a a ay z z x x y

∂ ∂⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂⎛ ⎞∇× = − + − + −⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠

( ) ( ) ( )2 2ˆ ˆ ˆ0 0 0x y za x y z a x z a= − + − + −

( ) ( )2 2ˆ ˆy zx y z a x z a= − −

2ˆ ˆ ˆ( ) sin cos zii Q a za z aρ φρ φ ρ φ= + +

ˆ ˆ ˆ z

z

a a a

zQ Q Q

Q

ρ φ

ρ φ

ρ

ρ φρ

∂×

∂ ∂∂ ∂ ∂

∇ =


Example 1

( )1 1ˆ ˆ ˆz zz

Q Q QQ QaQ a Q az zφ ρ ρ

ρ φ φρρ φ ρ ρ ρ φ

∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂− + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝

=⎠

∇ ×

( ) ( )2 21ˆ ˆ ˆsin 0 0 3 cos zz z a a z aρ φφ ρ ρ φρ ρ

⎛ ⎞−= − + − + −⎜ ⎟⎝ ⎠

( ) ( )31 ˆ ˆsin 3 cos zz a z aρφ ρ ρ φρ

= − + + −


ˆ ˆ ˆsin

sin

r

r

a ra r a

rT rT r T

T

θ φ

θ φ

θ

θ φθ

∂×

∂ ∂∂ ∂ ∂

∇ =


Example 1( )1 1 1 1ˆ ˆ ˆ( sin ) ( )

sin sinr r rT T T a T rT a rT T ar r r r rφ θ φ θ θ φθ

θ θ φ θ φ θ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞∇× = − + − + −⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠

1 ˆ(cos sin ) ( sin cos )sin rr a

rθ θ θ φ

θ θ φ⎛ ⎞∂ ∂

= −⎜ ⎟∂ ∂⎝ ⎠2

1 1 cos ˆ( cos )sin

r ar r r θ

θ θθ φ

⎛ ⎞∂ ∂⎛ ⎞+ −⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠

( )22

1 cos ˆsin cosr ar r r φ

θθ φθ

∂ ∂⎛ ⎞⎛ ⎞+ − ⎜ ⎟⎜ ⎟∂ ∂ ⎝ ⎠⎝ ⎠

( ) ( ) 21 1 1 sinˆ ˆ ˆcos2 sin sin 0 cos 2 sin cos

sin rr a a r ar r r rθ φ

θθ θ φ θ θ φθ

⎛ ⎞= + + − + +⎜ ⎟⎝ ⎠

3cos2 cos 1ˆ ˆ ˆsin 2cos

sin ra a ar r rθ φ

θ θφ φθ

⎛ ⎞ ⎛ ⎞= + − + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠


Stoke’s theoremStokes theorem states that the circulation of a vector field around a closed path L is equal to the surface integral of the curl of over the open surface S bounded by L provided that are continuous on S

AA

AA and ∇×

( )L S

A dl A dS⋅ = ∇× ⋅∫ ∫Surface SdS

dl

Closed path L


Stoke’s theorem

The surface S is subdivided in to a large number of cells.th

kIf the k cell has surface area ΔS and is bounded by the path L

k

k

LkL L

k k k

A dlA dl A dl S

S

⋅⋅ = ⋅ = Δ

Δ∫

∑ ∑∫ ∫


Stoke’s theoremThere is cancellation on every interior path. So the sum of line integrals around Lk’s is the same as the line integrals around the bounding curve L.Taking the limit of the above equation

By definition the quantity inside the brackets isAs

0lim k

kk

L

Sk k

A dlS

SΔ →

⋅

Δ

⎛ ⎞⎜ ⎟Δ⎜ ⎟⎝ ⎠

∑∫

A∇× thfor the k cell0 .kSΔ → the summation becomes integration over the whole surface

( ).L S

A dl A dS⋅ = ∇×∫ ∫ which is stoke's theorem


Example 1

0 Y

Y

2

2

5

5

6030

a

b

c

d

S

ˆ ˆcos sin , If a a evaluate A dlρ φρ φ φ+ ⋅∫ .around the path shown below' Verify Stoke s Theorem

S


Example 1b c d a

a b c dLA dl A dl⎡ ⎤⋅ = ⋅⎢ ⎥⎣

+ +⎦

+∫ ∫ ∫ ∫∫ˆ 2 Along ab dl d aφρ ρ φ= =

30

60sin

b

aA dl d

φρ φ φ

=⋅ =∫ ∫ [ ] ( )30

602 cos 3 1φ= − = − −

ˆ 30 Along bc dl d aρφ ρ= =

5

2cos

c

bA dl d

ρρ φ ρ

=⋅ =∫ ∫

52

2

21 3cos 302 4ρ⎡ ⎤

= =⎢ ⎥⎣ ⎦

ˆ 5 Along cd dl d aφρ ρ φ= =60

30sin

d

cA dl d

φρ φ φ

=⋅ =∫ ∫ [ ] ( )60

30

55 cos 3 12

φ= − = −


Example 1ˆ 60 Along da dl d aρφ ρ= =

2

5cos

a

dA dl d

ρρ φ ρ

=⋅ =∫ ∫

22

5

21cos 602 4ρ⎡ ⎤

= = −⎢ ⎥⎣ ⎦

( ) ( )21 3 5 213 1 3 14 2 4L

A dl − − + + − −⋅ =∫

4.941L

A dl⋅ =∫( ).

L SA dl A dS⋅ = ∇×∫ ∫Using stoke's theorem

( )1 1ˆ ˆ ˆz zz

A A AA AaA a A az zφ ρ ρ

ρ φ φρρ φ ρ ρ ρ φ

∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂− + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝

=⎠

∇ ×

ˆ zdS d d aρ φ ρ=


Example 1( ) ( ) ( )1ˆ ˆ ˆ0 0 0 0 1 sin zA a a aρ φ ρ φ

ρ∇ × = − + − + +

( )60 5

30 2

1 ˆ( ). 1 sin zSA dS d d a

φ ρρ φρ φ ρ

ρ= =∇× = +∫ ∫ ∫

( )1 ˆ1 sin zaρ φρ

= +

( )60 5

30 2sin 1d d

φ ρφ φ ρ ρ

= == +∫ ∫

[ ]52

60

302

cos2ρφ ρ

⎡ ⎤= − +⎢ ⎥

⎣ ⎦4.941=

( ). 4.941S L

A dS A dl∇× = = ⋅∫ ∫' Stoke s Theorem is thus verified


Laplacian of a scalar2 , The Laplacian of a scalar field V written as V is the divergence∇♣

. of the gradient of V It is another scalar field2In Cartesian coordinates,Laplacian V= V V∇⋅∇ = ∇

ˆ ˆ ˆ ˆ ˆ ˆ= x y z x y zV V Va a a a a a

x y z x y z⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂

+ + ⋅ + +⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

2 2 22

2 2 2V= V V Vx y z

∂ ∂ ∂∇ + +

∂ ∂ ∂

A scalar field V is said to be harmonic in a given region if its ♣ Laplacian vanishes in that region.

2V=0∇ Laplace's Equation⇒


Laplacian of a scalar in other coordinates

2 22

2 2 21 1V= V V V

zρ

ρ ρ ρ ρ φ⎛ ⎞∂ ∂ ∂ ∂

∇ + +⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠

22 2

2 2 2 2 21 1 1V= sin

sin sinV V Vr

r r r r rθ

θ θ θ θ φ∂ ∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞∇ + +⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

,In cylindrical coordinates

,In spherical coordinates


Laplacian of a vector2Laplacian of a vector A denoted as is defined asA∇

( ) ( )2 A A A∇ = ∇ ∇⋅ − ∇×∇×


Example 1Find the Laplacian of the scalar field given below. sin 2 coshzV e x y−=

2 2 22

2 2 2V= V V Vx y z

∂ ∂ ∂∇ + +

∂ ∂ ∂2 2 2

2 2 2= ( sin 2 cosh ) ( sin 2 cosh ) ( sin 2 cosh )z z ze x y e x y e x yx y z

− − −∂ ∂ ∂+ +

∂ ∂ ∂

= (2 cos2 cosh ) ( sin 2 sinh ) ( sin 2 cosh )z z ze x y e x y e x yx y z

− − −∂ ∂ ∂+ + −

∂ ∂ ∂

= 4 sin 2 cosh sin 2 cosh sin 2 coshz z ze x y e x y e x y− − −− + +

= 2 sin 2 coshze x y−−


Example 2Find the Laplacian of the scalar field given below.

2 cos2V zρ φ=2 2

22 2 2

1 1V= V V Vz

ρρ ρ ρ ρ φ

⎛ ⎞∂ ∂ ∂ ∂∇ + +⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠

2 22 2 2

2 2 21 1= ( cos2 ) ( cos2 ) ( cos2 )z z z

zρ ρ φ ρ φ ρ φ

ρ ρ ρ ρ φ⎛ ⎞∂ ∂ ∂ ∂

+ +⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠

( ) ( )2 22

1 1= 2 cos2 2 sin 2z zρ φ ρ φρ ρ ρ φ

∂ ∂+ −

∂ ∂

=4 cos2 4 cos2 0 z zφ φ− =


Classification of vector fields

A vector field is said to be solenoidal or divergenceless if A=0∇⋅♠

If A 0=0 V S

A dV A dS∇⋅ = ⋅ =∇⋅ ∫ ∫ implies that the flux lines of A entering a closed surfa e0 c

SA dS⋅ =∫

must also leave it.

:E x a m p le s1. Incompressible fluids 2. Magnetic fields

The field of curl F for any F is purely solenoidal be cause♠

( ) 0F∇⋅ ∇× =


Classification of vector fields A vector field is said to be irrotational or potential field if A=0∇×♠

( ) If A=0 0 S L

A dS A dl∇× ⋅ = ⋅ =∇× ∫ ∫ implies that the line integral of A is independent of th0 e

LA dl⋅ =∫

chosen path.

For this reason an irrotational field is also known as conservative field ♠:E x a m p le s

1. Electrostatic field 2. Gravitational field

The field of gradient V for any V is purely irrotational bec e aus♠

( ) 0V∇× ∇ =



0 0A A∇⋅ = ∇× = 0 0A A∇⋅ ≠ ∇× =

Solenoidal, Irrotational Non-solenoidal, Irrotational



0 0A A∇⋅ = ∇× ≠ 0 0A A∇⋅ ≠ ∇× ≠

Non-solenoidal, Rotational

Solenoidal, Rotational

EMT Electromagnetic Theory MODULE I

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