vector calculus

VECTOR CALCULUS

1.10 GRADIENT OF A SCALAR

1.11 DIVERGENCE OF A VECTOR

1.12 DIVERGENCE THEOREM

1.13 CURL OF A VECTOR

1.14 STOKES’S THEOREM

1.15 LAPLACIAN OF A SCALAR

1.10 GRADIENT OF A SCALAR

Suppose is the temperature at ,

and is the temperature at

as shown.

zyxT ,,1 zyxP ,,1

2P dzzdyydxxT ,,2

The differential distances are the components of the differential distance vector :

dzdydx ,,

zyx dzdydxd aaaL Ld

However, from differential calculus, the differential temperature:

dzz

Tdy

y

Tdx

x

TTTdT

12

GRADIENT OF A SCALAR (Cont’d)

But,

z

y

x

ddz

ddy

ddx

aL

aL

aL

So, previous equation can be rewritten as:

Laaa

LaLaLa

dz

T

y

T

x

T

dz

Td

y

Td

x

TdT

zyx

zyx

GRADIENT OF A SCALAR (Cont’d)

The vector inside square brackets defines the change of temperature corresponding to a vector change in position .

This vector is called Gradient of Scalar T.

LddT

GRADIENT OF A SCALAR (Cont’d)

For Cartesian coordinate:

zyx z

V

y

V

x

VV aaa

GRADIENT OF A SCALAR (Cont’d)

For Circular cylindrical coordinate:

zz

VVVV aaa

1

For Spherical coordinate:

aaa

V

r

V

rr

VV r sin

11

EXAMPLE 10

Find gradient of these scalars:

yxeV z cosh2sin

2cos2zU

cossin10 2rW

(a)

(b)

(c)

SOLUTION TO EXAMPLE 10

(a) Use gradient for Cartesian coordinate:

zz

yz

xz

zyx

yxe

yxeyxe

z

V

y

V

x

VV

a

aa

aaa

cosh2sin

sinh2sincosh2cos2

SOLUTION TO EXAMPLE 10 (Cont’d)

(b) Use gradient for Circular cylindrical

coordinate:

z

z

zz

z

UUUU

a

aa

aaa

2cos

2sin22cos2

1

2

SOLUTION TO EXAMPLE 10 (Cont’d)

(c) Use gradient for Spherical coordinate:

a

aa

aaa

sinsin10

cos2sin10cossin10

sin

11

2

r

rW

r

W

rr

WW

1.11 DIVERGENCE OF A VECTOR

Illustration of the divergence of a vector field at point P:

Positive Divergence

Negative Divergence

Zero Divergence

DIVERGENCE OF A VECTOR (Cont’d)

The divergence of A at a given point P is the outward flux per unit volume:

v

dS

div s

v

A

A A lim0

DIVERGENCE OF A VECTOR (Cont’d)

What is ?? s

dSA Vector field A at closed surface S

Where,

dSdSbottomtoprightleftbackfronts

AA

And, v is volume enclosed by surface S

DIVERGENCE OF A VECTOR (Cont’d)

For Cartesian coordinate:

z

A

y

A

x

A zyx

A

For Circular cylindrical coordinate:

z

AAA z

11A

DIVERGENCE OF A VECTOR (Cont’d)

For Spherical coordinate:

A

r

A

rAr

rrr sin

1sin

sin

11 22

A

DIVERGENCE OF A VECTOR (Cont’d)

EXAMPLE 11

Find divergence of these vectors:

zx xzyzxP aa 2

zzzQ aaa cossin 2

aaa coscossincos12

rr

W r

(a)

(b)

(c)

18

(a) Use divergence for Cartesian coordinate:

SOLUTION TO EXAMPLE 11

xxyz

xzzy

yzxx

z

P

y

P

x

P zyx

2

02

P

(b) Use divergence for Circular cylindrical

coordinate:

cossin2

cos1

sin1

11

22

Q

zz

z

z

QQQ z

SOLUTION TO EXAMPLE 11 (Cont’d)

SOLUTION TO EXAMPLE 11 (Cont’d)

(c) Use divergence for Spherical coordinate:

coscos2

cossin

1

cossinsin

1cos

1

sin

1sin

sin

11

22

22

W

r

rrrr

W

r

W

rWr

rrr

It states that the total outward flux of a vector field A at the closed surface S is the same as volume integral of divergence of A.

VV

dVdS AA

1.12 DIVERGENCE THEOREM

EXAMPLE 12

A vector field exists in the region

between two concentric cylindrical surfaces

defined by ρ = 1 and ρ = 2, with both

cylinders extending between z = 0 and z = 5.

Verify the divergence theorem by evaluating:

aD 3

S

dsD

V

DdV

(a)

(b)

SOLUTION TO EXAMPLE 12

(a) For two concentric cylinder, the left side:

topbottomouterinnerS

d DDDDSD

Where,

10)(

)(

2

0

5

01

4

2

0

5

01

3

z

zinner

dzd

dzdD

aa

aa

160)(

)(

2

0

5

02

4

2

0

5

02

3

z

zouter

dzd

dzdD

aa

aa

2

1

2

05

3

2

1

2

00

3

0)(

0)(

zztop

zzbottom

ddD

ddD

aa

aa

SOLUTION TO EXAMPLE 12 Cont’d)

Therefore

150

0016010

SDS

d

SOLUTION TO EXAMPLE 12 Cont’d)

SOLUTION TO EXAMPLE 12 Cont’d)

(b) For the right side of Divergence Theorem,

evaluate divergence of D

23 41

D

So,

150

4

5

0

2

0

2

1

4

5

0

2

0

2

1

2

zr

z

dzdddVD

1.13 CURL OF A VECTOR

The curl of vector A is an axial (rotational) vector whose magnitude is the maximum circulation of A per unit 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.

maxlim0

a

A

A A ns

s s

dl

Curl

Where,

CURL OF A VECTOR (Cont’d)

dldldacdbcabs

AA

CURL OF A VECTOR (Cont’d)

The curl of the vector field is concerned with rotation of the vector field. Rotation can be used to measure the uniformity of the field, the more non uniform the field, the larger value of curl.

For Cartesian coordinate:

CURL OF A VECTOR (Cont’d)

zyx

zyx

AAAzyx

aaa

A

zxy

yxz

xyz

y

A

x

A

z

A

x

A

z

A

y

AaaaA

z

z

AAAz

aaa

A1

z

zz

AA

z

AA

z

AA

a

aaA

1

1

For Circular cylindrical coordinate:

CURL OF A VECTOR (Cont’d)

CURL OF A VECTOR (Cont’d)

For Spherical coordinate:

ArrAArrr

r

sinsin

12

aaa

A

a

aaA

r

rr

A

r

rA

r

r

rAA

r

AA

r

)(1

sin

11sin

sin

1

EXAMPLE 13

zx xzyzxP aa 2

zzzQ aaa cossin 2

aaa coscossincos12

rr

W r

(a)

(b)

(c)

Find curl of these vectors:

SOLUTION TO EXAMPLE 13

(a) Use curl for Cartesian coordinate:

zy

zyx

zxy

yxz

xyz

zxzyx

zxzyx

y

P

x

P

z

P

x

P

z

P

y

P

aa

aaa

aaaP

22

22 000

(b) Use curl for Circular cylindrical coordinate

z

z

zzz

zz

z

z

y

Q

x

QQ

z

Q

z

QQ

aa

a

aa

aaaQ

cos3sin1

cos31

00sin

11

3

2

2

SOLUTION TO EXAMPLE 13 (Cont’d)

(c) Use curl for Spherical coordinate:

a

aa

a

aaW

22

2

cos)cossin(1

coscos

sin

11cossinsincos

sin

1

)(1

sin

11sin

sin

1

rr

r

r

r

rrr

r

r

W

r

rW

r

r

rWW

r

WW

r

r

r

rr

SOLUTION TO EXAMPLE 13 (Cont’d)

SOLUTION TO EXAMPLE 13 (Cont’d)

a

aa

a

aa

sin1

cos2

cossin

sin

2cos

sincossin2

1

cos01

sinsin2cossin

1

3

2

r

rr

rr

r

rr

r

r

r

1.14 STOKE’S THEOREM

The circulation of a vector field A

around a closed path L is equal to the

surface integral of the curl of A over

the open surface S bounded by L that A

and curl of A are continuous on S.

SL

dSdl AA

STOKE’S THEOREM (Cont’d)

EXAMPLE 14

By using Stoke’s Theorem, evaluate

for dlA

aaA sincos

EXAMPLE 14 (Cont’d)

SOLUTION TO EXAMPLE 14

Stoke’s Theorem,

SL

dSdl AA

where, and

zddd aS Evaluate right side to get left side,

zaA

sin11

SOLUTION TO EXAMPLE 14 (Cont’d)

941.4

sin11

0

0

60

30

5

2

aA zS

dddS

EXAMPLE 15

Verify Stoke’s theorem for the vector field

for given figure by evaluating: aaB sincos

(a) over the semicircular contour.

LB d

(b) over the surface of semicircular contour.

SB d

SOLUTION TO EXAMPLE 15

(a) To find LB d

321 LLL

dddd LBLBLBLB

Where,

dd

dzddd z

sincos

sincos

aaaaaLB

So

202

1

sincos

2

0

2

0

0

00,0

2

01

LB

zzL

ddd

4cos20

sincos

0

0,200

2

22

LB

zzL

ddd

SOLUTION TO EXAMPLE 15 (Cont’d)

202

1

sincos

0

2

2

00,0

0

23

r

zzL

ddd

LB

SOLUTION TO EXAMPLE 15 (Cont’d)

Therefore the closed integral,

8242 LB d

SOLUTION TO EXAMPLE 15 (Cont’d)

(b) To find SB d

z

z

z

zz

a

aaa

a

aa

aaB

11sin

sinsin1

00

cossin1

0cossin01

sincos

SOLUTION TO EXAMPLE 15 (Cont’d)

Therefore

82

1cos

1sin

11sin

0

2

0

2

0

2

0

0

2

0

aaSB

dd

ddd zz

1.15 LAPLACIAN OF A SCALAR

The Laplacian of a scalar field, V written as: V2

Where, Laplacian V is:

zyxzyx z

V

y

V

x

V

zyx

VV

aaaaaa

2

For Cartesian coordinate:

2

2

2

2

2

22

z

V

y

V

x

VV

For Circular cylindrical coordinate:

2

22

22 11

z

VVVV

LAPLACIAN OF A SCALAR (Cont’d)

LAPLACIAN OF A SCALAR (Cont’d)

For Spherical coordinate:

2

2

22

22

22

sin

1

sinsin

11

V

r

V

rr

Vr

rrV

EXAMPLE 16

Find Laplacian of these scalars:

yxeV z cosh2sin 2cos2zU

cossin10 2rW

(a)

(b)

(c)

You should try this!!

SOLUTION TO EXAMPLE 16

yxeV z cosh2sin22

02 U

2cos21

cos102 r

W

(a)

(b)

(c)

Yea, unto God belong all things in the heavens and on earth, and enough is

God to carry through all affairs

Quran:4:132

END