VECTOR CALCULUS

1. GRADIENT OF A SCALAR

2. DIVERGENCE OF A VECTOR

3. DIVERGENCE THEOREM

4. CURL OF A VECTOR

Dr. ANANT KUMAR SINHAASSOCIATE PROFESSORDEPTT. OF PHYSICSA.M. COLLEGE, GAYA

1. GRADIENT OF A

SCALARis the temperature atSuppose T1x, y, z P1x, y, z ,

and T2x dx, y dy, z dz is the temperature at P2

as shown.

The differential distances dx, dy,dz are the

components of the differential distance

vector dL :

dL dxax dya y dzaz

However, from differential calculus, the

differential temperature:

x y zdT T T

T dx

T dy

Tdz2 1

GRADIENT OF A SCALAR (Cont’d)

But, dx dL ax

dy dLay

dz dL az

So, previous equation can be rewritten as:

zyx

zyx

a dLz

T a

T

ya

x

T

z dL

Ta dL

x ydT

Ta dL

Ta

GRADIENT OF A SCALAR (Cont’d)

GRADIENT OF A SCALAR (Cont’d)

The vector inside square brackets defines the

change of temperature dT corresponding to a

vector change in position dL .

This vector is called Gradient of Scalar T.

For Cartesian coordinate:

zyxy z

Va

Va

xV

Va

zz

GRADIENT OF A SCALAR (Cont’d)

For Circular cylindrical coordinate:

V V

a 1 V

a V

a

For Spherical coordinate:

r r

V V

a rr sin

1 Va

1 Va

EXAMPLE

1

Find gradient of these scalars:

V e z sin 2x cosh y(a)

(a) U 2z cos2

(b) W 10r sin2 cos

SOLUTION TO EXAMPLE 1

(a) Use gradient for Cartesian coordinate:

x y z

2e z cos2x cosh yax e z sin 2x sinh ya y

e z sin 2x cosh yaz

Va

Va

x y zV

Va

SOLUTION TO EXAMPLE 1

(Cont’d)

z

2zcos2a 2zsin2a

2 cos2az

z

(b) Use gradient for Circular cylindrical

coordinate:

U U

a 1 U

a U

a

SOLUTION TO EXAMPLE 1

(Cont’d)

(c) Use gradient for Spherical coordinate:

10sin2 cosar 10sin 2 cosa

10sin sina

r

1 Wa

1 Wa

r r sin rW

Wa

1.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

A dS

div A A lim s

v0

DIVERGENCE OF A VECTOR

(Cont’d)

What is A dS ??s

Vector field A at

closed surface S

s

front back left right top bottom

A dSA dS

And, v is volume enclosed by surface S

DIVERGENCE OF A VECTOR

(Cont’d)

Where,

Az Ay

x y z A

Ax

For Circular cylindrical coordinate:

AzA

zA

1 1

A

DIVERGENCE OF A VECTOR

(Cont’d)

For Cartesian coordinate:

Ar A r

r sin r sin

1 A sin 121

r 2r A

DIVERGENCE OF A VECTOR

(Cont’d)

For Spherical coordinate:

EXAMPLE

1

Find divergence of these vectors:

P x2 yzax xzaz

r s in cosa cosar 2 r

(a)

(b) Q sina 2za zcosaz

(c) W 1

cosa

18

(a) Use divergence for Cartesian

coordinate:

SOLUTION TO EXAMPLE 1

Pz Py

x y z

x2 yz 0 xzx y z

2xyz x

PPx

coordinate:

1

2sin cos

1

Q

z

1 2 sin 1 2 z z cos

z

QQQ z

SOLUTION TO EXAMPLE 1

(Cont’d)

(b) Use divergence for Circular cylindrical

SOLUTION TO EXAMPLE 1

(Cont’d)

(c) Use divergence for Spherical coordinate:

1

2

2cos cos

1

W

r sin

1 cos 1 rsin2 cos

r sin

cos

r 2r

W1 W sin 1

r sin r sin r W

r 2rr

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.

A dS AdVV V

1.DIVERGENCE THEOREM

EXAMPLE

1

3A vector field D a 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:

D dsS

DdVV

(a)

(b)

SOLUTION TO EXAMPLE 1

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

D dS Dinner Douter Dbottom Dtop

S

Where,2 5

101

10 z0

2 5

4a ddz(a )

0 z0

Dinner 3a ddz(a )

1602

20 z0

2 5

4a ddz(a )0 z0

10

10

2 2

z5

z0

Dtop 3a dd(az ) 0

2 2

Dbottom 3a dd(az ) 0

SOLUTION TO EXAMPLE 1

Cont’d)

2 5

Douter 3a ddz(a)

SOLUTION TO EXAMPLE 1

Cont’d)

Therefore

D dS 10 160 0 0S

150

SOLUTION TO EXAMPLE 1

Cont’d)

(b) For the right side of Divergence Theorem,

evaluate divergence of D

D 1 3 42

2

5 2 2

150

5

4 2

0z0

r1

So, DdV 42dddz

z0 0 1

1. 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

when the area is oriented

the area

so as to

make the circulation maximum.

an max

sCurl AA lim s

A dl

s0

Where,

CURL OF A VECTOR (Cont’d)

s ab bc cd da

A dl A dl

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)

x y z

Ax Ay Az

ax a y az

A

zyx aa

Ax x y

Ay

z

Ax ax

Az z

Ay y

A Az

a a az

z A A Az

A 1

za

aa

Az

z z

A

A

A

1 A

A 1 Az

For Circular cylindrical coordinate:

CURL OF A VECTOR (Cont’d)

CURL OF A VECTOR (Cont’d)

For Spherical coordinate:

Ar rA r sin Ar2 sin r

a

ar a

1 A

a

a

a

r

rr

rA A

) Ar

1 (rA

r sin r

1 1 Ar A r sin

1 sinA

P x2 yzax xzaz

r s in cosa cosar 2 r

(a)

(b) Q sina 2za zcosaz

(c) W 1

cosa

EXAMPLE

1

Find curl of these vectors:

SOLUTION TO EXAMPLE 1

(a) Use curl for Cartesian coordinate:

zx y aa

0 0a x x2 y za y 0 x2 za z

x2 y za y x2za z

z x z

x y

Pz Px a

Px

Py

Py

y P

Pz

z

z

zaaa

2

1 z sin 3a 3z cosa

1 3 2z cosa

z

sin a 0 0a

x y

1 Q Q

Qz z

Q

Q

z Q

1 Qz

SOLUTION TO EXAMPLE 1

(Cont’d)

(b) Use curl for Circular cylindrical coordinate

(c) Use curl for Spherical coordinate:

a

a

a

aa

2 cos

cos11

r2 1 (r sin cos)

r r

r

rcos r2

r sin

1 cossin rsin cos rsin

) Wr

1 (rW

r

rr sin

W 11 sinW rW W rsin

ar

r

1 Wrr

SOLUTION TO EXAMPLE 1

(Cont’d)

SOLUTION TO EXAMPLE 1

(Cont’d)

3 2cos 1 sina

cosa

cos2 sin a

r

r

rsin

r212r sin cos

sin a

r

rcos 2 r sin sin a

1 0 cosa1

rsin

r

r