Vector Calculus

2

Line Integral of Vector Functions B

APath L

dl

P

C

B

APath L

dl

P

C

A

Path L

dl

P

C

cos ,B

L A

d dl P l P dl = dxi +dyj +dzk

For a closed loop, i.e. ABCA,

L

dP l = circulation of P around L

Line given by L(x(s), y(s), z(s)), s = parametric variable

,)(ds

ds

sdxdx ,

)(ds

ds

sdydy .

)(ds

ds

sdzdz

Always take the differential element dl as positive and insert the integral limits according to the paths!!!

3

Example For F = yi – xj, calculate the circulation of F along the two paths as shown below.

y

(1,2,0)

Cx

(1,0,0)

x Cy

Cz

(1,2,4)

C1 C2=Cx+Cy+Cz

dl = dxi +dyj +dzk

Solution:

d ydx xdy F l

Along path C2

2 x y zC C C C

d d d d F l F l F l F l

2

1 2 4

0 10 0 0

0 2y x

C

d ydx xdy dz

F l

4

Example - Continue

y

(1,2,0)

Cx

(1,0,0)

x Cy

Cz

(1,2,4)

C1 C2=Cx+Cy+Cz

Along path C1

Using x as the parametric variable, the path equations are given as

( ) , ( ) 2 , ( ) 4x x x y x x z x x

Therefore,

( ), 2 , 4

dx x dy x dz xdx dx dx dy dx dx dz dx dx

dx dx dx

and ( ) ( 2 4 )

2 2 2 0

d y x dx dx dx

ydx xdx xdx xdx

F l i j i j k

1

0C

d F l

5

Example - Continue

1 2

2 0C C

d d d F l F l F l

The vector field defined by F in a given domain is non-conservative. The line integral is dependent on the integration path!

Yes, because the work done when we move a charge from one point to another is independent of the path but determined by the potential difference between these two points.

C

dF l is work done on an object along path C if F = force !!

6

Surface Integral

Smooth

Surface S

dS

n̂

P

Surface integral or the flux of P across the surface S is

ˆ cosS S

dS dS P n P

n̂ is the outward unit vector normal to the surface.

For closed surface,

ˆS

dS P n = net outward flux of P.

7

ExampleIf F = xi + yj + (z2 – 1)k, calculate the flux of F across the surface shown in the figure.

z

x

y

(1, -1, 2) (1, 1, 2)

(-1, 1, 2)(-1, -1, 2)F

n̂

Solution:

1 1 2

21 1

1 1 2

21 1

1 1

1 1

ˆ

( ( 1) )

( 1)

3

12

S

z

z

dS

x y z dxdy

z dxdy

dxdy

F n

i j k k

8

Volume Integral x y zV V V V

dV F dV F dV F dV F i j k

Evaluation : choose a suitable integration order and then find out the suitable lower and upper limits for x, y and z respectively.

Example: Let F = 2xzi – xj + y2k. Evaluate VdVF

where V is the region bounded by the surface x = 0, x = 2, y = 0, y = 6, z = 0, z = 4.

9

Volume Integral Solution:

2 6 4 2

0 0 0

2 6 4 2 6 4

0 0 0 0 0 0

2 6 4 2

0 0 0

(2 )

2 ( )

192 48 576

x y z

V x y z

x y z x y z

x y z x y z

x y z

x y z

dV xz x y dzdydx

xzdzdydx x dzdydx

y dzdydx

F i j k

i j

k

i j k

In electromagnetic,

v

V

dV Q = Total charge within the volume

where v = volume charge density (C/m3)

10

Scalar Field

Every point in a region of space is assigned a scalar value obtained from a scalar function f(x, y, z), then a scalar field f(x, y, z) is defined in the region, such as the pressure in atmosphere and mass density within the earth, etc.

Partial Derivatives

0

( , ,...) ( , ,...)limx

f f x x y f x y

x x

Mixed second partials 2 f f

y x y x

11

Gradient

Del operator

x y z

i j k

Gradient

grad f f f

f fx y z

i j k

Gradient characterizes maximum increase. If at a point P the gradient of f is not the zero vector, it represents the direction of maximum space rate of increase in f at P.

12

Example Given potential function V = x2y + xy2 + xz2, (a) find the gradient of V, and (b) evaluate it at (1, -1, 3).

Solution:(a)

2 2 2(2 ) ( 2 ) 2

V V VV

x y z

xy y z x xy xz

i j k

i j k

(b) (1, 1,3)

( 2 1 9) (1 2) 6

8 6

V

i j k

= i j k

2 2 2

8 6 1ˆ (8 6 )

1018 ( 1) 6

i j ka i j k Direction of

maximum increase

13

Vector Field

Electric field: E = E(x, y, z), Magnetic field : H = H(x, y, z)

Every point in a region of space is assigned a vector value obtained from a vector function A(x, y, z), then a vector field A(x, y, z) is defined in the region.

31 2

l l l l

AA A

t t t t

A

i j k

R(t1, t2) = acos t1i + asin t1j + t2k 1 1

1

sin cosa t a tt

R

i j

2t

R

k

14

Divergence of a Vector FieldRepresenting field variations graphically by directed field lines - flux lines

A BP

P

P

15

Divergence of a Vector FieldThe divergence of a vector field A at a point is defined as the net outward flux of A per unit volume as the volume about the point tends to zero:

0

div lim S

v

d

v

A s

A

It indicates the presence of a source (or sink)! term the source as flow source. And div A is a measure of the strength of the flow source.

16

Divergence of a Vector FieldIn rectangular coordinate, the divergence of A can be calculated as

1 2 3

31 2

div ( ) ( )A A Ax y z

AA A

x y z

A A i j k i j k

For instance, if A = 3xzi + 2xyj – yz2k, then

div A = 3z + 2x – 2yz

At (1, 2, 2), div A = 0; at (1, 1, 2), div A = 4, there is a source; at (1, 3, 1), div A = -1, there is a sink.

17

Curl of a Vector FieldThe curl of a vector field A is a vector whose magnitude is the maximum net circulation of A per unit area as the area tends to zero and whose direction is the normal direction of the area.

0curl lim L

s

d

s

A l

A

It is an indication of a vortex source, which causes a circulation of a vector field around it.

Water whirling down a sink drain is an example of a vortex sink causing a circulation of fluid velocity.

If A is electric field intensity, then the circulation will be an electromotive force around the closed path.

18

Curl of a Vector FieldIn rectangular coordinate, curl A can be calculated as

1 2 3

3 32 1 2 1

curl x y z

a a a

A AA A A A

y z z x x y

i j k

A A

i j k

19

Curl of a Vector FieldExample:

If A = yzi + 3zxj + zk, then

curl

3

(3 ) ( ) (3 ) ( )

3 2

x y z

yz zx z

z zx yz z zx yz

y z z x x y

x y z

i j k

A A

i j k

i j k