Introduction to Vector Calculus (29)

SOLVED EXAMPLESQ.1 If vector ˆ ˆ ˆ ˆ ˆ ˆˆ ˆA 2i j 2k, B 2i j, C 2i 3j k

then find

(a) A B

(b) A B

(c) A . B C

(d) B. C A

(e) A B C

(f) a unit vector perpendicular to both B and C

(g) Component of A along B

.

Solution:

(a) A B

= ˆ ˆ ˆ ˆ ˆ2i j 2k 2i j

= ˆ ˆ ˆ4i 2 j 2k

(b) A B

= ˆ ˆ ˆ ˆ ˆ2i j 2b 2i j

= ˆ2k

(c) A. B C

=

2 1 22 1 02 3 1

= –2 + 2 – 8 = –8

(d) B. C A

=

2 1 02 3 12 1 2

= –10 + 2= –8

(e) A B C

= B A.C C A.B

= ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ2i j 2i j 2k . 2i 3j k

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ2i 3j k 2i j 2k . 2i j

= ˆ ˆ12i 3k

(30) Introduction to Vector Calculus

(f) n = 2 2 2

ˆ ˆ ˆ ˆ ˆ2i j 2i 3j kB CB C 1 2 4

=ˆ ˆ ˆi 2 j 4b

21

(g) AB = ˆA cos B

= 2

A.B Bˆ ˆA.B BB

as B = BB

=

22

ˆ ˆ ˆ ˆ ˆ ˆ2i j 2k . 2i j 2i j

2 1

=ˆ ˆ10i 5j5

= ˆ ˆ2i j

Q.2: Find the angle between ˆ ˆ ˆA 2i j k

and ˆ ˆ ˆB i j 3k

.

Solution :

as A.B

= A B cos

cos = 22 2 2 2 2

ˆ ˆ ˆ ˆ ˆ ˆ2i j k . i j 3kA.BA B 2 1 1 1 1 3

=1 4cos

6 11

= 60.5° approx.

Q.3: If ˆ ˆ ˆ ˆˆ ˆA 2i 3j 7k and B 2i j k

, then show that A and B

areperpendicular to each other.

Solution :

A.B

= ˆ ˆ ˆ ˆ ˆ ˆ2i 3j 7k . 2i j k

= – 4 – 3 + 7 = 0

so A.B

= A B cos 0

cos = 0

Introduction to Vector Calculus (31)

= 90°

so A is perpendicular to B

.

Q.4: Find the unit vector perpendicular to bothˆ ˆ ˆA 2i 3j 5k

and ˆ ˆ ˆB 2i 3j k

. Also find the angle between them.

Solution:

As A B

=

ˆ ˆ ˆi j k2 3 52 3 1

= ˆ ˆ12i 8j

unit vector perpendicular to both A and B

n =A BA B

= 22

ˆ ˆ ˆ ˆ12i 8j 12i 8j20812 8

Again sin =A B

A B

=208 208

4 9 25 4 9 1 38 14

=1 208sin

38 14

= 38.7° Approx.

Q.5: A particle is acted upon by two constant forces 1ˆ ˆ ˆF i 4j 3k

and

2ˆ ˆ ˆF 3i j k

due to which particle is displaced from ˆ ˆ ˆ ˆˆ ˆi 2j 3k to 4i 5j k . Calculate

the total work done.Solution:Displacement of the particle

r = ˆ ˆ ˆ ˆ ˆ ˆ4i 5j k i 2j 3k

(32) Introduction to Vector Calculus

= ˆ ˆ ˆ3i 3j 2k Hence total work done

= Total force . displacement

= 1 2F F .r

= ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆi 4 j 3k 3i j k . 3i 3j 2k

= ˆ ˆ ˆ ˆ ˆ ˆ4i 5j 4k . 3i 3j 2k 12 15 8

= 35 units.Q.6: A rigid body is rotating with angular velocity of 5 rad/s about an axis parallel

to ˆ ˆ3j k and passing through the point ˆ ˆ ˆi j 3k . Find the velocity vector of the

particle, when it is at the point ˆ ˆ ˆ2i 4 j k .

Solution : Suppose r is the position vector

then r = ˆ ˆ ˆ ˆ ˆ ˆ2i 4j k i j 3k

= ˆ ˆ ˆi 5j 4k angular velocity

=

ˆ ˆ3j k 5 ˆ ˆ5 3j kˆ ˆ 103j k

linear velocity

v = 5 ˆ ˆ ˆ ˆ ˆr 3j k i 5 j 4k10

=

ˆ ˆ ˆi j k5 0 3 110 1 5 4

= 5 ˆ ˆ ˆ7i j 3k10

units.

Q.7 : Calculate the torque of a force ˆ ˆ ˆ2i 2j 5k about the point ˆ8j acting

through the point ˆ ˆ ˆ6i 4j 2k .

Solution : Here

Introduction to Vector Calculus (33)

r = ˆ ˆ ˆ ˆ8j 6i 4j 2k

= ˆ ˆ ˆ6i 4 j 2k

torque = ˆ ˆ ˆ ˆ ˆ ˆr F 6i 4j 2k 2i 2j 5k

=

ˆ ˆ ˆi j k6 4 22 2 5

= ˆ ˆ ˆ24 i 34 j 4k

Q.8: A force vector ˆ ˆ ˆ10 i 25 j 35k passes through a point (2, 5, 7). Prove thatforce is also passing through the origin.

Solution: The position vector

r = ˆ ˆ ˆ2i 5 j 7k

and moment of the form about this point i.e. torque

= r F

= ˆ ˆ ˆ ˆ ˆ ˆ2i 5j 7k 10i 25j 35k

=

ˆ ˆ ˆi j k2 5 7

10 25 35

= 0As the moment is zero, which shows that forces is passing through the origin.

Q.9: A force ˆ ˆ ˆ4i 3j 2k passes through the point (–9, 2, 1). Find the componentof moment of the force about the axis of reference.

Sol.: Here

r = ˆ ˆ ˆ9i 2 j k

so moment of force i.e. torque

r F = ˆ ˆ ˆ ˆ ˆ ˆ9i 2j k 4i 3j 2k

(34) Introduction to Vector Calculus

=

ˆ ˆ ˆi j k9 2 1

4 3 2

= ˆ ˆ ˆ7i 22 j 19k

Hence components of moment of force are 7 unit, 22 units and 19 units in x, y and zdirection respectively.

Q.10: A proton is moving with velocity 108 cm/s along z-axis through an electricfield of intensity 3 × 104 volt/cm along x-axis and magnetic field of intensity 2000 gaussalong y-axis. Calculate the magnitude and direction of total force.

Solution: Intensity of electric field

E = 4 ˆ ˆ3 10 i volt / cm 100 i esu/cm

Proton charge = 19 101.6 10 C 4.8 10 esu

Magnetic field B = ˆ2000 j gauss

velocity v = 8 ˆ10 k cm / s

so total force acting on the proton

F =

v Bq EC

= 10 810

1ˆ ˆ ˆ4.8 10 100i 10 k 2000 j3 10

= 8 ˆ4.47 10 i dyne

Hence total force acting on the proton has magnitude +4.47 × 10–8 dyne along the +ve x-direction.

Q.11: Find the value of the constant p so that

ˆ ˆ ˆ ˆˆ ˆA 2i j 3k, B 2i 3j k

and ˆ ˆ ˆC 3i pj k

are coplanar..

Solution: We know that three vectors are said to be coplanar if A. B C 0

2 1 32 3 13 p 1

= 0

Introduction to Vector Calculus (35)

4p – 38 = 0or 4p = 38

p =38 9.54

Q.12: Evaluate A B C

where

ˆ ˆ ˆ ˆ ˆA 2i j, B i j k

and ˆ ˆ ˆC 5i 3j k

.

Solution:

A B C

= B A.C C A.B

= ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆi j k 2i j 0k . 5i 3j k

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ5i 3j k 2i j 0k . i j k

= ˆ ˆ ˆ ˆ ˆ ˆ7 i j k 1 5i 3j k

= ˆ ˆ ˆ2i 4j 8k

Q.13: If r si the position vector of any point (x, y, z) and A is a constant vector

then show that

(i) r .A . A 0

is the equation of a constant plane.

(ii) r A .r

is the equation of a sphere.

Also show that result of (i) is of the form Ax By Cz D 0 where

2 2 2D A B C and that of (ii) is of the from 2 2 2 2x y z r . [RU 2005]

Solution: (i) Suppose A = A, B, C and r x, y, z

r A . A

= x A A y B B z C C

= 2 2 2xA A yB B zC C

= 2 2 2xA yB zC A B C

= x y zA B C D

(36) Introduction to Vector Calculus

where D = 2 2 2A B C

so r A .A

= 0 x y zA B C D 0

which is an equation of a plane.

(ii) r A .A

= x A x y B y z C z

if r A .A

= 0 then

2 2 2x y zx y z A B C = 0

Which is the equation of sphere whose surface touches the origin.

Q.14: A particle moves on the curve 2 2x 2t , y t 4t, z 3t 5 where t is thetime. Find the components of velocity and acceleration at time t = 1 in the directionˆ ˆ ˆi 3j 2k .

Solution: Position vector

r = 2 2ˆ ˆ ˆ2t i t 4t j 3t 5 k

so velocity vector

drvdt

= 2 2d d dˆ ˆ ˆ2t i t 4t j 3t 5 kdt dt dt

= ˆ ˆ ˆ4t i 2t 4 j 3k

acceleration a =dvdt

= d d dˆ ˆ ˆ4t i 2t 4 j 3 kdt dt dt

= ˆ ˆ4i 2 j 0

at` t = 1, velocity ˆ ˆ ˆv 4i 2 j 3k

acceleration a = ˆ ˆ4i 2 j

and the component of ˆ ˆ ˆv along i 3j 2k

Introduction to Vector Calculus (37)

is

22 2

ˆ ˆ ˆ ˆ ˆ ˆ4i 2 j 3k . i 3j 2k

1 3 2

=16 8 14

714

and component of ˆ ˆa along i 3j 2k is

=

22 2

ˆ ˆ ˆ ˆ ˆ4i 2 j . i 3j 2k 2141 3 2

=14

7

Q.15: Calculate the unit vector, which is normal to the surface

= 2 2x y xy 3xyz at the point (1, 1, –1).

Solution : Here

= 2 2ˆ ˆ ˆi j k x y xy 3xyzx y z

= 2 2 2 2ˆ ˆx y xy 3xyz i x y xy 3xyz jx y

2 2 ˆx y xy 3xyz kz

= 2 2ˆ ˆ2xy y 3yz i x 2xy 3xz j 3xy k

At (1, 1, –1),

= ˆ ˆ ˆ ˆ2 1 3 i 1 2 3 j 3k 3k

so the unit vector normal to the surface at (1, 1, –1) is

2

ˆ3k

3 =

ˆ3k3

= k

Q.16: Find the direction derivative of 2 2x, y,z x y xy at the point (2, –1, –4)along the direction of the vector (1, 2, –1).

Solution: as

(38) Introduction to Vector Calculus

= 2 2x y xy

= ˆ ˆ ˆi j k x, y,zx y z

= 2 2 2 2 2 2ˆ ˆ ˆx y xy i x y xy j x y xy kx y z

= 2 2ˆ ˆ2xy y i x 2xy j

2, 1, 4

= ˆ3i

Position vector r = ˆ ˆ ˆi 2 j k

and unit vector along this position vector

n =ˆ ˆ ˆ ˆ ˆ ˆi 2 j k i 2 j k1 4 1 6

and direction derivative n

= ˆ ˆ ˆi 2j kˆ3i .

6

=3 6

26

Q.17: Find the equation of the tangent plane and normal line to the surface2 22x y 2z 3 at the point (2, 1, –3).

Solution :

Here x,y,z = 2 22x y 2z 3

x

= 2 22x y 2z 4xx

y = 2 22x y 2z 2y

y

z

= 2 22x y 2z 2z

so the components , andx y z at the point (2, 1, –3) will be

Introduction to Vector Calculus (39)

x

= 4 2 8, 2 1 2, 2y z

Hence the equation of the tangent plane to the surface at the point (2, 1, –3) is

X 2 8 Y 1 2 Z 3 2 = 0

or 4X + y + Z = 6so the equation of normal to the surface at (2, 1, –3) is

X 28

=Y 1 Z 3

2 2

orX 2

4

= Y – 1 = Z + 3

Q.18: Find the angle between the surfaces 2 2 2x y z 9 and 2 2x y z 3 atthe (1, 2, 2)

Solution:

Suppose 1 = 2 2 2 2 22x y z and x y z

so 1

= ˆ ˆ ˆ2x i 2y j 2z k

and 2 = ˆ ˆ ˆ2x i 2y j k

and 1 1,2,2

= ˆ ˆ ˆ2i 4 j 4k

2 1,2,2

= ˆ ˆ ˆ2i 4 j k

since 1 2and

are normal to 1 and 2

then 1 2.

= 1 2 cos

where is the angle between thesurfaces 1 and 2.

so =1 1 2

1 2

.cos

=1 14 16 4 16cos cos

36 21 6 21

= 54.41° approx.

Q.19 (i) Provle that 1 1 2 2ˆ ˆ ˆ ˆP cos i sin j and cos i sin j

are unit vectors in

(40) Introduction to Vector Calculus

the xy-plane respectively making 1 and 2 with the x-axis.

(ii) By means of dot product, obtain the formula for 2 1cos . by similarly

formulating P and Q, obtain the formula for 2 1cos .

(iii) If is the angle between P and Q find 1 P Q2

in terms of .

Solution:

(i) Given P = 1 1

ˆ ˆcos i sin j

Q

= 2 2ˆ ˆcos i sin j

y

1

2

Q

P

P

= 2 21 1cos sin 1

Q

= 2 22 2cos sin 1

hence P and Q

are unit vectors.

(ii) P.Q

= 2 1P Q cos

= 2 11.1 cos ...(1)

But P.Q

= 1 1 2 2ˆ ˆ ˆ ˆcos i sin j . cos i sin j

= 1 2 1 2cos cos sin sin ...(2)

so 2 1cos = 1 2 1 2cos cos sin sin

let 1P P

= 1 1ˆ ˆcos i sin j

and 1Q

= 2 2ˆ ˆcos i sin j

then 1 1P .Q

= 1 21.1 cos

Introduction to Vector Calculus (41)

= 1 2 1 2cos cos sin sin

(iii) 1 1P and Q

are unit vectors.

so 1 P Q2

= 2 21 Q 2PQ cos2

=1 1 1 2cos2

= 1 cos

= 22sin2

Q.20: A vector field is given as 2 2ˆ ˆ ˆW 4x y i 7x 2z j 4xy 2z k

(i) What is the magnitude of the field at point (2. –3, 4).(ii) At what point on z-axis is the magnitude of W equal to unity? [RU 2002]Solution: (i)

W = 2 2ˆ ˆ ˆ4x y i 7x 2z j 4xy 2z k

at P(2, –3, 4), W = 2 2ˆ ˆ ˆ4 2 3 i 7 2 4 2 j 4 2 3 2 4 k

= ˆ ˆ ˆ48i 22 j 8k

W

= 2 2 248 22 8 53.4

(ii) As the required point is on z-axis so x = 0, y = 0

W = 2 2ˆ2z j 2z k for that point.

W

= 22 2 2 42z 2z 4z 4z 1

so 4 24z 4z 1 = 0

z2 = 4 16 16 1 1 1

8 2 2

z2 = –1.207 and 0.207taking z as positive z2 = 0.207

z = ± 0.455Q.21: Calculate the differential volume to obtain the expression for volume of the(i) sphere of radius 'b'

(42) Introduction to Vector Calculus

(ii) Semispherical shell of inner radius 'a' and outer radius 'b'.(iii) Cylinder of radius 'b' and height 'h'.Solution:(i) Differential volume in spherical coordinates

dV = 2r sin dr d d

here r = 0 to b, = 0 to , = 0 to 2.So volume of sphere

V =b 2

2

V 0 0 0

dV r sin dr d d

=b 2

2

0 0 0

r dr sin d d

= b

2

0 0

2 r dr cos

= b

2

0

2 1 1 r dr

=3b4

3

so Vsphere =34 b

3

(ii) For semispherical shellr1 =a, r2 = b

so dV = 2r sin dr d d

here r = a to b, 0 to , 0 to

so V =b

2

a 0 0

r sin dr d d

Introduction to Vector Calculus (43)

=b

2

a 0 0

r dr sin d d

= b

2

a 0

r dr cos

=

b2

a

r23

= 3 32 b a3

(iii) Differential volume for a cylinder

dV = r dr d dz

here r = 0 to b, z = 0 to h, and = 0 to 2.

so V =V

dV

V =b h 2

a 0 0

r dr d dz

=b h 2

0 0 0

r dr dz d

=

b2

0

r2 .h .2

so cyl.V = 2b h

Q.22 : For positive x, y, z let = 40 xyz c/m3. Find the total charge within theregion bounded by x = 0, y = 0, 0 2x 3y 10 and 0 z 2 .

Solution : Here

Q =y5 z

0 0 0

40 xyz dx dydz

(44) Introduction to Vector Calculus

=

10 2x 25 2 23

0 0 0

y z40x dx2 2

= 5

2 3

0

40 100x 40x 4x dx9

=

52 3 4

0

40 100x 40x 4x9 2 3 4

Q = 925.926 CQ.23: Given point P in Cartesian coordinate system as P(1, 2, 3). Calculate its

coordinates in cylindrical system.Solution: As given x = 1, y = 2, z = 3

= 2 2 1 yx y tan , z zx

so = 2 21 2 5 2.236

= 1 2tan 63.431

z = 3so Pcyl. = (2.236, 63.43°, 3)Q.24: The cooridnate of a point P in cylindrical system is P(1, 45°, 2). find its

equivalent in cartesion system.Solution:Here = 1, = 45°, z = 2

and x = cos , y sin , z z

x =11. cos 45 0.7072

y = 0.707z = 2

so Pcart. = (0.707, 0.707,2)Q.25: Find the constant m such that the vector

Introduction to Vector Calculus (45)

= ˆ ˆ ˆx 3y i y 2z j x mz k is solenoidal.

Solution: The vector will be solenoidal if

so ˆ ˆ ˆ ˆ ˆ ˆi j k x 3y i y 2z j x mz k 0x y z

i.e. x 3y y 2z x mz 0x y z

or 1 + 1 + m = 0m = –2

Q.26: Find 3 3 3div F and curl F if F grad x y z 3xyz

. [WBUT 2001]

Solution: Here F = grad 3 3 3x y z 3xyz

= 3 3 3 3 3 3ˆ ˆx y z 3xyz i x y z 3xyz jx y

3 3 3 ˆx y z 3xyz kz

= 2 2 2ˆ ˆ ˆ3x 3yz i 3y 3xz j 3z 3xy k

div F

= 2 2 23x 3yz 3y 3xz 3z 3xyx y z

= 6 x y z

F =

2 2 2

ˆ ˆ ˆi j k

x y z

3x 3yz 3y 3xz 3z 3xy

= ˆ ˆ ˆ3x 3x i 3y 3y j 3z 3z k 0

Q.27 Show that curl grad f = 0 where f = x2y + 2xy + z2.

(46) Introduction to Vector Calculus

Solution: grad f =f f fˆ ˆ ˆi j kx y z

= 2ˆ ˆ ˆ2xy 2y i x 2x j 2z k

curl grad f =2

ˆ ˆ ˆi j k

x y z

2xy 2y x 2x 2z

= ˆ0 0 2x 2 2x 2 k 0

Q.28: If the scalar function 2x,y,z 2xy z ,. is its corresponding scalar fieldis solenoidal or irrotational?

Solution : Let

F = ˆ ˆ ˆ2y i 2x j 2zk

so .F

= 2y 2x 2zx y z

= 0 +0 + 2 = 2 0

So field is not solenoidal.

Now F

=

ˆ ˆ ˆi j k

x y z2y 2x 2z

= 0so field is irrotational.Q.29: Verify the divergence theorem for the vector function

F = 2ˆ ˆ ˆ4xz i y j yz k

taken over the cube bounded by x = 0, 1 y = 0, 1, z = 0, 1.[WBUT (math) 2002]

Solution.:

Introduction to Vector Calculus (47)

x

y

z

3 2

5

1

67

0

4

for face 4567; ˆn i and x = 1

1

ˆF . n ds =

1 1

0 0

4z dydz 2

2

ˆF . n ds = i dy dz 0 for 1230

for ˆ2561 n = j, y 1

3

ˆF. n ds = –1

and for 30744

ˆF. n ds = 0

for face 2345,5

ˆF. n ds =

12

and6

ˆF. n ds = 0 for 0167

totalS

ˆF. n ds = 1 32 0 1 0 0

2 2

Again F

= 24xz y yzx y z

= 4z – y

(48) Introduction to Vector Calculus

1 1 1

0 0 0

.F dV

= 1 1 1

0 0 0

4z y dx dy dz

=

11 1 2

0 0 0

4z yz dx dy2

=32

s

ˆF. n ds =

V

F dV

Hence divergence theorem is verified.

Q.30: Calculate the line integral of ˆ ˆA cos z sin z

around the edge L of thewedge defined by 0 4 , 0 30 , z 0.

Solution:

Given A = ˆ ˆcos z sin z

differential length

dl

= ˆˆ ˆd d dzz

x

y

0

(3)(2)

(1)

Circulation of A around path is

L

A.dl

=1 2 3

A.dl A.dl A.dl

for (1) 0, d 0, z 0, dz 0

1

A.dl

= 1

ˆˆ ˆ ˆcos zsin d d dz z

Introduction to Vector Calculus (49)

=

44 2

1 0 0

cos d d2

=16 82

for (2) 4, d = 0

z = 0, dz = 0

2

A. dl

=2

cos d zsin dz 0

for (3) / 6, d = 0, z = 0, dz = 0

3

A. dl

=3

cos d

=0

4

cos d6

=00 2

4 4

3 d 0.8662 2

= –6.93

So totalL

A. dl

= 8 + 0 – 6.93

= 1.07

Q.31 Given 2A x xy

, calculate A . ds

over the region y = x2, 0 < x < 2.

Solution:So y = x2, ds = dx dy

(50) Introduction to Vector Calculus

x

y

0

2

2

A. ds

= 2x xy dx dy

= 2 22 x 2 x

2

0 y 0 0 y 0

x dx dy xy dx dy

=2 2 5

4

x 0 x 0

xx dx dx2

=32 645 12

= 11.7

Q.32 For a scalar function zx ysin sin e2 3

. Calculate the magnitude

direction of maximum rate of increase of at the point (1, 1, 1).Solution: As gradient of a scalar function gives the magnitude and direction of max. rate

of change of that

So = ˆ ˆ ˆi j kx y z

=z zx y x yˆ ˆsin sin e i sin sin e j

x 2 3 y 2 3

zx y ˆsin sin e kz 2 3

=z zx x yˆ ˆe sin j sin sin e k

6 2 2 3

at (1, 1, 1)

Introduction to Vector Calculus (51)

1,1,1 =1 1ˆ ˆe sin j sin sin e k

6 2 2 3

= ˆ ˆ0.367 j 0.866 0.367 k6

= ˆ ˆ0.192 j 0.318 k

and 1,1,1 = 1/22 20.192 0.318

= 0.37

and j =ˆ ˆ0.192j 0.318k0.37

= ˆ ˆ0.52j 0.86k

Q.33 : Determine the divergence of the following vector fields at given points–

(i) ˆ ˆ ˆA yzi 4 x y j xyz k at 1, 2,1

(ii) ˆˆ ˆB z sin 5 z cos z z at 5, , 12

(iii) ˆ ˆˆC 2rsin cos r cos r at 1, ,3 3

Solution: In cartesion

(a) .A

= yz 4x 4y xyzx y z

= 0 + 4 + xy

.A

= 4 + xy

1, 2,1at 1, 2,1 , .A

= 4 + 1 × (–2)

= 2(b) In cylindrical

.B

= z1 1B B B

z

(52) Introduction to Vector Calculus

given B = 1 zz sin B 5 z cos , B z

.B

= 2 1z sin 5. z sin 1

= 1 3z sin

at 5, , 12

.B

= 1 – (3 × 1 × 1)

= –2(c) In spherical system

.C

= 2r2

C1 1 1r C sin Cr r sin rsinr

= 322 cossin cos r sin

r r sin rsinr

=cos cot6sin cos

r

at 1, ,3 3

.C

=cos cot

3 36sin cos3 3 1

= 2.6 + 0.288= 2.88

Q.34: Find the nature of the vector 2ˆ ˆ ˆF 30i 2xy j 5xz k

. [RU 2003]

Solution :

.F

= 30 2xy 5xzx y y

= 2x + 10xz 0

Introduction to Vector Calculus (53)

Curl F

= F =

2

ˆ ˆ ˆi j k

x y z

30 2xy 5xz

= 2 ˆ ˆ5z j 2yk

0

as .F 0

fields is not solenoidal

and F 0

field is rotational.

Q.35: Given the vector field 2ˆ ˆ ˆG 16xy 3 i 8x j xk

.

(i) Is G irrotational (or conservative)?(ii) Find the net flux of G over the cube 0<x, y, z < 1.(iii) Determine the circulation of G around the edge of the square z = 0, 0 < x,

y < 1. Assume anticlockwise direction. [RU 2003]Solution:

(i) G

=2

ˆ ˆ ˆi j k

x y z

16xy z 8x x

= ˆ ˆ ˆ0 i 1 1 j 16x 16x k 0

So G is irrotational.

(ii) Net flux of G over the cube

= V

. G dV

. G

= 216xy z 8x xx y z

= 16y 0 0 16y

so V

. G dV

=1 1 1

0 0 0

16y dx dydz 16 dx dz y dy

(54) Introduction to Vector Calculus

=

12

0

y16.1.1 82

(iii)0

G . dl

= y 0 x 1

1 12

x 0 y 0z 0 z 0

16xy z dx 8x dy

y 1 x 0

0 02

x 1 y 1z 0 z 0

16xy z dx 8x dy

= 021

01

x0 8 1 y 16 1 02

= 8 – 8 = 0.

SUMMARY• A vector is with magnitude and direction.• In space a quantity is specified by a function.• When the result of multiplication of two vectors is a scalar then it is called scalar

product or dot product.• When product is a vector then it is called vector product.

• Multiplication of three vectors can give scalar A . B C

or a vector A B C

.

• Vector differentiation is done using dal () operator the gradient of a scalar field is

, divergence as . A

and curl by A

and laplacian by 2A.

• In Cartesion coordinate system

dl

= ˆ ˆ ˆdx i dy j dz k, dV dx dy dz

Gradient = ˆ ˆ ˆi j kx y z

Divergences . A

=yx zAA A

x y z

Introduction to Vector Calculus (55)

Curl A

=x y z

ˆ ˆ ˆi j k

x x zA A A

Laplacian =2 2 2

2 2 2x y z

• In cylindrical system

dl

= ˆˆ ˆd d dz z, dV d d dz

gradient T =T 1 T Tˆˆ z

z

divergence . A

= zA A1 1Az

Curl A

=

z

ˆˆ z1

zA A A

Laplacian =2

2 2 21 1

z

• In spherical system

dl

= ˆ ˆˆdr r rd r sin d

dV = 2r sin d dr d

gradient T =T 1 T 1 Tˆ ˆrr r r sin

divergence

= 2r2

A1 1 1r A A sinr r sin rsinr

(56) Introduction to Vector Calculus

Curl A

=

r

ˆ ˆr r sin1

r sin rA rA rsin A

Laplacian =2

21 1r sin

r r r sin

2

2 2 21

sin

• Gauss Divergence theorem

s

A . ds

= V

. A dV

• Stoke's theorem

L

A . dl

= V

A . ds

• A vector field is solenoidal if

Irrotational or conservative if

• Triple products

A . B C B C A C A B

A B C B A . C C A . B

• Second derivatives

. A 0

f 0

2A . A A

EXERCISE

Introduction to Vector Calculus (57)

1. Give the basic concepts of transformation of one coordinate system to another. Derivenecessary relations for rectangular, cylindrical and spherical systems. [RU 2003]

2. Write short note on "Physical significance of curl, divergence and gradient".3. State-Gauss divergence theorem. Write its applications, advantages and limitations.

[RU 2002]4. State and prove stoke's theorem. [RU 2000]5. Explain how stoke's theorem enables us to obtain the integral form of ampere circuital law.6. Explain various types of vector fields.

(i) Solenoidal and irrotational fields.(ii) Irrotational but not solenoidal fields(iii) Solenoidal but not irrotaitonal fields(iv) Neither irrotational nor solenoidal fields.

7. For the vectors ˆ ˆ ˆ ˆ ˆA i 3k and B 5i 2 j 6k

calculate

(i) A B

(ii) A B

(iii) A . B

(iv) A B

(v) Angle between A and B

(vi) A unit vector parallel to 3A B

.

(vii) Length of the projection of A on B

.

Ans.: ˆ ˆ ˆ ˆ ˆ ˆˆ ˆi 6i 2j 3k ii 4i 2j 9k iii 13 iv 6i 21j 2k

ˆ ˆ ˆ8i 2j 3kv 60 vi vii 1.6m

77

8. Use the differential volume dV to find volume of region.

(i) 0 x 1, 1 y 2, 3 z 3 [Ans.: 6]

(ii) 2 5, , 1 z 43

[Ans.: 110]

9. Find area of the region 0 on the spherical shell of radius 'b'. [Ans. 2b2]

10. Evaluate the gradient of the following scalar fields

(58) Introduction to Vector Calculus

(a) zP e sin 2x [Ans.: z zˆ ˆP 2cos 2xe i sin2xe k ]

(b) 2q zcos 2 2ˆˆAns. : q 2 zcos zsin cos z

(c) s 20r sin cos ˆAns. : r 20sin cos r 20rcos

ˆ ˆcos 20rsin sin

11. If f = xy + yz + xz then(i) Find the magnitude and direction of the maximum rate of change of the function at

point (1, 2, 3)(ii) Find the rate of change of the function at the same point in the direction of the

vector.

Ans. : i f 14

ˆ ˆ ˆf 2i 3j kˆ ff 14

(ii) df f .dl 11

12. If ˆ ˆ ˆT 2x i 3y j 4z k and V = xyz evaluate . VT

. [Ans.: 2xyz]

13. If 2 2 2U xz x y y z find div (grad U). [Ans.: 2(–y + z2 + y2)]

14. Given 2 2 2 2 3ˆ ˆ ˆD 6xyz i 3x z j 6x y k C/m

Find the total charge lying within the region

bounded by 0 < x < 1, 1 < y < 2 and z 1 by separately evaluating each side of divergencetheorem. [Ans.: 6C]

15. If ˆ ˆ ˆA x y 1 i j x y k

then prove that A . 0

16. Prove that vector ˆ ˆ ˆA x y z i 2y zx j 2z xy k

is not solenoidal.[WBUT 2005]

17. If 2 2x y 2z find .

. [WBUT 2005]

18. Show that 2 2ˆ ˆ ˆB 2xyz i x z 2y j x yk

is irrotational. [WBUT 2005]

19. find a unit vector perpendicular to 2 2 2x y z 100 at (1, 2, 3). [WBUT 2007]

Introduction to Vector Calculus (59)

ˆ ˆ ˆi 2j 3kAns. :14

20. if 2 3 23x y y z find

at (1, –2, –1). [WBUT 2004]

ˆ ˆ ˆAns. : 12i 9j 16k

21. Show that 3 2 2ˆ ˆ ˆF 2xy z i x j 3xz k

is a conservative force field. Find also the scalar

potential. [WBUT 2006, 2003] 2 3Ans. : x y z x constant

22. Evaluates s

ˆF.n ds where 2ˆ ˆ ˆF 8x z i y j yzk

and s is the surface of the cube bounded

by x = 0, 1, y = 0, 1, z = 0, 1.