UNIT-II

VECTOR CALCULUS

Directional derivative

The derivative of a point function (scalar or vector) in a particular

direction is called its directional derivative along the direction.

The directional derivative of a scalar point function φ in a given

direction is the rate of change of φ in the direction. It is given by the

component of gradφ in that direction.

The directional derivative of a scalar point function

φ (x,y,z) in the direction of →a is given by

→

→∇

a

a.φ.

Directional derivative of φ is maximum in the direction of φ∇ .

Hence the maximum directional derivative is φφ grador∇

Unit normal vector to the surface

If φ (x, y, z) be a scalar function, then φ (x, y, z) = c represents

A surface and the unit normal vector to the surface φ is given by

φφ

∇∇

Equation of the tangent plane and normal to the surface

Suppose →a is the position vector of the point ),,( 000 zyx

On the surface φ (x, y, z) = c. If →→→→

++= kzjyixr is the position vector of

any point (x,y,z) on the tangent plane to the surface at→a , then the

equation of the tangent plane to the surface φ at a given point →a on it is

given by 0. =

−→→

φgradar

If →r is the position vector of any point on the normal to the surface

at the point →a on it. The vector equation of the normal at a given point

→a on the surface φ is 0=×

−→→

φgradar

The Cartesian form of the normal at ),,( 000 zyx on the surface

φ (x,y,z) = c is

z

zz

y

yy

x

xx o

∂∂−=

∂∂−=

∂∂−

φφφ00

Divergence of a vector

If ),,( zyxF→

is a continuously differentiable vector point function in

a given region of space, then the divergences of →F is defined by

z

Fk

y

Fj

x

FiFdivF

∂∂+

∂∂+

∂∂==∇

→→

→→

→→→→

.

=x

Fi

∂∂

→→

∑

If →→→→

++= kFjFiFF 321 ,then ).( 321

→→→→++∇= kFjFiFFdiv

i.e., z

F

y

F

x

FFdiv

∂∂+

∂∂+

∂∂=

→321

Solenoidal Vector

A vector →F is said to be solenoidal if 0=

→Fdiv (ie) 0. =∇

→F

Curl of vector function

If ),,( zyxF→

is a differentiable vector point function defined at each

point (x, y, z), then the curl of →F is defined by

→→

×∇= FFcurl

= z

Fk

y

Fj

x

Fi

∂∂×+

∂∂×+

∂∂×

→→

→→

→→

= x

Fi

∂∂×

→→

∑

If →→→→

++= kFjFiFF 321 ,then )( 321

→→→→++×∇= kFjFiFFcurl

321 FFF

zyx

kji

Fcurl∂∂

∂∂

∂∂=

→

=

∂∂−

∂∂+

∂∂−

∂∂−

∂∂−

∂∂ →→→

y

F

x

Fk

z

F

x

Fj

z

F

y

Fi 121323

Curl→F is also said to be rotation

→F

Irrotational Vector

A vector →F is called irrotational if Curl 0=

→F

(ie) if 0=×∇→F

Scalar Potential

If →F is an irrotational vector, then there exists a scalar function φ

Such that φ∇=→F . Such a scalar function is called scalar potential of

→F

Properties of Gradient

1. If f and g are two scalar point function that ( ) gfgf ∇±∇=±∇ (or)

( ) gradggradfgfgrad ±=±

Solution: ( ) ( )gfz

ky

jx

igf ±

∂∂+

∂∂+

∂∂=±∇

→→→

= ( ) ( ) ( )

±∂∂+±

∂∂+±

∂∂ →→→

gfz

kgfy

jgfx

i

= z

gk

z

fk

y

gj

y

fj

x

gi

x

fi

∂∂±

∂∂+

∂∂±

∂∂+

∂∂±

∂∂ →→→→→→

=

∂∂+

∂∂+

∂∂±

∂∂+

∂∂+

∂∂ →→→→→→

z

gk

y

gj

x

gi

z

fk

y

fj

x

fi

= gf ∇±∇

2. If f and g are two scalar point functions then ( ) fggffg ∇+∇=∇ (or)

ggradffgradgfggrad +=)(

Solution: ( ) =∇ fg ( )fgz

ky

jx

i

∂∂+

∂∂+

∂∂ →→→

= ( ) ( ) ( )

∂∂+

∂∂+

∂∂ →→→

fgz

kfgy

jfgx

i

=

∂∂+

∂∂+

∂∂+

∂∂+

∂∂+

∂∂ →→→

z

fg

z

gfk

y

fg

y

gfj

x

fg

x

gfi

=

∂∂+

∂∂+

∂∂+

∂∂+

∂∂+

∂∂ →→→→→→

z

fk

y

fj

x

fig

z

gk

y

gj

x

gif

= fggf ∇+∇

3. If f and g are two scalar point function then 2g

gffg

g

f ∇−∇=

∇ where

0≠g

Solution: =

∇g

f

∂∂+

∂∂+

∂∂ →→→

g

f

zk

yj

xi

= ∑

∂∂→

g

f

xi

= ∑

∂∂−

∂∂

→

2g

x

gf

x

fg

i

=

∂∂−

∂∂

∑ ∑→→

x

gif

x

fig

g 21

= [ ]gffgg

∇−∇2

1

4. If →→→→

++= kzjyixr such that rr =→

,prove that →

−=∇ rnrr nn 2

Solution: nn r

zk

yj

xir

∂∂+

∂∂+

∂∂=∇

→→→

=

∂∂+

∂∂+

∂∂ →→→

z

rk

y

rj

x

ri

nnn

= z

rnrk

y

rnrj

x

rnri nnn

∂∂+

∂∂+

∂∂ −

→−

→−

→111

=

++→→→

−

r

zk

r

yj

r

xinr n 1

=

++→→→−

kzjyixr

nr n 1

= →−

rr

nr n 1

5. Find a unit normal to the surface 422 =+ xzyx at (2,-2, 3)

Solution: Given that xzyx 22 +=φ

)2( 2 xzyxz

ky

jx

i +

∂∂+

∂∂+

∂∂=∇

→→→φ

= ( ) ( ) ( )xkxjzxyi 222 2→→→

+++

At (2,-2, 3)

( ) )4()4(68→→→

+++−=∇ kjiφ

= →→→

++− kji 442

63616164 ==++=∇φ

Unit normal to the given surface at (2,-2,3)

φφ

∇∇

=6

442

→→→

++− kji

=

++−→→→kji 22

3

1

6. Find the directional derivative of xyzxzyzx ++= 22 4φ at (1,2,3) in the

direction of →→→

−+ kji2

Solution: Given xyzxzyzx ++= 22 4φ

)4( 22 xyzxzyzxz

ky

jx

i ++

∂∂+

∂∂+

∂∂=∇

→→→φ

= ( ) ( ) ( )→→→+++++++ kxyxzyxjxzzxiyzzxyz 842 222

At (1, 2, 3)

→→→

++=∇ kji 28654φ

Given: →→→→

−+= kjia 2

6114 =++=∴→a

→

→

∇=∴a

aDD .. φ

= 6

22.28654

→→→→→→ −+

++ kjikji

= [ ] [ ]866

1286108

6

1 =−+

7. Find the angle between the surface 5222 =++ zyx and

52222 =−++ xzyx at (0,1,2)

Solution: Let 222

1 zyx ++=φ and xzyx 2222

2 −++=φ

zz

yy

xx

2,2,2 111 =∂∂=

∂∂=

∂∂ φφφ

zz

yy

xx

2,2,22 222 =∂∂=

∂∂−=

∂∂ φφφ

→→→

++=∇ kzjyix 2221φ

→→→

++−=∇ kzjyix 22)22(2φ

At (o,1,2)

→→

+=∇ kj 421φ

→→→

++−=∇ kji 4222φ

Cos644416

422.42.

21

21

+++

++−

+=

∇∇∇∇=

→→→→→kjikj

φφφφθ

2420

20

2420

164cos =+=θ

= −

2420

20cos 1θ

=

−

24

20cos 1

8. Find the angle between the surfaces 1log 2 −= yzx and zyx −= 22 at the

point (1,1,1)

Solution: let zxy log2

1 −=φ and zyx += 2

2φ

z

x

zy

yz

x−=

∂∂=

∂∂−=

∂∂ 111 ,2,log

φφφ

1,,2 2222 =∂∂=

∂∂=

∂∂

zx

yxy

x

φφφ

→→→−+−=∇ kz

kjyiz 2)log(1φ

→→−=∇ kj22φ

Cos65

1

11414

12.

21

21 =+++

−=∇∇∇∇=

φφφφθ

= −

65

1cos 1θ

9. Find ( )nr2∇

Solution: ( )nr2∇ = ( )nr∇∇.

= ( ) ( ) ( )nnn rz

kry

jrx

i∂∂+

∂∂+

∂∂ →→→

= z

rnrk

y

rnrj

x

rnri nnn

∂∂+

∂∂+

∂∂ −

→−

→−

→111

→→→→

++= kzjyixr

222 zyxrr ++==

→

2222 zyxr ++=

r

x

x

rx

x

rr =

∂∂

⇒=∂∂

22

r

y

y

ry

y

rr =

∂∂

⇒=∂∂

22

r

z

z

rz

z

rr =

∂∂

⇒=∂∂

22

( )=∇∴ nr2

++→→→

−

r

zk

r

yj

r

xinr n 1

=

++→→→

− kzjyixnr n 2

=

→−

rnrn 2

Since →→→

+∇=

∇ udivuu φφφ .

( )

∇=∇→

− rnrr nn 22

= ( ) →−

→− ∇+

∇ rnrrnr nn .. 22

++

∂∂+

∂∂+

∂∂=∇

→→→→→→→kzjyix

zk

yj

xir.

=1+1+1 = 3

( ) ( ) →−− ∇+=∇ rrnnrr nnn .3 222

= ( )( ) 242 .23 rrnnnr nn −− −+

= ( )( )22 23 −− −+ nn rnnnr

( ) [ ] ( ) 2222 1 −− +=+=∇ nnn rnnnnrr

10. If →→→→

++= kzjyixr and rr =→

.Prove that →rr n is solenoidal if 3−=n and

→rr n is irrotational for all vectors of n.

Solution: →rr n

→→→++= krjyrixr nnn

div ( ) ( ) ( )zrz

yry

xrx

rr nnnn

∂∂+

∂∂+

∂∂=

→…………………(1)

Now 2222 zyxr ++=

Differentiating partially w.r.to x,

r

x

x

rx

x

rr =

∂∂

⇒=∂∂

22

Similarly, r

y

y

ry

y

rr =

∂∂

⇒=∂∂

22

r

z

z

rz

z

rr =

∂∂

⇒=∂∂

22

Now ( ) ( ) nnn rx

rr

rxxr

x+

∂∂

∂∂=

∂∂

.

= x.nnn r

r

xr +−1

( ) nnn rynryry

+=∂∂ − 22

( ) nnn rznrzrz

+=∂∂ − 22

From (1) we have

( ) nnn rzyxnrrrdiv 32222 +++=

−→

= nn rnr 3+

= ( ) nrn 3+

The vector →rr n is solenoidal if

→rrdiv n

= 0

( ) 03 =+⇒ nrn

03 =+⇒ n

3−=⇒ n →

∴ rr n is solenoidal only if n = -3

Now

zryrxr

zyx

kji

rrcurl

nnn

n

∂∂

∂∂

∂∂=

→→→

→

= ( ) ( )∑

∂∂−

∂∂→

yrz

zry

i nn

= ∑

∂∂−

∂∂ −−

→

z

rynrz

y

rnri nn 11

= ∑

− −−→

r

zynrz

r

ynri nn 11

= ( )∑ −−→

− yznryznri nn 22

= 0

Curl (→rr n ) =

→→→++ kji 000 =0

Curl (→rr n ) = 0 for all values of n

Hence →rr n is irrotational for all values of n.

11. Prove that ( ) ( )→→→→

+−++= kxzjxyizxyF 232 34sin2cos is irrotational and

find its scalar potential

Solution:

232 34sin2cos xzxyzxy

zyx

kji

Fcurl

−+∂∂

∂∂

∂∂=

→→→

→

= [ ] [ ] [ ] 0cos2cos23300 22 =−+−−−→→→

xyxykzzji →

∴F is irrotational.

To Find φ such that φgradF =→

( ) ( )z

ky

jx

ikxzjxyizxy∂∂+

∂∂+

∂∂=+−++∴

→→→→→→ φφφ222 34sin2cos

Integrating the equation partially w.r.to x,y,z respectively

),(sin 1

32 zyfxzxy ++=φ

),(4sin 2

2 zxfyxy +−=φ

),(33 yxfxz +=φ

,4sin 32 Cyxzxy +−+=∴φ is scalar potential

12. Prove that ).().(→→→→→→

−=

× BcurlAAcurlBBAdiv

Proof : ).(→→→→

×∇=

× BABAdiv

= ∑

×∂∂ →→→

BAx

i

= ∑∑

×

∂∂+

∂∂×

→→

→→

→→B

x

Ai

x

BAi

= ∑∑

×

∂∂+

×

∂∂−

→→

→→→

→B

x

AiA

x

Bi

= →

→→→

→→

∂∂×+

∂∂×− ∑∑ B

x

AiA

x

Bi ..

= →→→→

+

− BAcurlABcurl ..

13.Prove that →→→

∇−

•∇∇=

FFFcurlcurl 2

Solution:

×∇×∇=

→→FFcurlcurl

By using →→→→→→→→→

−

=

×× cbabcacba ..

= ( )→→

∇∇−∇

∇ FF ..

= →→

∇−∇

∇ FF 2.

VECTOR INTEGRATION

Line, surface and Volume Integrals

Problems based on line Integral

Example 1:

If ( ) →→→→+−+= kxzjyziyxF 22 201463 Evaluate ∫

→

C

drF . from (0,0,0) to

(1,1,1) along the curve 32 ,, tztytx ===

Solution: The end points are (0, 0, 0) and (1, 1, 1)

These points correspond to t = 0 and t = 1

23,2, tdztdydtdx ===∴

∫→

C

drF . = ( )∫ +−+C

dzxzyzdydxyx 22 201463

= ( ) ( ) ( )∫ +−+1

0

27522 32021463 dttttdttdttt

= ( )∫ +−1

0

962 60289 dtttt

= ( )101073 643 ttt +−

= ( )[ ] 50643 =−+−

Example 2:

Show that →→→→

++= kzjyixF 222 is a conservative vector field.

Solution: If →F is conservative then 0=×∇

→F

Now 0000

222

=++=∂∂

∂∂

∂∂=×∇

→→→

→→→

→kji

zyx

zyx

kji

F

→∴F is a conservative vector field.

Surface Integrals

Definition: Consider a surface S. Let n denote the unit outward normal to the

surface S. Let R be the projection of the surface x on the XY plane. Let→f be

a vector valued defined in some region containing the surface S. Then the

surface integral of →f is defined to be dydx

kn

nfdsnf

RS

.

.

.. ∫∫∫∫ →

→

∧→∧→

=

Example 1;

Evaluate dsnfS

∫∫∧→. where

→→→→−+= kzyjxizF 2

and S is the surface of

the cylinder 122 =+ yx included in the first octant between the planes z = 0

and z = 2.

Solution: Given →→→→

−+= kzyjxizF 2

122 −+= yxφ

→→

+=∇ jyix 22φ

22 44 yx +=∇φ

=222 yx +

=2

The unit normal ∧n to the surface =

φφ

∇∇

= yjxiyjxi +=+

2

22

xyxzjyixkzyjxiznF +=

+

−+=→→→→→∧→

.. 2

INTEGRAL THEOREMS

(i) Gauss’s divergence theorem

(ii) Stoke’s theorem

(iii) Green’s theorem in the plane

Green’s Theorem

Statement:

If M(x,y) and N(x,y) are continuous functions with continuous

partial derivatives in a region R of the xy plane bounded by a simple closed

curve C, then

dxdyy

M

x

NndyMdx

Rc

∫∫∫

∂∂−

∂∂=+ , where C is the curve described in the

positive direction.

Verify Green’s theorem in a plane for the integral ( ) xdydxyxc

+−∫ 2

taken around the circle 422 =+ yx

Solution: Green’s theorem gives

dxdyy

M

x

NNdyMdx

Rc

∫∫∫

∂∂−

∂∂=+

Consider ( ) xdydxyxc

+−∫ 2

M = x – 2y N = x

1,2 =∂∂−=

∂∂

x

N

y

M

dxdyy

M

x

N

R

∫∫

∂∂−

∂∂∴

( ) ∫∫∫∫ =+RR

dxdydxdy 321

= 3[Area of the circle]

= 32rπ

=3. 4.π

= π12 ……………………(1)

Now ∫ + NdyMdx

We know that the parametric equation of the circle 422 =+ yx

x = 2 cosθ y = 2 sinθ

θθddx sin2−= , θθddy cos2=

( ) xdydxxxNdyMdx +−=+∴ 2

= ( )( ) ( ) θθθθθθθ dd cos2cos2sin2sin4cos2 +−−

= θθθθθ d22 cos4sin8sincos2 ++−

Where θ various from 0 to π2

( )∫∫ ++−=+∴π

θθθθ2

0

2 4sin4sincos2 dNdyMdxC

= ∫

+

−+−π

θθθ2

0

42

2cos142sin d

= ( )∫ −+−π

θθθ2

0

2cos262sin d

=

πθθθ 2

02

2sin26

2

2cos

−+

= ππ 122

112

2

1 =−+ …………………….(2)

∴From (1) and (2)

dxdyy

M

x

NNdyMdx

Rc

∫∫∫

∂∂−

∂∂=+

Hence Green’s Theorem is verified.

Example 2

Using Green’s theorems find the area of a circle of radius r.

Solution: By Green’s theorem we know that

Area enclosed by C = ∫ −C

ydxxdy2

1

The parametric equation of a circle of radius r is x = θθ sin,cos ryr =

Where πθ 20 ≤≤

∴Area of the circle = ∫ −−π

θθθθθ2

0

)sin(sin)cos(cos2

1drrrr

= ( ) θθθπ

drr∫ +2

0

2222 sincos2

1

= ∫π

θ2

0

2

2

1dr

= [ ] 22

0

2

2

1rr πθ π =

Example 3:

Evaluate ( )[ ]∫ −−c

xdydxyx cossin where c is the triangle with

vertices (0,0) ,( )0,2

π and )1,

2(π

Solution: Equation of OB is

02

0

01

0

−

−=−−

πxy

πx

y2=⇒

By Green’s theorem dxdyy

M

x

NNdyMdx

Rc

∫∫∫

∂∂−

∂∂=+

Here 1,sin −=∂

∂−=y

MyxM

N xx

Nx sin,cos =

∂∂−=

( )[ ] ( )dxdyxxdydxyxRC

∫∫∫ +=−−∴ 1sincossin

In the region R, x varies from x = 22

ππtoy

and y varies from y = 0 to y = 1

( ) =−−∴ ∫ xdydxyxC

cossin ( )∫ ∫ +1

0

2

2

1sin dxdyxy

π

π

= [ ]∫ +−1

0

2

2

cos

π

πyxx dy

= dyyy

∫

−+1

0222

cosπππ

=

1

0

2

422sin2

−+ yy

y ππππ

= 2

2

42

2 ππ

πππ

+=−+

Example 4

Verify Green’s theorem in the plane for

( ) ( )∫ −+−C

dyxyydxyx 6483 22 where C is the boundary of the region defined

by

X = 0 , y= 0, x + y =1

Solution: We have to prove that

dxdyy

M

x

NNdyMdx

Rc

∫∫∫

∂∂−

∂∂=+

M = xyyNyx 64,83 22 −=−

yx

Ny

y

M6,16 −=

∂∂−=

∂∂

By Green’s theorem in the plane

dxdyy

M

x

NNdyMdx

Rc

∫∫∫

∂∂−

∂∂=+

= ( )∫ ∫−1

0

1

0

10 dydxy

x

= ∫−

1

0

1

0

2

210

x

y

= ( )∫ −1

0

215 dxx

= ( )

3

5

3

15

1

0

3

=

−− x

Consider ∫∫∫∫ ++=+BOABOAc

NdyMdx

Along OA, y=0 , x varies from 0 to 1

[ ] 131

0

3

1

0

2 ===+∴ ∫∫ xdxxNdyMdxOA

Along AB, y = 1 - x dxdy −=⇒ and x varies from 1 to 0 .

( ) ( ) ( )[ ]dxxxxxxNdyMdxAB

∫∫ −+−−−−=+∴0

1

22 1614183

= ( ) ( ) 0

1

32

232

232

14

3

18

3

3

−+

−−−

−−− xx

xxx

= 3

82312

3

8 =+−−+

STOKE’S THEOREM

If S is an open surface bounded by a simple closed curve C and if a vector

function →F is continuous and has continuous partial derivatives in S and on

C, then ∫∫∫→→→→

=c

rdFdsnFcurl .. where →n is the unit vector normal to the

surface (ie) The surface integral of the normal component of →Fcurl is equal

to the integral of the tangential component of →F taken around C.

Example 1

Verify Stoke’s theorem for ( )→→→→

−−−= kzyjyziyxF 222 where S is the upper

half of the sphere 1222 =++ zyx and C is the circular boundary on z = 0

plane.

Solution: By Stoke’s theorem ∫∫∫→→→→

=sc

dsnFcurlrdF ..

( )→→→→

−−−= kzyjyziyxF 222

zyyzyx

zyx

kji

Fcurl

222 −−−∂∂

∂∂

∂∂=

→→→

→

= [ ] ( ) ( )→→→→

=++−−+− kkjyzyzi 100022

Here →→

= kn since C is the circular boundary on z = 0 plane

=∴ ∫∫S

area of the circle ∫∫=→→

S

dxdydsnFcurl .

= ππ =2)1( ……….(1)

ON z = 0, ∫∫∫→→→→

=sc

dsnFcurlrdF ..

On C, x = cos θθ sin, =y

θθθθ ddyddx cos,sin =−=

θ varies from 0 to π2

( )( ) θθθθπ

drdFc

sinsincos2.

2

0

−−=∴ ∫∫→→

= ( ) ∫∫ +−ππ

θθθθθ2

0

2

2

0

sinsincos2 dd

= ( ) ∫∫

−+−ππ

θθθθ2

0

2

02

2cos12sin dd

=

ππ θθθ 2

0

2

0 2

2sin

2

1

2

2cos

−+

−

= ππ =++−2

1

2

1 ……………(2)

∴From (1) and (2)

∫∫∫→→→→

=sc

dsnFcurlrdF ..

Hence stoke’s theorem is verified

Example 2

Verify stoke’s theorem for ( ) ( )→→→→

−+++−= kxzjyzizyF 42 where s is

the surface of the cube x = 0, x = 2, y = 0, y = 2, z = 0 and z = 2 above the xy –

plane.

Solution:

By Stoke’s theorem

∫∫∫→→→→

=sc

dsnFcurlrdF ..

Given ( ) ( )→→→→

−+++−= kxzjyzizyF 42

xzyzzy

zyx

kji

Fcurl

−++−∂∂

∂∂

∂∂=

→→→

→

42

= [ ] [ ] [ ]1010 −++−−−→→→kzjyi

= [ ]→→→

−−+− kzjiy 1

Hence Stoke’s theorem is verified.

Example 3:

Verify Stoke’s theorem for →→→→

++= kxjziyF where S is the upper half

surface of the sphere 1222 =++ zyx and C is its boundary.

Solution: By stoke’s theorem

∫∫∫→→→→

=sc

dsnFcurlrdF ..

Gauss Divergence theorem

Statement:

The surface integral of the normal component of a vector

function F over a closed surface S enclosing volume V is equal to the volume

integral of the divergence of F taken throughout the volume V ,

dvFdsnFVS

∫∫∫∫∫→∧→

∇= ..

Evaluate zdxdyxydzdxxdydzx 223 ++∫∫ over the surface bounded by z = 0 ,z

= h, 222 ayx =+

Solution:

16

3

22

1

4

3cos

2

0

4 ππθθ

π

==∫ d

2

3.

4adsnF

S

=∫∫→→