chapter

VECTOR CALCULUS
TRODUCTION
I quantity, which is a function of the
position of a
called a scalar
point function or a vector
according as the quantity is a scalar or
Aiht
function

vector.
pof velocity are respectively scalar and vector noint
inOn. When a point function is defined at every point of
tunregion of space, then that region is called afield.
r Differential Operator V(del)
V is defined V=7
Iieoperator
as
=ij+k
ôy oz
Gradient of ascalar point function
iafxV2) be defined and differentiable at each
point (x,y,z)
ta Grtain region of space. Then the gradient of f
ifr gradf is defined as
written as

ôy
-+kÔz

-+j +k
ôy Oz
sa scalar
point function, then Vð
Vf is a vector point
3 2 E n g i n e e r i n g M a t h e m a t i c s

Wector Calculus 3.3

Angle betubetween two surfaces
Result 3.
Level Surfaces =c, ¢, =c is given
field over a region R. The by cos -
scalar
firyz) be a fayz)
Let
equation of the type
satisfying an in three Divergence of a vector point function
constitute a family of surfaces
dimens Let be any given
constant)
called level Surfan Definition:
differentiable
sDace. The surfaces
of this family are
function. Then the divergence of V written as vector point
V.V or div
of a scalar point function ie defined as div V=V=|++Sz.
Directional derivatives Ox ôy
Let fx. y, z) define ascalar field in a region
Definition: í=v)+v7 +v,k then div?My,oy
region. Suppose is a pointi
let Pbe any point in this Solenoidal Vector
neighborhood of Pin the direction of a
region in the
Avector is said to be solenoidal if div j=0//
lim f(O)-f) if it exits is calle
unit vector a. Then 0’P PQ Curl of a vector point function
directional derivative of f,y, 2) at P in the direction ofi Definition: Let F be a vector point function. Then curl F
written as
Result 1. The directional derivative of a scalar field f
VxF is defined as curlF =)ix
point P(s, y, z) in the direction of a unit vector âis gie Ox

Vf å. The directional derivative is maximum i

direction of grad fand its maximum value is grad f|· If F-Fi
+E*+Ek then curlF =
ôx ôy ôz
Result 2. Unit normal to a surface f= c is given by
rrotational Vector
A Vector F is said to be irotational if VxF=0. In this case
we can
represent E= V¢, where is called scalar potential.
Note 1. Physical meaning of V
3.4 Engineering Mathematics
Vector Calculus 3,5
VAe-23), =i(-8+6)+j(49)+E(4)
If
r = y + + , 7 + y * is
a vector
point function
moving fluid
at
represet
the pointo =-2i +4j+4k
velocity ofa
instarntancous
per unit IV=V4+16+16 =6
the
). then VI
representsrate of
loss of fluid
volume
that point. 2,2,
+j+*
meaning of curlF 3 3 3
Note 2. Physical a y, z)
point (x,
inear velocity of the
the axis with Evample3
If F represents about a fixed COnsa
rotates
rigid body that that point represents 2 directional derivative of
angular velocity , then curlF at Findthe f(,y,z) =x'y +4x2 at the
point(1,-2,-1) in the direction of the
..a...taA.
vector 2i -j-2k.
Example 1 Solution: The required directional derivative is Vfà.
=ry+y'r+z', find yo at the
point (1, 1. I) Here
Ifoa. y.z) =2+-j-2k.

S+=iry+y'* +z') 2i-j-2k

Soluion: Let Vý= Ox So la =V4+1+4=3 and â=
la 3
=i(21y +y)+7(*' +2yx) +k(22)
Now gradf =Vf= ax -7yz+4az')
Vguy =i(2+I) +j1+2) +k(2)
-3+ +37 +2k =+(2ryz +4z') +j(x'z) +k(*'y+&xz)
Example 2. Hence (V/), 3 =i(8)+j(-I) +k(-10).
surface x´y + 2xz =4 .The directionalderivative
Find a unit normal vector to the level
the point (2,-2, 3).
=-(87-j-10) (2i -j-2k)_16,!20
37
V where = 3 3 3 3 3
Solution: The unit normal vector ' =
Example 4
2xz -4. Find the directional derivative of f(xy.z) =x-2y'+4* a
ne point (1, 1,-1)in the direction of 2+ +j -k.
Now Vo-E+-y+
ôx 2x)
Solution: The directional derivative = f.a.
=+(2.xy +22) +7(r') +k(2x)
 Vector Calculus 3.7
3.6 Engincering Mathem
r e q u i r e
directionalI
d
derivative
The
2i+j-k -47.+12 ). 4i-2j+ k
Here a = a=(27
V21
(ad/).

Now
+4*')+7-3y'+ 4s')+*S -2y' +42 28 2
2 1 - V32 1 .
V2721
v-2y'
-8E
=i(2r)+j(4y)+k(8:) and (VDa )=2i -4j Example6 .
Directional derivative direction from the point (1, 1,-1) is the
4-4+8
In
what

of f(x, y, z) = x? -2y +42 a maximum?

directional
=
derivative of Also find
ofthis maximum directional derivative.
8 the value

Solution:
We have grad f = 2xi -4yj +8zk =2i -4j-8k at
1,-1).
Example 5 the point (1,
Find the directional derivative of the Tunctig The directional derivative of fis a maximum in the direction
funcion
f(r.y.z) =r-y'+22 at the point P(1,2,3) in the direction of grad f =27 -4j-8k.
of the line PO, where Qis the point (3,0,4). The maximum value of this directional derivative
Solution: Given f(x,y, z) =x*-y' +22. =\grad f|=|27 -47 -sk
Hence gradf = J7k= 2x+ -2,yj +4zk =2i -4j +1) =J4+16+64
oy ôz
=2V21
at the point (1, 2, 3)
Also PO = Position vector of Q- Position vector of P Example 7
=(5+ +0j+4k)-(7 +27 +3k) =4+ -2j +k Show that the vector field F given by

If à be the unit vector in the direction of PO the =(-yz)i +(y* - zx)7 +(z? -xy)k is irrotational. Find a
a= 4i-2j+k 4i-2j+k scalar such that F=V¢
V16+4+1 W21
3.8 Engineering Mathematics

Vector Clculus 3.9

k
curiÅ = oz
Solution: We have Then curlF =
- xy\
ôy Oz
=(-rtr)i--(-y+y)j +(-2+g 'cos x+z 2ysin x-4 3xz+2

or+2)-ysin x-4)-7
irotational.

vector field
Fis -(3x2 +2)
The =i

(or)
Let F= V¢
+('-z)j
+(z²-y)k=ôx
i: +
+k
| ôx
-(2ysinx-4)
ôy +
(r-)7
y+f.z)
=+(0-0)-)i(32?-32') +k(2y cos x-2ycos x)
=r-yz whence =:3
Then =0
Fis irrotational.
o=y'-zx whence ¢= --xyz +f(*,z)
3
oy Ifo is scalar potential let F=V6.
='-y whence o=-xyz+j;(*,y)
3
requirements.
We choose ó satisfying all the above o' cosx+z)i +(2ysin x- 4)7 +(3xr' +2) =7P72,¡e
y
r+y'+ --xyz +C.
3 =y´ coS x+ z': =2ysin x-4; =3x2+2
ôy
Example 8
Integrating partially, we get,
Prove that the vector (y cosx+z)i +(2ysinx-4)j +(3 =y' sin r+z'x+fy,z)
is irotational and find its scalar potential. (SRM - Su
2010) =y» sinx-4y+ f;(x, z)

Solution: Let F=(y' cos x +z')+ +(2ysin x-4)7+(3x¢+ o= xz' +2z + f(x, y)

=y' sinx+z'x-4y +2z +c

 Engineering Mathematics

Vector Calculus 313

.F=i(2r) +j(2y)-F is -2(x - ) + (y t 3) + 3(z- 2) =
i4 It's
cquation

)
3-6=0
Now V4 =i -2.+ytJz
+2+

-k (or)

(2.-1.2) =47-2j Jz +,|=0.

(Tá) at
2 x - y -

(or)
and T4,= j , , -ôb - 2xi +(2y)j +(22)X Erample 3 . .
+j oz
y
c o n s t a n t s a,b,c, so that
47-27+4k. Findt
the
Ta2-1 = (r+2y+
az)i - z)j +(4x+cy +2z)k may be
+(bx-3y-z)j
cos0= F=
between g,, then Fori these values of a, b, c, find its scalar
If 8' is the
angle i r T O t a t i o n a l .

potential.
(47-27-)\47 -27 +4k) Solution: We know that F Fis irrotational if curlF =0.
cOsO VI6+4+1/16+4 +16
16+4-4
. VxF=0’

V21N6 3V21 (ieXCurtY=0) |=0

..0= cos
8 x+2y +4z bx-3y -z 4x+cy + 22|

Example l2 i4r+ey +22)- Oz (b-3y-z)}

Find the equation of the tangent plane to the surface
-i ôx
(4r+cy- 2z) - -(r+2y+az)}
x+x'y=z-lat the point (1,-3, 2).
Solution: Let ¢= x'+x'y-z+l=0. +k (br -3y-z)- ôy(+2y+ az)}=0
ie) i{c+l) - j(4-a)+ k(b-2) =0
=+(2 +2y)+7')+ k(2xz-1)
ie) c +1 = 0,4- a = 0, b-2 = 0
» (V0n-3 =-27 +j +3k. ie) a= 4, b= 2 and c = -1

Direction ratios of the normal to the plane is (-2, 1.3). Jsing these values in F, we get
Required tangent plane passes through (1,-3, 2). F=(r+2y +4z)+ +(2x-3y- z)j +(4x-y +22)k
316 Engneering Mathematics Vector Calculus 3.17
=l+1+1ls2
- ( r ) t ) + ( 2 )

Then dy F
Oz
=
url
gradg=Vx(VØ)

ax ôy oz

VxF = ôz ax ôy oz
Also curlF-

+k
Oxôz Ozôx Oxôy ôxôy
=0
Here F=xi+N +à
and V ( r ) = i ) - i m . F.Fi+Fj +F;k.
'=x'+y'+z'. ( )Let
r=r'ty' +z* =F (ie)
Or
2r=2x.
Differentiate partially w.r.to x, w get, Ôx curl F=(VxF)=
ôr_y ôr z
Similarly. F
y r'zr
Now grad(r) =V(r") OF, OF
=i +k
ôy 0z Ôx Ôz ôx Ôy

Now
-(É +yË +zk) =nrF OF,
diy curl F= V.(Vx F) = oz
+
J ôy ôz Oz ôx ôy
Example 16
Prove that (i) curl gradó = 0(i) div curl F-V(VxF) =0
(üi) io +) =¢divi +V¢.7 ôx-y ôxôz Oyôz Ôyâx ôzâx ôzêy
=0
Solution: )We know that grad = Vo=7+79. +k
ôy
S.18 EnqineeringMathematics

Vector Colculus 3.19

av F =)i. curl(øu) =Voxk+
(ii) We know that curl k.
that
k n o w

We
Vr"xF+r curl F
=nr(FxF) +r".0 (as
Fx?
ôx ôx =0 =0,curl T=9)
r fisiro curllr" F=0,1. e)
yalue of n. whatever be the
Ou
r"F
is solenoidal as swell as
ôx Hence

Axample 18 .
irrotational when n =-3.
=Vo. k+o div+ (Note: .(mb) =(ã).b Prove that f(r)rF is an irrotational vector.
Example 17..
Solution:

Find the value of n, if rF is both solenoidal and irrotatiog, Consider I(fr) F)=VVf(r)xF+fr)
curl
curl F (1)
when F=xÉ +yË +zk
Ox
Solution: Consider dir"F)=VF+r"divr

We know that div F=3, Vr"=nrF _W++yj+zh) =IV,

From (1) divrF)=(nrF. F)+r(3) Further curl F =0.
=nr+3"
From (1), curl(f(r)F) = Fx)+fr),0 =0+ 0=0.
=nr" +3r"
=(n+3)r" fr)r is an irrotational vector.
’ ris solenoidal means div(r" F) =
0, which is poss
only when n = -3.
 3.20 Engincering Mathematics Vector Colculus 3.21
Erample 1 9 ] . . . . as =x*yz
l)r"-2 whereeY=xi +yj+zk,
Prove that °=n(n + -z) +k(-»)
V= V(VO). =4i -j-4k
Solution: We know that

Hcnce V'=S(r) = V(nrF). - yz)

at (1,1,1)
Since V{oz) = Vé.k +Ùdiv k,
VnrF+nr div
we have V.(nrF) =
requiredi angle then
Now V(nr) =)ir=n(7-2) y I Oisthe.

, (47-j -4k)) (27-j-h

-)in (-2) r3 -Sn n-2) pd xi osl a7 VI6+1+16 V4+1+1
8+1+4 13
Now put (2) in (1), we have V33V6 V98
V(nr)= nln-2)r*F. F+nr2 3
13
. =cos
=nr(n-2+3) =n(n +I) 2 (3V22
Example 21
Example 20
Prove that 4=(2yfyz)i +(4y +zr)j -(6z-xy)k is solenoidal
Findthe angle between the normals to the surface x =
the points (1, 1, I)and (2, 4, 1). 8wellas irçeational. Also find the scalar potential of å.
Solution: We know that
Solution: Let ñ=unit nomal to x -yz =0 at (2, *,4, 1).
1bi:2r+ yz) +(4y +zx)+(6z+3y)
Ôz
(ie) + =
=2+4-6=0.
Ais solenoidal.
3.22 Engineering Mathematics
Vector Calculus 3.23
integrating (1), (2) and
(3) w. I. to
Artially
rspectivelY, weg e r x, y, z
constant independent of x

|2r+)* 4y+z* -6z+xy| d=2y+yZ+a constant independent of y

-iS)-r)-a-t+)-,
fz- +yz +a constant independent of z.
Hencea possible form of is =x*+2y' -3z +xyz +c.
Erample22
-ia-)-jy-y) +k(z-z)
Provethatthe directional derivative of =xy'z at (1, 2, 3) is
=0
along the direction f 9) +37+k. Also
maximum

find the
.. Ais irotational. maximum directional derivative.
scalar potential of Solution: Given =x'y'z.
Let y.z) be the

ôy ôz Hence Vp=| 7+j +-k

ax ôy ôz

Hence¡ e.;,¡e-(2x+= yz)+ +(4y +z)j -(6z

- -i -7y'z) +7(y')
ôy +y)
xy oz

Equating components by i,j,k -(3x'y'z)+ +(2r'yz)7 +(ry')½

=(3x4x3)i+12j +4k
= 2x+ yz (0)
ôx =36i +12j +4k =4(9) +3j +k)
We know that the directional derivative of is maximum
=4y+zx (2)
ôy along the direction Vo.
hence it is maximum along the direction 4(97 +37 +k).(. e)
=-62+xy (3
ôz along (9+ +3j +k).
Magnitude of this vector = 4/9² +3+1 =4/91 and this S
lIne maximum directional derivative.
3.24 Engincering Mathematics
Vector Calculus 3.25
EXERCISE

Questions)
Y=(r+y+I)7.
perpendicularto curl F
+j-(r+y)k, show that F is
Answer
PART- A(Short
surface Find the maximum directional derivative of
1. Findthe unit normal
to the
i-27+2k
42y'+z), |3.
at the point (1.1,1) |Ans:/ia)
=xyz
(1-1,2)| dns: 3 Findthe value of a, ififF=(ay-2i+(a-2)xj+-a)zk,
14.
2. IfVo= yz+ +zj +yk, find [Ans :¢ = is irrotational.

the point are irrotational, prove that (kxv)

If ø(r, y.z)=r'y-2y', find Vý at
15. If Tandy

3. solenoidal.
is
(1.-1.2), Ans:-27 +337 - 24Ã| that F= (sin y+ z)i +(rcos y-z)j
that the 16.
Show
+(*-y)k is
Show irrotational.

F=(r+3y)i +(y-32)j +(*-22)k is solenoidal When is said to

a vector sajd be (i) solenoidal, (ü)
the 17.
5. Show that irrotational?

Ä=3r'yi +(r - 2y2')j +(32-2y'z)k is irrotational 18.

Find the unit normal to the surface x.-3xyz +2 +l=0
If a is aconstant vector, find div(F x). Ans :0]
6.
at the point (1.1,1). :-j
Ans:

7. Determine the constant 'a' so that the vector

10 If r=F|,
19. where F=xi +yj +zk, prove that
T=(r+3y)i +(y-22)j +(r+ az)k is solenoid
r
[Ans:a =-2]
8. Evaluate ivf where 7= 2x'zi -y'j+jyi . Find gradf, where fis given by f=r'-y'+r' att tie
(Ans :2xz(2- y)]. point (1,-1,2) Ans :7+ -3j +4k|
9. If u=*-' +4z, show that V'u =0 at the
21. What is the greatest rate of increase of u =y
10. If F=ryi +xj +2yzk, verify that div curl F=0 point (1,0,3)? [Ans :9]
11. Find the equation of the tangent plane to the surfac 2. Prove that grad f(u) =f'(w) gradu
z=t+y' at the point (2,-1,5) (Ans :4x-2y-z=5]
3.26 Engineering Mathematics

Vector Calculus 3.27

vo= 21yz'i +x*' +3ryz'. find
PART- B
of the
If
px,y, z) given
Find the
directional
derivative
funs, 28.

that ol.-2,2) =4. |Ans ip=x'y +20]

23.
point (2,-1,1) in
¢='+yr at the
- y +.
+4=0 at
direcçon o Ifr=F|
where F isthe position vector of
the point (x,
nomal to the surface xlogz the püin 29.
with respect to the origin,
15 prove that
1.21)|Ans:nits. v'fr)=f(r)+

and b, so that the

24. Find the constants a suria 0. Findthe directional derivative of a=xy+ yz +zx
in the
5x'-2vz-9x=0 and ax'y +bz =4 may
direction offvector i++2j +2k at (1,2,0). Ans: 10
orthogonally atthe point (1,-1, 2). [Ans :a =4,b=) 3

Eind the angle of intersection at (4,-3, 2) of spheres

25. Find the angle between the surfaces r'-y' 31.
2'tz? = 29 andx +y' +z +4r-6y-8z -47 =(0.
and xy +yz zx = 18 at the point (6, 4, 3).
|Ans:cos V19/29
Ans : = cos -24
W5246) 32. Find Vo and (V when =(r+y'+z')eey' ty
26. Show that
[Ans :(2-r)e '7.(2-r)e "r|
k=(2r' +8ry'z)+ +(3x'y -3xy)+ -(4y'2' +2x'z)k s
3. If u=3x'y and v= xz- 2y, then find grad[ (grad
not solenoidal, but ¼= yz'7 is solenoidal.
w).(grad v)]
27. Find the directional derivative of =
2xy+z at t
point (1,-l, 3) in the direction of 7 +2j+) |Ans :(6y2-4r)i +6x2j +12xyzk|
14)
Ans:
328 EngineeringMathematics

Vector Calculus 3.29

and curli? where f= grad(x
Find diy?
34. e that e(2y+3z)i +2j +3k| is irrotational and
[Ans :diyf= 6(r+y +z), curf =0] 40.
find its
scalar potential. [Ans :e (2y +32)+c].
Findthe directional derivative of p=xy + yz'
35. vector field
in the direction of PO w Given the
point
P(2.-1,1)
where Qi 41.
F=(r'-y'+-2xz)
2xz)i +(xz - xy +yz)j +(z' +x)k. Find
point (3. 1, 3),|4ns:
Show that the vectors given by
curlD. curl v at
Find the values of A and u, if
36. the suri z(2,.-3) and R(2,3,12) are orthogonal.
r-yz =(2+2)x and 4xy+z'=4 cut orthogo
5 Afluid motion is given by V=(+)i +(z+)j +(rtyk. Is
at the point (1,-1,2). | Ans :d=. 42.
.his motion irrotational? If so, find the velocity
37. Prove that yz+ +(r2? -1)j +2(xyz -1)k is potential.(Ans :yes, (p =y +yz+z].
irrotatig
and find its scalar potential. Ans :qp=xy´-y-2+e)
43.
In what direction from (3, 1,-2) is the directional
38. Find the angle between the surfaces
xlog z =y'-la derivative p=xyz is maximum? Find the
*y=2-z at thepoint (1,1,1). Ans :cos
magnitude of this maximum.
39.
J30 [Ans :96/19|
Find the directional
derivative of o= 2x'+3y' +z
1, 3) in the al 44. Establish the relation: crl curlf= grad div f -V'.
direction of the
4
vector i
Ans
3.30 Engineering Mathematics
Vector Calculus 3.31
INTEGRATION OF VECTORs components form the above line integral
Therefore
án is
writtena s
Line Integral
vector function and a curve AB.
Let F(x,y.:) be a
Note1. lodr. where is ascalar point functions and [Fxur
integrals,

line
arealso

IfF F represents
the variable force acting on a particle
Note
2.
2. If
B

done = (Fd
along arc AB, then the total work 4
X

Fig. 1 a fY
Note 3. If
FaF is
F represents the velocity of aliquid then VFdr

Line integral of a vector function F along the curve 4 the circulation of F round the curve C.
called
defined as the integral component of F along the tangen
the curve AB. Note 4. If FdF =0. then the field F is called consecutive
Component of Falong atangent PT at P =Dot product ol (i. e) VxF= 0)
and unit vector along PT -F.{ dr is a unit vector B
ds ds alo
tangent PT; Note 5. If F.dr is said to be independent of path then
4

Line Integral =)F. d

=) F from Ato B
ds Balong the curv

Line Integral = dr
Surface Integral: Let F be a vector point function and S be
ds
C the given surface. Surface integral of a vector point function
Since F =xi +yj F over the surface S is defined as the integral ethe
+zk, therefore dr =
Fd+ =(F+ +F,j + dxi +dyj +dek component F along the normal to the surface.
=Fdr+ F,dy F,k)(dk+ +dyj + dzk )
+F,dz
 Vector Calculus 3.33

ds Nowalongthe C, :y=x’dy =2xd

Fig.2 x=0

Component of Falong the normal = F.n where nis 7

the +2x')dk = 3 8
normal vector to an element dS and ñ= gradf dS = dxdy o 12
gradf (nky Example 2 ....
Surface Integral of Fover S=XFâ= ||(Fi)ds
S
.F-(3+6y)i -14yzj +20xz´k ,evaluate FdF , whereC
Velume Integral C

Let Fbe a vector point function and volume straight line joining (0,0,0))to (1,1,1).
Venclosed isthe
closed surface. by
Solution: The equation of the straight line joining (0, 0, 0)
The volume integral =|Fdv and (1, I, 1)are
x-0_ y-0_2-0=t (say
Example l 1-0 1-0 1-0
Evaluate Fdf , Then along C, x =1, y=,z=t
where F=x++y7 and curve C is the arc Also Y=xi
of the parabola y=x +yj +zk dr =(+ +j +*)d1
in the x-y plane
from (0,0) to (1,1)
Solution: In the x y Also along C, F=(3 +6r)+ -147+20rk
dr - dzi +dyË plane, we have
have F= xi +) At (0, 0, 0), t = 0 and at (1, 1, ), t=1
Fdr =(+
+y)(dk) +dyj) =*dx +y'dy |FdF = [(3/ +6t)-14/ +20r`]dt
 3.34 Engineering Mathematics

Vector Calculus 3.35

14
=1+3-+5 42 2
Jo
3
-
=914 13
= 303
3 3

Erample4|.
Example 3
Y=(2xy +z')7T+i+3xz'k
+x*7. is a conservative force
Find the total work done
in moving a particle in aforce
that

the scalar potential. Find also the work done in

Show

given by F=3xy+ -527 +10xk along the

Find
in this field from
4
field.
an
object (1,-2, 1)to (3, 1, 4).
r=r+l. v= 2r'.z =t' from = lto t = 2 moving
i
The field F will be consecutive if VxF=0.
Solution.:

Solution: Let Cdenote the arc

of the given curve
from ts
to t= 2. We h a v e V x F =
ôx ôy ôz
Then the total work done F.dF
2.xy +2 3xz
- |(3xyi - 5zj +10xk ).(dki+dyj +dek)
C -(3xz)-(2ry +z)}
|ôy
- |(3ydr -5zdy +10xdz)
2
+k. ()2y+)}
ôy
dy -+10xdt
dt dt dt
=+(0-0)-j(3z*-3z) +k(2r-2x)
2 =0
-3 +1X2'(2)-(5X4) +10( +13) Therefore F is consecutive force field.
Let F= Vo » (2y +z)i +x*7+3xzk=7
2

- |(12r'+12 - 20r +30r ôx y

+30r)dt
2

=(12F +10 +12 -= 2.xy +z' whence ¢=x'y+z'x+ fo.:) (1)

+30)dt
 Engineering Mathematics

3.36 Vector Calculus 3.37

+x(1+ cos y)j 1-[dxi + dyË l
=x whence g=x´y +s(x, z)
Oy
x(1+ cos y)dy]
dp312² whence = xz'+f(r, y)
d) +xdyl
-V[(sin yd + x cos
Then (1), (2) and (3)
each represents o. These
agree it,
choose

f(r,y) = xy 2n
So.) = 0, f,(1,2) =z'x,
-Tala cost sin(asint)] + Jacosta cos tdt
.:ý=x'y+xz +c 2r

(3,1,4) (3,1,4) -facostsin(asin )l6"+a' fcos rdt

Work done = F.dr +
(1,-2,1) (1,-2,1) 2 2r

- 2
a+ cos 21)dt
i4)
(i. e) work done -xytx3 Ji,-2,1) -sin2
12
Jo

= (9+192)-(-2+1)

= 202. the curve

If F=2y) - j +xk, evaluate GFxdr along
Example 5 z = 2costfrom t =0 to t=
F=sin yi + x(1+ cos v)ì x= cost, y = sini,
2
A vector field is given by
Evaluate the line integral over the circular path given b
x +y' =d,z= 0. Fx dF =2y
Solution.: Let
Solution: The parametric equation of the circular path are r - dx dy dz
a cost, y=a sint, z = 0 where t varies from 0 to 2T. Since tht
particle moves in the xy plane (2 = 0), we zd)k
=(-zdx- xdy)+ + (xde- 2ydz)+ +(2ydz +
can take
F=xÉ + yË ,so that dr = dxi + dyj
 Engincer

3.38

t)d|7
In terms oft,
cost(cos ---
sin /)dt - 1)d]+
FxdF =[-2
cost(-2
sin dy = dx
t(-sin t) -
2 sin i(-2 )dt]k line y = x,
+[cos I(-sin sttraight

/(cost)dt +
2 cos
sin)7ldt
the

+[2sin sin´7- cost a l o n g

[Far
cos1)i +(4 (i)
=
-
-[(4cos1 sin
done
work
4sin-costt sin () Jdt o - y + y)cy-(2y'+y)dy

cos' )i +(4:
st sint- - Si n t

Sin
2
2 3
3
Jo

done is the same.

that the work
o b s e r v e

We
Example73
when a force
F=(x-y²+x)+
-(2xy +y) Example 8 being the
Find the work done along the parahak F =zi+xj + yk and C
plane from (0, 0)
Fdr , given that
moves particle in the xy different when the
path is Caluate
y' =x. Is the work done
r =
c o s i +sin tj + tk from t= 0to t = 2I.
the c u r v e
straight line y x? arc of
dr
-(2xy + y)j, in the xy plane: +sin tj + tk =-sinti +cost/ +k.
Solution: As F=(x-y' +x)+ Let F =cos ti dt
Solution:
= 0, F =xÉ +y

dF =d+ +dyj and F.dF = (*-y +x)dx-(2xy + y)dy .

(i)along the parabola x=y', dx =2ydy. 2

.Work done =W= |Far 7+ cosij + sintk).4-sint+ +cosij +k)}dt

C

1
2

- Jo-y'+y'X2ydy)-(2y' +y)dy = (sint+ cos' i+ sin ()dt

3.40 Engineering Mathemat1cs
Vector Calculus 3.41
2 2r

- (t sin tdt +Jcos' tdt +[sin ta, -sin!

+cos
3
tcos kit
21 -feos sintssin3 Jo
2 1+cos 2t
=-Jtd-cosi)+ | 0
2
dt +
0
|sin tdt =-1-| = -2

27
-2
done
=
Work

E a m p l e1 D . .

=27+1= 3n as sin2 = 0. cos2z = 1, s the

C iis
the curve x = -t,y= 2t, z = fromt =0
sino and
lod
cos) = 1. Ifd=
2yz*
lne integral
evaluate the
el.
Example9
d= 2xyz, x =r, y = 21, z = so that
Let
Find the work done by the force F=(*+y)i +(ß+zi Salution:

2r21.
=4
when it moves a particle along the upper half of the +(2dt)j +(3rdr)k
r'+y'=l from the point (-1, 0) to the point (1, 0).
Solution: As the circle is in the xy plane z = 0. Wem
x = cost, y = sint, 0<t<I - (s"7+8rj+
12" )dt l+12
dx
-=-sint, dy = cost, dr = dxi +dyi, bË47+k
4-
dt dt 11
F=(r'+y')) +(*+j+ yk and
FaY =(x+y )dr +x'dy 2 Example
FF , where Cis the
IF=(2r+y)+ +(3y-x)+, evaluate
Work done =|FaY = (-sin t+cos' tcos t)dt straight lines from (0.
C consisting of the
urve in the xy - plane
': cos3A=4cosA-3 cos d O)to (2, 0) and then to (3, 2).
\(-sint +cos' )dt 3 1 Chas been shown Ue
of integration
cos" A=-cos A+-cos34|| Soltion: The nath
4
ligure.
342 Engineering Mathematics

Vector Calculus 3.43

Erample.
work done
the
by the
Faet-2yy).i+(l-x*)7+(e +z)k when it force
Find

moves a particle
1-)to
(2, 3, 0) along any path.
(0
fom
the path of is integration
aSSume that the work done depends on endnot points
G o l u t i o n :S i n c e

given only,
we may
for

curlF = 0.
w h i c h

Fig. 3 check cif Fis conservative.

Wefirst
It consists of the straight lines OA and AB. We have
[Fa= (2x+y)+ +8y-n)+)\(àtÉ+dù)
C C ôy ôz

=|(2r+ yldr +(3y-x)d e'z-2xy I-** e+z

Now along the straight line OA, y = 0, dy = 0and =+(e'+z) - (e'+z)

oy
from 0to 2.
The equation of the straight line AB is
2-0
Ox La-r)-e-25)}
y-0-*-2)
3-2 (i. e) y=2x-4 =(0-0)7-(e-e')j+(-2x+2x)B
Along AB, y = 2x-4, dy = 2dx andx varies from 2 to 3 =0
2
F isconservative force.
3

[FaF =((2r+0),dz +0]-+ [(2r +2r-4)dr +(6x-12-x)2d:]

2
3
La F=Vo (i.e) (e'z-2xy)+ +((-j+(e'+)k J7
-2-(u4x-28J:
=e'z-2xy: =1-x';
= 4+7 = |1. ôy ôz
3.44 Engineering M a t h e m a t i c s

Vector Calculus 3.45

Integrating. -q(xi-j + 2yk ).dák+ +dyj t dÅ)y
o=e':-r'y+fo.z)

-(udt -záy +2yut)

C

Z=0, d: = )
p=y-r'y+f(,z)
y=X,
0sxS1,
G;,
on
o=e'z++f(r.y) 1
2

We choose p=e'z-xy+y ++c

2
onC, 0szSl, x=1. y=1, ,dydx = 0, dy = 0
C;, 0
on

Now work done, W= Far =(0)

Jo,1,-1) -lez-tyayn2
12I2 0,, y=z=x, dy = dz= dx
=(0-12+3)- -1+1+ on C;,

=-9.
1 19 W,= ((xd -xdx +2xd) =2 |xd =-1 1

2 2
Total work done =W, +W, +W,
1 ,3
=-+2-1=.
Find the work done by the force F=xi -j +2yk ins 2 2
displacement along the closed path C, consisting of segma
C, C; and C;, where Evample 14 |
on C,, 0sxsl, y= x, z =0 Evaluate (änds, where A=(x+y')+ -2yj +2yzk and Sis
on C, 0szsl, x=l, y=l the surface of the plane 2x+y+22 = 6 inthe first octant.
on C,, 12x20,y=2X.
Solution: A unit vector normal to the surface S is given by.
Solution: Total work done
C
346 Engineering Mathematics Vector Calculus 3.47
dxdy

n=
Ve where ¢= 2x+y+2z =6 Hence
S

36-2x
-[|2y(3-)dydk
J4+1+4 3 3 0 0

2 76-2x
3
dx

drdy where Ris the 3

projection of s . - f(3-xX6-2x)'de

xy - plane. The region Ris bounded by x axis, y - axis,

2x+y = 6, z =0. in: 3

=4|(3-x)'a

-4
4(-1) Jo

=-(0-81)=81.

Evample 15

Fig. 4 Evaluate (|Añds, where A=zi +j-3y'zk and S is the

Now Ä. ñ=|(*+y')+-2gj +2yzk 1, 2
surface of the cylinder x* +y' =16 included in the first octant
2
between z =0 and z=5.

2 4 (6-2x-y
6-2x-y Solution: A vector normal to the surface S is given by
4
P' +y') =2xÉ +2yj
0-)
348 Engineering Mathematics Vector Calculus 3.49
dydz dhydz
unit vector
normal to surface
ñ=a R
X

xi +yi 4*
2xi +2yj
54
4

00
- as x'+y'=16
4 5

then +4z)dz =
-(8+4 2
Let Rbe the projection of Son yz -plane,
dhydt
= 40+ 50

= 90

Examplel6.
B'

5 Eraluate |Fds , where F=yzi +zxj +xyk

and Sis the part
4
sphere +y'+z'=1 that lies in the first octant.
of the

Fig. 5 Solution: Let / = ||FRds

The region Ris OBB'O' enclosed by y = 0to y =4 and 2=l
to z = 5
where =x+y' +z' =1
1
Now i ñ=i-xi 2xi +2yj +2zk =xi+yË +zk
4
J4(r' +y' +z')
A.ñ=(zi +rj-3y' zk)x++j
4 4 as (, y, z) lies on S
1
72ty=x(y+2) 4
3.50 Engineering Mathematics
Vector Calculus 3.5I
2 4-2x 8-4x-2y
1= l(oz+ +zy +syB )\(a7 +yj +zk )ds =|3ryz:
S
S
S o l u t i o n :L e t x=0 y=0 =0
45x ydzdyd
2 4-2x
dudy where Ris the projection in the y- = 45
8-4x-2y
dydk
R x=0 y=0

2 4-2x
bounded by the circle
x'+y=] and lying in the
= 45
x=0 y=0
8-4x-2ykhydk
quadrant.

=45 (8-4r)-21
2
d:
3
22

B
y =45|(4-
3
2x}d =128

Example18.
X

2x-3z)i +2x}j -4xk, then evaluate fV.Fd

Fig. 6

where Vis bounded by the planes x = 0, y = 0,z = 0,and

dxdy +y+z = 4.
- 3ydaáy = dy
wlution: Let V =(2*-3z) +(-2.xy)+(-4r) =4x-2x=2x
Z
R 0 Jo

3 3|2 3
22 4 Jo 8
2 2-x 4-2x-2y

ExaDple 17 Hence ij.Fdv =||2xdayde =| 0 0 0

2rudzajd
Evaluate llpdv, where p
=45xy and Vis the closed origin 2 2-x 22-x
4-2x-2y
bounded by the planes 4x+2y+z = 8, x=0, y= 0, 2 =0. -|| 2x(2)%
0 0
=||2x(4-2r-2y)ldyd
00
3.52 Engineering Mathematics
Vector Calculus 3.53
22-x i e 3 y i - y ' j , ealuate
FdF , where Cis the
-||(4x(2-1)-4ryldsd
0 0

the parabola y = 2x from (0, 0) to (1. 2).

of
2 2-r arc

0
Ans:|6
2
4 IF=(4y - 3x'z')7 +2xj-2x'zk, show that FaF is
independent fthe path C.
2

Define
the eline integral ofof aa vector point function.
5
2 State the
neceSsary and sufficient condition for the line
=2(4x-4r +)d =2 2r2-4 4
6. BB

aporal .ar to be ndependent of the path of

A
8
=28-4
3 |3 integration.

What is meant by conservative vector field?

7.
EXERCISE
8. Evaluate (xdjy -yds around the circle r'+y=.
1. If fÖ)=i +(¢ -2)j+(3 +s), find the ra
[Ans:2r]
9, Find the circulation of F round the curve C, where
2. If Y=ti-j+{t-1)k and a=21+ +6tk , evalua F=y+ +zj +xk and C is the circle +y'=1.:=0.
ev

10.If d=(5xy -6x')+ +(2y - 4x)7, evaluate AdF where

CIs the curve y = in the xy - plane from the point
(.1) to (2,8). [Ans :35.