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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 in a given direction is the rate of change of in the direction. It is given by the component of grad in that direction. The directional derivative of a scalar point function
(x,y,z) in the direction of a is given by
a
a..
Directional derivative of is maximum in the direction of . Hence the maximum directional derivative is grador Unit normal vector to the surface
If (x, y, z) be a scalar function, then (x, y, z) = c represents A surface and the unit normal vector to the surface is given by
Equation of the tangent plane and normal to the surface
Suppose a is the position vector of the point ),,( 000 zyx
On the surface (x, y, z) = c. If
++= kzjyixr is the position vector of
any point (x,y,z) on the tangent plane to the surface ata , then the
equation of the tangent plane to the surface at a given point a on it is
given by 0. =
gradar
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 is 0=
gradar
The Cartesian form of the normal at ),,( 000 zyx on the surface
(x,y,z) = c is
z
zz
y
yy
x
xx o
=
=
00
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
+
+
==
.
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=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
CurlF 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
Such that =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
+
+
=
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= ( ) ( ) ( )
+
+
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
0g
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
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= 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 +=
)2( 2 xzyxz
ky
jx
i +
+
+
=
= ( ) ( ) ( )xkxjzxyi 222 2 +++ At (2,-2, 3)
( ) )4()4(68
+++= kji
=
++ kji 442
63616164 ==++= Unit normal to the given surface at (2,-2,3)
=6
442
++ kji
=
++kji 22
3
1
6. Find the directional derivative of xyzxzyzx ++= 22 4 at (1,2,3) in the
direction of
+ kji2 Solution: Given xyzxzyzx ++= 22 4
)4( 22 xyzxzyzxz
ky
jx
i ++
+
+
=
= ( ) ( ) ( ) +++++++ kxyxzyxjxzzxiyzzxyz 842 222 At (1, 2, 3)
++= kji 28654
Given:
+= kjia 2
6114 =++=a
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=a
aDD ..
= 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 ++= and xzyx 2222
2 ++=
zz
yy
xx
2,2,2 111 ==
=
zz
yy
xx
2,2,22 222 ==
=
++= kzjyix 2221
++= kzjyix 22)22(2 At (o,1,2)
+= kj 421
++= kji 4222
Cos644416
422.42.
21
21
+++
++
+=
=
kjikj
2420
20
2420
164cos =+=
= 2420
20cos 1
=
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 log21 = and zyx +=2
2
z
x
zy
yz
x=
=
=
111 ,2,log
1,,2 2222 ==
=
zx
yxy
x
+= kz
kjyiz 2)log(1
= kj22
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Cos65
1
11414
12.
21
21 =+++
==
= 65
1cos 1
9. Find ( )nr2 Solution: ( )nr2 = ( )nr. = ( ) ( ) ( )nnn r
zkr
yjr
xi
+
+
= 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
=
rnr
n 2
Since
+=
udivuu .
( )
=
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
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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
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=
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 such that gradF =
( ) ( )z
ky
jx
ikxzjxyizxy+
+
=+++
222 34sin2cos
Integrating the equation partially w.r.to x,y,z respectively
),(sin 132 zyfxzxy ++=
),(4sin 22 zxfyxy +=
),(33 yxfxz +=
,4sin 32 Cyxzxy ++= is scalar potential
12. Prove that ).().(
=
BcurlAAcurlBBAdiv
Proof : ).(
=
BABAdiv
=
BAx
i
=
+
B
x
Ai
x
BAi
=
+
B
x
AiA
x
Bi
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=
+
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.
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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. Letf 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 += yx
+= jyix 22
22 44 yx +=
=222 yx +
=2
The unit normal n to the surface =
= yjxiyjxi +=+
2
22
xyxzjyixkzyjxiznF +=
+
+=
.. 2
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INTEGRAL THEOREMS
(i) Gausss divergence theorem
(ii) Stokes theorem
(iii) Greens theorem in the plane
Greens 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.
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Verify Greens theorem in a plane for the integral ( ) xdydxyxc
+ 2
taken around the circle 422 =+ yx Solution: Greens 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]
= 32r
=3. 4. = 12 (1) Now + NdyMdx We know that the parametric equation of the circle 422 =+ yx x = 2 cos y = 2 sin
ddx sin2= , ddy cos2= ( ) xdydxxxNdyMdx +=+ 2
= ( )( ) ( ) dd cos2cos2sin2sin4cos2 + = d22 cos4sin8sincos2 ++ Where various from 0 to 2
( ) ++=+
2
0
2 4sin4sincos2 dNdyMdxC
=
+
+
2
0
42
2cos142sin d
= ( ) +
2
0
2cos262sin d
=
2
02
2sin26
2
2cos
+
= 122
112
2
1 =+ .(2)
From (1) and (2)
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dxdyy
M
x
NNdyMdx
Rc
=+
Hence Greens Theorem is verified.
Example 2
Using Greens theorems find the area of a circle of radius r.
Solution: By Greens theorem we know that
Area enclosed by C = C
ydxxdy2
1
The parametric equation of a circle of radius r is x = sin,cos ryr = Where 20
Area of the circle =
2
0
)sin(sin)cos(cos2
1drrrr
= ( )
drr +2
0
2222 sincos2
1
=
2
0
2
2
1dr
= [ ] 22022
1rr =
Example 3:
Evaluate ( )[ ] c
xdydxyx cossin where c is the triangle with
vertices (0,0) ,( )0,2
and )1,
2(
Solution: Equation of OB is
02
0
01
0
=
xy
x
y2=
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By Greens 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
toy
and y varies from y = 0 to y = 1
( ) = xdydxyxC
cossin ( ) +1
0
2
2
1sin dxdyxy
= [ ] +1
0
2
2
cos
yxx dy
= dyyy
+1
0222
cos
=
1
0
2
422sin2
+ yyy
= 2
2
42
2
+=+
Example 4
Verify Greens 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
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dxdyy
M
x
NNdyMdx
Rc
=+
M = xyyNyx 64,83 22 =
yx
Ny
y
M6,16 =
=
By Greens 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
[ ] 13 1031
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
+
xxxxx
= 3
82312
3
8 =++
STOKES THEOREM
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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 Stokes 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 Stokes 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 .
= =2)1( .(1)
ON z = 0,
=sc
dsnFcurlrdF ..
On C, x = cos sin, =y ddyddx cos,sin ==
varies from 0 to 2
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( )( )
drdFc
sinsincos2.
2
0
=
= ( ) +
2
0
2
2
0
sinsincos2 dd
= ( )
+
2
0
2
02
2cos12sin dd
=
2
0
2
0 2
2sin
2
1
2
2cos
+
= =++2
1
2
1 (2)
From (1) and (2)
=sc
dsnFcurlrdF ..
Hence stokes theorem is verified
Example 2
Verify stokes 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 Stokes theorem
=sc
dsnFcurlrdF ..
Given ( ) ( )
+++= kxzjyzizyF 42
xzyzzy
zyx
kji
Fcurl
++
=
42
= [ ] [ ] [ ]1010 ++kzjyi
= [ ]
+ kzjiy 1
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Hence Stokes theorem is verified.
Example 3:
Verify Stokes theorem for
++= kxjziyF where S is the upper half surface of the sphere 1222 =++ zyx and C is its boundary. Solution: By stokes theorem
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=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:
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16
3
22
1
4
3cos
2
0
4
== d
2
3.
4adsnF
S
=
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