Vector calculus
55
VECTOR CALCULUS 1.10 GRADIENT OF A SCALAR 1.11 DIVERGENCE OF A VECTOR 1.12 DIVERGENCE THEOREM 1.13 CURL OF A VECTOR 1.14 STOKES’S THEOREM 1.15 LAPLACIAN OF A SCALAR
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Transcript of Vector calculus
VECTOR CALCULUS
1.10 GRADIENT OF A SCALAR
1.11 DIVERGENCE OF A VECTOR
1.12 DIVERGENCE THEOREM
1.13 CURL OF A VECTOR
1.14 STOKES’S THEOREM
1.15 LAPLACIAN OF A SCALAR
1.10 GRADIENT OF A SCALAR
Suppose is the temperature at ,
and is the temperature at
as shown.
zyxT ,,1 zyxP ,,1
2P dzzdyydxxT ,,2
The differential distances are the components of the differential distance vector :
dzdydx ,,
zyx dzdydxd aaaL Ld
However, from differential calculus, the differential temperature:
dzz
Tdy
y
Tdx
x
TTTdT
12
GRADIENT OF A SCALAR (Cont’d)
But,
z
y
x
ddz
ddy
ddx
aL
aL
aL
So, previous equation can be rewritten as:
Laaa
LaLaLa
dz
T
y
T
x
T
dz
Td
y
Td
x
TdT
zyx
zyx
GRADIENT OF A SCALAR (Cont’d)
The vector inside square brackets defines the change of temperature corresponding to a vector change in position .
This vector is called Gradient of Scalar T.
LddT
GRADIENT OF A SCALAR (Cont’d)
For Cartesian coordinate:
zyx z
V
y
V
x
VV aaa
GRADIENT OF A SCALAR (Cont’d)
For Circular cylindrical coordinate:
zz
VVVV aaa
1
For Spherical coordinate:
aaa
V
r
V
rr
VV r sin
11
EXAMPLE 10
Find gradient of these scalars:
yxeV z cosh2sin
2cos2zU
cossin10 2rW
(a)
(b)
(c)
SOLUTION TO EXAMPLE 10
(a) Use gradient for Cartesian coordinate:
zz
yz
xz
zyx
yxe
yxeyxe
z
V
y
V
x
VV
a
aa
aaa
cosh2sin
sinh2sincosh2cos2
SOLUTION TO EXAMPLE 10 (Cont’d)
(b) Use gradient for Circular cylindrical
coordinate:
z
z
zz
z
UUUU
a
aa
aaa
2cos
2sin22cos2
1
2
SOLUTION TO EXAMPLE 10 (Cont’d)
(c) Use gradient for Spherical coordinate:
a
aa
aaa
sinsin10
cos2sin10cossin10
sin
11
2
r
rW
r
W
rr
WW
1.11 DIVERGENCE OF A VECTOR
Illustration of the divergence of a vector field at point P:
Positive Divergence
Negative Divergence
Zero Divergence
DIVERGENCE OF A VECTOR (Cont’d)
The divergence of A at a given point P is the outward flux per unit volume:
v
dS
div s
v
A
A A lim0
DIVERGENCE OF A VECTOR (Cont’d)
What is ?? s
dSA Vector field A at closed surface S
Where,
dSdSbottomtoprightleftbackfronts
AA
And, v is volume enclosed by surface S
DIVERGENCE OF A VECTOR (Cont’d)
For Cartesian coordinate:
z
A
y
A
x
A zyx
A
For Circular cylindrical coordinate:
z
AAA z
11A
DIVERGENCE OF A VECTOR (Cont’d)
For Spherical coordinate:
A
r
A
rAr
rrr sin
1sin
sin
11 22
A
DIVERGENCE OF A VECTOR (Cont’d)
EXAMPLE 11
Find divergence of these vectors:
zx xzyzxP aa 2
zzzQ aaa cossin 2
aaa coscossincos12
rr
W r
(a)
(b)
(c)
18
(a) Use divergence for Cartesian coordinate:
SOLUTION TO EXAMPLE 11
xxyz
xzzy
yzxx
z
P
y
P
x
P zyx
2
02
P
(b) Use divergence for Circular cylindrical
coordinate:
cossin2
cos1
sin1
11
22
Q
zz
z
z
QQQ z
SOLUTION TO EXAMPLE 11 (Cont’d)
SOLUTION TO EXAMPLE 11 (Cont’d)
(c) Use divergence for Spherical coordinate:
coscos2
cossin
1
cossinsin
1cos
1
sin
1sin
sin
11
22
22
W
r
rrrr
W
r
W
rWr
rrr
It states that the total outward flux of a vector field A at the closed surface S is the same as volume integral of divergence of A.
VV
dVdS AA
1.12 DIVERGENCE THEOREM
EXAMPLE 12
A vector field exists in the region
between two concentric cylindrical surfaces
defined by ρ = 1 and ρ = 2, with both
cylinders extending between z = 0 and z = 5.
Verify the divergence theorem by evaluating:
aD 3
S
dsD
V
DdV
(a)
(b)
SOLUTION TO EXAMPLE 12
(a) For two concentric cylinder, the left side:
topbottomouterinnerS
d DDDDSD
Where,
10)(
)(
2
0
5
01
4
2
0
5
01
3
z
zinner
dzd
dzdD
aa
aa
160)(
)(
2
0
5
02
4
2
0
5
02
3
z
zouter
dzd
dzdD
aa
aa
2
1
2
05
3
2
1
2
00
3
0)(
0)(
zztop
zzbottom
ddD
ddD
aa
aa
SOLUTION TO EXAMPLE 12 Cont’d)
Therefore
150
0016010
SDS
d
SOLUTION TO EXAMPLE 12 Cont’d)
SOLUTION TO EXAMPLE 12 Cont’d)
(b) For the right side of Divergence Theorem,
evaluate divergence of D
23 41
D
So,
150
4
5
0
2
0
2
1
4
5
0
2
0
2
1
2
zr
z
dzdddVD
1.13 CURL OF A VECTOR
The curl of vector A is an axial (rotational) vector whose magnitude is the maximum circulation of A per unit area tends to zero and whose direction is the normal direction of the area when the area is oriented so as to make the circulation maximum.
maxlim0
a
A
A A ns
s s
dl
Curl
Where,
CURL OF A VECTOR (Cont’d)
dldldacdbcabs
AA
CURL OF A VECTOR (Cont’d)
The curl of the vector field is concerned with rotation of the vector field. Rotation can be used to measure the uniformity of the field, the more non uniform the field, the larger value of curl.
For Cartesian coordinate:
CURL OF A VECTOR (Cont’d)
zyx
zyx
AAAzyx
aaa
A
zxy
yxz
xyz
y
A
x
A
z
A
x
A
z
A
y
AaaaA
z
z
AAAz
aaa
A1
z
zz
AA
z
AA
z
AA
a
aaA
1
1
For Circular cylindrical coordinate:
CURL OF A VECTOR (Cont’d)
CURL OF A VECTOR (Cont’d)
For Spherical coordinate:
ArrAArrr
r
sinsin
12
aaa
A
a
aaA
r
rr
A
r
rA
r
r
rAA
r
AA
r
)(1
sin
11sin
sin
1
EXAMPLE 13
zx xzyzxP aa 2
zzzQ aaa cossin 2
aaa coscossincos12
rr
W r
(a)
(b)
(c)
Find curl of these vectors:
SOLUTION TO EXAMPLE 13
(a) Use curl for Cartesian coordinate:
zy
zyx
zxy
yxz
xyz
zxzyx
zxzyx
y
P
x
P
z
P
x
P
z
P
y
P
aa
aaa
aaaP
22
22 000
(b) Use curl for Circular cylindrical coordinate
z
z
zzz
zz
z
z
y
Q
x
z
Q
z
aa
a
aa
aaaQ
cos3sin1
cos31
00sin
11
3
2
2
SOLUTION TO EXAMPLE 13 (Cont’d)
(c) Use curl for Spherical coordinate:
a
aa
a
aaW
22
2
cos)cossin(1
coscos
sin
11cossinsincos
sin
1
)(1
sin
11sin
sin
1
rr
r
r
r
rrr
r
r
W
r
rW
r
r
rWW
r
WW
r
r
r
rr
SOLUTION TO EXAMPLE 13 (Cont’d)
SOLUTION TO EXAMPLE 13 (Cont’d)
a
aa
a
aa
sin1
cos2
cossin
sin
2cos
sincossin2
1
cos01
sinsin2cossin
1
3
2
r
rr
rr
r
rr
r
r
r
1.14 STOKE’S THEOREM
The circulation of a vector field A
around a closed path L is equal to the
surface integral of the curl of A over
the open surface S bounded by L that A
and curl of A are continuous on S.
SL
dSdl AA
STOKE’S THEOREM (Cont’d)
EXAMPLE 14
By using Stoke’s Theorem, evaluate
for dlA
aaA sincos
EXAMPLE 14 (Cont’d)
SOLUTION TO EXAMPLE 14
Stoke’s Theorem,
SL
dSdl AA
where, and
zddd aS Evaluate right side to get left side,
zaA
sin11
SOLUTION TO EXAMPLE 14 (Cont’d)
941.4
sin11
0
0
60
30
5
2
aA zS
dddS
EXAMPLE 15
Verify Stoke’s theorem for the vector field
for given figure by evaluating: aaB sincos
(a) over the semicircular contour.
LB d
(b) over the surface of semicircular contour.
SB d
SOLUTION TO EXAMPLE 15
(a) To find LB d
321 LLL
dddd LBLBLBLB
Where,
dd
dzddd z
sincos
sincos
aaaaaLB
So
202
1
sincos
2
0
2
0
0
00,0
2
01
LB
zzL
ddd
4cos20
sincos
0
0,200
2
22
LB
zzL
ddd
SOLUTION TO EXAMPLE 15 (Cont’d)
202
1
sincos
0
2
2
00,0
0
23
r
zzL
ddd
LB
SOLUTION TO EXAMPLE 15 (Cont’d)
Therefore the closed integral,
8242 LB d
SOLUTION TO EXAMPLE 15 (Cont’d)
(b) To find SB d
z
z
z
zz
a
aaa
a
aa
aaB
11sin
sinsin1
00
cossin1
0cossin01
sincos
SOLUTION TO EXAMPLE 15 (Cont’d)
Therefore
82
1cos
1sin
11sin
0
2
0
2
0
2
0
0
2
0
aaSB
dd
ddd zz
1.15 LAPLACIAN OF A SCALAR
The Laplacian of a scalar field, V written as: V2
Where, Laplacian V is:
zyxzyx z
V
y
V
x
V
zyx
VV
aaaaaa
2
For Cartesian coordinate:
2
2
2
2
2
22
z
V
y
V
x
VV
For Circular cylindrical coordinate:
2
22
22 11
z
VVVV
LAPLACIAN OF A SCALAR (Cont’d)
LAPLACIAN OF A SCALAR (Cont’d)
For Spherical coordinate:
2
2
22
22
22
sin
1
sinsin
11
V
r
V
rr
Vr
rrV
EXAMPLE 16
Find Laplacian of these scalars:
yxeV z cosh2sin 2cos2zU
cossin10 2rW
(a)
(b)
(c)
You should try this!!
SOLUTION TO EXAMPLE 16
yxeV z cosh2sin22
02 U
2cos21
cos102 r
W
(a)
(b)
(c)
Yea, unto God belong all things in the heavens and on earth, and enough is
God to carry through all affairs
Quran:4:132
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