vector calculus - magadh university
TRANSCRIPT
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VECTOR CALCULUS
1. GRADIENT OF A SCALAR
2. DIVERGENCE OF A VECTOR
3. DIVERGENCE THEOREM
4. CURL OF A VECTOR
Dr. ANANT KUMAR SINHAASSOCIATE PROFESSORDEPTT. OF PHYSICSA.M. COLLEGE, GAYA
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1. GRADIENT OF A
SCALARis the temperature atSuppose T1x, y, z P1x, y, z ,
and T2x dx, y dy, z dz is the temperature at P2
as shown.
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The differential distances dx, dy,dz are the
components of the differential distance
vector dL :
dL dxax dya y dzaz
However, from differential calculus, the
differential temperature:
x y zdT T T
T dx
T dy
Tdz2 1
GRADIENT OF A SCALAR (Cont’d)
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But, dx dL ax
dy dLay
dz dL az
So, previous equation can be rewritten as:
zyx
zyx
a dLz
T a
T
ya
x
T
z dL
Ta dL
x ydT
Ta dL
Ta
GRADIENT OF A SCALAR (Cont’d)
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GRADIENT OF A SCALAR (Cont’d)
The vector inside square brackets defines the
change of temperature dT corresponding to a
vector change in position dL .
This vector is called Gradient of Scalar T.
For Cartesian coordinate:
zyxy z
Va
Va
xV
Va
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zz
GRADIENT OF A SCALAR (Cont’d)
For Circular cylindrical coordinate:
V V
a 1 V
a V
a
For Spherical coordinate:
r r
V V
a rr sin
1 Va
1 Va
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EXAMPLE
1
Find gradient of these scalars:
V e z sin 2x cosh y(a)
(a) U 2z cos2
(b) W 10r sin2 cos
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SOLUTION TO EXAMPLE 1
(a) Use gradient for Cartesian coordinate:
x y z
2e z cos2x cosh yax e z sin 2x sinh ya y
e z sin 2x cosh yaz
Va
Va
x y zV
Va
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SOLUTION TO EXAMPLE 1
(Cont’d)
z
2zcos2a 2zsin2a
2 cos2az
z
(b) Use gradient for Circular cylindrical
coordinate:
U U
a 1 U
a U
a
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SOLUTION TO EXAMPLE 1
(Cont’d)
(c) Use gradient for Spherical coordinate:
10sin2 cosar 10sin 2 cosa
10sin sina
r
1 Wa
1 Wa
r r sin rW
Wa
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1.DIVERGENCE OF A VECTOR
Illustration of the divergence of a vector
field at point P:
Positive
Divergence
Negative
Divergence
Zero
Divergence
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DIVERGENCE OF A VECTOR
(Cont’d)
The divergence of A at a given point P
is the outward flux per unit volume:
v
A dS
div A A lim s
v0
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DIVERGENCE OF A VECTOR
(Cont’d)
What is A dS ??s
Vector field A at
closed surface S
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s
front back left right top bottom
A dSA dS
And, v is volume enclosed by surface S
DIVERGENCE OF A VECTOR
(Cont’d)
Where,
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Az Ay
x y z A
Ax
For Circular cylindrical coordinate:
AzA
zA
1 1
A
DIVERGENCE OF A VECTOR
(Cont’d)
For Cartesian coordinate:
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Ar A r
r sin r sin
1 A sin 121
r 2r A
DIVERGENCE OF A VECTOR
(Cont’d)
For Spherical coordinate:
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EXAMPLE
1
Find divergence of these vectors:
P x2 yzax xzaz
r s in cosa cosar 2 r
(a)
(b) Q sina 2za zcosaz
(c) W 1
cosa
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18
(a) Use divergence for Cartesian
coordinate:
SOLUTION TO EXAMPLE 1
Pz Py
x y z
x2 yz 0 xzx y z
2xyz x
PPx
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coordinate:
1
2sin cos
1
Q
z
1 2 sin 1 2 z z cos
z
QQQ z
SOLUTION TO EXAMPLE 1
(Cont’d)
(b) Use divergence for Circular cylindrical
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SOLUTION TO EXAMPLE 1
(Cont’d)
(c) Use divergence for Spherical coordinate:
1
2
2cos cos
1
W
r sin
1 cos 1 rsin2 cos
r sin
cos
r 2r
W1 W sin 1
r sin r sin r W
r 2rr
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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.
A dS AdVV V
1.DIVERGENCE THEOREM
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EXAMPLE
1
3A vector field D a 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:
D dsS
DdVV
(a)
(b)
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SOLUTION TO EXAMPLE 1
(a) For two concentric cylinder, the left side:
D dS Dinner Douter Dbottom Dtop
S
Where,2 5
101
10 z0
2 5
4a ddz(a )
0 z0
Dinner 3a ddz(a )
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1602
20 z0
2 5
4a ddz(a )0 z0
10
10
2 2
z5
z0
Dtop 3a dd(az ) 0
2 2
Dbottom 3a dd(az ) 0
SOLUTION TO EXAMPLE 1
Cont’d)
2 5
Douter 3a ddz(a)
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SOLUTION TO EXAMPLE 1
Cont’d)
Therefore
D dS 10 160 0 0S
150
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SOLUTION TO EXAMPLE 1
Cont’d)
(b) For the right side of Divergence Theorem,
evaluate divergence of D
D 1 3 42
2
5 2 2
150
5
4 2
0z0
r1
So, DdV 42dddz
z0 0 1
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1. 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
when the area is oriented
the area
so as to
make the circulation maximum.
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an max
sCurl AA lim s
A dl
s0
Where,
CURL OF A VECTOR (Cont’d)
s ab bc cd da
A dl A dl
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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.
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For Cartesian coordinate:
CURL OF A VECTOR (Cont’d)
x y z
Ax Ay Az
ax a y az
A
zyx aa
Ax x y
Ay
z
Ax ax
Az z
Ay y
A Az
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a a az
z A A Az
A 1
za
aa
Az
z z
A
A
A
1 A
A 1 Az
For Circular cylindrical coordinate:
CURL OF A VECTOR (Cont’d)
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CURL OF A VECTOR (Cont’d)
For Spherical coordinate:
Ar rA r sin Ar2 sin r
a
ar a
1 A
a
a
a
r
rr
rA A
) Ar
1 (rA
r sin r
1 1 Ar A r sin
1 sinA
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P x2 yzax xzaz
r s in cosa cosar 2 r
(a)
(b) Q sina 2za zcosaz
(c) W 1
cosa
EXAMPLE
1
Find curl of these vectors:
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SOLUTION TO EXAMPLE 1
(a) Use curl for Cartesian coordinate:
zx y aa
0 0a x x2 y za y 0 x2 za z
x2 y za y x2za z
z x z
x y
Pz Px a
Px
Py
Py
y P
Pz
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z
z
zaaa
2
1 z sin 3a 3z cosa
1 3 2z cosa
z
sin a 0 0a
x y
1 Q Q
Qz z
Q
Q
z Q
1 Qz
SOLUTION TO EXAMPLE 1
(Cont’d)
(b) Use curl for Circular cylindrical coordinate
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(c) Use curl for Spherical coordinate:
a
a
a
aa
2 cos
cos11
r2 1 (r sin cos)
r r
r
rcos r2
r sin
1 cossin rsin cos rsin
) Wr
1 (rW
r
rr sin
W 11 sinW rW W rsin
ar
r
1 Wrr
SOLUTION TO EXAMPLE 1
(Cont’d)
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SOLUTION TO EXAMPLE 1
(Cont’d)
3 2cos 1 sina
cosa
cos2 sin a
r
r
rsin
r212r sin cos
sin a
r
rcos 2 r sin sin a
1 0 cosa1
rsin
r
r