2 The divergencen foravector function F=fvertfelettored is determined as follows: V · F = 1 [ahnhof 2) + dm₂h3F₂) + Xhaha 3) = /1_) (1²fr) + 1 a[sine fa) + 1 DF] h2 h₂h3d91 rsine do dhah = (1 + 2(1²71) + 1 a(sinofa) 892 893 do
2 The divergencen foravector function F=fvertfelettored is determined as follows: V · F = 1 [ahnhof 2) + dm₂h3F₂) + Xhaha 3) = /1_) (1²fr) + 1 a[sine fa) + 1 DF] h2 h₂h3d91 rsine do dhah = (1 + 2(1²71) + 1 a(sinofa) 892 893 do
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
How did they work out the divergence. I sent a picture of the previous steps.
![blas:
C traced out
LA
$
(FAR)
2
#2² +41² / 2
1
0
4
€
the course C-traced out
R
%
F
5
V
Zk, along the cune C track out
T
G
» Example: Spherical Polar Coordinates
Sor
Let us know determine the quantities above the most the standard coordinate system, sphorical polar Coordinates:
x=rsinecas &
y=rsingsing re[0,00, 0€ [0,1]],
[0,271).
Z=rcose
for a vector of the genued form : √=x[r₂0₂0) T+y(r₂0₂0) J+ z(r, 0, 0) F (₂)
Owe Start by Calculating:
a
V = Sine Cosør +Sines ing J Hosek
de = rosecoso I + rosesing J &-rsinsk
of = -rsingsing ++rsinecaseJ
2 The hame coefficients are therefore:
From these we can now
get the Lamé coefficients.
hr = |ar| = |Sinecospi+SinesindJHosek = 1
fal
he = 1301= | rose casar trosesingJ4#-rsinekl=r
hø=134|=|-rsinesing it rsine cospJ) = rsine
3 From these we get the whit vectors:
er=fror - Sinecosoi + sinesing5+ CoseR
1
le=no de = cosecose + cosesing J-Sinek
1 J
ef=h₂ 3g = -singi Hose J
hø
Owe can convince ourselves that these vectors are indeed an orthogonal curvilinear coordinate system, i...
ei es= Si holds for i, j=1,2,3 for example, as follows:
er.eo-Sinecosecastsinecosesino-sine Case=0
er-er-Sin²ecos²³p+ Sitesin²+ cos²e=Size+cos³0=1
15 Using our equations from earlier, the square of the line element is:
2
2
ds² = h²dq² th ² dq ² th 3d q3=dr²tr²de²tr²sin²edo ²
the area element on the r surface (??) is
do - h₂h3dq₂dq3 = r²sineded
and the volume element (??) is
2
(2)
dT = h₂ h₂h3dq1d2₂dqz = r²³sin @drdedø
16 The gradient for a scalar function
(r,0,0) is determined as follows:
Do=1 does + 1 do ² + 1 223, VV= dve +1 ve + 1 dved
h₂ 892
hz 892
из даз
rsine dø
dr
rdo
pg dn
enter
el)
E i5
en](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F12d366b4-7e59-4d81-8f2b-965d1a83d6f8%2Ffacc3427-6e13-417a-a534-dc86a3d6d3ef%2Fndwsawf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:blas:
C traced out
LA
$
(FAR)
2
#2² +41² / 2
1
0
4
€
the course C-traced out
R
%
F
5
V
Zk, along the cune C track out
T
G
» Example: Spherical Polar Coordinates
Sor
Let us know determine the quantities above the most the standard coordinate system, sphorical polar Coordinates:
x=rsinecas &
y=rsingsing re[0,00, 0€ [0,1]],
[0,271).
Z=rcose
for a vector of the genued form : √=x[r₂0₂0) T+y(r₂0₂0) J+ z(r, 0, 0) F (₂)
Owe Start by Calculating:
a
V = Sine Cosør +Sines ing J Hosek
de = rosecoso I + rosesing J &-rsinsk
of = -rsingsing ++rsinecaseJ
2 The hame coefficients are therefore:
From these we can now
get the Lamé coefficients.
hr = |ar| = |Sinecospi+SinesindJHosek = 1
fal
he = 1301= | rose casar trosesingJ4#-rsinekl=r
hø=134|=|-rsinesing it rsine cospJ) = rsine
3 From these we get the whit vectors:
er=fror - Sinecosoi + sinesing5+ CoseR
1
le=no de = cosecose + cosesing J-Sinek
1 J
ef=h₂ 3g = -singi Hose J
hø
Owe can convince ourselves that these vectors are indeed an orthogonal curvilinear coordinate system, i...
ei es= Si holds for i, j=1,2,3 for example, as follows:
er.eo-Sinecosecastsinecosesino-sine Case=0
er-er-Sin²ecos²³p+ Sitesin²+ cos²e=Size+cos³0=1
15 Using our equations from earlier, the square of the line element is:
2
2
ds² = h²dq² th ² dq ² th 3d q3=dr²tr²de²tr²sin²edo ²
the area element on the r surface (??) is
do - h₂h3dq₂dq3 = r²sineded
and the volume element (??) is
2
(2)
dT = h₂ h₂h3dq1d2₂dqz = r²³sin @drdedø
16 The gradient for a scalar function
(r,0,0) is determined as follows:
Do=1 does + 1 do ² + 1 223, VV= dve +1 ve + 1 dved
h₂ 892
hz 892
из даз
rsine dø
dr
rdo
pg dn
enter
el)
E i5
en
![-
Ⓒ
V · F = 4
"The divergence pra vector function F=Frert fele+ Forex is determined as follows:
minife) + (toks/2) + Xhorof=) = (1, 2(13fr) + 1 ) (Sinefa) + 3 F
Chahaf3)
1
h2 h₂h3 Ldq1
1
rsine dø
892 893
dr
Isine do
h1 h₂h3
8) The court of F is determined from our earlier equation:
VXF= 1
ez
d
h3e3
2 d
892 193
= 1
091
1F₂h1 F₂h2 F3h3|
Fr
= [0(SineFo) - Ofe ] er + 1 Jfr - Xrfd) eo + X(Ifo) - dfr e ed
Feat
dørsine sine dø orr
dr
r²sino
et re. rsinded
ad
or de dø
Fr vfe rsinoFø
2
The Laplace Operator
Finally we can determine that the Laplace operator acting on Hr, e, Dis:
3²√√² = 1
(h₂h3
d h₂h3 √ √ + (hihs JV + 2
av
h1h₂h3/09₁ h₁ 821 092 h22 192 193 h3 623
h2₂
hibe dy
біз
= 1 [3 (Prince)) + 2 (sing (√) +) (1 52
sino
av
izcing or
sine do
р
ente](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F12d366b4-7e59-4d81-8f2b-965d1a83d6f8%2Ffacc3427-6e13-417a-a534-dc86a3d6d3ef%2F3lkhlof_processed.jpeg&w=3840&q=75)
Transcribed Image Text:-
Ⓒ
V · F = 4
"The divergence pra vector function F=Frert fele+ Forex is determined as follows:
minife) + (toks/2) + Xhorof=) = (1, 2(13fr) + 1 ) (Sinefa) + 3 F
Chahaf3)
1
h2 h₂h3 Ldq1
1
rsine dø
892 893
dr
Isine do
h1 h₂h3
8) The court of F is determined from our earlier equation:
VXF= 1
ez
d
h3e3
2 d
892 193
= 1
091
1F₂h1 F₂h2 F3h3|
Fr
= [0(SineFo) - Ofe ] er + 1 Jfr - Xrfd) eo + X(Ifo) - dfr e ed
Feat
dørsine sine dø orr
dr
r²sino
et re. rsinded
ad
or de dø
Fr vfe rsinoFø
2
The Laplace Operator
Finally we can determine that the Laplace operator acting on Hr, e, Dis:
3²√√² = 1
(h₂h3
d h₂h3 √ √ + (hihs JV + 2
av
h1h₂h3/09₁ h₁ 821 092 h22 192 193 h3 623
h2₂
hibe dy
біз
= 1 [3 (Prince)) + 2 (sing (√) +) (1 52
sino
av
izcing or
sine do
р
ente
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