SF• 25 = √ aw and 11 (x, 2x, 3x) • (-2) Ix dy -√3x dx dy 3х бабу =0 D 41 (BY SYMMETRY) x (0,0,b) a y
SF• 25 = √ aw and 11 (x, 2x, 3x) • (-2) Ix dy -√3x dx dy 3х бабу =0 D 41 (BY SYMMETRY) x (0,0,b) a y
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
not a graded question. how do you get the flux of the first
Transcribed Image Text:5.0 Solution continued
√√ F•£$ = £F·43 + [F• 18 + SF • 45
ow
aw
3₂W
W
√ 7.25 = = √ (²
S F.
·S(x, 2x, 3x) • (-2) St dy
-√3x
3х бабу =0
(BY SYMMETRY)
and
&: DC R² → 2₂ WCR ³₂
$(0,₁z) = (acno, a sim 0, Z)
3 = (-asino, a cao, 0)
= (0,0,1)
숟
→ ||
1 x $ $ = a-sin core o
0 1
= a(
a (coso, sin 0, 0)
Egl Batchelor Math 03 HOME-WORK (for F.E.)
X
b-
Z
(0,0,6)
y
D=
a
D.
y
2TC
Points away from interior of W
as is required.
Page 24 of (40)
Transcribed Image Text:5.0 5.0: Solution
DIVERGENCE THEOREM: V.Fav = √ F•d$
aw
GIVEN: W is the right circular cylindrical solid W
of radius, a.
height, b
bottom disk on the xy-plane
top disk in the plane: z = b
and axis of symmetry is the z-axis.
The field F = (x, 2x, 3x)
SHOW: Divergence theorem is true
for this field, F and solid, W.
X
Hence,
Z
(0,0,b)
a
a₂W
Ə3 W
= bottom disk
Φ
Egl Batchelor Math 03 HOME-WORK (for F.E.)
=
=
lateral cylindical surface
top disk
2₂ W = lateral surface of W
Q: DCR²
-a₂W CR³
Þ(0, z) = (a cos , a sin, z)
F = (x, 2x, 3x) ⇒>D•F
Z
b
11
11
£•Flv = √(₁1&v = V() =
W
D
35 + 25 + 35
F3
əx
dy
1 + 0 + 0
1
πa² b
2π
(EASY)
Next, we will show: [ Ƒ•d$ = ♫a²b
Jaw
Page 23 of (40)
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