GIVEN: a>0, The surface, : x+y+z = a, x ≥ 0, y ≥ 0, z ≥ 0. Consider the 3D field: F = (x, 2x, - x) Orient the surface so that its normal vector has a positive z - component. FIND: The flux of F through with the given orientation. x A Ω F = A = (a,0,0) B = (0,a,0) C = (0,0,a) B (x, 2x, - x) Y
GIVEN: a>0, The surface, : x+y+z = a, x ≥ 0, y ≥ 0, z ≥ 0. Consider the 3D field: F = (x, 2x, - x) Orient the surface so that its normal vector has a positive z - component. FIND: The flux of F through with the given orientation. x A Ω F = A = (a,0,0) B = (0,a,0) C = (0,0,a) B (x, 2x, - x) Y
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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for the first image attach please do the calculations similar to the second image attach
please please answer everything correctly I would really appreciate if you would answer
this is not a graded question
![[10] (2) GIVEN: a>0, The surface, Q: x+y+z = a, x ≥ 0, y ≥ 0, z ≥ 0.
Consider the 3D field: F
(x, 2x, - x)
=
Orient the surface so that its normal vector has a positive z component.
FIND: The flux of F through Q with the given orientation.
x A
2
Ω
A = (a,0,0)
B = (0, a, 0)
C = (0,0,a)
B
F = (x, 2x,- x)
Y](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe343d170-4423-4dcd-9c75-f5f0118e0ff9%2F95195c9f-e4ae-4de2-ad31-b323fd62449c%2F3xw7n5t_processed.png&w=3840&q=75)
Transcribed Image Text:[10] (2) GIVEN: a>0, The surface, Q: x+y+z = a, x ≥ 0, y ≥ 0, z ≥ 0.
Consider the 3D field: F
(x, 2x, - x)
=
Orient the surface so that its normal vector has a positive z component.
FIND: The flux of F through Q with the given orientation.
x A
2
Ω
A = (a,0,0)
B = (0, a, 0)
C = (0,0,a)
B
F = (x, 2x,- x)
Y
![[20] (2)
GIVEN: a > 0, The surface, Q2: x + y + z = a, x ≥ 0, y ≥ 0, z ≥ 0.
Consider the 3D field: F (x, 2x, 3x)
=
Orient the surface so that its normal vector has a positive z- component.
FIND: The flux of F through with the given orientation.
Parameterization of D. :
D= PROJ, :D
xy
= {(5₂₂) |05*{a-x}
y
Q
а-х-у
$(x,y)=(x, y, a-z-y)
⇒ = (1,0,-1)
y = (0,1,-1)
⇒x=(1,1,1)
canonical
||
3
a
X
parameterization
3
= 6 [−¾a³+ ½a³]
A
Z
Ω
A = (a,0,0)
B = (0, a, 0)
C = (0,0,a)
B
F = (x, 2x, 3x)
D=
FLUX = √ F·Ã$ = √₂ (2,22,3x)-(1,1,1) dedy
= √₂ 6x4 dy = 65° 1² 727144
бу
dy
y=a-x
.a
= 6 √ ²x (a-x) & = 6 √ ² x ² + ax de
→L
Y](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe343d170-4423-4dcd-9c75-f5f0118e0ff9%2F95195c9f-e4ae-4de2-ad31-b323fd62449c%2Fjvpitth_processed.png&w=3840&q=75)
Transcribed Image Text:[20] (2)
GIVEN: a > 0, The surface, Q2: x + y + z = a, x ≥ 0, y ≥ 0, z ≥ 0.
Consider the 3D field: F (x, 2x, 3x)
=
Orient the surface so that its normal vector has a positive z- component.
FIND: The flux of F through with the given orientation.
Parameterization of D. :
D= PROJ, :D
xy
= {(5₂₂) |05*{a-x}
y
Q
а-х-у
$(x,y)=(x, y, a-z-y)
⇒ = (1,0,-1)
y = (0,1,-1)
⇒x=(1,1,1)
canonical
||
3
a
X
parameterization
3
= 6 [−¾a³+ ½a³]
A
Z
Ω
A = (a,0,0)
B = (0, a, 0)
C = (0,0,a)
B
F = (x, 2x, 3x)
D=
FLUX = √ F·Ã$ = √₂ (2,22,3x)-(1,1,1) dedy
= √₂ 6x4 dy = 65° 1² 727144
бу
dy
y=a-x
.a
= 6 √ ²x (a-x) & = 6 √ ² x ² + ax de
→L
Y
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