4 For SSXF).ds S (4.) Verify Stokes' Theorem for the field F = (xz, xy, 3xz) over theportion of the plane 2x + y + z = 2 lying in the first octant. LHS ①F =<1-+, 2+, o> FCF)=<(+)10, (1-+)2+, 3(1-t) .0) = <0,2+-2+², σ> 5' = <-1, 2,0> F.F'= 4t-4t² ⇒ SF₁dr₁ = Set-t²)d= 0 - 4 [€] = 4 (4-3)=4(+2= 3/ F F= <0, 2-2+, 2+> > F(*) = <0,0,0> SF.& = 0 F = <+, 0,2-²+> 05+61 F(F) = <+ (2-2+), t.0, 3+ (2-2+)) * <2+-2+², 0, 6+-6+²> F = <1, 0, -2> Fir' = 2t-2+² - 12t+12€² = 100+10t 1 [Foli = j*+tQoL = - 110 (½ ½ - ½ 2 ) = 10 (- ½ 7 = 10 SF ·√r = 0 + 0 + 0 C = +0+ 513 RHS: √xF= 24 dz xz ху 3x2 S = <0, x-37, Y> Z-2-2x-Y ⇒ ˜¯xÊ > < 0, × -3 (2-2x-4), Y) -> = <0, x-6+6x+34, Y> = <0, 7 x + 34-6, 47 1 = (2x, z₁, 1) = (<2,1,1> PXF-ñ (2·0) + (1-(7x+34-6)) + (1.4) =7x+44-6 70-2-24 = SSEXF.d F 12-2x (7X+44-6)dy dx 0 2-27 dx = So (7x+9 (2-2x) +2 (2-2x)² ]ox = 5'' (-6x² + 18x1024) dx = > = -2x² + 5 x² - 4x1'; =-2+5-4=-1
4 For SSXF).ds S (4.) Verify Stokes' Theorem for the field F = (xz, xy, 3xz) over theportion of the plane 2x + y + z = 2 lying in the first octant. LHS ①F =<1-+, 2+, o> FCF)=<(+)10, (1-+)2+, 3(1-t) .0) = <0,2+-2+², σ> 5' = <-1, 2,0> F.F'= 4t-4t² ⇒ SF₁dr₁ = Set-t²)d= 0 - 4 [€] = 4 (4-3)=4(+2= 3/ F F= <0, 2-2+, 2+> > F(*) = <0,0,0> SF.& = 0 F = <+, 0,2-²+> 05+61 F(F) = <+ (2-2+), t.0, 3+ (2-2+)) * <2+-2+², 0, 6+-6+²> F = <1, 0, -2> Fir' = 2t-2+² - 12t+12€² = 100+10t 1 [Foli = j*+tQoL = - 110 (½ ½ - ½ 2 ) = 10 (- ½ 7 = 10 SF ·√r = 0 + 0 + 0 C = +0+ 513 RHS: √xF= 24 dz xz ху 3x2 S = <0, x-37, Y> Z-2-2x-Y ⇒ ˜¯xÊ > < 0, × -3 (2-2x-4), Y) -> = <0, x-6+6x+34, Y> = <0, 7 x + 34-6, 47 1 = (2x, z₁, 1) = (<2,1,1> PXF-ñ (2·0) + (1-(7x+34-6)) + (1.4) =7x+44-6 70-2-24 = SSEXF.d F 12-2x (7X+44-6)dy dx 0 2-27 dx = So (7x+9 (2-2x) +2 (2-2x)² ]ox = 5'' (-6x² + 18x1024) dx = > = -2x² + 5 x² - 4x1'; =-2+5-4=-1
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.3: Lines
Problem 30E
Related questions
Question
Please review and solve the following problem. The screenshot listed already has some work done and the correct answer listed. Please solve the problem and include an explanation of how the work was solved. Also, Please make sure to double check the answer provided matches up with the screenshot and the work is properly formatted so I am able to follow along. Thanks :)
![4
For SSXF).ds
S
(4.) Verify Stokes' Theorem for the field F = (xz, xy, 3xz) over theportion of the plane 2x + y + z = 2
lying in the first octant.
LHS ①F =<1-+, 2+, o>
FCF)=<(+)10, (1-+)2+, 3(1-t) .0)
= <0,2+-2+², σ>
5' = <-1, 2,0>
F.F'= 4t-4t² ⇒ SF₁dr₁ =
Set-t²)d=
0
- 4 [€] = 4 (4-3)=4(+2= 3/
F
F= <0, 2-2+, 2+>
> F(*) = <0,0,0> SF.& = 0
F = <+, 0,2-²+>
05+61
F(F) = <+ (2-2+), t.0, 3+ (2-2+))
* <2+-2+², 0, 6+-6+²>
F = <1, 0, -2>
Fir' = 2t-2+² - 12t+12€²
=
100+10t
1
[Foli = j*+tQoL
=
-
110 (½ ½ - ½ 2 ) = 10 (- ½ 7
= 10
SF ·√r = 0 + 0 + 0
C
=
+0+
513
RHS: √xF=
24
dz
xz
ху
3x2
S
= <0, x-37, Y>
Z-2-2x-Y
⇒ ˜¯xÊ > < 0, × -3 (2-2x-4), Y)
->
= <0, x-6+6x+34, Y>
= <0, 7 x + 34-6, 47
1 = (2x, z₁, 1) = (<2,1,1>
PXF-ñ (2·0) + (1-(7x+34-6)) + (1.4)
=7x+44-6
70-2-24
=
SSEXF.d
F
12-2x
(7X+44-6)dy dx
0
2-27
dx
= So (7x+9 (2-2x) +2 (2-2x)² ]ox
= 5'' (-6x² + 18x1024) dx
=
>
=
-2x² + 5 x² - 4x1';
=-2+5-4=-1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4effdf1a-67d0-42fb-b95c-7c4b54ad0624%2Fce18c992-1001-403e-83da-b481613f57ac%2Fzit9tra_processed.png&w=3840&q=75)
Transcribed Image Text:4
For SSXF).ds
S
(4.) Verify Stokes' Theorem for the field F = (xz, xy, 3xz) over theportion of the plane 2x + y + z = 2
lying in the first octant.
LHS ①F =<1-+, 2+, o>
FCF)=<(+)10, (1-+)2+, 3(1-t) .0)
= <0,2+-2+², σ>
5' = <-1, 2,0>
F.F'= 4t-4t² ⇒ SF₁dr₁ =
Set-t²)d=
0
- 4 [€] = 4 (4-3)=4(+2= 3/
F
F= <0, 2-2+, 2+>
> F(*) = <0,0,0> SF.& = 0
F = <+, 0,2-²+>
05+61
F(F) = <+ (2-2+), t.0, 3+ (2-2+))
* <2+-2+², 0, 6+-6+²>
F = <1, 0, -2>
Fir' = 2t-2+² - 12t+12€²
=
100+10t
1
[Foli = j*+tQoL
=
-
110 (½ ½ - ½ 2 ) = 10 (- ½ 7
= 10
SF ·√r = 0 + 0 + 0
C
=
+0+
513
RHS: √xF=
24
dz
xz
ху
3x2
S
= <0, x-37, Y>
Z-2-2x-Y
⇒ ˜¯xÊ > < 0, × -3 (2-2x-4), Y)
->
= <0, x-6+6x+34, Y>
= <0, 7 x + 34-6, 47
1 = (2x, z₁, 1) = (<2,1,1>
PXF-ñ (2·0) + (1-(7x+34-6)) + (1.4)
=7x+44-6
70-2-24
=
SSEXF.d
F
12-2x
(7X+44-6)dy dx
0
2-27
dx
= So (7x+9 (2-2x) +2 (2-2x)² ]ox
= 5'' (-6x² + 18x1024) dx
=
>
=
-2x² + 5 x² - 4x1';
=-2+5-4=-1
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps
Recommended textbooks for you
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage