2 outside; f = < 2 cos(u) cos(x), 2 (u)(u)sic (v), Fr <-2 Sin (4) sin(v), a sin (4) cos(v), (2.) Evaluate the outward flux of the vector field F: given by 1x2 + y² + z² < 4 (a.) both directly fff.d Parametrically = 20 (x, y, z) √x² + y² + z² & F (F(u,v)) • (Fox Tv ) dA- across the boundary of the region in U-vspace ř (u,v) = <2sin(0) cos(V), 2 sin(u) sin (v), in cus(u)> -25(4)> ต הלטVi Vio72π ° > 2cosusiv -25 (✔) = (4 sin² 4 cosy, 4 sin ² usin(v), 4 Sivcusu cos² + 4 sucusu si?. -2.5inu sinn 2 sinusv all + = <4 S² Cosv, 48siv, 4 Six(u) cos(V)) → outwarde, = SSřids'= ss Asin(u) dvdw 217.4. -603(431 ---81 (-1-1)=161 * Sin³ (4) cus(v), - 4 sin³ (4) sin(x); 4 sin()cus()) normal is rutru = Same as on outside Felf Since is nor smaliging F · ds = Sin(y) => 1 20 [ Fids = sss.ucu) drdo 0 0 Fids = 1617-417 17-11 S 412 دا (2.) continued (b.) and using the divergence theorem. Which do you prefer? 2 x²+22 = x²+y²+z² (x²+ y²+2²)³ 31 * Q·F= 2(x²+4'+2²) (x²+y²+2²) = √x² + y²+2² Div Thu: ff.d's - SSS Gi dv - = R 2r 2 sss 0 = G ' " 2 E p p³sin (4) dp do dd) § Sprinces dpudel 0 412 12 12. (-cos (4))" = · (2) = 1217." 20 3
2 outside; f = < 2 cos(u) cos(x), 2 (u)(u)sic (v), Fr <-2 Sin (4) sin(v), a sin (4) cos(v), (2.) Evaluate the outward flux of the vector field F: given by 1x2 + y² + z² < 4 (a.) both directly fff.d Parametrically = 20 (x, y, z) √x² + y² + z² & F (F(u,v)) • (Fox Tv ) dA- across the boundary of the region in U-vspace ř (u,v) = <2sin(0) cos(V), 2 sin(u) sin (v), in cus(u)> -25(4)> ต הלטVi Vio72π ° > 2cosusiv -25 (✔) = (4 sin² 4 cosy, 4 sin ² usin(v), 4 Sivcusu cos² + 4 sucusu si?. -2.5inu sinn 2 sinusv all + = <4 S² Cosv, 48siv, 4 Six(u) cos(V)) → outwarde, = SSřids'= ss Asin(u) dvdw 217.4. -603(431 ---81 (-1-1)=161 * Sin³ (4) cus(v), - 4 sin³ (4) sin(x); 4 sin()cus()) normal is rutru = Same as on outside Felf Since is nor smaliging F · ds = Sin(y) => 1 20 [ Fids = sss.ucu) drdo 0 0 Fids = 1617-417 17-11 S 412 دا (2.) continued (b.) and using the divergence theorem. Which do you prefer? 2 x²+22 = x²+y²+z² (x²+ y²+2²)³ 31 * Q·F= 2(x²+4'+2²) (x²+y²+2²) = √x² + y²+2² Div Thu: ff.d's - SSS Gi dv - = R 2r 2 sss 0 = G ' " 2 E p p³sin (4) dp do dd) § Sprinces dpudel 0 412 12 12. (-cos (4))" = · (2) = 1217." 20 3
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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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 :)
Transcribed Image Text:2
outside;
f = < 2 cos(u) cos(x), 2 (u)(u)sic (v),
Fr <-2 Sin (4) sin(v), a sin (4) cos(v),
(2.) Evaluate the outward flux of the vector field F:
given by 1x2 + y² + z² < 4
(a.) both directly
fff.d
Parametrically
=
20
(x, y, z)
√x² + y² + z²
& F (F(u,v)) • (Fox Tv ) dA-
across the boundary of the region
in U-vspace
ř (u,v) = <2sin(0) cos(V), 2 sin(u) sin (v), in cus(u)>
-25(4)>
ต
הלטVi
Vio72π
° >
2cosusiv
-25 (✔)
=
(4 sin² 4 cosy, 4 sin ² usin(v), 4 Sivcusu cos² + 4 sucusu si?.
-2.5inu sinn 2 sinusv
all +
= <4 S² Cosv, 48siv, 4 Six(u) cos(V))
→ outwarde,
=
<Sin (v) cos(x), Six (u) six (V), cup(u))
F(F(u,v)) = < 2 Sin (4) Cos(V), 2 Sin (4) sin (V), 2 cos(u))
† (ŕ(u,v)) = (Po׳v) =
2
4 Sin³ ³ (u) cos² (v) + 4 sin³ (4) sin² (x) + 4 sin(u) cos² 14)
4 sin³ 14) + 4 sin(u) cos² (4)
2R
17
on
S, (inside)
outwards
³ (ÿ þµ‚× ) ) =
-=)
= 4 sin(es) (sin² (4) + cos² (4)) = 4 sin(u)
=>
SSřids'= ss Asin(u) dvdw
217.4.
-603(431
---81 (-1-1)=161
* Sin³ (4) cus(v), - 4 sin³ (4) sin(x); 4 sin()cus())
normal is rutru
=
Same as on outside
Felf
Since is
nor
smaliging
F · ds =
Sin(y)
=>
1 20
[ Fids = sss.ucu) drdo
0 0
Fids
=
1617-417
17-11
S
412
دا
Transcribed Image Text:(2.) continued (b.) and using the divergence theorem. Which do you prefer?
2
x²+22
=
x²+y²+z²
(x²+ y²+2²)³
31
* Q·F= 2(x²+4'+2²)
(x²+y²+2²)
=
√x² + y²+2²
Div Thu: ff.d's - SSS Gi dv
-
=
R
2r
2
sss
0
=
G
'
"
2
E
p p³sin (4) dp do dd)
§ Sprinces dpudel
0
412 12 12. (-cos (4))"
=
· (2)
=
1217."
20
3
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