Evaluate the surface integral SS F * dS for the given vector field F and the oriented surface S. For closed surfaces, use the positive (outward) orientation. F(x,y,z) = -xi - yj +z^3k. S is the part if the cone z = (x^2+y^2)^1/2 between the planes z = 1 and z =3 with a downward orientation. I attempted to solve this with an (r, theta) parametrization. I'm wondering if I did this properly, or where I went wrong since the final answer differs from the solved solutions on this website
Evaluate the surface integral SS F * dS for the given vector field F and the oriented surface S. For closed surfaces, use the positive (outward) orientation. F(x,y,z) = -xi - yj +z^3k. S is the part if the cone z = (x^2+y^2)^1/2 between the planes z = 1 and z =3 with a downward orientation. I attempted to solve this with an (r, theta) parametrization. I'm wondering if I did this properly, or where I went wrong since the final answer differs from the solved solutions on this website
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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Question
Evaluate the surface
F(x,y,z) = -xi - yj +z^3k. S is the part if the cone z = (x^2+y^2)^1/2 between the planes z = 1 and z =3 with a downward orientation.
I attempted to solve this with an (r, theta) parametrization. I'm wondering if I did this properly, or where I went wrong since the final answer differs from the solved solutions on this website
( https://www.bartleby.com/solution-answer/chapter-137-problem-24e-essential-calculus-early-transcendentals-2nd-edition/9781337692991/evaluate-the-surface-integral-s-f-ds-for-the-given-vector-field-f-and-the-oriented-surface-s-in/09e1d51f-05ec-4023-aa9e-a016378d0dc7 ).
Transcribed Image Text:[Chapter 13.7, Problem 24E
Evaluate the surface integral JS, F. ds for the given vector field F and the oriented surface S. In other words, find the
flux of F across S. For closed surfaces, use the positive (outward) orientation.
x² + y2 between the planes z = 1 and z= 3 with downward
24. F(x, y, z) = -xi-yj + z³ k, S is the part of the cone z =
orientation
![13,7 (Surface Integrals): 21, 22, 24, 27, 28, 30%
(8)
24) F(x, y, z) = -xi - yj + z³ K; 5 is part of the cone 2=√x² + y² between
z = r
Planes
downward orientation.
Z=1
and Z=3 W/
Let X=rcasa; y=rsins, z = f(y₁ z) = r
R(r, e) = <rcose, sine, r
R₁ = <cos e, sine, 1)
Ro= <-rsine, +rease, o)
j
Rr X Ro= cose) sine 1
1-r sine rcose
-rcosei- (o + rsine);
(rcose + rsin's) K
+
21T
[8 [424] 10 = 277 [424] =
de
1≤r≤3
05032TT
84817
3
d #
Brx Ro = <-rcoso, -rsine, r
3
So S. <-(trcose), - (rsine), (1)³). <-rcose, -rsine, r> (rarde)
Sea S² (r² cos²o + r² sin ²6 +5²) Farde
.274,3
सीं
5
Sent S² (r² +r²4) rdrd@ = 5" 5₁ r²³²+r³ ardo
SO" [ +₁5² + + ²)² = (+ (0)² + + (3) ²) - (-) = +¹-(2)
+ 21
+
729
6
1
У
bot
=
Z
JA
тов
у
X
r²sin²@+ 1² cos²@
= r²(Sin²e + cos²e)
r² (1) = r²](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F93013f87-eae9-408f-8367-c564e6d22e0c%2Fff1a1a9c-27f8-4784-a48d-87c4bd16541b%2Fm9pe6c7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:13,7 (Surface Integrals): 21, 22, 24, 27, 28, 30%
(8)
24) F(x, y, z) = -xi - yj + z³ K; 5 is part of the cone 2=√x² + y² between
z = r
Planes
downward orientation.
Z=1
and Z=3 W/
Let X=rcasa; y=rsins, z = f(y₁ z) = r
R(r, e) = <rcose, sine, r
R₁ = <cos e, sine, 1)
Ro= <-rsine, +rease, o)
j
Rr X Ro= cose) sine 1
1-r sine rcose
-rcosei- (o + rsine);
(rcose + rsin's) K
+
21T
[8 [424] 10 = 277 [424] =
de
1≤r≤3
05032TT
84817
3
d #
Brx Ro = <-rcoso, -rsine, r
3
So S. <-(trcose), - (rsine), (1)³). <-rcose, -rsine, r> (rarde)
Sea S² (r² cos²o + r² sin ²6 +5²) Farde
.274,3
सीं
5
Sent S² (r² +r²4) rdrd@ = 5" 5₁ r²³²+r³ ardo
SO" [ +₁5² + + ²)² = (+ (0)² + + (3) ²) - (-) = +¹-(2)
+ 21
+
729
6
1
У
bot
=
Z
JA
тов
у
X
r²sin²@+ 1² cos²@
= r²(Sin²e + cos²e)
r² (1) = r²
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