7. Let S = {(x, y, z): 0 ≤ x ≤ 1, 1 ≤ y ≤ 3, z = 3 - ry}. That is, S is the piece of the surface z = 3 - xy that lies above the rectangle [0, 1] x [1, 3]. a) Circle the integral whose value would give the area of S: 13 jj√(3)² + (ry)² dy dz 01 13 [/v 01 -x-y+1 dy dx S: I Z 13 HVITE [[√1 + x² + y² dy dr 0 1 b) Find the upward flux of the field F = (z,y², x) through S. That is, evaluate where n is the upward-oriented unit normal vector for S. [[F.n S F.ndS, Y
7. Let S = {(x, y, z): 0 ≤ x ≤ 1, 1 ≤ y ≤ 3, z = 3 - ry}. That is, S is the piece of the surface z = 3 - xy that lies above the rectangle [0, 1] x [1, 3]. a) Circle the integral whose value would give the area of S: 13 jj√(3)² + (ry)² dy dz 01 13 [/v 01 -x-y+1 dy dx S: I Z 13 HVITE [[√1 + x² + y² dy dr 0 1 b) Find the upward flux of the field F = (z,y², x) through S. That is, evaluate where n is the upward-oriented unit normal vector for S. [[F.n S F.ndS, Y
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
6th Edition
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
ChapterA: Appendix
SectionA.2: Geometric Constructions
Problem 10P: A soda can has a volume of 25 cubic inches. Let x denote its radius and h its height, both in...
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Help with the following question.
![7. Let S = {(x, y, z) : 0 ≤ x ≤ 1, 1 ≤ y ≤ 3, z = 3 - xy}. That is, S is the piece of the surface
z = 3 - xy that lies above the rectangle [0, 1] × [1,3].
4] a) Circle the integral whose value would give the area of S:
13
liv
][√−x−y+1 dy dz
0 1
1 3
[[√(3)² + (xy)² dy da
01
S:
jjv
[[√1 + x² + y² dy dx
0 1
b) Find the upward flux of the field F = (z,y², x) through S. That is, evaluate
where n is the upward-oriented unit normal vector for S.
JfF.
S
F.n dS,
Y](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcef3f1b2-56cd-4f45-949e-fc7a38ef994c%2F749b896e-0707-4193-b015-d0b47f0942e6%2F2ct36v_processed.jpeg&w=3840&q=75)
Transcribed Image Text:7. Let S = {(x, y, z) : 0 ≤ x ≤ 1, 1 ≤ y ≤ 3, z = 3 - xy}. That is, S is the piece of the surface
z = 3 - xy that lies above the rectangle [0, 1] × [1,3].
4] a) Circle the integral whose value would give the area of S:
13
liv
][√−x−y+1 dy dz
0 1
1 3
[[√(3)² + (xy)² dy da
01
S:
jjv
[[√1 + x² + y² dy dx
0 1
b) Find the upward flux of the field F = (z,y², x) through S. That is, evaluate
where n is the upward-oriented unit normal vector for S.
JfF.
S
F.n dS,
Y
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