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5) Let F(x, y, a vector field and S the surface with equation x In (y − 1) described in the = Following figure. When taking n (e*, -1, 0) then the = normal vector unitary that determines the orientation of S is N = The double integral that allows to calculate he flux integral J.F. F. Nds is: 6 cln 5 A) 1² h (x² − 2) dx dz -6 pln 5 z)=2xi+xyj+x²zk be In 5 -In(y-1) N S

5) Let F(x, y, a vector field and S the surface with equation x In (y − 1) described in the = Following figure. When taking n (e*, -1, 0) then the = normal vector unitary that determines the orientation of S is N = The double integral that allows to calculate he flux integral J.F. F. Nds is: 6 cln 5 A) 1² h (x² − 2) dx dz -6 pln 5 z)=2xi+xyj+x²zk be In 5 -In(y-1) N S

Advanced Engineering Mathematics
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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6) Let F(x, y, z) = 2xi+xyj+x²zk be a vector field and S the surface with equation x = In (y - 1) described in the following figure. = When taking n (e, 1, 0) then the normal vector unitary that determines the orientation of S is N = The double integral that allows to calculate the flux integral F F. Nds is: -6 5 A) [² m² (re² − x) dxdz 6 cln 5 B) (x − xe²) dx dz 0 0 6 In 5 xe. -x C) [ L™³ dx dz √e²+1 -6 5 D) Looth ² (re* – x) dx dz In 5 In(y-1) S
Transcribed Image Text:6) Let F(x, y, z) = 2xi+xyj+x²zk be a vector field and S the surface with equation x = In (y - 1) described in the following figure. = When taking n (e, 1, 0) then the normal vector unitary that determines the orientation of S is N = The double integral that allows to calculate the flux integral F F. Nds is: -6 5 A) [² m² (re² − x) dxdz 6 cln 5 B) (x − xe²) dx dz 0 0 6 In 5 xe. -x C) [ L™³ dx dz √e²+1 -6 5 D) Looth ² (re* – x) dx dz In 5 In(y-1) S
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