How do I evaluate a vector surface integral over a cylindrical surface?

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Homework Help Overview

The discussion revolves around evaluating a vector surface integral over a cylindrical surface defined by the equation x² + y² = 9, with z ranging from -3 to 3. The vector field in question is F(x,y,z) = . Participants are exploring the correct approach to compute the integral ∫∫S F · dS, particularly focusing on the parameterization and the orientation of the normal vector.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to parameterize the surface using cylindrical coordinates and calculates the normal vector. They express concern over a discrepancy between their computed integral result and the expected answer.
  • Some participants question whether the outward normal vector has been correctly identified and whether the problem requires consideration of the closed surface of the cylinder, including the top and bottom surfaces.
  • Others suggest reconsidering the use of the divergence theorem as a potential approach to the problem.
  • There is a discussion about the implications of using inward versus outward normals and how that affects the integral's sign.

Discussion Status

The discussion is active, with participants providing guidance on checking the orientation of the normal vector and clarifying the definition of the surface in the problem statement. Multiple interpretations of the problem setup are being explored, particularly regarding whether the entire cylindrical surface or just its lateral area is to be considered.

Contextual Notes

Participants note that the problem as stated may not clearly indicate whether the integral should include the top and bottom surfaces of the cylinder, leading to potential confusion about the setup. There is also mention of imposed homework rules regarding the use of calculators for evaluating integrals.

musicmar
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Homework Statement


Compute ∫∫SF.dS

F(x,y,z)=<y,x2y,exz> over x2+y2=9, -3<=z<=3, outward pointing normal.

The Attempt at a Solution



I parameterized the surface in cylindrical coordinates:
Φ(z,θ)=<3cosθ,3sinθ,z>.
The normal vector of this surface is n(z,θ)=<0,0,1>x<-3sinθ,3cosθ,0>=
<-3cosθ,-3sinθ,0>.

∫∫SF.dS
=∫∫F(Φ(z,θ)).n(z,θ)dzdθ

F(Φ(z,θ))=<3sinθ,9cos2θ,e3zcosθ>

I took the dot product of that and the normal, then evaluated the integral with z from -3 to 3 and θ from 0 to 2π and got an answer of 0. I know the answer should be (243π)/2

Some help in finding my mistake would be much appreciated. And I know it wasn't from evaluating the integral, as I used my calculator(we are allowed to).
 
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A couple of questions for you.

1. Did you check that you have the outward normal?

2. Did the problem specify the closed surface of the cylinder? If so what about the top and bottom surfaces?

3. And if so, have you considered using the divergence theorem?
 
1. Wouldn't the inward normal be the reverse cross product and thus negative?
2. The problem is exactly as I typed it. If -3<=z<=3 and 0<=theta<=2pi, that includes the entire cylinder, not just the outside. I don't know if vector surface integral was the most appropriate title for this question.
 
musicmar said:
1. Wouldn't the inward normal be the reverse cross product and thus negative?
Yes, that is exactly why what you have, <-3cosθ,-3sinθ,0>, is the inward normal, not the outward normal. Of course, that would only change the sign on the integral.

2. The problem is exactly as I typed it. If -3<=z<=3 and 0<=theta<=2pi, that includes the entire cylinder, not just the outside. I don't know if vector surface integral was the most appropriate title for this question.
The problem, as you typed it, says you are trying to integrate over the surface of the cylinder. That does not include "the entire cylinder". The "-3<=z<=3 and 0<=theta<=2pi" define the cylinder, not its surface. I would interpret what you wrote, the surface of the figure given, as including the two ends. That is, you need to include the integral over the two circles of radius 3 with z= 3 and z= -3. The "outward normal" for the first is <0, 0, 1> and for the second <0, 0, -1>.
 

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