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

[musicmar]

Homework Statement

Compute ∫∫_SF.dS

F(x,y,z)=<y,x²y,e^xz> 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,e^3zcosθ>

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).

 

[LCKurtz]

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?

 

[musicmar]

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.

 

[HallsofIvy]

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>.