# Vector surface integral

## 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).

LCKurtz
Homework Helper
Gold Member
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.

HallsofIvy