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:
The normal vector of this surface is n(z,θ)=<0,0,1>x<-3sinθ,3cosθ,0>=
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).