Find the flux for the vector field F=< xz, yz, 5> over the 
surface S (with outward pointing normal) given parametric
form r(s,t) = < cos s, sin s, s + t> for 0 <= s <= pi and
1 <= t <= 3



Answer.

partial r
---------   = <-sin s, cos s, 1>
partial s

partial r
---------   = <     0,     0, 1>
partial t


->
dA = < cos s, sin s, 0>

On S
F = < (s+t)cos s, (s+t)sin s, 5> 

                   ->    ->
so the dot product F dot dA = (s + t) cos^2 s + (s + t) sin^2 s 
= (s + t).  The integral runs for s=0..Pi and t=1..3 so

The inner integral yields
           |3
(st + t^/2)|   = 2s + 4
           |t=1

The outer integral yields
          | pi
(s^2 + 4s)|      = pi^2 + 4pi
          | s = 0