Find the flux for the vector field F=< x/2, -3y, z> over the surface S
(with upward pointing normal) given by the portion of z = x^2 + 2y over
the region D in the xy-plane between the curves y = x and y = x^2.



Answer.
->
dA = < -f_x, -f_y, 1>


->
dA = < -2x, -2, 1>

On S
F = < x/2, -3y, x^2+ 2y> 

                   ->    ->
so the dot product F dot dA = -x^2 +6y + x^2 +2y
= 8y.  The integral runs for y=x^2..x and x=0..1 so

The inner integral yields
      |x
(4y^2)|   = 4 x^2 - 4 x^4
      |y=x^2

The outer integral yields
               | 1
(4x^3/3-4x^5/5)|      = 4/3 - 4/5 = (20 - 12)/15 = 8/15
               | x = 0