Find the flux integral

  / /
 | |   ->    ->
 | |   F dot dA
 | |
/ /
    S

When F = < 2x, y, z > and S is the upward oriented
piece of the plane z = 3x + 2y over the rectangle
D = {(x,y) | 0 <= x <= 1, 0 <= y <= 2}


Answer:
This is set up for the form z = f(x,y) = 3x + 2y, upward normal.

->
dA = < -f_x, -f_y, 1> dA = < -3, -2, 1> dy dx

F on the surface is < 2x, y, 3x + 2y> so the dot product is
  -6x -2y + 3x + 2y = -3x

	1    2
   /    /
  |    |
  |    |   -3 x dy dx = -3.
  |    |
 /    /
   0    0