->     ^          ^
Compute the flux of the vector field F =  y i + (1 + z)j through the part of
the xy-plane oriented downward between x>=0, y>=0 and 3x + 3y + z = 3

Answer z = f(x,y) form, namely z = 0. Let T be 3x  + 3y <= 3 in the 1st
quadrant, or equivalently x + y <= 1. This is a triangle.
->
dA = < 0, 0, -1>

->
F  = < y, 0, 1 + 0>


flux = integral over T of -1 dA    = -1 area(T) = -1/2


[If you did limits it would be integral of x from 0 to 1 of the
integral of y from 0 to 1 - x of -1 dy dx.]