Nevermind, after the TA and i stared at this for a while i finally figured it out.. um if anyone wants to know lemme know and i'll post how.
I really don't even know where to start on this problem.
I found this formula in the book:
The surface integral of F over S is:
Dbl Integral (F dot dS) = Dbl Integral (F dot n dS)
This is also called the flux of F across S.
Dbl Integral (F dot dS) = Dbl Integral(F dot (r sub u X r sub v) dA
Here is the problem:
Evaluate: Dbl Integral (A dot n dS) over S in each of the following cases:
(a) A = yi + 2xj - zk and S is the surface of the plane 2x + y = 6 in the first octant cut off by the plane z = 4.
(b) A = (x + y^2)i - 2xj + 2yzk and S is the surface of the plane
2x + y +2z = 6 in the first octant.
The answers are (a) 108. (b) 81.
Thanks a lot! It'd be great if someone could help me out on here by morning or I'm going to have to go talk to a TA for help.