$z = 1 - x - y$
$z_x = -1$
$z_y = -1$
$D = \sqrt{1 + z_x + z_y} = \sqrt{3}$
$I = \displaystyle \int_S f(1-x-y)\cdot D = \int_S 4(1-x-y)^2 \cdot \sqrt{3}$
The integral over the surface is
$I = 4\sqrt{3} \displaystyle \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}~(1-x-y)^2~dy~dx$
In polar coordinates
$I = \displaystyle 4\sqrt{3}\int_0^{2\pi}\int_0^1 (1-r \cos(\theta)-\sin(\theta))^2\cdot r ~dr ~d\theta$
I'll let you decide which way you want to actually integrate it.
The answer is $I = 6\sqrt{3}\pi$
Is it z=-x-y you mean? I think the 1 you have in the z function is not suppose to be there. Would that change the answer?
Update: I got the answer with your help. I just changed up the function inside the integral . Thanks!