Answer
$ \dfrac{ -\sqrt 2}{2}$
Work Step by Step
The surface integral can be found as: $dS= \sqrt {(f_x)^2 +(f_y)^2 +1} \ dy \ dx=\sqrt {(-1)^2+0^2+1} \ dz \ dy =\sqrt 2 $
We set up the integral and solve the surface integral as follows:
$\iint_{S} G (x,y,z) \ dS=\int_0^1 \int_{0}^{1} (1-2y-z) \sqrt {2} \ dz \ dy$
or, $=\sqrt 2 \int_0^1 [z-2yz -\dfrac{z^2}{2}]_0^1 \ dy$
or, $=\sqrt 2 \int_0^1 [1-2y(1) -\dfrac{1}{2}(1^2-0)] \ dy$
or, $=\sqrt 2 \times \int_0^1 [\dfrac{1}{2}-2y] \ dy$
or, $= \sqrt 2[\dfrac{y}{2}-y^2]_0^1$
or, $= \dfrac{ -\sqrt 2}{2}$
