Answer
$$\dfrac{ 3 \sqrt 6- \sqrt 2}{3}$$
Work Step by Step
The surface integral can be found as:
$dS= \sqrt {(f_x)^2 +(f_y)^2 +1} \ dy \ dx$
or, $dS=\sqrt {(2x)^2 +(1)^2 +1} \ dy \ dx =\sqrt {(4x^2+2)} \ dy \ dx $
Now, $$I=\iint_{S} G (x,y,z) \ dS=\int_0^1 \int_{-1}^{1} x \sqrt {(4x^2+2)} \ dy \ dx \\=\int_0^1 x \times \sqrt {(4x^2+2)} [y]_{-1}^1 \ dx \\=(2)(\dfrac{1}{8}_(\dfrac{2}{3}) [(4x^2+2)^{3/2}]_0^1 \\ \dfrac{6 \sqrt 6-2 \sqrt 2}{6} \\=\dfrac{ 3 \sqrt 6- \sqrt 2}{3}$$
