## Thomas' Calculus 13th Edition

$\dfrac{ 3 \sqrt 6- \sqrt 2}{3}$
The surface integral can be found as: $dS= \sqrt {(f_x)^2 +(f_y)^2 +1} \ dy \ dx=\sqrt {(2x)^2 +(1)^2 +1} \ dy \ dx =\sqrt {(4x^2+2)} \ dy \ dx$ We set up the integral and solve the surface integral as follows: $\iint_{S} G (x,y,z) \ dS=\int_0^1 \int_{-1}^{1} x \sqrt {(4x^2+2)} \ dy \ dx$ or, $=\int_0^1 x \sqrt {(4x^2+2)} [y]_{-1}^1 \ dx$ or, $= \int_0^1 x(4x^2 +2)^{1/2} \ dx$ or, $=2 \times \dfrac{1}{8} \times \dfrac{2}{3} [(4x^2+2)^{3/2}]_0^1$ or, $= \dfrac{6 \sqrt 6-2 \sqrt 2}{6}$ or, $= \dfrac{ 3 \sqrt 6- \sqrt 2}{3}$