Answer
$-\sqrt{\dfrac{2}{3}}$
Work Step by Step
The surface ares is given by:
$S=\int_0^1 \int_0^1 [(u+v)(u-v)-v^2] \sqrt 6 du dv =\sqrt 6 \int_0^1 u^2-2v^2 du dv$
This implies that
$S=\sqrt 6 \int_0^1 [\dfrac{u^3}{3}-2uv^2]_0^1 dv$
Thus, $\sqrt 6[\dfrac{v}{3}-\dfrac{2v^3}{3}]_0^1=-\sqrt{\dfrac{2}{3}}$
