Answer
$\dfrac{\sqrt 2}{3}+\ln(\sqrt 2+\sqrt 3)$
Work Step by Step
$Surface \ Area=\iint_{D} \sqrt {2+4x^2} dy dx=4 \times \int_{0}^1 \sqrt {\dfrac{1}{2}+x^2} dx-\int_{0}^1 2x \sqrt {2+4x^2} dx$
Set $2+4x^2 =a $ and $8x dx =da$
Now, $Surface Area=\sqrt 6+\ln(\sqrt 2+\sqrt 3) -\dfrac{1}{4} \int_2^6 a^{1/2} da \\=\sqrt 6+\ln(\sqrt 2+\sqrt 3) -\dfrac{1}{4} [\dfrac{2}{3}a^{3/2}]_2^6 $
$\implies =\sqrt 6+\ln(\sqrt 2+\sqrt 3)-\sqrt 6-\dfrac{\sqrt 2}{3}$
$\implies Surface \ Area=\dfrac{\sqrt 2}{3}+\ln(\sqrt 2+\sqrt 3)$
