Answer
$\dfrac{\sqrt 2}{3}+\ln(\sqrt 2+\sqrt 3)$
Work Step by Step
Surface area is given by:
$A(S)=\iint_{D} \sqrt {2+4x^2} dy dx$
or, $=4 \int_{0}^1 \sqrt {\dfrac{1}{2}+x^2} dx-\int_{0}^1 2x \sqrt {2+4x^2} dx$
Consider,
$a=2+4x^2 \implies da=8x dx$
$A(S)=\sqrt 6+\ln(\sqrt 2+\sqrt 3) -\dfrac{1}{4} \int_2^6 \sqrt a da $
or, $=\sqrt 6+\ln(\sqrt 2+\sqrt 3) -\dfrac{1}{4} [\dfrac{2a^{3/2}}{3}]_2^6 $
or, $=\sqrt 6+\ln(\sqrt 2+\sqrt 3)-\sqrt 6-\dfrac{\sqrt 2}{3}$
or, $=\dfrac{\sqrt 2}{3}+\ln(\sqrt 2+\sqrt 3)$
