Answer
$2\pi(2-\dfrac{3\sqrt 2}{4})$
Work Step by Step
We are given that $x=t^2+(1/2t) \implies \dfrac{dx}{dt}= 2t-\dfrac{1}{2t^2}$ and $y=4\sqrt t \implies \dfrac{dy}{dt}= \dfrac{2}{\sqrt t}$
The surface area is given as follows:
$S=\int_{1/\sqrt 2}^{1} 2\pi (t^2+(1/2t)) \sqrt { (2t-\dfrac{1}{2t^2})^2 +(\dfrac{2}{\sqrt t})^2} dt$
Then, $S=(2 \pi)[\dfrac{1}{2}t^4+\dfrac{3}{2} t-\dfrac{1}{8}t^{-2}]_{1/\sqrt 2}^{1}$
Thus, $S=2\pi(2-\dfrac{3\sqrt 2}{4})$
