Answer
$2\pi(2-\dfrac{3\sqrt 2}{4})$
Work Step by Step
Here, $ \dfrac{dx}{dt}= 2t-(\dfrac{1}{2t^2}) ;y=4\sqrt t$
This implies that $ \dfrac{dy}{dt}= \dfrac{2}{\sqrt t}$
Surface area: $S=\int_{(1/\sqrt 2)}^{1} 2\pi [t^2+(\dfrac{1}{2t})] \sqrt { [(2t-\dfrac{1}{2t^2})^2 +(\dfrac{2}{\sqrt t})^2}] dt$
Then, $S=(2 \pi)[(\dfrac{1}{2)}t^4+(\dfrac{3}{2}) t-(\dfrac{1}{8})t^{-2}]_{(1/\sqrt 2)}^{1}$
Hence, $S=2\pi(2-\dfrac{3\sqrt 2}{4})$
