Answer
$\dfrac{28 \pi}{9}$
Work Step by Step
Since, $dS=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$
Thus, $dS=\sqrt{(t^{1/2})^2+(t^{-1/2})^2}=\sqrt{\dfrac{t^2+1}{t}}$
or, $S=\int_{0}^{\sqrt 3} (2 \pi) x ds=\int_{0}^{\sqrt 3} (2 \pi) (2/3t^{3/2}) (\sqrt{\dfrac{t^2+1}{t}}) dt$
or, $S=\dfrac{4 \pi}{3}\int_{0}^{\sqrt 3} t\sqrt{t^2+1} dt$
Suppose $p=t^2+1 \implies dp=2 dt$
$S=\dfrac{2 \pi}{3}\int_{1}^{4} \sqrt{p} dp=\dfrac{28 \pi}{9}$
