Answer
$S=\left.16 \pi\left[\frac{8}{5} t^{\frac{5}{2}}-2 t^{-0.5}\right]\right|_{t=1} ^{t=3}=16 \pi(23.7868+0.4)=1215.7611$
Work Step by Step
$\begin{gathered}\frac{\mathrm{d} x}{\partial t}=4 t-\frac{1}{t^2} \\ \frac{\mathrm{d} y}{\partial t}=\frac{4}{\sqrt{t}} \\ S=2 \pi * \int_1^3 8 \sqrt{t} \sqrt{\frac{16}{t}+\left(4 t-\frac{1}{t^2}\right)^2} d t \\ S=16 \pi \int_1^3 \sqrt{16+t\left(4 t-\frac{1}{t^2}\right)^2} d t \\ =16 \pi \int_1^3 \sqrt{16+t\left(16 t^2+\frac{1}{t^4}-\frac{8}{t}\right)} d t \\ =16 \pi \int_1^3 \sqrt{8+16 t^3+\frac{1}{t^3}} d t\end{gathered}$
$\begin{gathered}=16 \pi \int_1^3 \sqrt{\left[8+\frac{16 t^6+1}{t^3}\right]} d t \\ =16 \pi \int_1^3 \sqrt{\frac{8 t^3+16 t^6+1}{t^3}} d t=16 \pi \int_1^3 \sqrt{\frac{\left(4 t^3+1\right)^2}{t^3}} d t=16 \pi \int_1^3\left(4 t^3+1\right) * t^{\frac{-3}{2}} d t \\ S=16 \pi \int_1^3\left(4 t^3+1\right) * t^{\frac{-3}{2}} d t=16 \pi \int_1^3\left(4 t^{\frac{3}{2}}+t^{\frac{-3}{2}}\right) d t\end{gathered}$
$S=\left.16 \pi\left[\frac{8}{5} t^{\frac{5}{2}}-2 t^{-0.5}\right]\right|_{t=1} ^{t=3}=16 \pi(23.7868+0.4)=1215.7611$
