Answer
$\frac{1152\pi\sqrt{3}-128\pi}{5}+\frac{96\pi-32\pi\sqrt{3}}{3}$
Work Step by Step
Recall: The area of the surface obtained by rotating the curve $x=f(t)$ and $y=g(t)$ for $a\leq t\leq b$ about the x-axis is formulated by the integral $S=\int_a^b2\pi yds$ where $ds=\sqrt{(dx/dt)^2+(dy/dt)^2}dt$.
Given: $f(t)=2t^2+1/t$ and $g(t)=8\sqrt{t}$ for $1\leq t\leq 3$
Find $dx/dt$ and $dy/dt$:
$dx/dt=\frac{d}{dt}(2t^2+1/t)=4t-\frac{1}{t^2}$
$dy/dt=\frac{d}{dt}(8\sqrt{t})=8\cdot \frac{1}{2\sqrt{t}}=\frac{4}{\sqrt{t}}$
Find $ds$:
$ds=\sqrt{(4t-\frac{1}{t^2})^2+(\frac{4}{\sqrt{t}})^2}$
$ds=\sqrt{16t^2-\frac{8}{t}+\frac{1}{t^4}+\frac{16}{t}}dt$
$ds=\sqrt{16t^2+\frac{8}{t}+\frac{1}{t^4}}dt$
$ds=\sqrt{(4t+\frac{1}{t^2})^2}dt$
$ds=(4t+\frac{1}{t^2})dt$
Evaluate the area of the surface:
$S=\int_1^32\pi (8\sqrt{t})(4t+\frac{1}{t^2})dt$
$S=\int_1^364\pi t\sqrt{t}+\frac{16\pi}{t\sqrt{t}}dt$
$S=[\frac{128\pi t^2\sqrt{t}}{5}-\frac{32\pi}{\sqrt{t}}]_1^3$
$S=(\frac{128\pi 3^2\sqrt{3}}{5}-\frac{32\pi}{\sqrt{3}})-(\frac{128\pi 1^2\sqrt{1}}{5}-\frac{32\pi}{\sqrt{1}})$
$S=(\frac{1152\pi\sqrt{3}}{5}-\frac{32\pi\sqrt{3}}{3})-(\frac{128\pi}{5}-\frac{96\pi}{3})$
$S=\frac{1152\pi\sqrt{3}-128\pi}{5}+\frac{96\pi-32\pi\sqrt{3}}{3}$
Thus, the area of the surface is $\frac{1152\pi\sqrt{3}-128\pi}{5}+\frac{96\pi-32\pi\sqrt{3}}{3}$.
