Answer
The area of the surface obtained by rotating the curve about the x-axis is:
$$
\begin{aligned}
S&=\int 2 \pi y d s \\
&=\int_{\frac{1}{2}}^{1} 2 \pi (\frac{x^{3}}{6}+\frac{1}{2x} ) (\frac{x^{2}}{2}+\frac{1}{2x^{2}}) dx\\
&=\frac{263}{256} \pi.
\end{aligned}
$$
Work Step by Step
$$
y= \frac{x^{3}}{6}+\frac{1}{2x}, \quad \quad \frac{1}{2} \leq x \leq 1
$$
$\Rightarrow$
$$
y^{\prime} =\frac{dy}{dx}= \frac{x^{2}}{2}-\frac{1}{2x^{2}}
$$
$\Rightarrow$
$$
\begin{aligned} ds &= \sqrt {1+\left(y^{\prime}\right)^{2}} dx \\
&=\sqrt{1+\left(\frac{x^{2}}{2}-\frac{1}{2x^{2}} \right)^{2}} dx\\
&=\sqrt{\frac{x^{4}}{4}+\frac{1}{2}+\frac{1}{4x^{4}}}dx \\
&=\sqrt{(\frac{x^{2}}{2}+\frac{1}{2x^{2}})^{2}} dx\\
&= (\frac{x^{2}}{2}+\frac{1}{2x^{2}})dx
\end{aligned}
$$
So, the area of the surface obtained by rotating the curve about the x-axis is:
$$
\begin{aligned}
S&=\int 2 \pi y d s \\
&=\int_{\frac{1}{2}}^{1} 2 \pi (\frac{x^{3}}{6}+\frac{1}{2x} ) (\frac{x^{2}}{2}+\frac{1}{2x^{2}}) dx\\
&=2 \pi \int_{\frac{1}{2}}^{1} (\frac{x^{5}}{12}+\frac{x}{12} +\frac{x}{4}+\frac{1}{4x^{3}} ) dx\\
&=2 \pi \int_{1 / 2}^{1}\left(\frac{x^{5}}{12}+\frac{x}{3}+\frac{x^{-3}}{4}\right) d x\\
&=2 \pi\left[\frac{x^{6}}{72}+\frac{x^{2}}{6}-\frac{x^{-2}}{8}\right]_{1 / 2}^{1}\\
&=2 \pi\left[\left(\frac{1}{72}+\frac{1}{6}-\frac{1}{8}\right)-\left(\frac{1}{64 \cdot 72}+\frac{1}{24}-\frac{1}{2}\right)\right]\\
&=2 \pi\left(\frac{263}{512}\right) \\
&=\frac{263}{256} \pi.
\end{aligned}
$$
