Answer
The area of the surface obtained by rotating the curve about the y-axis
$$
\begin{aligned}
S&=\int 2 \pi x d s \\
&=\int_{1}^{2} 2 \pi (x)(\frac{x}{2}+\frac{1}{2x})dx\\
&= \frac{10}{3}\pi
\end{aligned}
$$
Work Step by Step
$$
y= \frac{1}{4}x^{2}-\frac{1}{2}\ln x, \quad \quad 1 \leq x \leq 2
$$
$\Rightarrow$
$$
y^{\prime} =\frac{dy}{dx}= \frac{x}{2}-\frac{1}{2x}
$$
$\Rightarrow$
$$
\begin{aligned} ds &= \sqrt {1+\left(y^{\prime}\right)^{2}} dx \\
&=\sqrt{1+\left(\frac{x}{2}-\frac{1}{2x} \right)^{2}} dx\\
&=\sqrt{1+\frac{x^{2}}{4}-\frac{1}{2}+\frac{1}{4x^{2}}}dx \\
&=\sqrt{\frac{x^{2}}{4}+\frac{1}{2}+\frac{1}{4x^{2}}}dx \\
&=\sqrt{(\frac{x}{2}+\frac{1}{2x})^{2}} dx\\
&= (\frac{x}{2}+\frac{1}{2x})dx
\end{aligned}
$$
So, the area of the surface obtained by rotating the curve about the y-axis
$$
\begin{aligned}
S&=\int 2 \pi x d s \\
&=\int_{1}^{2} 2 \pi (x)(\frac{x}{2}+\frac{1}{2x})dx\\
&=2 \pi \int_{1}^{2} x(\frac{x}{2}+\frac{1}{2x})dx \\
&= \pi \int_{1}^{2} (x^{2}+1)dx \\
&= \pi\left[ \frac{1}{3}x^{3}+x\right]_{1 }^{2}\\
&= \pi\left[ (\frac{8}{3}+2)-(\frac{1}{3}+1)\right]\\
&= \frac{10}{3}\pi
\end{aligned}
$$
