Answer
$$\dfrac{1179 \pi}{256} $$
Work Step by Step
Let us consider that $f(x)$ is a non-negative function which defines a continuous first derivative on the interval $[a,b]$. The area of the surface obtained when the graph of a function $f(x)$ on the interval $\left[ {a,b} \right]$ is revolved about the $x$- axis can be expressed as:
$$S = \int_a^b 2\pi f(x) \sqrt {1 + [f'(x)]^2} dx $$
We have: $f(x)=y=\dfrac{x^4}{8}+\dfrac{1}{4x^2} \implies f'(x)=\dfrac{x^3}{2}-\dfrac{1}{2x^3}$
Then $$S=\int_{1}^{2} 2\pi (\dfrac{x^4}{8}+\dfrac{1}{4x^2}) \sqrt {1 + (\dfrac{x^3}{2}-\dfrac{1}{2x^3})^2} dx \\=2 \pi \int_{1}^{2} (\dfrac{x^4}{8}+\dfrac{1}{4x^2}) \sqrt {\dfrac{x^6}{4}+\dfrac{1}{4x^6} +\dfrac{2x^3}{4x^3}} \ dx \\=2\pi \int_1^2 (\dfrac{x^4}{8}+\dfrac{1}{4x^2}) \sqrt {(\dfrac{x^3}{2}+\dfrac{1}{2x^3})^2}\ dx \\=2 \pi [\dfrac{x^8}{(16)(8)}+\dfrac{3x^2}{32}+\dfrac{x^{-4}}{-32}]_1^2\\=\dfrac{1179 \pi}{256} $$
