Answer
$$\dfrac{275 \pi}{32} $$
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: $f(x)=y=\dfrac{x^3}{3}+\dfrac{1}{4x} \implies f'(x)=x^2-\dfrac{1}{4x^2}$
Then $$S=\int_{1/2}^{2} 2\pi (\dfrac{x^3}{3}+\dfrac{1}{4x}) \sqrt {1 + (x^2-\dfrac{1}{4x^2})^2} dx \\=2 \pi \int_{1/2}^{2} (\dfrac{x^3}{3}+\dfrac{1}{4x}) (x^2+\dfrac{1}{4x^2}) \ dx \\=2\pi \int_{1/2}^2 (\dfrac{x^5}{5}+\dfrac{x}{3}+\dfrac{x^{-3}}{16}) \ dx \\=2 \pi [\dfrac{x^6}{18}+\dfrac{x^2}{6}-\dfrac{x^{-2}}{-32}]_{1/2}^2\\=\dfrac{275 \pi}{32} $$
