Answer
$$\dfrac{1256001}{1024}\pi$$
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: $f(y)=x=9x^{2/3} -x^{2/3}/32 \implies f'(y)=\dfrac{6}{x^{1/3}}-\dfrac{x^{1/3}}{24}$
Use formula: $\int x^n dx=\dfrac{x^{n+1}}{n+1}$
Then $$S=\int_{1}^{8} 2\pi (9x^{2/3} -x^{2/3}/32) \sqrt {1 + (\dfrac{6}{x^{1/3}}-\dfrac{x^{1/3}}{24})^2} dy \\=2 \pi \int_{1}^{8} (54x^{1/3}+\dfrac{3}{8} x-\dfrac{3}{16}x -\dfrac{x^{5/3}}{(32)(24)} \ dx \\=\dfrac{1256001}{1024}\pi$$
