Answer
$$\dfrac{\pi}{9} (17^{3/2}-1)$$
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:
$$S = \int_a^b 2\pi f(x) \sqrt {1 + [f'(x)]^2} dx $$
Rewrite the function as: $y=(3x)^{1/3} \implies f(y)=\dfrac{y^3}{3}$
We have: $f(y)=\dfrac{y^3}{3} \implies f'(y)=y^2$
Then $$S=\int_{0}^{2} 2\pi (\dfrac{y^3}{3}) \sqrt {1 + (y^2)^2} dx \\=\dfrac{2 \pi}{3} \int_{0}^{2} y^3 \sqrt {1+y^4} \ dy$$
Plug $a=1+y^4 \implies 4y^3 =da$
Now, $$S=\dfrac{2\pi}{3} \int_{0}^2 (\sqrt a) \dfrac{\ da}{4} \\=\dfrac{2\pi}{12} [\dfrac{a^{3/2}}{2}]_0^2\\=\dfrac{\pi}{9} (17^{3/2}-1)$$
