Answer
$$ \frac{\pi}{27}(145 \sqrt{145}-1)$$
Work Step by Step
Given $$y= x^3, \ \ \ 0 \leq x \leq 2$$
Since the surface area of the surface of revolution
formed by revolving the graph of $f (x)$ around the $x-$axis is given by
\begin{align*}
S_x&=\int_{a}^{b}2\pi f(x)\sqrt{1+[f'(x)]^2}dx\\
&=\int_{0}^{2}2\pi x^3\sqrt{1+9x^4}dx
\end{align*}
Let $ u=1+9x^4\ \ \Rightarrow \ \ du =36x^3dx$, at $x=0\to u=1$, $x=2\to u=145$, then
\begin{align*}
S_x&=\frac{1}{18}\int_{1}^{145} \pi u^{1/2}du\\
&=\frac{\pi}{27} u^{3/2} \bigg|_{1}^{145}\\
&=\frac{\pi}{27}(145 \sqrt{145}-1)\ \ \text{unit}^2
\end{align*}
