#### Answer

$$ 3.43653$$

#### Work Step by Step

Since
$$ y=x^{4}\to y'=4x^{3}$$
Then
\begin{align*}
\text{Surface area =}& 2\pi \int_{a}^{b} f(x) \sqrt{1+f'^2(x)}dx\\
&=2 \pi \int_{0}^{1}x^4 \sqrt{1+16x^6 } dx\\
& \text{Using a calculator gives:} \\
&\approx 3.43653
\end{align*}