#### Answer

$$24\sqrt{2}\pi$$

#### Work Step by Step

Given $$ y=x+1, \quad[0,4]$$
\begin{align*}
y'&= 1\\
1+y'^2&= 2
\end{align*}
Then the surface area is given by
\begin{align*}
S&=2\pi \int_{a}^{b} y\sqrt{1+y'^2}dx\\
&=2\pi \int_{0}^{4}\sqrt{2}(x+1)dx\\
&=\sqrt{2}\pi (x+1)^2 \bigg|_{0}^{4}\\
&=24\sqrt{2}\pi
\end{align*}