Chapter 9 - Further Applications of the Integral and Taylor Polynomials - Chapter Review Exercises - Page 496: 7

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

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*}