Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.1 Arc Length and Surface Area - Exercises - Page 469: 44

$$ 3.43653$$

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*}