Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.1 Arc Length and Surface Area - Exercises - Page 469: 43

$$ 7.603062807$$

Since $$ y=x^{-1}\to y'=-x^{-2}$$ Then \begin{align*} \text{Surface area =}& 2\pi \int_{a}^{b} f(x) \sqrt{1+f'^2(x)}dx\\ &=2 \pi \int_{1}^{3} \frac{1}{x} \sqrt{1+\left(-\frac{1}{x^{2}}\right)^{2}} d x \\ &=2 \pi \int_{1}^{3} \frac{1}{x} \sqrt{1+\frac{1}{x^{4}}} d x \\ & \text{Using a calculator gives:} \\ &\approx 7.603062807 \end{align*}