Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.1 Arc Length and Surface Area - Exercises - Page 469: 45

$$ 8.222695606$$

Since $$ y=e^{-x^2/2}\to y'=-xe^{-x^2/2}$$ Then \begin{align*} \text{Surface area }& =2\pi \int_{a}^{b} f(x) \sqrt{1+f'^2(x)}dx\\ &=2 \pi \int_{0}^{2} e^{-x^{2} / 2} \sqrt{1+\left(-x e^{-x^{2} / 2}\right)^{2}} d x\\ &=2 \pi \int_{0}^{2} e^{-x^{2} / 2} \sqrt{1+x^{2} e^{-x^{2}}} dx \\ & \text{Using a calculator gives:} \\ &\approx 8.222695606 \end{align*}