#### Answer

$$ 8.222695606$$

#### Work Step by Step

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*}