1. **State the problem:** We need to find the volume of the solid formed by rotating the region enclosed by the curve $y = e^{-x^2}$, the x-axis, and the vertical lines $x = -1$ and $x = 1$ about the x-axis. 2. **Set up the integral for volume:** The volume $V$ of the solid of revolution about the x-axis is given by the disk method formula: $$V = \pi \int_{a}^{b} [f(x)]^2 \, dx$$ where $f(x) = e^{-x^2}$, $a = -1$, and $b = 1$. 3. **Write the integral:** $$V = \pi \int_{-1}^{1} (e^{-x^2})^2 \, dx = \pi \int_{-1}^{1} e^{-2x^2} \, dx$$ 4. **Simplify the integral:** The integrand is $e^{-2x^2}$. 5. **Evaluate the integral:** The integral $$\int_{-1}^{1} e^{-2x^2} \, dx$$ is an even function, so $$\int_{-1}^{1} e^{-2x^2} \, dx = 2 \int_{0}^{1} e^{-2x^2} \, dx$$ 6. **Express the integral in terms of the error function:** The integral of $e^{-ax^2}$ from 0 to $t$ is $$\int_0^t e^{-ax^2} \, dx = \frac{\sqrt{\pi}}{2\sqrt{a}} \operatorname{erf}(t\sqrt{a})$$ where $a=2$ and $t=1$. 7. **Calculate the integral:** $$\int_0^1 e^{-2x^2} \, dx = \frac{\sqrt{\pi}}{2\sqrt{2}} \operatorname{erf}(\sqrt{2})$$ 8. **Substitute back to volume:** $$V = \pi \times 2 \times \frac{\sqrt{\pi}}{2\sqrt{2}} \operatorname{erf}(\sqrt{2}) = \pi \frac{\sqrt{\pi}}{\sqrt{2}} \operatorname{erf}(\sqrt{2})$$ 9. **Final answer:** $$\boxed{V = \frac{\pi^{3/2}}{\sqrt{2}} \operatorname{erf}(\sqrt{2})}$$ This is the exact volume of the solid of revolution formed by rotating the region about the x-axis.