Chapter 6 - Inverse Functions - 6.3* The Natural Exponential Function - 6.3* Exercises - Page 454: 98

$V= \pi (1-e^{-1})=1.986$

The volume of the solid obtained by rotating the region under the curve $y=e^{-x^{2}}$ From 0 to 1 about the x-axis is equal to $V=\int_{0}^{1} 2\pi xe^{-x^{2}} dx$ Consider $x^{2}=t$ $2xdx=dt$ $dx=\frac{1}{2x}dt$ Then $V=\pi \int_{0}^{1} e^{-t} dt$ $V=\pi (e^{0}-e^{-1})$ Hence, $V= \pi (1-e^{-1})=1.986$