Answer
$$V = \left( {\frac{{1 - {e^{ - 2}}}}{2}} \right)\pi $$
Work Step by Step
$$\eqalign{
& y = {e^{ - x}},{\text{ }}y = 0,{\text{ }}x = 0,{\text{ }}x = 1 \cr
& {\text{From the graph shown below, we can apply the disk method}} \cr
& V = \pi \int_0^1 {{{\left[ {{e^{ - x}}} \right]}^2}} dx \cr
& V = \pi \int_0^1 {{e^{ - 2x}}} dx \cr
& {\text{Integrate}} \cr
& V = - \frac{1}{2}\pi \left[ {{e^{ - 2x}}} \right]_0^1 \cr
& V = - \frac{1}{2}\left[ {{e^{ - 2\left( 1 \right)}} - {e^{ - 2\left( 0 \right)}}} \right] \cr
& V = - \frac{1}{2}\left[ {{e^{ - 2}} - 1} \right] \cr
& V = \left( {\frac{{1 - {e^{ - 2}}}}{2}} \right)\pi \cr} $$
