Answer
$V = \sqrt {2\pi } \left( {1 - \frac{1}{{\sqrt e }}} \right)$
Work Step by Step
$$\eqalign{
& y = \frac{1}{{\sqrt {2\pi } }}{e^{ - {x^2}/2}},{\text{ }}y = 0,{\text{ }}x = 0,{\text{ }}x = 1 \cr
& {\text{Applying the shell method about the }}y{\text{ - axis}} \cr
& V = 2\pi \int_a^b {x\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr
& {\text{From the graph}} \cr
& {\text{Let }}f\left( x \right) = \frac{1}{{\sqrt {2\pi } }}{e^{ - {x^2}/2}}{\text{ and }}g\left( x \right) = 0{\text{ on the interval }}\left[ {0,1} \right],{\text{ then}} \cr
& V = 2\pi \int_0^1 {x\left[ {\frac{1}{{\sqrt {2\pi } }}{e^{ - {x^2}/2}} - 0} \right]} dx \cr
& V = \sqrt {2\pi } \int_0^1 {x{e^{ - {x^2}/2}}} dx \cr
& V = \sqrt {2\pi } \int_0^1 {{e^{ - {x^2}/2}}} \left( { - x} \right)dx \cr
& {\text{Integrating}} \cr
& V = - \sqrt {2\pi } \left[ {{e^{ - {x^2}/2}}} \right]_0^1 \cr
& V = - \sqrt {2\pi } \left[ {{e^{ - {{\left( 1 \right)}^2}/2}} - {e^{ - {{\left( 0 \right)}^2}/2}}} \right] \cr
& {\text{Simplifying}} \cr
& V = - \sqrt {2\pi } \left( {{e^{ - 1/2}} - 1} \right) \cr
& V = \sqrt {2\pi } \left( {1 - {e^{ - 1/2}}} \right) \cr
& V = \sqrt {2\pi } \left( {1 - \frac{1}{{\sqrt e }}} \right) \approx 0.986 \cr} $$
