# `y = e^(-x^2/2)/sqrt(2pi) , y = 0 , x = 0 , x = 1` Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis. Using the shell method we can find the volume of the solid generated by the given curves,

`y = e^(-x^2/2)/sqrt(2pi), y = 0 , x = 0 , x = 1`

Using the shell method the volume is given as

`V= 2*pi int _a^b p(x) h(x) dx`

where p(x)...

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Using the shell method we can find the volume of the solid generated by the given curves,

`y = e^(-x^2/2)/sqrt(2pi), y = 0 , x = 0 , x = 1`

Using the shell method the volume is given as

`V= 2*pi int _a^b p(x) h(x) dx`

where p(x) is the function of the average radius  `=x`

and

h(x) is the function of height =  `e^(-x^2/2)/sqrt(2pi)`

and the range of x is given as 0 to 1

So the volume is  `= 2*pi int _a^b p(x) h(x) dx`

`= 2*pi int _0^1 (x) (e^(-x^2/2)/sqrt(2pi)) dx`

`=(2*pi)/(sqrt(2pi)) int _0^1 (x*e^(-x^2/2)) dx`

`=(2*pi)/(sqrt(2pi)) int _0^1 (x*e^(-x^2/2)) dx`

let us first solve

`int (x*e^(-x^2/2)) dx`

let `u = x^2/2`

`du = 2x/2 dx = xdx`

so ,

`int (x*e^(-x^2/2)) dx`

`= int  (e^(-u)) du`

`= -e^(-u) = -e^(-x^2/2)`

So,  `V=(2*pi)/(sqrt(2pi)) int _0^1 (x*e^(-x^2/2)) dx`

`=(2*pi)/(sqrt(2pi)) [-e^(-x^2/2)]_0^1`

=`(2*pi)/(sqrt(2pi)) [[-e^(-(1)^2/2)]-[-e^(-0^2/2)]]`

=`(2*pi)/(sqrt(2pi)) [[-e^(-1/2)]-[-e^(0)]]`

=`(sqrt(2pi)) [1-[e^(-1/2)]] `

=` 0.986`

is the volume

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