`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.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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`


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

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial