`y = cos2x , y = 0 , x = 0 , x = pi/4` Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis.

Expert Answers

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

Given

`y=cos(2x), y=0 x=0,x=pi/4`

so the solid of revolution about x-axis is given as

`V = pi * int _a ^b [R(x)^2 -r(x)^2] dx`

here

`R(x) =cos(2x)`

`r(x)=0` and the limits are `a=0 ` and` b=pi/4`

so ,

`V = pi * int _a ^b [R(x)^2 -r(x)^2] dx`

= `pi * int _0 ^(pi/4) [(cos(2x))^2 -0^2] dx`

=`pi * int _0 ^(pi/4) [(cos(2x))^2 ] dx`
 
as we know `cos^2(x) = (1+cos(2x))/2`
so ,
`cos^2(2x) = (1+cos(4x))/2`
now
 
=`pi * int _0 ^(pi/4) [(1+cos(4x))/2 ] dx`
 
=`pi *  (1/2) int _0 ^(pi/4) [(1+cos(4x))] dx`
 
=`pi *  (1/2)  [(x+(1/4)sin(4x))]_0 ^(pi/4) `
 
=`pi/2 [pi/4 +(1/4)(sin(pi))-[0+0]]`
 
= (`pi/2)[pi/4]`
 
=`pi^2/8`

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