`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.

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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...

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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

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