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.

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