# Find exact volume of the solid obtained by rotating about `x=pi/2` the region bounded by `y=sin^2x, y=sin^4x, x in[0,pi/2]` Note: WA can be used Please keep in mind that `c in[a,b]...

Find exact volume of the solid obtained by rotating about `x=pi/2`

the region bounded by `y=sin^2x, y=sin^4x, x in[0,pi/2]`

Note: WA can be used

Please keep in mind that `c in[a,b] ->a<=c<=b`

### 1 Answer | Add Yours

The volume generated by the curves is

`V=int_0^(pi/2)pi(sin^2(x)-sin^4(x))^2dx`

Now let simplify the problem

`sin^2(x)-sin^4(x)=sin^2(x)cos^2(x)`

`(sin^2(x)-sin^4(x))^2=(sin(x)cos(x))^4=(sin^4(2x))/16`

`=(1/16)((1-cos(4x))/2)^2`

`=(1/64)(1+cos^2(4x)-2cos(4x))`

`=(1/64)(1+(cos(8x)+1)/2-2cos(4x))`

`=(1/128)(2+cos(8x)+1-4cos(4x))`

`=(1/128)(3+cos(8x)-4cos(4x))`

`` Thus

`V=(pi/128)int_0^(pi/2)(3+cos(8x)-4cos(4x))dx`

`=(pi/128){3x+(sin(8x))/8-sin(4x)}_0^(pi/2)`

`=(pi/128)(3pi)/2`

`=(3pi^2)/256` cubic unit