Question Number 104703 by abony1303 last updated on 23/Jul/20

Commented byabony1303 last updated on 23/Jul/20

pls help too :)

Answered by bemath last updated on 23/Jul/20

vol =π∫_0 ^a  {(sin x)^2 −(((2x)/π))^2 }dx  = π ∫_0 ^a  {(1/2)−(1/2)cos 2x−((4x^2 )/π^2 )} dx  = π { (x/2)−((sin 2x)/4)−((4x^3 )/(3π^2 ))}_0 ^a   = π{(a/2)−((sin 2a)/4)−((4a^3 )/(3π^2 ))}  where a is solution from  ⇒sin x = ((2x)/π)