From the standard surface of revolution formula we have,
Area=2piintf[x]2π∫f[x]sqrt [1+[f'x]^2dx, ...........[1], let y =f[x]
So Area=2piint[x^3]sqrt[1+[dy/dx]^2dx and now e need to differentiate x^3 and dy/dx[x^3]= 3x^2 and so [dy/dx]^2=9x^4.
Now we can Re-state[1] as follows Area=2piintx^3sqrt[[1+9x^3]]dx.............[2]
Let u=1 +9x^4......[2] .Differentiating wrt# x...[du]/dx]=36x^3# and so #[du]/[36x^3]=dx#.
substituting for dx and u in [2] we obtain.........
Area=2piintx^3sqrtu[du]/[36x^3],and the terms in x^3 will cancel after taking the1/36 constant outside the integral leaving Area=pi/18intsqrtudu=pi/18{2/3sqrt[u^3]}= pi/27[sqrt[u^3]] .......[3]
Substituting, when x=1, u =10 , and when x=0,u=1from
[2] into [3] as the limits of integration,will yield the required answer, above.
