From the standard surface of revolution formula we have,
Area=#2piintf[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^3##sqrt[[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 the#1/36# 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=1#from
[#2#] into [#3#] as the limits of integration,will yield the required answer, above.
