The standard formula for the surface area of revolution is ,

#SA="2piintf[x]sqrt[1+[[f'x]]^2dx#, Let #y=f[x]=x^3#.....so in this case having taken the #1/3# outside the integrand we obtain the following,

#SA=2pi/3intx^3sqrt[1+[d/dx[x^3]]^2dx#

#d/dx[x^3]=3x^2# and squaring..we have #9x^4#

Therefore #SA=2pi/3intx^3sqrt[1+9x^4] dx#.... Now let #u =1+9x^4#..........#[1]#

Differentiating ......#[1]#, #[du]/[36x^4]=dx# and so the integral now becomes , #2pi/3intx^3sqrt[udu]/[36x^3]# The terms in #x# will now cancel and #1/36 #can also be taken outside the integrand leaving

#SA=[2pi]/[108]intsqrt[udu#= #[2pi]/108intu^[1/2]du#

After integration, #SA=[4pi]/324sqrt[u^3]#............ From......#[1] #, #u=1+9x^4# and changing the bounds of integration

When #x=1#, #u=10#, and when #x=0#, #u=1#

Evaluated this becomes #SA#= #[4pi]/[324]##[10sqrt10-1]#units^2. Hope this is helpful.