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.
