How do you find the area of the surface generated by rotating the curve about the y-axis #x=t^2, y=1/3t^3, 0<=t<=3#?

1 Answer
Jul 30, 2018

For infinitesimal arc length, #ds#:

#ds = sqrt(dx^2 + dy^2 )#

  • # {(x = t^2, dx = 2 t\ dt),(y = 1/3 t^3, dy = t^2\ dt):}#

#:. ds = sqrt(4t^2 + t^4 ) \ dt = t sqrt(4 + t^2 ) \ dt#

Spinning round the y-axis, creating infinitesimal surface area #dS#, where:

  • #dS = 2 pi \ x \ ds#

#= 2 pi \ t^3 \ sqrt(4 + t^2 ) \ dt#

#:. S =2 pi int_0^3 \ dt qquad t^3 sqrt(4 + t^2 ) #

#= (2 pi)/15 (64 + 247 sqrt13)#

I have skipped the mechanical integration steps. you can start with a sub:

  • #u = 4 + t^2 qquad du = 2 t \ dt#

#:. S =2 pi int_4^13 \ (du)/(2t) qquad t^3 \ sqrtu#

#= pi int_4^13 \ du qquad (u - 4) \ sqrtu#

#= pi int_4^13 \ du qquad u^(3/2) - 4u^(1/2) #

Which is trivial but protracted