What is the surface area of the solid created by revolving #f(t) = ( t^3-t, t^3-t, t in [2,3]# around the x-axis?

Question

Scarlett Allison · Accepted Answer

not including the area of the end caps, S = # 540 \sqrt(2) \ \pi#, for the end caps add in # pi (6^2 + 18^2)# also

if you look at the parameterisation, #x = y = t^3 - t# so this is a straight line. and the interval specified corresponds in Cartesian to #(6,6) o (24, 24)# maybe there is a typo here as bothering with the calculus seems overkill but we can proceed anyways. starting with the simple idea of arc length, viz that #ds^2 = dx^2 + dy^2# we can say that #(\frac{ds}{dt})^2 = (\frac{dx}{dt})^2 + (\frac{dy}{dt})^2# and so #ds =\sqrt { (\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt# the elemental surface area of revolution is therefore # dS = 2 \pi \ y \ ds# so Surface Area is #S = 2 \pi \int y(t) \sqrt { (\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \ dt # and as the params are the same, #\frac{dy}{dt} = \frac{dx}{dt} = 3t^2 -1# so we have #S = \sqrt{2} \ 2 \pi \int_{t = 2}^{3} \ (t^3 - t) (3t^2 - 1) \ dt # # = 540 \sqrt(2) \ \pi# This is what you would get from the formula for the surface are of a truncated cone #S = pi ( s (R+r))# where s is slant height #s = \sqrt{(R-r)^2 + h^2}# NB this does not include the area of the end caps.