Question

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?

Answer 1

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

Explanation:

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) \to (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.