Question

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`?

Answer 1

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