Find the volume of the solid obtained by rotating the region under the graph of the function f(x) = x^(13/12) about the x-axis over the interval [1,3].
Thanks!
This is pretty straightforward. Just for a sanity check, f(x) is very nearly x, so the volume is quite close to a truncated cone of height 2 and base radius 3, or 26π/3 = 8.67π
Using discs,
v = ∫[1,3] πr^2 dx
where r = y = x^(13/12)
v = π∫[1,3] x^(13/6) dx
= π (6/19 x^(19/6)) [1,3]
= 6π/19 (27 * 3^(1/6) - 1)
= 9.92π
Using shells, we have
x = y^(12/13), so
v = ∫[1,3^(13/12)] 2πrh dy
where r=y and h=(3-x) = (3-y^(12/13))
v = 2π∫[1,3^(13/12)] y(3-y^(12/13)) dy
= 2π (3/2 y^2 - 13/38 y^(38/13)) [1,3^(13/12)]
= 7.92π
Hmm. They should agree. Better check my algebra.