Using the washer method, the volume is given by the integral
[tex]\displaystyle\pi\int_{-\pi/3}^{\pi/3}\bigg((3-1)^2-((1+\sec x)-1)^2\bigg)\,\mathrm dx=2\pi\int_0^{\pi/3}(4-\sec^2x)\,\mathrm dx[/tex]
where 3 - 1 = 2 is the distance from y = 3 to the axis of revolution, and similarly (1 + sec(x)) - 1 = sec(x) is the distance from y = 1 + sec(x) to the axis. The integrand is symmetric about x = 0, so the integral "folds" in on itself, and the integral from -π/3 to π/3 is twice the integral from 0 to π/3.
So the volume is
[tex]\displaystyle2\pi\int_0^{\pi/3}(4-\sec^2x)\,\mathrm dx=2\pi(4x-\tan x)\bigg|_0^{\pi/3}=\boxed{\dfrac{8\pi^2}3-2\pi\sqrt3}[/tex]
