Question

Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = 1 + sec(x), −π 3 ≤ x ≤ π 3 , y = 3; about y = 1

Answer 1

Using the washer method, the volume is given by the integral

$\displaystyle\pi\int_{-\pi/3}^{\pi/3}\bigg((3-1)^2-((1+\sec x)-1)^2\bigg)\,\mathrm dx=2\pi\int_0^{\pi/3}(4-\sec^2x)\,\mathrm dx$

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

$\displaystyle2\pi\int_0^{\pi/3}(4-\sec^2x)\,\mathrm dx=2\pi(4x-\tan x)\bigg|_0^{\pi/3}=\boxed{\dfrac{8\pi^2}3-2\pi\sqrt3}$