Rotational Area?

Use the washer method and see graph
https://www.desmos.com/calculator/u4vljrotwy
Rotating about y = 2
Outer radius = 4-2 = 2
Inner radius = f(x)-2 = 2+secx-2 = secx
As the curve is symmetrical we can use bounds 0 to π/3 and double the result.
V = volume = 2π∫[2^2-sec^2(x)]dx [0,π/3]
= 2π[4x-tanx] [0,π/3]
= 2π[4π/3-√3]
≈ 15.44

Note that sec(pi/3) = sec(-pi/3) = 2, and sec(0) = 1,
so the curve y = 2 + sec(x) meets y = 4 at the boundaries,
and the given region is something like a tongue hanging
down from the line y=4, with its tip at the point (0,3).

The method of "washers" is probably a little easier than the method of shells, since I don't feel like trying to integrate [2*(4-y)*arcsec(y-2)] dy.

The outer radius of a washer is 4 and the inner radius is 2 + sec(x).
The area of the washer is pi*[16 - 4 - 4*sec(x) - sec^2(x)]
= pi*[12 - 4*sec(x) - sec^2(x)].
Then this will be integrated with respect to x; the indefinite integral is
pi*[12x - 4*ln|sec(x) + tan(x)| - tan(x)].
At x = pi/3, the value is
pi*[4pi - 4*ln|2 + sqrt(3)| - sqrt(3)];
at x = -pi/3, the value is
pi*[-4pi - 4*ln|2 - sqrt(3)| + sqrt(3)],
so the volume is
8pi^2 - (4pi)*ln[1] - 2pi*sqrt(3) = 8pi^2 - 2pi*sqrt(3) = about 68.

This seems possible, as the volume of the enclosing cylinder is
16*pi*(2*pi/3) = about 105.