Question

Using disk washers, find the volume of the solid obtained by rotating the region bounded by the curves y=sec(x), y=1, x=−1, and x=1 about the x-axis.

Answer 1

General integral for volume using washer method:
$V = \pi \int_a^b (R^2 - r^2) dx$

First determine limits along x-axis.
region is bounded by x=1 and x=-1, so the limits are also -1 and 1.

Next determine R(x) and r(x)
R is the long radius, which is distance from x-axis to upper bound.
The upper bound of region is y = sec(x).
R = sec(x)

r is the short radius, which is distance from x-axis to lower bound.
The lower bound is y = 1.
r = 1

Sub into integral:
$V = \pi \int_{-1}^1 (sec^2 x - 1) dx$

Integrate
$V = \pi |_{-1}^1 (tan x - x)$

Evaluate
$V = \pi[(tan(1) -1) - (tan(-1) +1)] \\ \\ V = 2\pi(tan(1) -1) \\ \\ V = 3.502$