I am expected to use Mathematica to complete this task.
Given the functions $f(x)=\arctan x$ and $g(x)=2\cos x$, find the volume of the solid of revolution when the region bounded by $f$ and $g$, within the interval $[-\pi, \pi]$, is rotated about the line $y=-2$.
On paper, I sketched the solid, and found it to be a donut with a widened hole at one side.
The inner radius of the donut, then, was $IR=-2-\arctan(x)$, because that is the function on the inside of the donut. The outer radius was $OR=-2-2\cos(x)$, because that is the function on the outside of the donut.
Using Mathematica, I found the intersection points within the required interval to be $1.1331$ and $-2.17747$.
According to the formula for solid volume, I should integrate as follows:
$$V=\int \pi(OR^2−IR^2)dx,$$ which translates to:
$$V=\pi \int ((-2-2\cos x)^2−(-2-\arctan x)^2)dx,$$ using a definite integral from the intersection point $1.1331$ to $-2.17747$.
Mathematica evaluates this integral and provides the answer $94.5322 - 1.11022*10^{-16}i$.
It has been a while since I've used calculus; what am I misunderstanding or doing incorrectly? Any and all help is appreciated.