Originally Posted by

**HallsofIvy** The region bounded by those curves is rather small! And the simplest way to find the volume rotated around the x-axis is to do it as two integrals. The volumes bounded above by y= sec(x), below by y= 0, on the left by x= -1, and on the right by x= 1, rotated around the x-axis, is $\displaystyle \pi\int_{-1}^1 sec^2(x) dx$. The volume bounded above by y= 1, below by y= 0, on the left by x= -1, and on the right by x= 1, rotated around the x-axis, is $\displaystyle \pi\int_{-1}^1dx= 2\pi$ (actually the volume of a cylinder with radius 1 and height 2). The volume you want is the **difference** between those two. That is the same as $\displaystyle \pi \int_{-1}^1 (sec^2(x)- 1) dx$.

That is based on the fact that the volume of a cylindrical "donut" with inner radius r, outer radius R, and height h, is $\displaystyle \pi(R^2- r^2)$ which is the same as the volume of the whole cylinder, $\displaystyle \pi R^2h$ minus the volume of the inner cylinder, $\displaystyle \pi r^2h$.