# Math Help - Solid of revolution

1. ## Solid of revolution

I tried to understand the solution to this problem but one thing I don't get:

the problem states: find the volume of the solid obtained by rotating the region bounded by the curves y= 1 + sec(x), y = 3, about y = 1;

so, the inner radius is (1 + sec(x)) - 1 = sex(x); the outer radius is 3-1 = 2; and we have the area A(x) = pi(4 - sec^2(x))

and it is integrated from -pi/3 to pi/3; how does they get these x-axis coordinates of pi/3?

thank you

2. ## Re: Solid of revolution

Set the two equations equal. You get $1+\sec(x) = 3$ or $\sec(x) = 2$. That happens when $\cos(x) = \dfrac{1}{2}$. Since $\cos\left(\pm \dfrac{\pi}{3} \right) = \dfrac{1}{2}$, those are your x-coordinates. Now, since $\sec(x)$ is periodic, there are actually an infinite number of those regions bounded by that curve and the line, so I must assume they only want you to look in a specific range, such as $[-\pi,\pi]$.