Determining the volume of the solid of revolution formed by y=1+sec x and y=3 rotated around y=1

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The instructions for this question verbatim are: "Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer."

I believe the domain is $-\pi/2$ to $\pi/2$

I've been having a really hard time visualizing the functions for the radius of the inner and outer rings. I've graphed the function and the lines but I'm still stumped. I thought that the formula for the outer radius would be $\pi(2)^2$ because the outer radius is always 2 units away from the line of rotation, and the inner radius $2-(1+\sec x)$. With these equations I end up having to find the anti-derivative of 2 sec x which we haven't done in class. This makes me think I'm wrong about one of the expressions - likely the inner radius. Do you have any advice for understanding how the expressions for the inner and outer slices are affected by the point of rotation?

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It would probably help to first graph the functions so you can visualize the region you are rotating. You can do this via online graphing programs or a graphing calculator if you have one. Does the question specify which quadrant/interval to look at? I ask because $y=1 + sec(x)$ intersects $y=3$ at many points.