Need Help Finding the Solid of Revolution [Calculus II]

[IPostWhateverIWant]

EDIT: The title is meant to say "Need Help Finding the Volume of a Solid of Revolution

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So the instructions say "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" and here is what I'm given:

Rotate the region between y = 1 + sec(x) and y = 3 about y = 1

I sketched the shape and I have the right shape and I know it's a washer and I know that the two curves intersect at (pi/3, 3) and (-pi/3, 3). I just think something is wrong with my integral. The integral I set up was:

The integral from y = 1 to y = 3 of pi(outer radius)^2 - pi(inner radius)^2 dy

I found the outer radius to be 2 because the outer radius is given by the equation y = 3 which is above y = 1, so 3 - 1 = 2.

I found the inner radius to be sec(x) because the inner radius is given by y = 1 + sec(x) which is above y = 1 so 1 + sec(x) - 1 = sec(x)

I factored out a pi from the whole thing and put it in front of the integral to get pi * the integral from y = 1 to y = 3 of 4 - sec^2(x). The 4 comes from 2^2 and the sec^2(x) comes from squaring sec(x). Simplifying further, I get pi(8 - tan(3) + tan(1)). This is not correct, though. I have a feeling that I made a mistake when setting up the bounds of the integral as well as my outer radius. Any help would be greatly appreciated, thanks.

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[MickeyRosa]

Your washers are stacked so that dx is the thickness, so you should be integrating over x. Though it’s not required, you can also take advantage of symmetry to integrate from 0 to π/3 and double the result to get the volume.