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I was working through a problem about finding the volume of a solid of revolution using the shell method, and I got a different answer than the one in the textbook. I set up my integral as ∫ 2π x (f(x)) dx over the interval, but I think my error might be in how I determined the radius of the cylindrical shell.

If you’re rotating around the y axis and you’re using vertical shells, the radius is just x. So 2π x times the shell height is the right setup provided the height is the vertical distance between the top and bottom boundaries (top minus bottom). If the region isn’t simply f(x) above the axis, you need to replace the height with the appropriate difference.

I remember thinking the same thing once: the distance to the axis changes things. If you rotate about a vertical line other than the y axis, use distance = |x minus the axis value|. And if your region is between two curves, height is top minus bottom, not just f(x).

I once did this and the numbers seemed off. I graphed the region and realized I was using f(x) as the height when the actual height was the difference between two curves; that mismatch made the volume off. I chased a couple of tiny algebra slips and even second-guessed the axis a bit, only to find it was the height I’d mis-specified.

Are you rotating around the y axis or a vertical line?

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