What approach helps when the axis isn't a boundary in a solid of revolution?

[Justin_H]

I’m working through a calculus problem where I need to find the volume of a solid of revolution, but I’m stuck on setting up the integral correctly. I understand the disk method, but when the axis of rotation isn’t a boundary of the region, I get confused about the radius expression.

[John91]

Yeah, I ran into this too. I ended up treating the cross sections as washers, and I kept two radii. The outer radius is the distance from the axis to the far boundary of the region at that slice, the inner radius is the distance to the near boundary. Then I wrote the integral as pi times the difference of squares, integrated over the slice’s range. It sounds simple, but you have to get the two distance functions right and keep them as functions of the same variable.

[Matthew46]

One concrete thing I learned the hard way: don’t forget the inner radius when there is a hole. I once wrote R(y) but forgot r(y) and got nonsense. If the axis sits inside the region’s horizontal span, you’ll likely have R(y) bigger than r(y).

[Isabella_H]

I keep a rough sketch handy and label distances from the axis, even if it feels pedantic. If the axis is not a boundary, the radius isn’t just the max distance; you’re measuring two boundaries. Sometimes the algebra looks nicer with shells, but then you’re flipping variables and the limits shift.

[Charles53]

Do you know which boundary is closest to the axis for each cross-section? If not, you might be calculating the wrong inner radius.