Why do shell and disk methods give different volumes for the same solid?

[PenelopeWL]

I'm working through a problem about finding the volume of a solid of revolution using the shell method, but I keep getting a different answer than when I use the disk method for the same region. I've checked my bounds and my integrals several times, and I'm starting to wonder if my understanding of the radius for the cylindrical shells is off.

[ElizabethFT]

I had the same doubt once. With shells the radius is the distance from the axis of revolution to the shell. If you rotate about the y axis and slice with dx, the radius is x. The height should be the vertical extent of the region at that x, basically top y minus bottom y. If you’re using only one function, make sure you’re accounting for where that function sits relative to the axis.

[Adam.H]

I found I was mixing up height. I wrote the height as f(x) but the region actually sits between two curves, so the height is f_top(x) - f_bot(x). Small mistake like that makes the numbers diverge fast, even though the setup looks right.

[Mark_M]

Could be the bounds. Sometimes the x limits come from where the region starts and ends in x, not just the intersection; with shells you may have to project the region onto the axis of rotation. I once swapped in the wrong x-limits and the numbers didn't match the disk method.

[Robert_T]

I keep doubting whether the problem's real issue is something else, like the axis being misread in the diagram. I’d plot the region and do a quick sanity check by approximating the volume with a bunch of tiny shells by hand and see if the trend matches the disk result, even if the exact value is off.