What did i mess up to get a negative volume with the shell method?

[LunaL]

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[Nora_G]

I'm working through a problem about finding the volume of a solid of revolution using the shell method, and I've hit a wall with the integration step. My setup seems correct, but when I evaluate the definite integral from the bounds I established, I'm getting a negative result, which can't be right for a volume. I'm wondering if my bounds are reversed or if I made an error in the integrand's simplification.

[ScottYR]

Yep, that happens a lot. If the integral comes out negative, I first check the bounds. With shells around the y axis the volume is 2 pi times integral of radius times height. If you swapped the limits, the sign flips. Try swapping the limits and see if the answer goes positive.

[DonaldM]

I had a case where I defined height as y_top minus y_bottom, but I accidentally labeled the lower curve as top. The numbers looked right but the sign was wrong until I swapped them.

[Oliver_M]

I once dropped a minus sign while simplifying the inner expression. The height looked positive on one part of the interval but negative on another, and the net integral came out negative. Re-deriving the height step by step fixed it, but it was sneaky.

[Madison.W]

Another angle: if your region straddles the axis or your axis is changing, you might need to split the integral. A single interval can hide a sign issue if the height changes sign over the range.

[ZacharyGR]

Could the problem be that the radius should be the distance to the axis, so if you're rotating around the y axis and x crosses zero you should use |x|? If your x-interval includes negative and positive, that sign can bite you even though the volume is positive.

[Gary32]

I remember tracing the curves on paper, then shading the shell heights. Seeing the top clearly above the bottom made the sign issue obvious even if the algebra stayed fuzzy for a while.