What mistake can cause negative volume in shell method for solids of revolution?

[StephenA]

I’m working through a calculus 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 right, but when I evaluate the definite integral from the bounds, I’m getting a negative result for volume, which can’t be correct. I’m not sure if my radius or height function is wrong, or if I messed up the antiderivative.

[Jonathan.T]

I’ve run into this exact snag with the shell method. If you’re getting a negative result, the usual culprit is the height function h being negative somewhere on the interval. The radius is fine as long as you’re measuring from the axis of rotation; the fix is to set h(x) = y_top − y_bottom so it’s nonnegative over [a, b], or just take |h| in the integral. Then the 2π ∫ r h dx should come out positive and match the volume.

[Chloe.G]

One time I swapped the order of the functions and the bounds without noticing, and the definite integral came out negative even though the region was fine. Rewritten with h = top − bottom and kept r positive, and it cleared up.

[BrandonCS]

I also checked the axis: around y-axis, the radius is x, not y. If you flip the axis without flipping r, you get wrong signs.

[NoraUJ]

Are you sure about the axis of revolution? Sometimes the real issue isn’t the antiderivative but the axis choice that makes radius sign flip in the middle of the interval.