Why is my washer method volume negative, could it be the limits or squaring?

[Stella31]

I'm working through a problem about finding the volume of a solid of revolution using the washer method, and I've set up my integral with the correct outer and inner radii. But when I evaluate it, I keep getting a negative result, which doesn't make sense for a volume. I must be messing up the limits of integration or the order of the functions when I square them.

[AdamDM]

I had this exact thing last semester. I kept getting a negative volume until I realized I was squaring the difference of radii instead of squaring each radius separately. The correct setup is pi times (R^2 minus r^2). If you do (R - r)^2, you’ll see negative values wherever R < r on part of the interval.

[Aurora70]

Another pitfall is the limits. If you’re integrating with respect to x but your washers are built from y-values, you might be mixing things up and get a sign flip. Make sure the bounds reflect the interval for the axis and that R(x) and r(x) really match what’s on that interval.

[Luna_H]

I tried a quick numeric check: picked a point, computed R and r, and R^2 - r^2 came out negative there when I expected positive. It helped to switch to absolute distances to the axis: R = |f(x)|, r = |g(x)|, then square. It’s ugly but it made the sign consistent.

[Luna_W]

Is the axis of rotation really the one you used to define the radii, or could the problem be that you’re rotating around the y-axis while you wrote the radii as distances to the x-axis?