What did i mess up with the shell method radius in a solid of revolution?

[JamesMM]

I was working through a problem about finding the volume of a solid of revolution using the shell method, and I got a different answer than the one in the book. I set up my integral as ∫ 2π * x * f(x) dx over the interval, but my constant factor is off. I’m wondering if I made a mistake in identifying the correct radius for the cylindrical shell.

[FrankT]

Radius is distance to the axis. If you rotate about the y axis, the shell radius is x. If the axis is x = c, use |x - c| as the radius.

[JohnFS]

The height of a shell is the vertical thickness of the region. If your region sits between y = f(x) and y = g(x), the height is f(x) - g(x). If it’s between y = 0 and y = f(x), height is f(x).

[William6]

Check your bounds and axis—sometimes a sign slip or mixing left and right limits changes the result, even if you kept r = x.

[Stella85]

Could the axis be a line other than the y axis, like x = 2? Then the radius would be |x - 2|, which changes the constant.

[AddisonL]

I tried this once and still felt unsure; I compared to a washer setup and the numbers lined up only after I reinterpreted the height; still not confident.