How to set up disk method for region between y=x^2 and y=4 about the x-axis?

[Ryan.D]

I'm working through a problem about finding the volume of a solid of revolution using the disk method, but I'm getting stuck on setting up the integral for the region bounded by y = x² and y = 4. My limits of integration seem off because when I solve for x I get ±2, but I'm not sure if that's correct for revolving around the x-axis.

[AubreyM]

On revolving around the x axis the washers have outer radius 4 and inner radius x^2, so the cross sectional area is pi(4^2 - (x^2)^2) = pi(16 - x^4). The x limits come from where the curves meet: x^2 = 4, so x = -2 to 2. So the volume is integral from -2 to 2 of pi(16 - x^4) dx, which evaluates to 256 pi / 5.

[Ethan.W]

I messed with it last night and initially tried 0 to 2 because of symmetry, but the region sits on both sides of the origin, so you do -2 to 2.

[Noah.J]

I once forgot the subtraction and ended up with pi∫ (4^2) dx; realized later the hole matters because y=x^2 becomes the inner radius.

[Mark89]

Maybe you're accidentally mixing axes—the same region rotated about the x-axis gives washers; about the y-axis would be shells. Is it possible you meant around the y-axis?