You are revolving about the x-axis from x=1- to x=1
Using washers:
Which simplifies to
i'll give the problem then explain my issue with it.
the solid lies between planes perpendicular to the x axis at x=-1 and x=1. The cross sections perp. to the x axis are circular disks whose diameters run from the parabola y=x^2 to the parabola y=2-x^2
seems pretty simple but for the life of me i can't figure it out. I figured to use the washer method, with y=2-x^2 being R and y=x^2 being r. so i set that up and integrate it, and get volume = pi(4x-2x^2/3) . (There were two other functions, but one was positive, one was negative, so they cancel out, leaving this). I need to evaulate that from -1 to 1. but doing that results in 0, which obviously can't be the answer. where did i go wrong?
??? I see no reference in the problem to "rotating" anything around the x-axis.
Imagine dividing the solid into thin sheets of thickness " ". Each sheet can be approximated by a disk with upper boundary at and lower boundary at for some x in that section of the x axis. Thus, each disk has diameter and so radius . The area of each disk is and so the volume is . The approximate area, the sum of those disks, is . That "Riemann sum", in the limit, becomes the exact area .
The integral galactus gives is correct for the region between and , rotated around the x-axis but that does not appear to be the problem as stated: "The cross sections perp. to the x axis are circular disks whose diameters run from the parabola y=x^2 to the parabola y=2-x^2". I would certainly not call the line between inner and outer boundaries of a washer a "diameter".