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Math Help - volume of a solid

  1. #1
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    volume of a solid

    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?
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  2. #2
    Eater of Worlds
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    You are revolving about the x-axis from x=1- to x=1

    Using washers:

    {\pi}\int_{-1}^{1}[x^{4}-(2-x^{2})^{2}]dx

    Which simplifies to 4{\pi}\int_{-1}^{1}(1-x^{2})dx
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  3. #3
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    ??? I see no reference in the problem to "rotating" anything around the x-axis.

    Imagine dividing the solid into thin sheets of thickness " \Delta x". Each sheet can be approximated by a disk with upper boundary at y= 2- x^2 and lower boundary at y= x^2 for some x in that section of the x axis. Thus, each disk has diameter 2- x^2- x^2= 2- 2x^2 and so radius 1- x^2. The area of each disk is \pi(1- x^2)^2= \pi(1- 2x^2+ x^4) and so the volume is \pi(1- 2x^2+ x^4)\Delta x. The approximate area, the sum of those disks, is \sum\pi(1- 2x^2+ x^4)\Delta x. That "Riemann sum", in the limit, becomes the exact area \pi\int_{x= -1}^1 (1- 2x^2+ x^4)dx.

    The integral galactus gives is correct for the region between y= 2- x^2 and y= x^2, 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".
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  4. #4
    Eater of Worlds
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    Oh, I see. I quickly assumed it was a S.O.R. I see now what you mean, HOI. Sorry 'bout that. I was at the wheel
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  5. #5
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    Sorry about the confusion, my mistake.

    Thanks very much for all your help, both of you. Much appreciated
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