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Find the volume of the solid formed when the region bounded by y=3x^2 and y=36-x^2 is revolved about the x-axis.

  • Calculus - ,

    I see several posts made by you dealing mainly with these integrations problems, without any indication by you showing any steps or attempts.
    I will do this one. Try to follow these steps to solve the others
    Make a sketchl
    First you need the intersection to see the region that you are revolving.
    y = 3x^2 and y = 36-x^2

    3x^2 = 36-x^2
    4x^2 = 36
    x = ± 3
    Because of the symmetry, I will boundaries from 0 to 3, then double our volume

    in general,
    Volume = π∫y^2 dx from left boundary to right boundary
    the y , the height or radius of our rotation, is
    36-x^2 - 3x^2 = 36-4x^2

    V = 2π∫ (36-4x^2)^2 dx from 0 to 3
    = 2π∫ (1296 - 288x^2 + 16x^4) dx from 0 to 3
    = 2π [ 1296x - 96x^3 + (16/5)x^5] from 0 to 3
    = 2π (3888 - 2592 + 3888/5 - 0 - 0 - 0)
    = (20736/5)π = appr. 13028.8

  • Calculus - ,

    I use this to check my answers. I apologize for not realizing that I should have proven to you that I was doing them by myself at first.

  • Calculus - ,

    Fair enough.

    In that case, you should post the answer you got, that way I can just do some fast scribbling on a piece of scrap paper and either confirm or reject your answer.
    It would save us a lot of unnecessary typing of the whole solution.

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