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Math Help - Volume By Slicing

  1. #1
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    Volume By Slicing

    The question is read as follows: Solid lies between planes perpendicular to the x-axis at x=1 and x=-1 . The cross sections are perpendicular to the x-axis are circular disk.s whose diamters run from the parabola y=x^2 to the parabola y=2-x^2 FInd the volume of the solid.

    I tried doing this by the given integral - v= int (a to b) A(x) dx =int (a to b) pi R(x)^2 dx
    I ve determined that the radius should equal to 2 . I dont know where to go from here to determine the limits of integration (a and b values) and such. Any help is appreciated
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  2. #2
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    Quote Originally Posted by dreamx20 View Post
    The question is read as follows: Solid lies between planes perpendicular to the x-axis at x=1 and x=-1 . The cross sections are perpendicular to the x-axis are circular disk.s whose diamters run from the parabola y=x^2 to the parabola y=2-x^2 FInd the volume of the solid.

    I tried doing this by the given integral - v= int (a to b) A(x) dx =int (a to b) pi R(x)^2 dx
    I ve determined that the radius should equal to 2 . I dont know where to go from here to determine the limits of integration (a and b values) and such. Any help is appreciated
    1. Draw a sketch of the situation.

    2. The midpoints of the disks are placed on the line y = 1. Thus the radius of the disks can be calculated by

    r = 1-x^2

    3. The limits of integration are obviously given by the x-coordinates of the points of intersection between the two graphs.

    4. I've got a volume of V=\frac{16}{15} \pi \ vol.units
    Attached Thumbnails Attached Thumbnails Volume By Slicing-scheiben_inparab.png  
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  3. #3
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    Since, as earboth says, the radius of each disk is 1- x^2, the area is \pi(1- x^2)^2= \pi(1- 2x^2+ x^4). The thickness of each disk is along the x-axis and so is "dx". The volume of a disk is "area times thickness", \pi(1- 2x^2+ x^4)dx.

    "Add" the thicknesses of all the disks. In the limit, that becomes the integral \pi\int_{-1}^1 1- 2x^2+ x^4 dx.
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