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Homework Help: Volume of a solid of revolution

  1. Feb 16, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the volume of the solid that results when the region bounded by the curves y=16-x[tex]^{2}[/tex] and y=16-4x is revolved around x=8.

    If you could show me how to do it with both the shell method and washer method it would be greatly appreciated.

    2. Relevant equations

    3. The attempt at a solution

    When trying the shell method i tried to shift the graph to the left 8 units and then revolving it about the y axis and got:
    y=16-(x+8)[tex]^{2}[/tex] and y = 16-4(x+8) which gave me

    2[tex]\pi[/tex]*[tex]\int^{-4}_{-8}[/tex]x(16-(x+8)[tex]^{2}[/tex] - (16-4(x+8))) dx

    and that gave me a negative answer so I know that can't be right
  2. jcsd
  3. Feb 17, 2010 #2


    Staff: Mentor

    I would suggest working with the graphs as they are. Your way could work, but seems prone to errors that might not otherwise occur, and doesn't really simplify things that I can see.

    If you use washers you will need to solve each of your equations for x in terms of y. The volume of a typical volume element is
    pi(R^2 - r^2)dy

    For R, the larger radius, measure from the vertical line to the x value on the slanted line. For r, the smaller radius, measure from the vertical line to the x value on the parabola. For example, R = 8 - <x value on slanted line>. And similar for r, the smaller radius. Since R > r, then R^2 > r^2, so you'll get a positive value for V.

    Using shells, the volume of the typical volume element is
    2*pi*R*(y val on parabola - y val on slanted line)*dx. R is the radius of the shell, measured from the line x = 8. If you measure it the right way you should get a positive value for V.
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