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Homework Help: Volumes of Solid of Revolution

  1. Mar 17, 2010 #1
    I have a volume problem that I has been bothering me for a while now, as I have just not been able to figure it out.

    The question involves finding the volume of the solid generated by revolving the region bounded by y = x, y = 0, and y = 4 around the line x = 6. I tried doing [tex]\int[/tex]
    [tex]^{4}_{0}[/tex]16 - (4-y)[tex]^{2}[/tex] dy + [tex]\int[/tex][tex]^{4}_{0}[/tex]36-16 dy but that did not work out. I am fairly sure that the part of it that is giving me problems is the little extra area between x = 6 and x = 4, as the y = x and x =6 do not intersect due to the y = 4 boundary...

    There is also another question like this that has been giving me problems, involving fining the volume of the solid generated by x = y^2 and x = 4 around the line x = 6; would this be done the same way?

    Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 17, 2010 #2
    By what you wrote, I would think the solid is being revolved around x=6; therefore, you need to use the formula V= pi* the integral of [R(x)]^2-[r(x)]^2 and little r is the distance from the solid to the axis of revolution. So the shape meets at (4,4) as the end point with a difference of 2 units to x=6.

    I hope this helps.
     
  4. Mar 17, 2010 #3
    Oops, I seem to have left out the pis in the original post...

    So, now I tried using (6-x) for R(x) and 2 for r(x) and integrating from 0 to 6, but I end up with 48 pi... The answer is supposed to be 208(pi)/3... What am I doing wrong?
     
  5. Mar 17, 2010 #4
    Which one is 208 pi/3? The first or second? Also, the shell method maybe easier to use.
     
  6. Mar 17, 2010 #5
    Maple is telling me the answer is (224/3)*pi
     
  7. Mar 20, 2010 #6
    The first one is supposed to be 208(pi)/3, or at least that is what it says in my textbook.
     
  8. Mar 20, 2010 #7

    Mark44

    Staff: Mentor

    I don't believe you have posted the problem correctly. What you have described is the region between two parallel lines (y = 0 and y = 4), cut somehow by the line y = x. This does not determine a finite region.

    Also, for future reference, use one pair of [ tex] and [ /tex] tags for the entire integral. You are putting way too many pairs of them, which results in a garbled-up mess. Here is your first integral (which probably has nothing to do with this problem):

    [tex]\int_{0}^{4} 16 - (4-y)^{2} dy[/tex]

    Double-click the integral to see the LaTeX script that generates this integral.
     
  9. Mar 22, 2010 #8
    Well, I have rechecked the problem's statement in my textbook, and that is exactly what it says. I was assuming though that it meant for me to revolve a triangular area (bounded by y = x, y = 0, y= 4, and x=6) around x = 6.

    And thank you for the help with the typing of the integrals :]. Goodness knows I needed it.
     
  10. Mar 22, 2010 #9

    Mark44

    Staff: Mentor

    In that case, I think your book has a typo. The region bounded by y = x, x = 0, and y = 4 is probably what they meant.

    Probably the most natural way to do this is to use circular washers, disks with holes in them. The volume of a typical volume element is
    [tex]\Delta V = \pi (R^2 - r^2) \Delta x[/tex]

    Here, R is the radius from the axis of rotation (the line y = 6) to the line y = x, and r is the radius from the axis of rotation to the line y = 4.
     
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