1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?

    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


    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


    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook