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Calculating volumes by shell and disc method (Looking for Professional advice)

  1. Mar 9, 2009 #1
    1. The problem statement, all variables and given/known data

    The function is y = 2 - x. The region is bounded by x = 2 and x = 4. Calculate its volume by the shell method by rotating it by the x axis.


    3. The attempt at a solution

    This problem has been consuming my mind. I calculated it by the disc method and shell method but I got 2 different answers.

    Disc method:

    ∫ pi (y^2) dx...Evaluated from 2 to 4
    ∫ pi(2 - x)^2 dx

    Result : 8pi/3

    Shell method:

    ∫ 2pi*y*x*dy......Evalluated from -2 to 0.
    ∫ 2pi(y^2) dy

    Result : 16pi/3
     
  2. jcsd
  3. Mar 9, 2009 #2

    lanedance

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    I don't think your shell volume element is set up correctly....

    can you show you steps in getting to each element

    ie for the shell case
    r(y) = ?
    L(y) = ?
    A = 2.pi.r(y).L(y)
    dV = A.dy

    Also this problem can be simplified by shifting to the equivalent problem of y=x rotated about x axis with x from 0 to 2, which makes it a little easier, i did it this way and get 8.pi/3 for both
     
  4. Mar 9, 2009 #3

    lanedance

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    disclaimer: this advice is in no way professional
     
  5. Mar 9, 2009 #4
    Hi again

    Alright, kinda tired at this point (its 2am here) but I'm willing to show all steps

    For the shell the method the way I thought about it was folding it into a crown in "3 d." The circumference is just 2*pi*y times height x thickness dy which in your terms would be 2*pi*y*(2 - x) dy. The reason why I ended up with 2 - x is because the crown is bigger at one end yet thins out on the other side.

    r(y) = y
    L(y) = 2 - x

    So this was a strange problem for me, because 2 - x actually just reduces to y. No matter how hard I tried making it looking different. So yeah, I got 2pi*y^2 dy. Evaluated from the limits -2 to 0. I haven't thought about the function being shifted but I'll test that idea. We'll see

    Edit #2: Ok I got the same thing here, 8pi/3. You were right about shifting it. Although your method seems not so intuitive at first.

    Last question though, does the shell method always give the same answer as the disc method in your experience? I get kinda thrown off by using each methods sometimes
     
    Last edited: Mar 9, 2009
  6. Mar 9, 2009 #5

    lanedance

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    the shell method & disc method will always give the same result for well behaved functions revolved around an axis - you're calculating the volume, which is invariant of what method you use.

    So the shell method should give the same result here, regardless of the shift, so I'm not too sure what happend in your calc

    Generally they can both get pretty complicated & its easy to lose some arithmetic along the way. So the trick usually is to choose shell/disc based on which one is easiest to set up, and exploit whatever symmetry there is in the problem to make it as easy as possible
     
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