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Question about volume calculation using disks or shells

  1. Dec 25, 2009 #1
    Hello everyone,

    I was going through the Single Variable Calculus lectures on the MIT Opencourseware site and looking at calculating volumes in 3D using the method of shells and disks.

    In both these methods, we sweep the infinitesmal region in 3D sweeping and sweep a disk. My question is actually about setting up the limits for the definite integral to calculate the final volume.

    In the example, they were looking at calculating the volume of a couldron where the body was modelled like a parabola and path sweeped by the disk was modelled as a shell. Now the parabola was symmetric around the origin and when taking the limit, they only considered one side of the parabola (so, the lower limit was 0 and the upper limit was [tex]\sqrt{a}[/tex].

    My question is that should't the limit be from [tex]\sqrt{-a}[/tex] to [tex]\sqrt{+a}[/tex]. There was brief discussion about this but I could not follow the reasoning which was along the lines that the sweeping dx region took care of the other half...

    Would be vert grateful if someone can clarify this for me.

    And a merry xmas and new year to all!


  2. jcsd
  3. Dec 25, 2009 #2


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    They don't both sweep a disk. A shell is not the same as a disk.
    You mean [itex]-\sqrt{a}[/itex] not [itex]\sqrt{-a}[/itex], but no, you don't want the negative limit. The dV element of the shell is generally

    [tex] ( 2\pi x) (y_u-y_l)dx[/tex]

    for revolution about the y axis. The [itex]( 2\pi x)[/itex] is the circumference of the shell at position x and the y-upper - y-lower is the height of the shell. You get both sides of the x axis because the circumference goes all the way around.
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