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When to use shells, when to use disks?

  1. Feb 10, 2012 #1
    I have a test tomorrow on finding the volume, but I can never seem to use the right method.
    (easier, one-step method).

    What indicates if the disks method will be easier?
    What indicates if the shells method will be easier?
  2. jcsd
  3. Feb 10, 2012 #2


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    Hey esploded and welcome to the forums.

    This is really a hard question to answer and doesn't have a black/white do this/do that answer.

    It depends on the function you are trying to use for a volume and also what type of integral ends up being what you have to solve.

    If for example one integral type (say disk) is harder (or maybe is just too hard to integrate using normal techniques), then it would make sense to use the shell method because the disk method might just be a pain to deal with.

    This is pretty much an important realization for mathematics: we can end up coming at the same answer in many different ways and one way isn't necessarily the best way to do things. These kind of roadblocks happen a lot and when you have a lot of different techniques at your disposal (that people had to figure out before you!) then it makes your job easier because if one thing doesn't work, you have other things that you can try.
  4. Feb 10, 2012 #3

    Simon Bridge

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    Symmetry... and what, exactly, you have to achieve.
    It is usually easier to use the disks - or solid of rotation.
    But they are usually much the same. Can you provide two examples where you feel one or the other method was better so we know what you mean?
  5. Feb 13, 2012 #4
    Sadly, I believe that my answer will be too late for your test...

    Any way, I use shells when a function is being rotated around the y-axis or any x= equation. I would use disks (or rings if needed) for any function being rotated around the x-axis or any y= equation. This is so that I would not have to alter f(x) (assuming that it is f(x) and not f(y)) much at all. It would simply be a "plug and chug" situation. If the function given is f(y), just switch the situations that each method are used.

    These choices are purely selected from preference at which I find easier. I know some people that would rather alter the f(x) function to become f(y) if they need to rotate around the y-axis because they do not understand shells. So in the end, it is up to your preference. It is also important to point out that shells will need to be used if f(x) cannot be integrated without strenuous techniques.
  6. Feb 13, 2012 #5

    Simon Bridge

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    Don't see how the rotation axis changes anything ...
  7. Feb 14, 2012 #6
    The rotational axis doesn't actually change anything, but I find it pushes me away from certain methods.

    If f(x)=x2+5 and was rotated around the y-axis, I would use shells. This is because the function is in terms of x, and shells uses a cut that is parallel to the axis of rotation, in this case, dx. Since both are in terms of x, I do not need to modify any formulas.
    If I wanted to use disks (or rings), I would need to put the function in terms of y so it matches the dy needed for the disks formula at this rotational axis. To me it is just extra work that I do not want to do.

    I hope this clarifies what I meant.
  8. Feb 14, 2012 #7

    Simon Bridge

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    Oh I see, you just don't want to have to say that the disk between y and y+dy has volume [itex]\pi(y-5)dy[/itex] where the volume of the "ring" between x and x+dx is [itex]2\pi (x^2+5)xdx[/itex] ...

    Fair enough - the moral is to pick the method you are most comfortable with: that's the one with the least chance of a mistake.

    For the pedants:
    I know that the volumes above are not in same regions, the former is inside the curve and the latter under it - the example does not state which region the volume is to be found for. If, for instance, the idea is that it is a wine glass, and you want to know how much wine fills it to a height h above the table (at y=0)... why then the the two methods should provide:
    [tex]V\qquad=\pi \int_5^h (y-5)dy \qquad = \pi(h-5)h - 2\pi \int_0^{\sqrt{h-5}}(x^2+5)xdx[/tex]

    If, however, the volume is what remains after lathing a hole in a cylinder of glass (to make a shot-glass, say) h high and outer radius R, to the above specs
    ... the the glass that remains is given be each method as:
    [tex]V\qquad=\pi R^2h-\pi\int_5^h (y-5)dy \qquad
    = 2\pi\int_0^\sqrt{h-5}(x^2+5)xdx + 2\pi R(R-\sqrt{h-5}) \qquad :\;\; R > \sqrt{h-5}[/tex]

    Of course, in a math course you often have things laid out ... like, find the volume between f(x) rotated about the y axis and the x-z plane so that [itex]x^2+z^2 \leq r^2[/itex] then you get:
    [tex]V\qquad = \pi(r^2+5)r^2 -

    \pi\int_5^{r^2+5}(y-5)dy \qquad = 2\pi\int_0^r(x^2+5)xdx[/tex]
    I know also that in each case the second one gives the more complicated integral ... but it does not have to be the case - that's just this particular example.
    (I'm also pretty sure I have made a mistake or dozen through there...)

    So: pick the method that makes the math simplest and, when in doubt, use the one you personally are most comfortable with.
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