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When to use Washer, Shell or Disk

  1. Feb 21, 2005 #1
    I'm curious, when am I supposed to use Washer, Shell or Disk method when trying to answer questions involving integrals and volume? Is there something specific I should look out for?
    I just can't tell the difference.
    Any help is appreciated.
    Thanks.
     
  2. jcsd
  3. Feb 21, 2005 #2
    You can use whichever one you want. You can integrate using any shape. If you're doing the solids of revolution problems, I always thought that the shell method was easier.
     
  4. Feb 21, 2005 #3
    Oh, I thought you weren't able to choose them on your own and each problem reverted to a specific one or something?
     
  5. Feb 21, 2005 #4

    HallsofIvy

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    No, its a matter of how you find it easiest to visualize the solid.
     
  6. Feb 21, 2005 #5

    shmoe

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    Depending on how the solid is described, you'll sometimes find the shell method easier to integrate than the washer method and vice versa. If you set up the integral one way and you're finding it hard to evaluate, try using the other method. Try doing the same problems using both methods, it's a good way to get a feel for when one is preferable. Of course sometimes either way will be just as easy (or hard).
     
  7. Feb 21, 2005 #6
    Oh good question! I've been wondering the same thing! Disk is usually pretty obvious b/c it's just a volume of solid revolution, usually there's no hole in the center or whatever the situation usually is.

    Deciding between washer or shell can be tricky, and as posted above a lot of the time either method will work, but one might be easier. We actually had a homework assignment on the washer method, then the next homework assignment was to redo all the problems using the shell method! Talk about getting confused!

    Some things to keep in mind is when using the shell method, you can think of a tiny segment rotating parallel to the axis of rotation (forming a cylinder). This integral takes the form of: integral(circumference*height*delta r).

    When using washer, the segment is rotating perpendicular to the axis of rotation. This yields the integral where you have integral [(pi R^2)-(pi r^2)], where R is the radius of the larger disk or washer, and little r is (you guessed it!) the radius of the smaller or inner disk.

    Happy integrating!
     
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