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Evaluating triple integral

  1. Mar 13, 2009 #1
    1. The problem statement, all variables and given/known data

    http://img12.imageshack.us/img12/7181/integral.th.jpg [Broken]

    2. Relevant equations



    3. The attempt at a solution

    Well my first attempt is to convert this to a cylindrical coordinate first, which I believed to be:

    [tex] \int_0^1 \int_0^{2\pi} \int_0^1 1 \, dr \,d\theta \,dz [/tex]

    is this correct?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 13, 2009 #2
    Yep, that's the cylindrical coordinate version of the stated problem.
     
  4. Mar 13, 2009 #3
    and when I evaluate this I got 2*pi*r, but the answer doesn't seem to be correct...
     
  5. Mar 13, 2009 #4
    Keep in mind that after getting the antiderivative of the function you need to evaluate it at the upper and lower limits of integration and subtract each time. It looks like you did this for [itex]\theta[/itex] and z, but what about r? Since r is not a limit of integration for either [itex]\theta[/itex] or z, you're right in thinking that it shouldn't appear in the answer.

    In addition, we can figure out what the answer will be without actually doing the integral if we think about it as the volume of a solid object. If you can figure out what sort of object is described by these limits of integration, then you'll be able to check your solution by comparing it to the formula for the volume of whatever shape this is.
     
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