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Triple integral under a cone

  1. Sep 7, 2008 #1
    1. The problem statement, all variables and given/known data
    Triple integral of 1+z inside the cone z=2sqrt(x^2+y^2) above the xy plane and
    bounded by z=6



    2. Relevant equations



    3. The attempt at a solution
    when z=6, 6=2sqrt(r^2) so r=3 limits of integration are
    z=6 to z=2r r=3 to r=-3 theta=2pi to theta=0
    Just want to make sure my limits are right. Thanks!
     
  2. jcsd
  3. Sep 8, 2008 #2

    Hootenanny

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    You may want to rethink your radial limits, can a radius ever be negative? You should also be careful with the order of your limits.
     
  4. Sep 8, 2008 #3
    Good point. So the radius does not go from 3 to -3 it is 3 to 0. Oh and
    I have MAJOR issues with order. Without having to go through the problem and
    realize that the answer is negative, lol can someone PLEASE tell me how
    to determine the order of the limits?
     
  5. Sep 8, 2008 #4

    Defennder

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    As H00t mentioned, the order is very important. The limits of z can be gotten by imagining a line parellel to the z-axis passing through the required volume element from bottom to top. Where it enters denotes the lower limit, where it exits denotes the upper limits. In this case, it enters the solid from the cone so that's what it's lower limit should be and exits at the plane z=6, so that is the upper limits.

    The limits for the other integration variables can be obtained by similar observations. For r, just imagine a radial line originating from the z axis directed outwards radially on any plane parallel to the x-y plane. For theta, just think of a radial line sweeping across the entire volume element. Remember to specify the order. If you're expecting a numerical answer, the limits of your last integration variable cannot contain any variables.
     
  6. Sep 8, 2008 #5
    Whoa! I see your point1 For z, I drew the diagram of a simple cone with z-6 as the plane on top (ignoring the rest of the cone, I only draw the area we focus on). Then I imagined a simple line (1+z) with respect to the xz axis (or yz axis, its arbitrary) and at the origin of the line just before it gets "above" the xy plane from the -xy plane (x=0, y=0, z=1) and here the line is already inside of the cone, obviously it did not touch the z=6 plane yet but it certainly did cross into the cone at that point.
     
    Last edited: Sep 8, 2008
  7. Sep 8, 2008 #6

    Defennder

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    The technique I posted earlier on was for how to determine the limits of integration. So, you can't really just take any simple line (z+1, which isn't a line anyway) to get the limits. I learnt the technique from some calc textbook.
     
  8. Sep 9, 2008 #7
    Oh wait lol the line has to be parallel to the z axis no matter what and it should be read from bottom to top, so since z=6 is on top of the region we are integrating, the line does enter the cone before it touches z=6 if read from bottom to top, which determines the oirder of the limits. And from r=0 to r=3 that is how the limits should be read since r is read from 0 to 3 and
    theta sweeps around the surface but starting from the origin.
     
  9. Sep 9, 2008 #8

    Defennder

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    You still haven't specified the limits explicitly and the order of integration. Write out everything clearly so we can identify errors and correct them. As it stands your limits for z isn't stated, and the limits for r make sense only in one order which you did not specify.
     
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