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Triple integral help

  1. May 15, 2012 #1
    Hello everyone. I need help regarding putting boundaries in triple integrals using spherical coordinates. The figure attached below is an ice cream cone. I want someone to explain to me how to put the boundaries in the order of dr dz dθ? I had understood this in class but unfortunately i forgot the explanation.
     

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    Last edited by a moderator: May 15, 2012
  2. jcsd
  3. May 15, 2012 #2

    sharks

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    Hi sevag00
    Your question contradicts itself. You want to express the integrals in spherical coordinates, but the boundaries in terms of drdθdz are cylindrical coordinates! :smile:

    What is the whole question? Are you required to find the volume of the cone including the section of paraboloid on top?
     
    Last edited: May 15, 2012
  4. May 15, 2012 #3
    lol. I meant the volume in cylindrical coordinates.
     
  5. May 15, 2012 #4

    LCKurtz

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    There's no reason to use spherical coordinates on this problem. You want cylindrical coordinates. It is best not to integrate the dr variable first; better would be ##r\, dz dr d\theta##. The dz limits will be z on the lower surface and z on the upper surface, both expressed in terms of r. Once you have those z values in terms of r, you can set them equal to see what r is at their intersection.
     
  6. May 15, 2012 #5
    Yeah, i know it's easy integrating in order dzdrdθ . But the question in the book is asking drdzdθ.
     
  7. May 15, 2012 #6

    HallsofIvy

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    Well, "[itex]drdzd\theta[/itex]" is the hard way! I recommend doing it as two separate integrals. For z from 0 to 1, we have the cone, z= r, so, for each z, r goes from 0 to z. For z= 1 to 2, we have the paraboloid, [itex]z= 2- r^2[/itex] so, for each z, [itex]r^2=2-z[/itex] and r, being positive, goes from 0 to [itex]\sqrt{2- z}[/itex].
     
    Last edited: May 15, 2012
  8. May 15, 2012 #7

    sharks

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    The answer will be the sum of two triple integrals; volume of the cone + volume of paraboloid.
     
  9. May 15, 2012 #8
    Is this in order dzdrdθ?
     
  10. May 15, 2012 #9

    sharks

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    For the second triple integral, i think it should be from z=1 to ##z=2-r^2##
    The radius of the projection is r=1.
     
  11. May 15, 2012 #10

    LCKurtz

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    Is what in that order? Use the quote button to reply to a message so we know what you are replying to.
     
  12. May 15, 2012 #11
    Actually, i know the answer. But i didn't get how r is from 0 to z + 0 to sqrt(2-z). Answer is attached.
     

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  13. May 15, 2012 #12

    LCKurtz

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    Doing dr first, on the top part (the paraboloid) r goes from 0 to the r on the paraboloid. The equation of the paraboloid is ##z = 2-r^2##. Just solve that for ##r## to get ##\sqrt{2-z}##.
     
  14. May 15, 2012 #13
    Isn't the dr supposed to be from the x-axis to y-axis?
     
  15. May 15, 2012 #14

    LCKurtz

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    That doesn't make any sense. Think of starting on the z axis up at the level of the paraboloid. r is perpendicular to the z axis and it goes from r = 0 out till it hits the paraboloid. What direction it goes depends on ##\theta##.
     
  16. May 15, 2012 #15
    Aha. I think i get it.

    EDIT: Yes, i get it. Thanks for the help.
     
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