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Triple integral of sphere

  1. Mar 6, 2009 #1
    1. The problem statement, all variables and given/known data

    The figure below shows part of a spherical ball of radius 5 cm. Write an iterated triple integral which represents the volume of this region.

    http://img19.imageshack.us/img19/9237/sphereu.th.jpg [Broken]

    http://img19.imageshack.us/img19/1699/inte.jpg [Broken]


    2. Relevant equations



    3. The attempt at a solution

    I already got A, B, C and D correctly. However when I try E and F to be -3 and -sqrt(25-x^2-y^2) respectively and vice versa, the answer is wrong.. why is this.. what should be E and F then?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 6, 2009 #2

    gabbagabbahey

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    Is -3 really the minimum value of z for that volume?:wink:
     
  4. Mar 6, 2009 #3
    I guess so... z = -3 is just a plane right... other than that I really don't know what it is
     
    Last edited: Mar 6, 2009
  5. Mar 6, 2009 #4

    gabbagabbahey

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    Sure, but the region your interested in lies below the plane z=-3, so z=-3 should be the maximum not the minimum......likewise, the region of interest lies above the spherical surface z=-sqrt(25-x^2-y^2), so that should be the minimum value of z.

    Did you try E=-sqrt(25-x^2-y^2) and F=-3?
     
    Last edited: Mar 6, 2009
  6. Mar 6, 2009 #5
    yes I did and as I said on my original posting "vice versa" and it doesn't work as well..
     
  7. Mar 6, 2009 #6

    gabbagabbahey

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    That's odd; assuming that E and F are the lower and upper limits on z, it should work.....how do you know that it doesn't work? Are you imputing these values into an online submission? Or are you just getting an incorrect final answer for the volume?
     
  8. Mar 6, 2009 #7
    yes, into an online submission
     
  9. Mar 6, 2009 #8
    and shouldn't it be -sqrt(25-x^2-y^2) instead of -sqrt(25-x^2-z^2)
     
  10. Mar 6, 2009 #9

    gabbagabbahey

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    Hmmm, what were your A,B,C and D?

    Are you using the convention that [tex]\int_A^B\int_C^D\int_E^F dzdydx[/tex] means first integrate over z from E to F, then over y from C to D, and finally over x from A to B? Or are you using some other convention?
     
  11. Mar 6, 2009 #10
    A = -4
    B = 4
    C = -sqrt(16-x^2)
    D = sqrt(16-x^2)
     
  12. Mar 6, 2009 #11

    gabbagabbahey

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    Sorry, I meant to write -Sqrt(25-x^2-y^2)
     
  13. Mar 6, 2009 #12

    gabbagabbahey

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    That looks fine; you are either making a mistake/typo when typing in your E and F values, or the online submission program has a bug
     
  14. Mar 6, 2009 #13
    hmm...that's what I thought as well.. but are you sure that E=-sqrt(25-x^2-y^2) and F=-3?
     
  15. Mar 6, 2009 #14

    gabbagabbahey

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    Yes, why wouldn't that be correct? (Unless the volume is supposed to also include the cone that subtends that section of the sphere)
     
  16. Mar 6, 2009 #15
    Just making sure, as that's what I had in mind as well.. I'll have to ask my instructors about this then.. thanks for confirming
     
  17. Mar 7, 2009 #16
    Hmm.. there's a hint to this question, not sure if it helps

    Hint: If you flip this solid upside down, you can view it as an upper portion of a sphere that is centered at the origin.
     
  18. Mar 7, 2009 #17

    gabbagabbahey

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    Hmmm.... maybe they are expecting you to do that (even though it is clearly unnecessary); if so, then z would go from +3 to +Sqrt(25-x^2-y^2), so you could try those as your E and F values
     
  19. Mar 7, 2009 #18
    well..it makes no sense at all though... I only have one more chance to get this right
     
  20. Mar 8, 2009 #19

    gabbagabbahey

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    Why doesn't it make any sense? The volume of the spherical section is the same no matter where it is right?...If the problem suggests that you reflect it over the xy plane, then you might as well...
     
  21. Mar 8, 2009 #20
    yeah, I just realized that the volume is the same.. thanks for pointing that out.. now the answer is correct
     
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