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Triple Integral

  1. Feb 23, 2014 #1

    dwn

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    1. The problem statement, all variables and given/known data

    Image attached of problem.

    2. Relevant equations

    Not sure how this applies.

    3. The attempt at a solution

    ∫(o to ∞) ∫(0 to 5) ∫(0 to 4) 16xy dydxdz
    Below, I have completed the integrations in order (dydxdyz):

    ∫dy : 8xy2|(0 to 4) = 128 x

    ∫dx : 64x2 = 64(25) = 1600

    ∫dz : 1600z| (o to ∞)

    Do I set a limit on the integration dz, or set it at infinite?
     

    Attached Files:

  2. jcsd
  3. Feb 23, 2014 #2

    tiny-tim

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    hi dwn! :smile:
    the limits of z depend on x and y

    so you must integrate wrt z first (from 0 to … ? :wink:)
     
  4. Feb 23, 2014 #3

    Dick

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    If you want to set this up as a triple integral, then since you are trying to find volume, your integrand should be 1. And I would integrate over dz first. And the upper value of z will correspond to its value on the boundary surface. Then integrate over x and y.
     
  5. Feb 23, 2014 #4

    dwn

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    If I want to set this up as a triple integral? Just out of curiosity...what other method? I am going to calculate this using the triple, but I would like to know for my own selfish reasons.

    How do I go about finding ;) ? When you say that it depends on x and y....I can't just make it any constant, can I?
     
  6. Feb 23, 2014 #5

    Dick

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    You can also find the volume by integrating (height)dxdy. It's not really a different method, you just aren't writing the z integration out explicitly. Same thing. And the limit of z isn't any constant. It's the value of z on the bounding surface as a function of x and y. Don't think too hard about this.
     
  7. Feb 23, 2014 #6

    dwn

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    #smackforehead : just set the integral dz from 0 to z....
     
  8. Feb 23, 2014 #7

    Dick

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    .... means just set the integral dz from z=0 to z=16xy, right?
     
  9. Feb 23, 2014 #8

    dwn

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    exactly. So does this "thickheaded" thinking go away at some point? :)
     
  10. Feb 23, 2014 #9

    Dick

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    As far as this topic goes it should be going away already. You seem to be pretty clear on how obvious the answer is.
     
  11. Feb 24, 2014 #10

    tiny-tim

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    hi dwn! :smile:

    (just got up :zzz:)

    when you perform the first integration (in this case, z) in a triple integral to find a volume,

    you are essentially finding the volume of a tall thin square slice of the volume, of sides dx and dy …

    its volume obviously is (approximately) dxdy times the height,

    ie dxdy times [z(x,y) - zo(x,y)]​

    where z = z(x,y) is the equation of the top surface, and z = zo(x,y) is the equation of the bottom surface (in this case, the constant z = 0) :wink:

    btw, you will notice that the limits of integration for all variables except the last in a multiple integral will generally not be constants, but will generally depend on the later variables!
     
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