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Triple integral.

  1. Dec 19, 2011 #1
    1. The problem statement, all variables and given/known data

    find the volume between

    z +x^2 + y^2 = 4 and x^2+y^2+z^2 = 6

    2. Relevant equations



    3. The attempt at a solution

    what i did first was solve for the intersection points of z
    i got 2 and -1.

    then you get two equations for x^2 + y^2

    x^y+y^2 = 5 and x^2+y^2=2

    so then i used polars and i figure that r goes from 2 to 5, z goes from 2 to -1 and pi goes from 0 to 2pi

    is that right?
     
  2. jcsd
  3. Dec 20, 2011 #2

    ehild

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    The bounds for z are correct, but the bounds of r=sqrt(x2+y2) change with z:
    [tex]\sqrt{4-z}<r<\sqrt{6-z^2}[/tex]

    ehild
     
  4. Dec 20, 2011 #3

    vela

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    I've attached a plot of the two surfaces. Do you also have to take into account the cap?
     

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  5. Dec 20, 2011 #4
    No, just the portion in between. Thanks for the plot.
     
  6. Dec 20, 2011 #5

    vela

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    The cap is in between the two surfaces.
     
  7. Dec 23, 2011 #6
    but what if i want to make z variable doesnt r become constant???
    and one more thing when we you are finding the domain in cartesian is polar coordinates also considered to be in the cartesian way ??
     
  8. Dec 23, 2011 #7

    vela

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    Yes, you can let r run between constant values and have the limits for z depend on r, if you want. Ehild's suggestion, however, will make the math simpler.
     
  9. Dec 23, 2011 #8
    i get it 3 domains for z if i look at the figure i get it thanks
     
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