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Triple integral in cylindrical coordinates

  1. Feb 10, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the volume of the solid that lies between

    z=x2+y2 and
    x2+y2+z2=2

    2. Relevant equations

    z=r2
    z=√(2-r2)


    3. The attempt at a solution

    So changing this into cylindrical coordinates, I get

    z goes from r2 to √(2-r2)
    r goes from 0 to √2
    theta goes from 0 to 2π

    so we get [itex] \int_0^{2π} \int_0^{\sqrt{2}} \int_{r^2}^{\sqrt{2-r^2}} r dzdrdt[/itex]

    I dunno... something feels off about this. I think this way it would take into account the negative area on the outside of the curve... How else could I set up the integral?
     
  2. jcsd
  3. Feb 10, 2013 #2

    Dick

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    I would reconsider your statement that r goes from 0 to sqrt(2).
     
  4. Feb 10, 2013 #3
    How so? I was thinking of splitting it into 2 parts and having it go from 0 to the function in terms of z, but then I would have to integrate r first, and I run into the same problem...

    ie: z from r^2 to 1, then from 1 to √(2-r^2), and r from 0 to √z and then from 0 to √(2-z^2)...

    Is there a way to write it in terms of theta???
     
  5. Feb 10, 2013 #4

    Dick

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    The integral you have is fine. Except that the range of r is off. r^2 is x^2+y^2, not x^2+y^2+z^2. You need to figure out the value of r where r^2=z and r^2+z^2=2 intersect.
     
  6. Feb 10, 2013 #5
    Awesome, r from 0 to 1, and z from r^2 to √(2-r^2).
    Thank you so much, I see it perfectly now.
     
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