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Volume using Triple Integrals

  1. Nov 17, 2008 #1
    1. The problem statement, all variables and given/known data
    A volcano fills the volume between the graphs z = 0 and z =1/(x^2+y^2)^24, and outside the cylinder x^2+y^2=1 Find the volume of this volcano.


    2. Relevant equations
    This is a triple integral to be evaluated in cylindrical coordinates.



    3. The attempt at a solution
    Alright, so, this is what I've done. I don't understand what I'm doing wrong. I set up a triple integral, the limits of integration as follows:
    z goes from 0 to 1
    theta goes from 0 to 2pi
    and r goes from 1 to z^-1/48 (outside the unit circle, but between the other graphs)

    I first integrate with respect to r. I get r^2/2, evaluated between z^-1/48 and 1. This gives the integrand (z^-1/2304)/2-1/2. Then, I integrated with respect to z. This gives the answer ((2304z^(2303/2304)/2303)/2)-z/2. Then, this is evaluated from 0 to 1. This gives (2304/2303)/2-1/2, or, (1/2303)/2. This is integrated with respect to theta, giving (theta/2303)/2. Evaluated at 2pi, this results in (2pi/2303)/2. However, my homework software has told me this is wrong multiple times. Where am I going wrong? Thank you for any assistance.
     
  2. jcsd
  3. Nov 17, 2008 #2

    HallsofIvy

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    z does NOT go from 0 to 1. It goes from 0 up to 1/r48. r goes from 1 to infinity:
    [tex]\int_{\theta= 0}^{2\pi}\int_{r= 0}^\infty\int_{z= 0}^{1/r^{48}} r dz dr d\theta[/tex]
     
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