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Volume of volcano

  1. Apr 9, 2006 #1
    A volcano fills the volume between the graphs z=0 and [tex]\frac{1}{(x^2+y^2)^3}[/tex], and outside the cylinder [tex]x^2+y^2=1[/tex]

    so I found the z height to be from 0 to 1, the radius from 1 to infinity, and theta to be from 0 to 2pi

    [tex]\int_0 ^{2 \pi} \int_1 ^{\inf}\int _0 ^ {1} r dzdrd \theta [/tex]

    I know that this is not correct but I dont know how to set this integral up. any ideas?
    Last edited by a moderator: Apr 9, 2006
  2. jcsd
  3. Apr 9, 2006 #2


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    That is, your "volcano" is the region between the vertical line r= 1 and the curve z= 1/(x2+ y2)3[/sup]= 1/r6 rotated around the z-axis. Yes, [itex]\theta[/itex] goes from 0 to [itex]2\pi[/itex] and z goes from 0 to 1 but, for each z, r does not go from 1 to infinity, it goes from the boundary r= 1 to z= r6 or r= z-1/6. Your integral is
    [tex]\int_{\theta= 0}^{2\pi}\int_{z= 0}^1\int_{r=1}^{z^{-1/6}}rdrdzd\theta[/tex]
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