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Volume enclosed by two paraboloids

  1. Nov 6, 2008 #1

    This is a question in which I'm getting the right answer, I just want to make sure what I'm doing its legitimate, as it seems a bit off to me.

    The volume enclosed by
    [tex]z=16[x^{2}+y^{2}], z=8-16[x^{2}+y^{2}][/tex]

    I set my triple integral up as follows:
    [tex]\int^{?r}_{?r}\int^{2\pi}_{0}\int^{8-16r^{2}}_{16r^{2}} r dz d\vartheta dr[/tex]

    So, my height is the bottom function to the upper, and obviously completing a full circle.
    Now here is what I'm unsure about.

    I know I need to either do two integrals for r and add them together 0 to 1 and 0 to -1,
    or -1 to 1 so that I'm only left with 2/pi (the answer) but what is the logic behind getting the limits for r?

    Thanks in advance
  2. jcsd
  3. Nov 6, 2008 #2


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    Hello Damascus Road! :smile:
    I don't understand what your two integrals for r are :confused:

    r cannot be negative.

    You're taking all possible cylindrical slices, so they're from r = … to … ?

    (though it might be easier to take horizontal slices, from z = … to … )
  4. Nov 6, 2008 #3


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    That's a good start.
    It should be obvious that the origin, r=0, is contained in the volume so that is your lower limit on r. Now, where do the two graphs, z= 16r^2 and z= 8- 16r^2 intersect? That will give the upper limit on the r integral.
  5. Nov 6, 2008 #4
    Ok, thanks HallsofIvy.

    Now to make sure I understand it.... in my head it seems to make sense that maybe you have to take two integrals wrt r, since its technically moving through two different functions isn't it?

    Or does taking the integral of the height first take care of that, providing a "smooth" boundary?
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