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Voulme of an ice cream cone bound by a sphere

  1. Dec 4, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the volume of an ice cream cone bounded by the sphere x^2+y^2+z^2=1 and the cone z=sqrt(x^2+y^2-1)


    2. Relevant equations
    The two simultaneous equations yield x^2+y^2=1


    3. The attempt at a solution

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    Last edited: Dec 4, 2007
  2. jcsd
  3. Dec 5, 2007 #2

    HallsofIvy

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    z=sqrt(x^2+y^2-1) is NOT the equation of a cone- it is a hyperboloid.

    z= sqrt{x^2+ y^2) would be (the upper nappe of) a cone with vertex at the origin with sides making angle [itex]\pi/4[/itex] with the xy-plane.
     
  4. Dec 5, 2007 #3
    Thanks for your response, so is it logical to re-arrange the integral limits such that it becomes:
    Volume of cone =integral(limits theta= 0 to pi/4)integral(limits r=0 to 1/sqrt2)[sqrt((1-r^2)-r)dr d theta.
     
  5. Dec 5, 2007 #4

    HallsofIvy

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    First you are going to have to define the cone part! If it is z= sqrt{x^2+ y^2}, then yes, you take, in polar coordinates, [itex]\theta= 0[/itex] to [itex]\pi/4[/itex]. However, r goes from 0 to 1, not [itex]1/\sqrt{2}[/itex] because you are going up to the spherical cap.
     
  6. Dec 5, 2007 #5
    The "cone" part is given as z= sqrt{x^2+ y^2-1} which I agree is not an equation for a cone but a hyperboloid as you mentioned above. What is troubling me is how to deal with the (-1) inside the sqrt.
     
  7. Dec 6, 2007 #6
    Any ideas please, I am stuck.
     
  8. Apr 21, 2008 #7
    Did you ever get this figured out?
    I am working on the same problem with the exact same issue .. the -1.
     
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