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Spherical Coordinates

  1. Nov 27, 2011 #1
    1. The problem statement, all variables and given/known data

    Set up the triple integral for the volume of the given solid using spherical coordinates:

    The solid bounded below by the sphere ρ=6cosθ and above by the cone z=sqrt(x2+y2)
    2. Relevant equations

    3. The attempt at a solution

    I thought i had this set up right where ρ goes from 6cosθ to sqrt(3)/2 ( how i got sqrt(3)/2 was by z=sqrt(x2+y2 is th esame as ρcos∅=ρsin which means ∅=pi/4 plugging that into ρ=6cosθ, gives me sqrt3/2) i don't know if this is right but for ∅ it goes from 0 to pi/4 and for θ it goes from 0 to 2pi

    thanks in advance
  2. jcsd
  3. Nov 27, 2011 #2


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    What is ρ, and what is θ?

    How can be ρ=6cosθ the equation of a sphere?

  4. Nov 27, 2011 #3
    ρ is the distance from P to the origin

    ∅ is the angle that the line makes with the positive x axis

    θ is the angle from the cylindrical coordinates (0<θ<2pi)

    i know that the equation for a sphere is when ρ= constant, in this case she told us it was cos∅
  5. Nov 28, 2011 #4


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    Last edited: Nov 28, 2011
  6. Nov 28, 2011 #5


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    is that the volume you nee to find?

    http://img811.imageshack.us/img811/4920/conesphere.jpg [Broken]

    Last edited by a moderator: May 5, 2017
  7. Nov 28, 2011 #6
    i couldnt get a picture but that is how the sphere looks with a radius of six and the cone is just the top half as looking as any cone would look
  8. Nov 28, 2011 #7


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    Is the volume as shown in the picture in my previous post? If so, you have to integrate the volume element dV=r^2sin(∅)dr d∅ dθ from 0 to 2pi for θ,
    from 0 to pi/4 for ∅ (it is the angle of the cone it makes with the positive y axis) and for r from the top of the cone to the radius of the sphere r=ρ=6.

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