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Describing a Solid in Spherical Coordinates

  1. Nov 10, 2008 #1
    1. The problem statement, all variables and given/known data
    A solid lies above the cone [tex]z=\sqrt{x^2+z^2}[/tex] and below the sphere [tex]x^2+y^2+z^2=z[/tex]. Describe the solid in terms of inequalities involving spherical coordinates.

    2. Relevant equations
    In spherical coordinates, [tex]x=\rho\sin\phi\cos\theta[/tex], [tex]y=\rho\sin\phi\sin\theta[/tex], and [tex]z=\rho\cos\phi[/tex]

    3. The attempt at a solution
    I have no idea how to do this problem. My attempts have involved converting the two given equations to spherical coordinates, at which point everything is very messy and I don't know where to go next.

    I've attached a couple of 3D graphs to help with visualization.

    The answer is supposed to be [tex]0\leq\phi\leq\frac{\pi}{4}[/tex] and [tex]0\leq\rho\leq\cos{\phi}[/tex], but this doesn't make much sense to me.

    Any help would be great. Thanks!

    Attached Files:

    Last edited: Nov 10, 2008
  2. jcsd
  3. Nov 10, 2008 #2


    Staff: Mentor

    Shouldn't the equation of your cone be [tex]z = \sqrt{x^2 + y^2}[/tex]?
  4. Nov 10, 2008 #3


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    Staff Emeritus
    Science Advisor

    My recommendation is that you go back and read the problem again!

    You say "the cone [itex]z= \sqrt{x^2+ z^2}[/itex] is a complicated cylinder, not a cone. That may be why "everything is messy".

    I suspect it was really [itex]z= \sqrt{x^2+ y^2}[/itex]. In that case, the equation in spherical coordinates is [itex]r cos(\phi)= \sqrt{\rho^2 cos^2(\theta)sin^2(\phi)+ \rho^2 sin^2(\theta)sin^2(\phi)}= r sin(\phi)[/itex] which reduces to [itex]cos(\phi)= sin(\phi)[/itex]. It should be obvious that that reduces to [itex]\phi= \pi/4[/itex].
  5. Nov 10, 2008 #4
    Yes, this is correct. Thanks Mark!
  6. Nov 10, 2008 #5
    I didn't see how that reduced but I do now. Thanks!
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