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Question about equations of 3D solids

  1. Feb 16, 2016 #1
    I was working on a problem on this domain:
    [tex]
    E=[x,y,z)\:s.t. \: \sqrt{x^2+y^2}\leq z\leq \sqrt{3x^2+3y^2},\: x^2+y^2+z^2\leq 2]
    [/tex]
    and at some point I wanted to find the intersection of the internal cone(##\sqrt{3x^2+3y^2}=z##) with the sphere of radius ##\sqrt{2}## to find the height z of the circle. I did so by equating the two in cylindrical coordinates as follows:
    [tex]
    3x^2+3y^2-z^2=x^2+y^2+z^2-2\\x=rcos\theta,\: y=rsin\theta,\: z=z\\2r^2-2z^2+2=0 \rightarrow z^2=r^2+1
    [/tex]
    This result confuses me. The intersection of a cone along the z-axis and a sphere centered in (0,0,0) should be a circle around the z-axis(and so with a fixed z). The radius(##\sqrt{x^2+y^2}## is of course also fixed. Then why do I get a function of z and r? Isn't the intersection of a cone and a sphere a circle?
     
  2. jcsd
  3. Feb 16, 2016 #2

    BvU

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    That's not the way it works for an intersection. Both sides should be zero, so you have two equations:$$
    3x^2+3y^2-z^2=0 \\x^2+y^2+z^2-2 = 0 $$ and from this you have to find your way out....
     
  4. Feb 16, 2016 #3
    You are a genius. :D thank you!
     
  5. Feb 16, 2016 #4

    BvU

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    You're welcome !
     
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