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Homework Help: Parametric Equation of a sphere

  1. Jan 15, 2010 #1
    1. The problem statement, all variables and given/known data
    Find parametric equations for the part of sphere x2+y2+z2=9 that lies between the planes y=1 and y=2.


    2. Relevant equations



    3. The attempt at a solution

    Okay knowing that the p=3 I wrote the parametric equations for a sphere as
    x=3sin[tex]\phi[/tex]cos[tex]\theta[/tex] y=3sin[tex]\phi[/tex]sin[tex]\theta[/tex]
    z=3cos[tex]\phi[/tex] now the phi bound is 0<[tex]\phi[/tex]<[tex]\pi[/tex]
    but I'm not sure what to write for the [tex]\theta[/tex] bound.
     
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  3. Jan 15, 2010 #2

    Dick

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    It would be easier if they said between z=2 and z=1, right? Then you would just have to restrict phi. There's no reason why you can't interchange say, y and z in your parametrization.
     
  4. Jan 15, 2010 #3
    Oh okay so I can say z=3cos[tex]\phi[/tex]
    and from there plug in the values to find the [tex]\phi[/tex] bound to be cos(1/3)-1<[tex]\phi[/tex]<cos(2/3)-1 and the [tex]\theta[/tex] bound should just be 0< [tex]\theta[/tex]<2[tex]\pi[/tex]
     
  5. Jan 15, 2010 #4

    Dick

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    Right. But they did say y=1 and y=2. You'll also have to tweak your parametrization so y=3*cos(theta) instead of z=3*cos(theta).
     
  6. Jan 15, 2010 #5
    oh so then phi bound should be 0<[tex]\phi[/tex]<[tex]\pi[/tex] and since y=3cos[tex]\theta[/tex] the bounds for [tex]\theta[/tex] are
    cos(1/3)-1<[tex]\theta[/tex]<cos(2/3)-1
     
  7. Jan 15, 2010 #6

    Dick

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    Sorry, sorry!!! I meant to say your parametrization has z=3*cos(phi) and you want to change it so that y=3*cos(phi). Leave theta as it is.
     
  8. Jan 15, 2010 #7
    alright i think i got it now. So bound for theta 0<[tex]\theta[/tex]<2[tex]\pi[/tex]
    y=3cos[tex]\phi[/tex] the bounds for phi are then
    cos(1/3)-1<[tex]\phi[/tex]<cos(2/3)-1

    and the parametric equation for x=3sin[tex]\phi[/tex]cos[tex]\theta[/tex]
    and z=3sin[tex]\phi[/tex]sin[tex]\theta[/tex]
     
  9. Jan 15, 2010 #8

    Dick

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    That looks good to me.
     
  10. Jan 15, 2010 #9
    alright thank you for all the help, its much appreciated.
     
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