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Describe the cylinder

  1. Nov 4, 2012 #1
    1. The problem statement, all variables and given/known data

    Describe the cylinder x2+y2=9
    in terms of parameters theta and phi, where (p,theta,phi) are spherical coordinates of a point on the surface.

    I'm a bit confused I know that spherical coords are

    x=Psin(phi)cos(theta)=3sin(phi)cos(theta)

    y=Psin(phi)sin(theta)=3sin(phi)sin(theta)

    z= Pcos(phi)=3cos(phi)

    But the answer is way off it's
    x = 3 cos(theta) ; y = 3sin(theta) ; z = 3 cot(phi) 



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 4, 2012 #2

    jedishrfu

    Staff: Mentor

    In your third equation you're thinking that P is the radius of the circle.

    Slice the the cylinder with a plane containing X and Z axes and you'll see two vertical lines so a point on the cylinder is a vector from the origin to the point which makes a triangle with the P being the bottom side and the z value being the vertical side .

    Now describe Z in terms of spherical coordinates phi angle.
     
  4. Nov 4, 2012 #3
    Not really sure what you mean?
     
  5. Nov 4, 2012 #4

    jedishrfu

    Staff: Mentor

    your eqn is a cylinder along the z axis, right?

    So if you slice the cylinder with a plane containing the X and Z axes you'll see two vertical lines at x=3 and x=-3, right?
     
  6. Nov 4, 2012 #5
    Indeed
     
  7. Nov 4, 2012 #6

    haruspex

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    jedishrfu is pointing out that there is some confusion regarding p (or P).
    In the problem statement, p is the distance from the origin to a point on the cylinder. You have used P as the radius of the cylinder (=3). A point on the cylinder may be further than 3 from the XY plane, so z cannot be given by 3 cos(phi).
     
  8. Nov 5, 2012 #7

    HallsofIvy

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    [itex]\rho[/itex], what you are calling "P", is the straight line distance from the origin to the point, (x, y, z). It is NOT [itex]\sqrt{x^2+ y^2}[/itex], the distance, in the xy-plane, to the projection (x, y, 0).

    From [itex]x= \rho cos(\theta)sin(\phi)[/itex] and [itex]y= \rho sin(\theta)sin(\phi)[/itex], you can get [itex]x^2+ y^2= \rho^2 cos^2(\theta) sin^2(\phi)+ \rho^2sin^2(\theta)sin^2(\phi)= \rho^2sin^2(\phi)[/itex].
     
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