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Spherical coordinates equation

  1. Jul 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Identify surface whose equation in spherical coordinates is given
    p = sin(theta)*sin(fi)

    3. The attempt at a solution

    I know that y = r*sin(theta)*sin(fi). and thus, y = rp.

    This yields y = (x2 + y2)0.5*(x2 + y2+z2)0.5

    However, this is rather ugly. The answer is supposed to be "a sphere with radius 0.5, center (0. 0.5, 0)" however, I don't see how that results from this expression.
     
    Last edited: Jul 10, 2009
  2. jcsd
  3. Jul 9, 2009 #2

    Office_Shredder

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    What is p supposed to be?
     
  4. Jul 10, 2009 #3
    p is ro. p = (x^2 + y^2 + z^2)^0.5
     
  5. Jul 10, 2009 #4
    I'm guessing that the expression should read something like

    [tex]
    r = \sin\theta\sin\phi
    [/tex]

    where [tex]\theta[/tex] and [tex]\phi[/tex] are your polar angles. First, look in the x-y plane, where [tex]\theta=90^\circ[/tex]. This simplifies to

    [tex]
    r = \sin\phi.
    [/tex]

    Now, apply [tex]r[/tex] to the equations linking polar and Cartesian coordinates, for [tex]\theta=90^\circ[/tex], as

    [tex]
    x = r\cos\phi
    [/tex]

    [tex]
    y = r\sin\phi.
    [/tex]

    You'll see they come out to be

    [tex]
    x = \sin\phi\cos\phi
    [/tex]

    [tex]
    y = \sin^2\phi.
    [/tex]

    Next, recall the double-angle formulas that [tex]\sin(2x) = 2\sin x\cos x[/tex] and [tex] \cos(2x) = 1 - 2 \sin^2 x[/tex]. I'll leave it as an exercise to substitute these back into the expressions for [tex]x[/tex] and [tex]y[/tex]. What you should notice is that your values should now look like

    [tex]
    x = A \sin(2\phi)
    [/tex]

    and

    [tex]
    y = A \cos(2\phi) + B
    [/tex]

    where [tex]A[/tex] and [tex]B[/tex] are numbers. You should recognize this as the parametric representation of a circle, at the coordinate [tex](0,B)[/tex].

    Now, this doesn't exactly answer your question, but it should hopefully get you to visualize how the surface should be a sphere. Particularly, if you repeat this exercise in the y-z plane, where [tex]\phi=90^\circ[/tex], you should find find another case where the result is a circle, offset from the origin by some distance [tex]B[/tex].

    I think that, once you've identified what [tex]B[/tex] is, you can look into the coordinate transformation of

    [tex]
    x' = x, \quad y' = y + B, \quad z' = z
    [/tex]

    and then determine [tex]r' = \sqrt{x'^2 + y'^2 + z'^2}[/tex] which should be a constant. If this is the case, it shows that this surface is a sphere centered at [tex](0,B,0)[/tex].
     
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