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Rectangular to Spherical Coordinate conversion...

  1. Jul 9, 2017 #1
    1. The problem statement, all variables and given/known data
    Convert from rectangular to spherical coordinates.
    (-(sqrt3)/2 , 3/2 , 1)

    2. Relevant equations
    We know the given equations are
    ρ = sqrt(x^2 + y^2 + z^2)
    tan theta = y/x
    cos φ = z / ρ

    3. The attempt at a solution
    My answer was (2, -pi/3, pi/3)
    It should be a simple plug and go... Am I doing simple math wrong? The only part I think I could get wrong is the y/x but -3 / 2 / sqrt3 / 2 should be sqrt3 right?


    I'm using webassign if that matters at all. I'm sorry for posting such a simple question but I don't understand how that solution is wrong... I guess I'm just having a huge brain fart?
     
    Last edited: Jul 9, 2017
  2. jcsd
  3. Jul 9, 2017 #2

    Mark44

    Staff: Mentor

    No, since ##x = \frac{-\sqrt 3}{2}##. You are ignoring the sign, so your value for ##\theta## will be wrong.
     
  4. Jul 9, 2017 #3
    Sorry I must have deleted the negative sign by accident while I was editing. I've edited it back in and I had counted that into my calculations and that's why my theta is negative pi / 3 instead of positive.
     
  5. Jul 9, 2017 #4

    Mark44

    Staff: Mentor

    Your answer of (2, -pi/3, pi/3) looks fine to me. Is it possible that WebAssign doesn't recognize "pi" and wants you to enter a decimal value?
    Sometimes these programs are brain-dead...
     
  6. Jul 9, 2017 #5
    Yeah it's really frustrating especially considering it's such an easy question. I've tried decimal already and the simple "pi" or whatever, is correct for all the other choices...
     
  7. Jul 9, 2017 #6

    Mark44

    Staff: Mentor

    The second and third equations above are wrong, according to this wiki article (https://en.wikipedia.org/wiki/Spherical_coordinate_system).
    ##\theta = \arccos(z/r)## Here r is the same as your ##\rho##.
    ##\phi = \arctan(y/x)##
     
  8. Jul 9, 2017 #7
    There's no way that can be right... I've already done a few questions and gotten them all right except for this one... Also, I might trust the textbook more than wikipedia. I hope you don't take this as an insult or anything because I do appreciate the help.
     
  9. Jul 9, 2017 #8

    Mark44

    Staff: Mentor

    No, I don't take it as an insult at all. The wiki formulas seem weird to me, as well, and that's why I qualified my answer.

    I checked with wolframalpha, which gives an answer of (2, pi/3, 2pi/3) -- http://www.wolframalpha.com/input/?i=spherical+coordinates(-sqrt(3)/2,+3/2,+1).
    Not all books use the notation in the same way. For a point ##(\rho, \theta, \phi)##, some books call ##\theta## the inclination (measured away from the z-axis), and others call ##\phi## the inclination. These differences make your formulas correct as far as some books are concerned and incorrect in others.
     
  10. Jul 9, 2017 #9
    Ahh I see and understand now haha. Thank you for your help and it completely went over my head about how it could be more than one value of "n * pi / 3" for theta to be the value. I got the correct answer! Thank you so so much.
     
  11. Jul 9, 2017 #10

    Orodruin

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    Staff Emeritus
    Science Advisor
    Homework Helper
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    The typical convention in physics is that ##\theta## is the polar angle and ##\varphi## the azimuthal angle. In mathematics, the convention is usually the other way around. The Wikipedia page shows both conventions.
     
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