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Spherical Coordinates

  1. Dec 1, 2013 #1
    So, I was curious about this and found more or less what I was looking for here: http://electron9.phys.utk.edu/vectors/3dcoordinates.htm

    Except, something is bothering me about those equations. At the very bottom, the equation for Theta in a spherical coordinate system; shouldn't it be
    [itex]\theta = {tan^{-1}}( \frac{\sqrt{{x^{2}}+{y^{2}}}}{z})[/itex]
    instead of
    [itex]\theta = {tan^{-1}}( \frac{z}{\sqrt{{x^{2}}+{y^{2}}}})[/itex]

    (The image in question)

    Because [itex]{tan^{-1}}( \frac{opposite}{adjacent}) = \theta [/itex] , and looking at angle [itex] \theta [/itex] , the line opposite of it is exactly equal to [itex] \sqrt{{x^{2}}+{y^{2}}} [/itex] , and the line adjacent to it equal to [itex]z[/itex].

    So I'm wondering if I'm in error (and how so if I am) or if the linked page is.
  2. jcsd
  3. Dec 1, 2013 #2


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    Staff: Mentor

    You are correct. That equation on that page is in error.
  4. Dec 1, 2013 #3
    It appears that page is wrong. You're right, the correct relation is

    ##\theta = \tan ^{-1}\left( \frac{\sqrt{x^2 +y^2 }}{z}\right)##.

    You could also write it as ##\theta = \cos ^{-1} \left(\frac{z}{\sqrt{x^2 +y^2 +z^2 }}\right)##

    Edit: Darn you, jtBell, you beat me by seconds!
  5. Dec 1, 2013 #4

    Simon Bridge

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    Science Advisor
    Homework Helper

    yep... it sure looks like they've got their sides mixed up.
    However, it is more usual to use ##\theta = \cos^{-1}(z/r):r=\sqrt{x^2+y^2+z^2}##
  6. Dec 3, 2013 #5
    Thanks very much for all the replies! Are the equations for the x, y, and z components from a spherical coordinate system correct on that page, though?

    And why is the ##\theta = \cos ^{-1} \left(\frac{z}{\sqrt{x^2 +y^2 +z^2 }}\right)## approach more common? The [itex]\theta = {tan^{-1}}( \frac{\sqrt{{x^{2}}+{y^{2}}}}{z})[/itex] one has less terms. Something to do with higher mathematics?
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