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Spherical, cylinder and polar coordinates

  1. May 9, 2010 #1
    I can't really understand something in spherical and cylinder coordinates

    let me start with polar coordinates first

    if we have for example x^2 + y^2 = 4 this is a circle with center (0,0) and radius 2

    in polar coordinates x and y will be

    x = rcosφ
    y = rsinφ
    0<=r<=2
    0<=φ<=2π

    here φ is from 0 to 2π but sometimes i see in problems they take φ to be from -π/2 to π/2

    for example in a circle like this where you want to find the area

    [PLAIN]http://img576.imageshack.us/img576/333/87706111.jpg [Broken]

    why we have to take here r from -π/2 to π/2? I mean, the definition says from 0 to 2π right?

    this is my first question about polar coordinates

    let's go to spherical

    let's say we have this sphere x^2 + y^2 + z^2 = 1

    x = rsinθcosφ
    y = rsinθsinφ
    z = rsinθ
    0<=r<=1

    now sometimes i see this kind of definition about θ and φ

    -π/2<=θ<=π/2
    0<=φ<=2π

    or this

    -π<=θ<=π
    0<=φ<=π

    how can i know which one to take in different kind of problems? also i can't understand the difference between the both, aren't they the same thing? and why don't we have θ in polar coordinates?

    in cylinder we have the same definitions with polar coordinates except one thing, we have z too

    i would appreciate any explanation provided, i cant find anything online, and our teacher didnt explain these stuff at all

    thanks in advance
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 9, 2010 #2

    tiny-tim

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    Hi kliker! :smile:
    (you meant φ of course :wink:)

    φ always has to go from 0 to 2π (or from -π to π, or any other interval of length 2π).

    In this case, however, there are no values for π/2 < φ < 3π/2, so we can leave them out.
    You can use either system, there is no difference.

    (I've read somewhere that physicists prefer one system, and mathematicians prefer the other. :rolleyes:)

    Use whichever you prefer (in an exam, I can't see how it would matter). :smile:
    Polar coordinates usually do use θ, and cylindrical coordinates usually use φ (I expect to avoid confusion).
     
  4. May 9, 2010 #3
    Hi tiny-time, thanks for the answer I think that I get it right now, but let's look on this example

    suppose that we want to calculate I where

    gif.latex?I%20=%20\int_{D}%20\int%20\int%20z%20dxdydz.gif

    where D = {(x,y,z) ER^3: x^2+y^2+z^2<=1}

    using spherical coordinates we have

    x = rcosθsinφ
    y = rsinθcosφ
    z = ρcosθ

    also

    0<=ρ<=1
    0<=φ<=2π
    -pi/2<=θ<=p/2

    now If I try to calculate the following integral

    2}\int_{0}^{2pi}\rho%20cos\theta%20\rho%20^2%20sin\theta%20d\phi%20d\thetad\rho.gif

    I find result 0, but the correct result is π/6

    i cant understand what im doing wrong

    also as for the cylinder, you're right it's φ

    but what's the difference between φ and θ? I can't really get it, why we use two different letters if they are the same thing?

    i found this somewhere

    this is for ρ

    [PLAIN]http://www.math.montana.edu/frankw/ccp/multiworld/multipleIVP/spherical/rho.gif [Broken]

    this is for φ

    [PLAIN]http://www.math.montana.edu/frankw/ccp/multiworld/multipleIVP/spherical/phi.gif [Broken]

    this is for θ

    [PLAIN]http://www.math.montana.edu/frankw/ccp/multiworld/multipleIVP/cylindrical/theta.gif [Broken]

    thanks in advance
     
    Last edited by a moderator: May 4, 2017
  5. May 9, 2010 #4

    vela

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    Your limits for θ are wrong. θ runs from 0 to π.

    With spherical coordinates, you have the radial distance and two angles. There are various conventions on how these angles are defined and what they are called, and that's what's messing you up. For the equations you wrote,

    [tex]x = r \sin \theta \cos \phi[/tex]
    [tex]y = r \sin \theta \sin \phi[/tex]
    [tex]z = r \cos \theta[/tex]

    θ, the angle measured from the +z-axis, goes from 0 to π, and ϕ, the angle measured from the +x axis to the projection onto the xy-plane, typically runs from 0 to 2π. Sometimes people use the interval [-π,π] for ϕ.
     
  6. May 9, 2010 #5
    i think i made a mistake in my previous integral

    2}\int_{0}^{2pi}\rho%20cos\theta%20\rho%20^2%20sin\theta%20d\phi%20d\theta%20d\thetad\rho.gif

    here θ goes from -pi/2 to pi/2, isnt this the same as 0 to π?
    and φ from 0 to 2pi
     
  7. May 9, 2010 #6

    LCKurtz

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    You have φ and θ interchanged from the way most calculus texts do it. And, even so, your equations don't jibe with the usual notation. They also don't jibe with your animations below. Those animations illustrate θ being the usual polar coordinate angle in the xy plane going from 0 to 2π and φ being the angle between the z axis and the spherical ρ. φ goes from 0 to π in the animations.

     
    Last edited by a moderator: May 4, 2017
  8. May 9, 2010 #7

    vela

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    No, it's not. You would never be able to describe a point where z<0 if you used that range.

    The answer to your original problem should be 0, not π/6. You're essentially calculating the average of the z-coordinate over a sphere, which, by symmetry, you should be able to see will be 0. Are you sure you stated the problem correctly?
     
  9. May 9, 2010 #8
    im totally confused right now, the book says

    φ goes from 0 to 2pi, and θ from -pi/2 to pi/2 then on some other page, it says θ goes from 0 to pi and now from the shapes, φ must be from 0 to pi?

    @vela, I checked the problem statement again, you're right, It says z should be >=0
     
  10. May 9, 2010 #9

    vela

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    I suspect there's some missing info here that would clear up these apparent contradictions. How is θ defined in your book (for the first case you mention)?

    It helps to think about what points on the sphere the various values of θ and ϕ correspond to. They're similar to latitude and longitude for locating a point on the globe. Different values, like θ=π/4, Φ=0 and θ=-π/4, Φ=π can correspond to the same point, namely (1/√2, 0, 1/√2), but usually, you restrict the values of θ and ϕ so that this doesn't happen.

    In this case, the integral should evaluate to π/4, not π/6.
     
  11. May 10, 2010 #10
    yes it's pi/4

    θ for the polars is from 0 to 2pi

    i have another question

    is -pi to +pi the same as 0 to 2pi?

    or the same as -pi/2 to pi/2?

    in the first shape, if the circle has a center that is not 0,0

    in polars we have to use θ where θ goes from 0 to 2pi right? but this works if the center is on 0,0 only, why is that?
     
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