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Converting cartesian unit vectors to spherical unit vectors

  1. Jan 27, 2009 #1
    1. The problem statement, all variables and given/known data
    Well, it's all in the title. I just need to show that Gauss's theorem applies to this fluid flow and have converted all my (x,y,z) components to their respective (r,theta,phi) versions, but I can't remember the spherical counterparts of [tex]\hat{x}[/tex],[tex]\hat{y}[/tex],[tex]\hat{z}[/tex].
     
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  3. Jan 28, 2009 #2

    Dick

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  4. Jan 28, 2009 #3
    Wouldn't it have been easier to look at the page you linked to and see that there is no direct conversion there than to assume that I didn't already do that and reply to my post?... maybe not. Thanks though.
     
  5. Jan 28, 2009 #4

    Dick

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    I thought you wanted [tex]\hat{r}, \hat{\theta}, \hat{\phi}[/tex]. I would call those the "spherical counterparts" of the cartesian basis vectors. What do you want?
     
  6. Jan 28, 2009 #5
    You're right, I didn't word it very well. Here's what I was trying to compute:

    [tex] \vec{u} \cdot d\vec{A}[/tex]

    where

    [tex] \vec{u} = (2xy^{2}+2xz^{2})\hat{x} + (x^{2}y)\hat{y} + (x^{2}z)\hat{z}[/tex]

    and

    [tex] d\vec{A} = r^{2}sin\theta d\theta d\phi \hat{r} [/tex].

    I first converted the magnitudes of [tex]\vec{u}[/tex] to spherical coordinates, and what I intended to do was convert [tex] \hat{x}, \hat{y}, \hat{z} [/tex] to spherical as well. Does that make sense?

    Anyway, I ended up using [tex] \hat{r} = sin\theta cos\phi \hat{x} + sin\theta sin\phi \hat{y} + cos\theta \hat{z}[/tex], leaving the unit vecors in the cartesian basis.

    But then when I tried to compute the scalar product I ended up with a virtually unintegratable expression. The left side of the equation (remember I was trying to show that Gauss' theorem holds over a spherical region [tex]a^{2} = x^{2} + y^{2} + z^{2}[/tex]) came out to [tex]\frac{8}{3} \pi a^{5}[/tex].

    Do you think I was setting it up right? I was thinking that I probably skrewed up some algebra while computing the dot product.
     
  7. Jan 28, 2009 #6

    Dick

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    If you are integrating over a sphere of radius a, then the normal vector is (x*xhat+y*yhat+z*zhat)/a. Does that suggest any simplifications? Sorry, it's kind of late here so I haven't really thought this through.
     
  8. Jan 29, 2009 #7

    Dick

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    Another tip. Once you have your x,y,z expression for the integrand. You can choose the axis of the spherical coordinate system to point along any cartesian axis.
     
    Last edited: Jan 29, 2009
  9. May 10, 2009 #8
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