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Integration of a vector field in polar coordinates

  1. Apr 12, 2012 #1
    Hello all,

    I am trying to understand how to integrate a vector field in polar coordinates. I am not looking to calculate flux here, just the sum of all vectors in a continuous region. However, there is something I am not doing properly and I am a bit lost at this point. Any help would be greatly appreciated.

    1. The problem statement, all variables and given/known data

    Integrate the following two-dimensional vector field defined in polar coordinates: [tex]f=\frac{1}{\rho}\mathbf{e}_\theta[/tex]

    Here the unit vectors are: [tex]\mathbf{e}_\rho \mbox{ and } \mathbf{e}_\theta[/tex]
    for the radial and angular components, respectively.

    The region of integration is the two-dimensional circular region centered on the origin with radius ε, this is:

    [tex]\rho\rightarrow 0\ldots\epsilon,\; \theta\rightarrow 0\ldots 2\pi[/tex]

    2. Relevant equations

    Vector field (polar coordinates):
    [tex]f=\frac{1}{\rho}\mathbf{e}_\theta[/tex]

    Cartesian/polar conversion equations:

    [tex]x=\rho cos(\theta),\\
    y=\rho sin(\theta), \\
    \mathbf{i}=cos(\theta)\mathbf{e}_\rho - sin(\theta) \mathbf{e}_\theta,\\ \mathbf{j}=sin(\theta)\mathbf{e}_\rho + cos(\theta) \mathbf{e}_\theta
    [/tex]
    [tex]
    \rho = \sqrt{x^2+y^2},\\
    \theta = tan^{-1}(y/x), \\
    \mathbf{e}_\rho = cos(\theta) \mathbf{i} + sin(\theta) \mathbf{j}\\
    \mathbf{e}_\theta = -sin(\theta) \mathbf{i} + cos(\theta) \mathbf{j}
    [/tex]

    3. The attempt at a solution

    The first thing I did was to calculate the value of the integral in Cartesian coordinates, first, by representing the field in Cartesian coordinates as follows:

    [tex]\frac{1}{\rho}\mathbf{e}_\theta = \frac{1}{\rho}(-sin(\theta),cos(theta)) = \frac{1}{\rho^2}(-\rho sin(\theta),\rho cos(theta)) = (\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2})[/tex]

    The velocity field I want to integrate is similar to the one in the figure (I had to smooth the function to have a proper plot). So basically inside the region centered at the origin, the sum of all vectors should be zero. This can be verified by integrating the field in cartesian coordinates as follows:

    [tex]\int_{-\epsilon}^{\epsilon}\int_{-\sqrt{\epsilon^2-x^2}}^{\sqrt{\epsilon^2-x^2}}(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}) dxdy[/tex]

    The result is (0,0), which is correct. Now my problem is integrating this field in polar coordinates. My first impulse was to integrate the angular component of the vector field (as the radial is zero), over the circle defined in polar coordinates, this is:

    [tex]\int_{0}^{2\pi}\int_{0}^{\epsilon}f_{\theta} \rho d\rho d\theta = \int_{0}^{2\pi}\int_{0}^{\epsilon}\frac{1}{\rho} \rho d\rho d\theta = \int_{0}^{2\pi}\int_{0}^{\epsilon}d\rho d\theta=2\pi\epsilon[/tex]

    Which is incorrect, as the vector field I would obtain is:
    [tex]
    2\pi\epsilon \mathbf{e}_\theta
    [/tex]

    Then I attempted integrating:

    [tex]\int_{0}^{2\pi}\int_{0}^{\epsilon}\frac{1}{\rho} \mathbf{e}_\theta \rho d\rho d\theta[/tex]

    Converting eθ to Cartesians would add sin and cos functions that would make the integral go to zero, as a matter of fact, by using the conversions above, I wrote:

    [tex]\int_{0}^{2\pi}\int_{0}^{\epsilon}\frac{1}{\rho}(-sin(\theta), cos(\theta)) \rho d\rho d\theta[/tex]

    Here the result is the zero vector (0,0), which is the final result I was looking for. But, is this way of calculating the polar integral correct? I am afraid I am missing something conceptual here and the whole procedure may be flawed. Thanks in advance for any assistance, advice or help on this issue.

    Regards,
    M.
     
  2. jcsd
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