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I Infinitesimal area element in polar coordinate

  1. May 2, 2016 #1
    We know, that the infinitesimal area element in Cartesian coordinate system is ##dy~dx## and in Polar coordinate system, it is ##r~dr~d\theta##. This inifinitesimal area element is calculated by measuring the area of the region bounded by the lines ##x,~x+dx, ~y,~y+dy## (for polar coordinate ##r, r+dr, \theta, \theta + d\theta##).

    Now, I tried to calculate it in a different way. We know,
    ## x = r \cos {\theta} ; y= r\sin{\theta}; ##
    Hence,
    ## dx = - r \sin{\theta}~ d\theta + \cos{\theta}~ dr##
    & ## dy = r \cos{\theta}~ d\theta + \sin{\theta} ~dr##

    Now, we have, infinitesimal area element,
    ## dA = dx ~dy = - r^2 \cos {\theta} \sin {\theta} ~d\theta ^2 + (\cos^2 {\theta}-\sin^2{\theta} ) r ~dr~ d\theta + \sin{\theta}\cos{\theta}~ dr^2##
    Where did I make the mistake?
     
  2. jcsd
  3. May 2, 2016 #2

    Orodruin

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    This is not how coordinate transformations of area elements work. You cannot just take the infinitesimals and multiply them together. You need to compute the area element spanned by two infinitesimal displacements.
     
  4. May 2, 2016 #3
    It is not clear to me why it is incorrect to substitute for ##dx## and ##dy##. Could you please explain elaborately?
     
  5. May 2, 2016 #4

    PeroK

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    This is really about a change of variables, rather than comparing a specific small area in two coordinate systems.

    As you have seen, if you have a small rectangle defined by ##dx, dy## and the area of this rectangle is ##dA## then ##dA \ne rdrd\theta##. In fact, it's not even clear what ##dr## and ##d\theta## would be in this case. To see this, draw the rectangle and try to figure out what ##dr, d\theta## would be.

    Instead, if you imagine integrating in polar coordinates, you will take the limit of small areas, defined by ##dr, d\theta##. These are not rectangles (although they are close). Each has an area of approximately ##dA_{p} = rd\theta dr ##.

    This leads to:

    ##\int dxdy = \int rdrd\theta##

    But, there is not a direct relationship between the area of a specific small rectangular area in Cartesian coordinates and a different specific small (almost) rectangular area in polar coordinates.
     
    Last edited: May 2, 2016
  6. May 2, 2016 #5

    Orodruin

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    Because this is not what an area element means. You need to relate the displacement vectors corresponding to a change in each coordinate to the spanned area. This is done through the Jacobian determinant, not by multiplying infinitesimals together.

    You can multiply the infinitesimals if you use an anti-symmetric tensor product which defines an area form. For example:
    \begin{align}
    dx \wedge dy &= [\cos(\theta) dr - r \sin(\theta) d\theta]\wedge [\sin(\theta) dr + r \cos(\theta) d\theta] \nonumber \\
    &= \cos(\theta)\sin(\theta) (dr \wedge dr - r^2 d\theta \wedge d\theta) - r \sin^2(\theta) d\theta \wedge dr + r \cos^2(\theta) dr \wedge d\theta \nonumber \\
    &= r\, dr \wedge d\theta. \nonumber
    \end{align}

    Edit: Updated LaTeX display for the anti-symmetric tensor product.
     
    Last edited: May 3, 2016
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