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Nonsensical (lack of) relation between area and arc-length of polar curves

  1. Mar 14, 2012 #1
    It is known that the area of a sector of a polar curve is

    [itex]\frac{1}{2}\int r^{2} d \theta[/itex]

    This of course comes from the method of finding the area of an arc geometrically, by multiplying the area of the circle by the fraction we want

    [itex]\frac{\theta}{2\pi}\pi r^{2}[/itex]

    Today I learned how to calculate the arc length of a polar curve. The method is similar to the Cartesian method (by integrating [itex]ds[/itex]), where

    [itex]ds = \sqrt{r^{2}+ \left(\frac{dr}{dθ}\right)^{2}}[/itex]

    I found this odd, considering the parallels between area in calculus and geometry. I figured it would be based on the geometric arc length formula, where the circumference is multiplied by the fraction of the total circle

    [itex]\frac{\theta}{2\pi} 2 \pi r[/itex]

    Thus giving [itex]θr[/itex]. In order to integrate small pieces of arc with respect to θ as defined by [itex]r(θ)[/itex] (analogous to summing the area of small sectors), we have

    [itex]\int r dθ[/itex]

    However this doesn't work, and I don't know why. The geometric formulas both integrals are derived from are correct, but this formula doesn't give you the arc of a polar curve. Does anyone know why it doesn't work? Better yet is there a way to fix it?
  2. jcsd
  3. Mar 14, 2012 #2


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    welcome to pf!

    hi thetasaurus! welcome to pf! :smile:
    but that only works for constant r …

    you need the dr/dθ term for the extra length when r is changing :wink:
  4. Mar 14, 2012 #3
    Re: welcome to pf!

    I would agree with you yet we are able to calculate the area using this method with a non-constant r as a function of θ. My question is why doesn't this work for arc length.

    Technically r even is constant... the tiny sliver of the curve represented by dθ is such a small angle/sector/arc that it may as well be constant for any one value of θ.
  5. Mar 15, 2012 #4


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    Hey thetasaurus and welcome to the forums.

    Just out of curiosity do you or your teacher/lecturer derive this formula from first principles (going from pythagoras theorem to final derivation) or did you not?

    If you didn't it probably doesn't make sense since you are trying to probably see a pattern in terms of the formula rather than what it is based on and I think if you derived it yourself the confusion would probably go away.
  6. Mar 18, 2012 #5
    He derived it for me (as he did for the Cartesian formula as well) and I agree that it makes perfect sense. However I don't understand why the way I did it doesn't work. I know that the Pythagorean method is valid, but it seems to me like the geometric way I came up with should work as well.
  7. Mar 18, 2012 #6


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    The only comment I can make is not to try and use your intuition for the final result but instead to use it for the principle it was based on which is the pythagorean theorem.

    The pythagorean theorem expresses the relationship when we have cartesian co-ordinate systems and in that context it makes sense. Intuitively you can draw a two-dimensional right angle triangle or even a three-dimensional box to get the pythagorean theorem and when you look at it, it makes intuitive sense because you intuitively know cartesian type geometry.

    Don't fall into the trap of trying to make one formula 'look like' another because it will at some point (like this time) work against you.

    If you ever do probability and/or statistics you will have to do the same thing: get an intuition for the principles themselves and not the complex formulas that are derived from these.
  8. Mar 18, 2012 #7
    I'd love it if both formulas were the same, then they'd be all orderly and look nice. Evidently you're right that my look-alike formula is working against me, and as of now I have (reluctantly) accepted this fact. However I'm still curious if anyone can prove why it doesn't work, or where I went wrong, aside from the fact that it just doesn't work.
  9. Mar 18, 2012 #8


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    The best way that I can answer your question is to start with with the pythagorean form of distance where in two dimensions: ds^2 = dx^2 + dy^2 and from that use the relationships between (x,y) and (r,theta). Your ds is subsequently ds = SQRT(dx^2 + dy^2) and from this ds/dx = SQRT(dx^2/dx^2 + (dy/dx)^2) = SQRT(1 + (dy/dx)^2).

    Now from here you have to use the chain-rule. We have dy/dx for cartesian but we are looking at (r,theta) as opposed to (x,y). This means we need to find the transformation from (x,y) to (r,theta). For the 2D case we have r = SQRT(x^2 + y^2) and theta = arctan(y,x). This means that we need to find dy/dx = dy/dr dr/dx + dy/d(theta) d(theta)/dx.

    Using the relationship between (r,theta) and (x,y) can you prove the general formula for the 2D arclength in polar coordinates?
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