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Integrating a polar equation

  1. Sep 9, 2010 #1
    It seems to me that integrating a polar equation should give you the arc length of the curve, rather than the area under it. This is my reasoning:

    A polar equation is in the form of:

    (1) [tex]
    r = f(\theta)
    [/tex]

    The arc length of a segment of a circle where the radius is constant is given by [tex]s = r\theta
    [/tex]

    If you let [tex] \theta -> 0[/tex] then r essentially becomes constant over the interval [θ , θ +dθ]

    So, multiplying eq. (1) by [tex]d\theta[/tex] gives [tex]rd\theta = f(\theta)d\theta[/tex] and gives an arc length of zero width.

    Now integrate with respect to [tex] \theta [/tex]:

    [tex] \int{f(\theta)d\theta} = s[/tex] and you have the length of the curve.

    Is my reasoning correct?
     
  2. jcsd
  3. Sep 9, 2010 #2

    Mark44

    Staff: Mentor

    Nope, otherwise there would be no need for this formula for the arc length of a polar curve r = f(theta).
    [tex]\int_{\theta = \alpha}^{\beta}\sqrt{(f(\theta))^2 + (f'(\theta))^2 }d\theta[/tex]

    You can't say that r is constant on the interval [itex][\theta, \theta + d\theta][/itex]. That's the problem.
     
  4. Sep 13, 2010 #3
    could you explain why you cant? if the angle is approaching zero, then the radius should be approaching some constant value.
     
  5. Sep 13, 2010 #4

    Mark44

    Staff: Mentor

    That's true for a circle, but not true in general. To approximate the arc length between the points (r, [itex]\theta[/itex]) and (r + [itex]\Delta r[/itex], [itex]\theta + \Delta \theta[/itex]), you have to incorporate the fact that both r and [itex]\theta[/itex] are changing.

    Let's look at this in a different context, with f(x) = 2x, on an interval [x0, x0 + h]. By your logic, if h approaches zero, f(x) approaches some constant value, so the arc length would be just the horizontal distance along this interval. If this were true, the line segment from (x0, 2x0) to (x0 + h, 2(x0 + h)) would become flatter and flatter as h got smaller. Instead, what happens is that no matter how small an interval you use, the slope of the segment between the two points doesn't change, and never gets close to zero.

    If you look at two points (x0, 2x0) and (x0 + h, 2(x0 + h)), it can be seen that the distance between the two points is h[itex]\sqrt{5}[/itex].

    Jumping back to arc length in polar coordinates, let's say we have two points (r, [itex]\theta[/itex]) and (r + [itex]\Delta r[/itex], [itex]\theta + \Delta \theta[/itex]). We can approximate the distance between them using a right triangle. One of the sides has a length of [itex]r \Delta \theta[/itex], and the other side has a length of [itex]\Delta r[/itex]. The length of the hypotenuse is [tex]\sqrt{(\Delta r)^2 + (r \Delta \theta)^2}[/tex]. If you do some algebraic manipulations on this, you get [tex]\sqrt{(\frac{\Delta r}{\Delta \theta})^2 + (r)^2} \Delta \theta[/tex].

    Passing to the limit gives us the familiar form of arc length for polar curves.
     
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