- #1

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So I start with the arclength formula in Cartesian coordinates:

$$ ds = \sqrt{1 + \left( \frac{dy}{dx} \right)^2 } dx $$

With ##x= r \cos \theta, y = r \sin \theta ##

I start by making a change of variables to ##r##:

$$ \left( \frac{dy}{dx} \right)^2 = \left( \tan \theta \right)^2 $$

for the differential I get:

$$ dx = \left[ k \cos \theta - r \sin \theta \right] d\theta $$

this goes to:

$$ ds = \sqrt{1 + \left( \tan \theta \right)^2 } \left[ k \cos \theta - r \sin \theta \right] d\theta $$

If Iv'e done this correctly up to this point I see that it is integrable by researching it and I can get ##s## as a funtion of ##\theta##. But I realize its not really what I want... I want to measure the radius in the field.

I guess its true by definition of the Archimedean spiral that ##\theta## ( the upper limit of the integration ) is just the ##\frac{r}{T }2 \pi ##?

So if that is true, Is there a reasonable measure of some radius I can take in the field that would approximate the integral?