It seems to me that integrating a polar equation should give you the(adsbygoogle = window.adsbygoogle || []).push({}); arc lengthof 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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Integrating a polar equation

**Physics Forums | Science Articles, Homework Help, Discussion**