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[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?