Why Can't We Use \int^{β}_{α} rdθ for Polar Curve Arc Length?

[Bipolarity]

If we divide the polar curve into infinitely thin sectors, the arc length of a single sector can be approximated by ds = \frac{dθ}{2π}2πr = rdθ. So why can't we model the arc length of the curve as \int^{β}_{α} rdθ

It turns out that the correct formula is actually
\int^{β}_{α}\sqrt{r^{2}+(\frac{dr}{dθ})^{2}} \ dθ
I know how the correct formula is derived, I just can't figure out why the reasoning for the first formula is incorrect.

BiP

 

[haruspex]

By Pythagoras, ds² = (rdθ)²+dr². You can't go ignoring the dr term in general.

 

[Bipolarity]

By Pythagoras, ds² = (rdθ)²+dr². You can't go ignoring the dr term in general.

My question was, why can't you use the arc length of the sector?

BiP

 

[haruspex]

My question was, why can't you use the arc length of the sector?

BiP

Because in general the radial movement is of comparable magnitude to the tangential movement. In the extreme case, part of the curve might move directly away from the origin, so there dθ would be zero and you would only have dr.

 

[arildno]

A straight line radiating from the origin must also have an arc length formula in polar coordinates, even though the angular coordinate remains a constant along the whole curve.