why is the length of arc in polar coordinate like this?

Question

I learned $L=\int_a^b{\sqrt{r^2+\frac{dr}{d\theta}^2}d\theta}.$ But, Why do I calculate like $L=\int_a^b{rd\theta}$ ?
In my opinion, considering riemann sum like $\lim_{n\to\infty}{\sum_{i=1}^{i=n}{r(\theta_i)\triangledown\theta}}$ is better because the formula of arc in circle is $L=r\theta$.
Why doesn't this work?

Answer 1

Here is an exercise to test your formula.

Consider the straight line from polar coordinates $r=\sqrt2,\theta=\frac\pi4$ to polar coordinates $r=1,\theta=\frac\pi2.$ You may recognize this as the line segment from $(1,1)$ to $(0,1)$ in Cartesian coordinates. The length of the curve is $1.$

The polar equation of the line through these points is $r = \dfrac{1}{\sin\theta}.$

So according to your formula we should calculate the length of the line as $$ \int_{\pi/4}^{\pi/2} \frac{1}{\sin\theta}\, \mathrm d\theta \approx 0.8814. $$

To make a Riemann sum for the length of a curve in polar coordinates you can draw many rays from the origin intersecting your curve, including a ray through the start of the curve and one through the end, so that you partition the curve into smaller pieces. So far, this is a correct approach. The formula you have given is equivalent to the assumption that the length of each piece of the curve under this partition is approximately $r \Delta \theta$ where $r$ is somewhere between the minimum and maximum distance from the origin along that part of the curve and $\Delta \theta$ is the angle between the rays. This assumption is not a terrible one when $\mathrm dr/\mathrm d\theta$ is small, but it works very poorly when $\mathrm dr/\mathrm d\theta$ is large. Even for the example above, where $\mathrm dr/\mathrm d\theta = -\sqrt2$ at one end of the segment, you have a substantial overall error.

It is fundamentally the same mistake you would make by assuming that the length of the part of a curve between two vertical lines in Cartesian coordinates is simply the distance between the lines.

Answer 2

normally people say: $$dL=\sqrt{(dy)^2+(dx)^2}=\sqrt{\left(\frac{dy}{dx}\right)^2+1}\,dx$$ using the fact that $y=r\cos\theta=\sqrt{1-x^2}$ work out $dy/dx$ and $dx$ to derive $dL$ then integrate.