# Why is the formula $\ Arc length = \int r \times d\theta$ not correct?

The formula for calculating the area of a curve in a polar graph is $\large \rm \frac{1}{2}\int r^2~ d\theta$ and is adapted from $$\large \rm Area = \frac{1}{2}r^2\times \theta$$ But the formula to calculate the arc length is very different from $\large \rm Length =\int r ~d\theta$ which should've been adapted from $$\large \rm Length=r\times \theta$$ Why is the formula $\large \rm Length =\int r ~d\theta$ not correct to calculate the arc length of a sector in a polar graph?

• The true formula for length is $ds=\sqrt{dr^2 + r^2 d\theta^2}$. The problem is what if instead of curling around your curve diverges along a line? You need to take into account both degrees of freedom. – Hamed Aug 15 '16 at 15:10

Say we zoom in very close to the curve so that it looks like a straight line:

The range of $\theta$ values shown in the second picture will be taken to be $d\theta$, and the distance from the origin $r$. For the red line, we can see $rd\theta$ is a good approximation for the length. However, for the blue line, we can see that no matter how far we zoom in, $rd\theta$ will be off of the true length of the line by a factor of about $\sqrt{\frac{1}{m^2}+1}$, where $m$ is the slope of the blue line.

In fact, calculating $m$ would give us the formula for arc length as $rd\theta\sqrt{1+\frac{1}{m^2}}$.

• Oh thanks. That provides me a vague perception of the truth. One question, if we allow our value of r to change, would then rdθ be sufficient? – mathnoob123 Aug 15 '16 at 17:23
• If $rd\theta$ ever works it's only by coincidence. – florence Aug 15 '16 at 22:27

You are wrong. The formula for calculating the area enclosed by an arbitrary arc $A=\frac{1}{2}r^2\theta$ comes from the integration $\int_0^r r \, dr \int_0^\theta d\theta$.

The reason behind this is that the differential lengths along $r$ and $\theta$ axes in polar co-ordinates are $dr$ and $rd\theta$, just like in Cartesian co-ordinates, the differential lengths along $x$ and $y$ axes are $dx$ and $dy$.

A spiraling arc has a radial $dr$ component and circumferential $r d\theta$component. So with only the former considered you are indirectly still on a circle. The Pythagorean sum of the two as Hamed commented above, is the diagonal differential arc length.

Let $C$ be a smooth curve described parametrically by

\begin{align} \vec r&=\vec r(\theta)\\\\ &=\hat r(\theta)r(\theta) \end{align}

for $\theta \in [\alpha,\beta]$, where $\hat r(\theta)$ is the radial unit vector.

The arc length $s$ of the curve $C$ can be approximated by the sum of incremental displacements $\left|\Delta\vec r(\theta_n)\right|=\left|\vec r(\theta_n)-\vec r(\theta_{n-1})\right|$ for $n=1,\cdots ,N$ where $\theta_0=\alpha$ and $\theta_N=\beta$. Then, we can write heuristically

$$s\approx \sum_{n=1}^N \left|\vec r(\theta_n)-\vec r(\theta_{n-1})\right|$$

Letting $N\to \infty$ while $\max_{n\in [1,N]}\left(\left|\Delta\vec r(\theta_n)\right|\right)\to 0$, we have

\begin{align} s&=\int_\alpha^\beta \left|\frac{d\vec r(\theta)}{d\theta}\right|\,d\theta\\\\ &=\int_\alpha^\beta \left|\frac{d\hat r(\theta)r(\theta)}{d\theta}\right|\,d\theta\\\\ &=\int_\alpha^\beta \left|\hat r(\theta)\frac{dr(\theta)}{d\theta}+r(\theta)\frac{d\hat r(\theta)}{d\theta}\right|\,d\theta\\\\ &=\int_\alpha^\beta \left|\hat r(\theta)\frac{dr(\theta)}{d\theta}+r(\theta)\hat \theta(\theta)\right|\,d\theta\\\\ &=\int_\alpha^\beta \sqrt{\left(\frac{dr(\theta)}{d\theta}\right)^2+r^2(\theta)}\,d\theta\\\\ &\ne \int_\alpha^\beta r(\theta)\,d\theta\\\\ \end{align}

unless $\left(\frac{dr(\theta)}{d\theta}\right)^2=0$. This implies that $r(\theta)=\text{constant}$, which describes a segment of circle. Therefore, unless the contour $C$ is a segment on a circle, then the arc length $s$ is not given by the formula questioned in the OP.

• I am sorry. I am not very good at understanding mathematical notations. Can you just describe the logical flaw which occurs when $\ \int r d\theta$ is used to find the arc length? – mathnoob123 Aug 15 '16 at 17:17
• The flaw is that it does not account for the non-zero derivative $\frac{dr(\theta)}{d\theta}$. – Mark Viola Aug 15 '16 at 17:40
• I am still in dark. Why is $\ \frac{dr}{d\theta}$ significant if we allow our value of r to change frequently? By changing its value, I mean $\ r= r+dr$ (the equation is defined to be sensible in programming languages) – mathnoob123 Aug 15 '16 at 17:45
• It's not $dr$, it's $|d\vec r|=\sqrt{(dr)^2+r^2(d\theta)^2}$. – Mark Viola Aug 15 '16 at 18:35
• Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark – Mark Viola Aug 19 '16 at 4:11