Arc Length in Polar
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 7.4: “Arc Length of a Curve and Surface Area” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/7-4-arc-length-of-a-curve-and-surface-area |
| Textbook used in class | Stewart, Calculus, Section 10.4: “Areas and Lengths in Polar Coordinates” |
Opening Scenario
A spiral $r = \theta$ winds outward from the origin. How long is the spiral from $\theta = 0$ to $\theta = 2\pi$? The curve is not $y = f(x)$, so the Cartesian arc length formula does not apply directly. Instead, write the spiral as a parametric curve (using $\theta$ as the parameter) and apply the parametric arc length formula.
Quick Reference
Arc length of a polar curve $r = f(\theta)$, $\alpha \leq \theta \leq \beta$: $$L = \int_\alpha^\beta \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta = \int_\alpha^\beta \sqrt{[f(\theta)]^2 + [f'(\theta)]^2}\,d\theta.$$
Key Concepts
1. Derivation from Parametric Arc Length
For $r = f(\theta)$, write $x = r\cos\theta$, $y = r\sin\theta$ (parametric with $\theta$ as parameter). Then:
$\dfrac{dx}{d\theta} = r'\cos\theta - r\sin\theta$, $\quad \dfrac{dy}{d\theta} = r'\sin\theta + r\cos\theta$.
$\left(\dfrac{dx}{d\theta}\right)^2 + \left(\dfrac{dy}{d\theta}\right)^2 = (r'\cos\theta - r\sin\theta)^2 + (r'\sin\theta+r\cos\theta)^2$.
Expand: $(r')^2\cos^2\theta - 2r'r\cos\theta\sin\theta + r^2\sin^2\theta + (r')^2\sin^2\theta + 2r'r\sin\theta\cos\theta + r^2\cos^2\theta$.
$= (r')^2(\cos^2\theta+\sin^2\theta) + r^2(\sin^2\theta+\cos^2\theta) = (r')^2 + r^2$.
So $\sqrt{(dx/d\theta)^2+(dy/d\theta)^2} = \sqrt{r^2+(r')^2}$, giving $L = \int_\alpha^\beta\sqrt{r^2+(r')^2}\,d\theta$.
2. The Integrand $\sqrt{r^2 + (r')^2}$
Think of $r\,d\theta$ as a tiny arc of a circle (circumferential step) and $dr = r'\,d\theta$ as a radial step. The arc length element is: $$ds = \sqrt{(r\,d\theta)^2 + (dr)^2} = \sqrt{r^2+(r')^2}\,d\theta.$$
This Pythagorean picture is exact in the limit.
Worked Example
Find the arc length of the cardioid $r = 1 + \cos\theta$ from $\theta = 0$ to $\theta = 2\pi$.
$r' = -\sin\theta$.
$r^2 + (r')^2 = (1+\cos\theta)^2 + \sin^2\theta = 1 + 2\cos\theta + \cos^2\theta + \sin^2\theta = 2 + 2\cos\theta = 2(1+\cos\theta)$.
Use $1+\cos\theta = 2\cos^2(\theta/2)$:
$r^2+(r')^2 = 4\cos^2(\theta/2)$.
$\sqrt{r^2+(r')^2} = 2|\cos(\theta/2)|$.
For $\theta \in [0,\pi]$: $\cos(\theta/2) \geq 0$, so $|\cos(\theta/2)| = \cos(\theta/2)$. For $\theta \in [\pi, 2\pi]$: $\cos(\theta/2) \leq 0$, so $|\cos(\theta/2)| = -\cos(\theta/2)$.
By symmetry (the cardioid is symmetric about $\theta = 0$): $$L = 2\int_0^\pi 2\cos(\theta/2)\,d\theta = 4\left[2\sin(\theta/2)\right]_0^\pi = 4\cdot 2 = 8.$$
Boxed answer: $L = 8$.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Using $\int r\,d\theta$ for arc length | Writing $L = \int(1+\cos\theta)\,d\theta$ | The polar arc length formula is $\int\sqrt{r^2+(r')^2}\,d\theta$; $r\,d\theta$ gives arc of a circle, not general arc length |
| Forgetting the absolute value on $\cos(\theta/2)$ | Getting a negative contribution for $\theta \in [\pi, 2\pi]$ | $\sqrt{4\cos^2(\theta/2)} = 2|\cos(\theta/2)|$; use absolute value or split at $\theta = \pi$ |
| Using the Cartesian arc length formula $\int\sqrt{1+(dy/dx)^2}\,dx$ without converting | Trying to express the cardioid as $y = f(x)$ | Convert to parametric and use $\int\sqrt{r^2+(r')^2}\,d\theta$ directly |
Common Misconceptions
polar arc length is $\int_\alpha^\beta r\,d\theta$.
This is the multiplicative-not-additive error. The quantity $\int r\,d\theta$ computes the area of a circular sector swept by the radius, not the length of the curve. Arc length requires the Pythagorean theorem applied to infinitesimal steps in the plane: the tangential step is $r\,d\theta$ and the radial step is $dr = r'\,d\theta$, giving hypotenuse $\sqrt{r^2 + (r')^2}\,d\theta$. The squared terms and the square root are essential; omitting them produces an expression for area rather than length.
Leveled Practice
Level 1 -- Circle
Problem 1. Verify that the circumference of the circle $r = a$ is $2\pi a$ using the polar arc length formula.
Show answer
$r = a$, $r' = 0$.
$L = \displaystyle\int_0^{2\pi}\sqrt{a^2+0}\,d\theta = a\int_0^{2\pi}d\theta = 2\pi a$. Confirmed.
Level 2 -- Spiral
Problem 2. Find the length of the Archimedean spiral $r = \theta$ from $\theta = 0$ to $\theta = 1$.
Show answer
$r = \theta$, $r' = 1$.
$r^2+(r')^2 = \theta^2+1$.
$L = \displaystyle\int_0^1\sqrt{\theta^2+1}\,d\theta$.
This is a standard trig-substitution integral: $\theta = \tan\phi$, $d\theta = \sec^2\phi\,d\phi$, $\sqrt{\theta^2+1} = \sec\phi$.
$L = \displaystyle\int_0^{\pi/4}\sec^3\phi\,d\phi = \frac{1}{2}[\sec\phi\tan\phi + \ln|\sec\phi+\tan\phi|]_0^{\pi/4}$.
$= \frac{1}{2}[\sqrt{2}\cdot 1 + \ln(\sqrt{2}+1)] = \frac{\sqrt{2}+\ln(\sqrt{2}+1)}{2} \approx \frac{1.414+0.881}{2} \approx 1.148$.
Boxed answer: $L = \dfrac{\sqrt{2}+\ln(\sqrt{2}+1)}{2}$.
Mastery Checklist
Mental Model
Arc length in polar coordinates is the same integral-of-speed idea as for parametric curves. The speed is $\sqrt{(x')^2+(y')^2} = \sqrt{r^2+(r')^2}$ in polar form. Each infinitesimal step along the curve has a tangential component $r\,d\theta$ (spinning around) and a radial component $r'\,d\theta = dr$ (moving in or out). The step length is the hypotenuse: $\sqrt{(r\,d\theta)^2+(dr)^2} = \sqrt{r^2+(r')^2}\,d\theta$.
Connections
Looking back
- Parametric arc length (Section 10.2): The polar formula is derived from the parametric formula.
- Polar area (Section 10.4): Area uses $\frac{1}{2}r^2\,d\theta$; arc length uses $\sqrt{r^2+(r')^2}\,d\theta$.