Arc Length of Polar Curves

The arc length of a polar curve r = f(θ) from θ = α to θ = β is L = ∫ √(r² + (dr/dθ)²) dθ with limits α and β. It comes from the parametric arc length formula with x = r cos θ and y = r sin θ. As with area, choose limits that trace the curve exactly once.

Before this lesson: Arc Length of Parametric Curves, Derivatives of Polar Functions and Area in Polar Coordinates

The idea: two ways the point moves

As θ increases, a point on a polar curve moves in two ways at once. It slides outward or inward at the rate dr/dθ, and it swings around the origin. Swinging through a small angle Δθ at distance r covers an arc of length about rΔθ.

Those two motions are at right angles, so the Pythagorean theorem combines them: each small piece has length about √(r² + (dr/dθ)²) Δθ. Adding the pieces gives the integral.

Polar arc length formula

L = \int_{\alpha}^{\beta} \sqrt{r^2 + \left(\dfrac{dr}{d\theta}\right)^2}\,d\theta

Where the formula comes from

\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 = r'^2\cos^2\theta - 2rr'\sin\theta\cos\theta + r^2\sin^2\theta

\left(\dfrac{dy}{d\theta}\right)^2 = r'^2\sin^2\theta + 2rr'\sin\theta\cos\theta + r^2\cos^2\theta

\left(\dfrac{dx}{d\theta}\right)^2 + \left(\dfrac{dy}{d\theta}\right)^2 = r'^2 + r^2

So the parametric formula ∫ √((dx/dθ)² + (dy/dθ)²) dθ becomes the polar formula.

1. Find dr/dθ.
2. Compute r² + (dr/dθ)² and simplify. Look for sin² + cos² = 1 or a half-angle identity.
3. Take the square root, keeping absolute values where the expression can be negative.
4. Integrate over limits that trace the curve once. If there is no antiderivative, use a calculator.

Arc length of polar curves examples

Example 1: A circle centered at the origin

Find the length of r = 5.

r^2 + \left(\dfrac{dr}{d\theta}\right)^2 = 25 + 0 = 25

L = \int_0^{2\pi} 5\,d\theta = 10\pi

Answer: 10π, the circumference 2π(5).

Example 2: A circle through the origin

Find the length of r = 6 sin θ.

\dfrac{dr}{d\theta} = 6\cos\theta

36\sin^2\theta + 36\cos^2\theta = 36

L = \int_0^{\pi} 6\,d\theta = 6\pi

Answer: 6π. The curve is a circle of radius 3, and 2π(3) = 6π agrees.

Example 3: A cardioid

Find the length of the cardioid r = 1 + cos θ.

\dfrac{dr}{d\theta} = -\sin\theta

(1 + \cos\theta)^2 + \sin^2\theta = 1 + 2\cos\theta + \cos^2\theta + \sin^2\theta

1 + 2\cos\theta + 1 = 2 + 2\cos\theta

2 + 2\cos\theta = 4\cos^2\dfrac{\theta}{2}

L = 2\int_0^{\pi} 2\cos\dfrac{\theta}{2}\,d\theta

L = 2\left[4\sin\dfrac{\theta}{2}\right]_0^{\pi} = 2(4 - 0) = 8

Answer: 8. Integrating 2 cos(θ/2) straight from 0 to 2π would give 0, because cos(θ/2) is negative for θ between π and 2π. The absolute value matters.

Example 4: An exponential spiral

Find the length of r = e^θ for 0 ≤ θ ≤ 1.

\dfrac{dr}{d\theta} = e^{\theta}

e^{2\theta} + e^{2\theta} = 2e^{2\theta}

\sqrt{2e^{2\theta}} = \sqrt{2}\,e^{\theta}

L = \int_0^1 \sqrt{2}\,e^{\theta}\,d\theta = \sqrt{2}(e - 1)

Answer: √2(e − 1) ≈ 2.430. This is the same spiral as x = eᵗ cos t, y = eᵗ sin t in arc length of parametric curves, and the answer matches.

Example 5: A watch spring (word problem)

A flat watch spring follows the spiral r = 0.2θ (in centimeters) for three full turns, 0 ≤ θ ≤ 6π. How long is the spring?

\dfrac{dr}{d\theta} = 0.2

(0.2\theta)^2 + 0.2^2 = 0.04(\theta^2 + 1)

L = \int_0^{6\pi} 0.2\sqrt{\theta^2 + 1}\,d\theta

\int \sqrt{\theta^2 + 1}\,d\theta = \tfrac{1}{2}\left[\theta\sqrt{\theta^2 + 1} + \ln\left(\theta + \sqrt{\theta^2 + 1}\right)\right]

\tfrac{1}{2}\left[355.81 + 3.630\right] \approx 179.72

L \approx 0.2(179.72) \approx 35.94

Answer: about 35.9 cm. A quick check: three circles with the average radius, about 1.9 cm, would be about 35.5 cm, so the answer is reasonable.

Common mistakes

Related methods

Parametric arc length

The polar formula is arc length of parametric curves with θ as the parameter. If you forget the polar formula, convert to x and y and use the parametric one.

Area inside polar curves

Area in polar coordinates and area between polar curves use the same limits, so the work of finding where a curve starts and stops is shared.

Arc length of graphs

For y = f(x), the formula is ∫ √(1 + (dy/dx)²) dx, from arc length of a curve. All three formulas add up tiny straight pieces.

Practice problems

1. Find the length of r = 3.
2. Find the length of r = 4 cos θ.
3. Find the length of the cardioid r = 2 + 2 cos θ.
4. Find the length of r = θ² for 0 ≤ θ ≤ √5.
5. Find the length of r = e²ᶿ for 0 ≤ θ ≤ ln 2.
6. Find the length of r = 1 − cos θ.
7. Find the length of r = sec θ for 0 ≤ θ ≤ π/4, and check it with a sketch.
8. Set up, then estimate with a calculator, the length of r = 2 + sin θ.
9. Why does integrating the arc length formula for r = 6 sin θ from 0 to 2π give 12π?
10. A coiled rope lies in the spiral r = 0.5θ meters for one turn, 0 ≤ θ ≤ 2π. How long is the rope?

L = \int_0^{2\pi} 3\,d\theta = 6\pi

16\cos^2\theta + 16\sin^2\theta = 16

L = \int_{-\pi/2}^{\pi/2} 4\,d\theta = 4\pi

(2 + 2\cos\theta)^2 + 4\sin^2\theta = 8 + 8\cos\theta = 16\cos^2\dfrac{\theta}{2}

L = 2\int_0^{\pi} 4\cos\dfrac{\theta}{2}\,d\theta = 2\left[8\sin\dfrac{\theta}{2}\right]_0^{\pi} = 16

\theta^4 + (2\theta)^2 = \theta^2(\theta^2 + 4)

L = \int_0^{\sqrt{5}} \theta\sqrt{\theta^2 + 4}\,d\theta

L = \left[\tfrac{1}{3}(\theta^2 + 4)^{3/2}\right]_0^{\sqrt{5}} = \tfrac{1}{3}(27 - 8)

e^{4\theta} + 4e^{4\theta} = 5e^{4\theta}

L = \int_0^{\ln 2} \sqrt{5}\,e^{2\theta}\,d\theta = \dfrac{\sqrt{5}}{2}\left(e^{2\ln 2} - 1\right)

L = \dfrac{\sqrt{5}}{2}(4 - 1) = \dfrac{3\sqrt{5}}{2}

(1 - \cos\theta)^2 + \sin^2\theta = 2 - 2\cos\theta = 4\sin^2\dfrac{\theta}{2}

L = \int_0^{2\pi} 2\sin\dfrac{\theta}{2}\,d\theta = \left[-4\cos\dfrac{\theta}{2}\right]_0^{2\pi} = 8

\sec^2\theta + \sec^2\theta\tan^2\theta = \sec^2\theta(1 + \tan^2\theta) = \sec^4\theta

L = \int_0^{\pi/4} \sec^2\theta\,d\theta = \tan\dfrac{\pi}{4} - \tan 0 = 1

(2 + \sin\theta)^2 + \cos^2\theta = 5 + 4\sin\theta

L = \int_0^{2\pi} \sqrt{5 + 4\sin\theta}\,d\theta \approx 13.36

\int_0^{2\pi} 6\,d\theta = 12\pi

L = \int_0^{2\pi} 0.5\sqrt{\theta^2 + 1}\,d\theta

L = 0.25\left[\theta\sqrt{\theta^2 + 1} + \ln\left(\theta + \sqrt{\theta^2 + 1}\right)\right]_0^{2\pi}

L \approx 0.25(39.976 + 2.537) \approx 10.63

Polar arc length FAQ

What is the arc length formula for polar curves?

L = ∫ √(r² + (dr/dθ)²) dθ from θ = α to θ = β, where the curve is traced once on that interval.

Where does the polar arc length formula come from?

Write x = r cos θ and y = r sin θ and use the parametric formula. When you square and add the derivatives, the cross terms cancel and leave r² + (dr/dθ)².

Why does the formula need r² and not just dr/dθ?

Even when r stays the same, the point moves along a circle as θ changes. The r² term measures that swinging motion, and dr/dθ measures the in-and-out motion.

What if the integral can’t be done by hand?

Many polar lengths, like the limaçon r = 2 + sin θ, have no simple antiderivative. Use a calculator or a numerical method such as Simpson’s rule.

Summary

• Polar arc length: L = ∫ √(r² + (dr/dθ)²) dθ.
• It is the parametric formula with x = r cos θ and y = r sin θ.
• Simplify with sin² + cos² = 1 and half-angle identities.
• Keep absolute values after a square root, or use symmetry.
• Choose limits that trace the curve once.

Previous lesson: Area Between Polar Curves

Next lesson: Convergence of Sequences

Back to the Calculus 2 Index