Arc Length of Cardioids and Limacons

Introduction

A cardioid or limaçon can have a smooth outline, a cusp, or an inner loop. Finding its length begins with the polar arc-length formula, but the resulting integral has a useful structure: symmetry reduces the interval, and a trigonometric identity leads either to an elementary integral or to a standard elliptic integral.

The main family considered here is $r=a+b\cos\theta$, with $a,b\geq 0$. The same length results apply to $r=a+b\sin\theta$, since changing cosine to sine rotates the curve. The cases $a=b>0$ give a cardioid, while other positive choices give limaçons.

Set Up the Arc Length

For a polar curve $r=r(\theta)$ traced as $\theta$ ranges from $\alpha$ to $\beta$, its arc length is the integral of the speed of its parametrization. In polar coordinates that speed is $\sqrt{r(\theta)^2+(r'(\theta))^2}$. The interval must trace the portion of the curve whose length is being measured; for the standard limaçon with $a>0$, the interval from $0$ to $2\pi$ traces the complete curve.

Polar arc length: If $r$ is differentiable on $[\alpha,\beta]$, the length of the parametrized curve on that interval is $$ L=\int_{\alpha}^{\beta}\sqrt{r(\theta)^2+\left(r'(\theta)\right)^2}\,d\theta. $$

For $r=a+b\cos\theta$, differentiation gives $r'(\theta)=-b\sin\theta$. Combining the squared terms simplifies the integrand:

\begin{aligned} r^2+(r')^2 &=(a+b\cos\theta)^2+b^2\sin^2\theta\\ &=a^2+2ab\cos\theta+b^2\cos^2\theta+b^2\sin^2\theta\\ &=a^2+b^2+2ab\cos\theta\\ &=(a+b)^2-4ab\sin^2\left(\frac{\theta}{2}\right). \end{aligned}

Thus, for $a>0$, the full length is set up as

L=\int_0^{2\pi} \sqrt{(a+b)^2-4ab\sin^2\left(\frac{\theta}{2}\right)}\,d\theta.

The last form is especially useful because the expression under the square root depends on $\sin^2(\theta/2)$. For the full curve, the integrand is symmetric about the polar axis, so the interval can be reduced to $[0,\pi]$ and the result doubled.

Why a Standard Elliptic Integral Appears

For general positive $a$ and $b$, the square root in the length integral does not usually have an elementary antiderivative. Its standard form is the complete elliptic integral of the second kind. Here the word “complete” means that the integral runs from $0$ to $\pi/2$.

Definition: For $0\leq k\leq 1$, the complete elliptic integral of the second kind is $$ E(k)=\int_0^{\pi/2}\sqrt{1-k^2\sin^2 u}\,du. $$ The number $k$ is called the modulus.

The limaçon’s parameters determine the modulus $k=\frac{2\sqrt{ab}}{a+b}$. This lies between $0$ and $1$, since $(a+b)^2-4ab=(a-b)^2\geq 0$.

Theorem: For $a>0$, $b\geq 0$, the full length of the curve $r=a+b\cos\theta$, traced once as $\theta$ runs from $0$ to $2\pi$, is $$ L=4(a+b)E\left(\frac{2\sqrt{ab}}{a+b}\right). $$ This includes the cardioid case $a=b>0$ and the circle case $b=0$.

Use symmetry.
The arc-length integrand is even on $[-\pi,\pi]$, so the full length is twice its integral from $0$ to $\pi$.

Change variables.
In the reduced integral, set $u=\theta/2$. Then $d\theta=2\,du$, and the interval $0\leq\theta\leq\pi$ becomes $0\leq u\leq\pi/2$.

Match the definition of $E$.
Factor out $a+b$ from the square root. The remaining integral is $E$ with modulus $2\sqrt{ab}/(a+b)$.

In full, the calculation is

\begin{aligned} L &=2\int_0^\pi \sqrt{(a+b)^2-4ab\sin^2\left(\frac{\theta}{2}\right)}\,d\theta\\ &=4\int_0^{\pi/2} \sqrt{(a+b)^2-4ab\sin^2 u}\,du\\ &=4(a+b)\int_0^{\pi/2} \sqrt{1-\left(\frac{2\sqrt{ab}}{a+b}\right)^2\sin^2u}\,du\\ &=4(a+b)E\left(\frac{2\sqrt{ab}}{a+b}\right). \end{aligned}

This derivation explains both the symmetry reduction and the appearance of the special function: after the substitution, the integral matches its defining form exactly.

Cardioids and Circles: Elementary Cases

When $a=b>0$, the modulus is $k=1$. In this case the defining integrand simplifies on the interval of integration: $\sqrt{1-\sin^2u}=\cos u$ for $0\leq u\leq\pi/2$. Therefore $E(1)=1$, and the cardioid length is $8a$. When $b=0$, $k=0$, so $E(0)=\pi/2$; the curve is a circle of radius $a$, with length $2\pi a$.

Worked Example: Length of a Cardioid

Find the length of $r=2(1+\cos\theta)$. Here $a=b=2$, and the complete curve is traced for $-\pi\leq\theta\leq\pi$. First, $r'= -2\sin\theta$, so

\begin{aligned} r^2+(r')^2 &=4(1+\cos\theta)^2+4\sin^2\theta\\ &=4(1+2\cos\theta+\cos^2\theta+\sin^2\theta)\\ &=8+8\cos\theta\\ &=16\cos^2\left(\frac{\theta}{2}\right). \end{aligned}

On $[-\pi,\pi]$, $\cos(\theta/2)\geq 0$. Taking the square root therefore gives $4\cos(\theta/2)$, and symmetry yields

\begin{aligned} L &=2\int_0^\pi 4\cos\left(\frac{\theta}{2}\right)\,d\theta\\ &=8\left[2\sin\left(\frac{\theta}{2}\right)\right]_0^\pi\\ &=8(2-0)=16. \end{aligned}

Check: The general formula gives $4(2+2)E(1)=16\cdot1=16$, in agreement with the direct integration.

Worked Example: Length of a Circle in Polar Form

Find the length of $r=3$ over $0\leq\theta\leq2\pi$. Since $r'=0$, the speed is $\sqrt{3^2+0^2}=3$ at every parameter value. Hence

L=\int_0^{2\pi}3\,d\theta =3[\theta]_0^{2\pi} =3(2\pi-0)=6\pi.

Check: The curve is a circle of radius $3$, whose circumference is $2\pi\cdot3=6\pi$. The general formula also gives $4(3)E(0)=12(\pi/2)=6\pi$.

Limaçons With and Without Inner Loops

For $a,b>0$, the comparison between $a$ and $b$ describes an important geometric difference. If $a>b$, then $a+b\cos\theta$ remains positive, and the curve has no inner loop. If $a=b$, it reaches the pole at the cardioid’s cusp. If $0<a<b$, the radius is negative for some angles, and the curve has an inner loop.

The full-curve arc-length integral still runs through the entire parameter interval $0\leq\theta\leq2\pi$ in all these cases. In particular, when $0<a<b$, the resulting total length includes both the inner loop and the outer loop. A calculation of one loop alone would require the parameter bounds for that loop, not the full-period formula.

Worked Example: A Limaçon With an Inner Loop

Find the total length of $r=1+3\cos\theta$. Here $a=1$, $b=3$, so $0<a<b$ and the curve has an inner loop. Its derivative is $r'=-3\sin\theta$. The squared speed is

\begin{aligned} r^2+(r')^2 &=(1+3\cos\theta)^2+9\sin^2\theta\\ &=1+6\cos\theta+9\cos^2\theta+9\sin^2\theta\\ &=10+6\cos\theta\\ &=16-12\sin^2\left(\frac{\theta}{2}\right). \end{aligned}

The integrand is even on $[-\pi,\pi]$, so reduce to $[0,\pi]$ and double. With $u=\theta/2$, the integral becomes

\begin{aligned} L &=2\int_0^\pi\sqrt{16-12\sin^2\left(\frac{\theta}{2}\right)}\,d\theta\\ &=4\int_0^{\pi/2}\sqrt{16-12\sin^2u}\,du\\ &=16\int_0^{\pi/2}\sqrt{1-\frac34\sin^2u}\,du\\ &=16E\left(\frac{\sqrt{3}}{2}\right). \end{aligned}

Numerically, $E(\sqrt{3}/2)\approx1.211056$, so the total length is approximately $16(1.211056)=19.376896$. Check: The theorem gives $4(1+3)E(2\sqrt{3}/4)=16E(\sqrt{3}/2)$, exactly the same expression. This is the length of both loops together, not just the inner loop.

A Common Pitfall: Dropping the Absolute Value

Taking a square root of a squared trigonometric expression requires care. In the cardioid example, the speed simplified to $\sqrt{16\cos^2(\theta/2)}=4|\cos(\theta/2)|$, not $4\cos(\theta/2)$ on every possible interval.

Common mistake: On $0\leq\theta\leq2\pi$, replacing $4|\cos(\theta/2)|$ by $4\cos(\theta/2)$ gives $$ \int_0^{2\pi}4\cos\left(\frac{\theta}{2}\right)\,d\theta =8\left[\sin\left(\frac{\theta}{2}\right)\right]_0^{2\pi} =0. $$ A length cannot be zero here. The error is that $\cos(\theta/2)$ is negative on $\pi<\theta<2\pi$. The correct speed is $4|\cos(\theta/2)|$, whose integral over $0\leq\theta\leq2\pi$ is $16$, as found by integrating over $[-\pi,\pi]$ where the cosine is nonnegative.

Interpreting the Formula and Its Limits

The elliptic-integral expression is useful because it gives a precise, standard form for the length even when no elementary antiderivative is available. It also makes the special cases transparent: $a=b$ produces the cardioid value, while $b=0$ produces the circle value. For general positive parameters, numerical evaluation of $E(k)$ may be needed.

There is one parameter edge case to treat separately. If $a=0$ and $b>0$, then $r=b\cos\theta$ describes a circle of radius $b/2$, but the full parameter interval traces that circle twice. The arc-length integral over $0\leq\theta\leq2\pi$ consequently counts the circumference twice. The geometric circle’s length is $\pi b$, so the one-trace interval or the circle circumference should be used rather than interpreting the full-period integral as a once-traced curve.

Takeaway: For $r=a+b\cos\theta$ with $a>0$, the full length is $4(a+b)E(2\sqrt{ab}/(a+b))$. The cardioid and circle cases reduce to elementary values; general limaçons are naturally expressed using $E$.

Check Your Understanding

Use the polar arc-length formula and the symmetry or special-function ideas from this tutorial to answer the following.

For $r=4+4\cos\theta$, what is the full curve’s length, and which elementary value of $E$ applies?
For $r=5+2\cos\theta$, write the full length using $E$, including the modulus.
Why does the full length of $r=1+3\cos\theta$ include both loops?
What is the error in replacing $\sqrt{\cos^2(\theta/2)}$ with $\cos(\theta/2)$ over the entire interval $0\leq\theta\leq2\pi$?
What geometric curve does $r=b\cos\theta$ trace, and why does its full $2\pi$ parameter interval count its length twice?