Polar Curves

Explore polar curves: r as a function of angle θ.

Worked example — Polar Curves

Equation	1+cos(θ)
r(theta)	0.0000
A	4.7124
(x, y)	(-0.000, -0.000)
theta	3.142 rad (180.0 deg)

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

Odd k gives k petals, even k gives twice as many 🖖

A rose r = cos(kθ) does not have k petals, as you might expect. Odd k does: cos(3θ) draws 3 petals and finishes in half a revolution — carry on to 2π and the curve simply retraces the petals it already drew, because the negative r values land on top of the positive ones. Even k has nowhere to hide them, so those extra passes become new petals in the gaps: cos(4θ) ends up with 8. That is exactly why this tool’s two rose presets stop at different places, π for the 3-petal rose and a full 2π for the 8-petal one.

Distance and direction, not left and right 🖖

Polar coordinates pin down a point with two everyday questions: how far from the center, and in which direction? That is why shapes with a natural center — spirals, flower-like roses, orbits — get short, clean equations here, while they balloon into clumsy formulas on an x-y grid. Try the spiral preset: r = θ/π just means "the radius grows steadily as you turn," yet the Cartesian version is a tangle. The takeaway: choose coordinates that match a problem's symmetry and the math gets simpler.

The spiral Bernoulli wanted on his grave 🖖

The logarithmic spiral r = a·e^(bθ) so captivated Jakob Bernoulli that he called it spira mirabilis — "the marvelous spiral" — because scaling, rotating, or reflecting it returns the very same curve. He asked for one on his tombstone with the motto Eadem mutata resurgo ("though changed, I rise again the same"). The Basel stonemason, however, carved an Archimedean spiral (r = aθ) instead — a completely different curve, and the mistake is still visible today.

POLAR CURVES — WHICH FAMILY IS THIS, AND HOW MANY PETALS?

Which Polar Curve Case Are You In?

In polar form the shape is decided by the formula for r, and by how far θ has to run before the curve closes. Roses come from cos nθ, and the petal count depends on whether n is odd or even. Cardioids and limaçons come from a + b·cos θ, and the ratio a/b decides whether there is a dimple, a point, or a loop inside. Recognise the form and you know the picture.

a = b — the cardioid, with a single point at the origin a = b ⇒ r(π) = 0

cos nθ with n odd — n petals, drawn in half a turn n = 2k+1 ⇒ # = n

cos nθ with n even — 2n petals, and the full turn is needed n = 2k ⇒ # = 2n

b > a — a limaçon with an inner loop b > a ⇒ ∃θ: r < 0

a = b — the cardioid, with a single point at the origin

What you know: r = a + b·cos θ with a and b equal. The radius runs from 2a down to exactly zero, so the curve touches the origin once and comes to a cusp there.

Form: a = b ⇒ r(π) = 0

Worked example: r = 1 + cos θ: r = 2 at θ = 0°, r = 1 at 90°, and r = 0 at 180°. The area enclosed is 3π/2 ≈ 4.712.

Open this case: Cardioid

a = b — the cardioid, with a single point at the origin. The radius shrinks to nothing at 180° and the curve closes in a cusp. r = a + b·cos θ with a and b equal. The radius runs from 2a down to exactly zero, so the curve touches the origin once and comes to a cusp there. — The radius shrinks to nothing at 180° and the curve closes in a cusp.

cos nθ with n odd — n petals, drawn in half a turn

What you know: An odd n gives exactly n petals, and the whole rose is complete after θ has run from 0 to π. Carrying on to 2π retraces the same petals.

Form: n = 2k+1 ⇒ # = n

Worked example: r = cos 3θ over 0 to π: three petals, each spanning 60° of θ, with the first centred on θ = 0

Open this case: Rose (3)

cos nθ with n odd — n petals, drawn in half a turn. Three petals, complete after half a turn; the rest of the circle repeats them. An odd n gives exactly n petals, and the whole rose is complete after θ has run from 0 to π. Carrying on to 2π retraces the same petals. — Three petals, complete after half a turn; the rest of the circle repeats them.

cos nθ with n even — 2n petals, and the full turn is needed

What you know: An even n gives twice as many petals as you might expect, and all of them need θ to run the full 0 to 2π.

Form: n = 2k ⇒ # = 2n

Worked example: r = cos 4θ over 0 to 2π: eight petals, each spanning 45° of θ

Open this case: Rose (8)

cos nθ with n even — 2n petals, and the full turn is needed. Eight petals: four from positive r, four more from the intervals where r goes negative. An even n gives twice as many petals as you might expect, and all of them need θ to run the full 0 to 2π. — Eight petals: four from positive r, four more from the intervals where r goes negative.

b > a — a limaçon with an inner loop

What you know: r = a + b·cos θ with b larger than a. The radius goes negative over part of the turn, and that stretch draws a loop inside the main curve.

Form: b > a ⇒ ∃θ: r < 0

Worked example: r = 1 + 2cos θ: r = 3 at θ = 0°, and r < 0 for θ between 120° and 240°, which is the inner loop

Open this case: Limacon

b > a — a limaçon with an inner loop. The stretch where r is negative folds back through the origin as an inner loop. r = a + b·cos θ with b larger than a. The radius goes negative over part of the turn, and that stretch draws a loop inside the main curve. — The stretch where r is negative folds back through the origin as an inner loop.

Problem solved in full

Area of 4.7124 enclosed by the cardioid r = 1 + cos θ 5 steps
The cardioid r = 1 + cos θ encloses an area of 4.7124. Derive it, and compare it with the circle it is built from.
1. Area in polar coordinates is not ∫y dx. A thin wedge of angle dθ is a triangle of area ½r²dθ, and integrating those wedges is the whole formula.
  
  $A = \frac{1}{2} \int_{0}^{2 π} r^{2} d θ, r = 1 + cos θ$
2. So the integrand is the square of the radius, which for this curve expands into three terms.
  
  $(1 + cos θ)^{2} = 1 + 2 cos θ + cos^{2} θ$
3. Each integrates over a full turn without difficulty. The middle one vanishes — cosine averages to zero — and the last uses the standard result that cos² averages to a half.
  
  $\int_{0}^{2 π} 1 = 2 π, \int_{0}^{2 π} 2 cos θ = 0, \int_{0}^{2 π} cos^{2} θ = π$
4. Adding them gives 3π/2, which is 4.7124.
  
  $A = \frac{1}{2} (2 π + 0 + π) = \frac{3 π}{2} = 4.7124$
5. The curve touches the origin exactly once, at θ = π, where 1 + cos θ = 0. That is the cusp, and it is the reason the cardioid is not a circle displaced.
  
  $r (π) = 1 + cos π = 0 (the cusp), \frac{3 π /2}{π} = 1.5$
Answer

The tool prints an area of 4.7124, with r = 0 at θ = 3.142. The comparison is the payoff: the unit circle encloses π, so the cardioid holds exactly 1.5 times as much, despite reaching out only to r = 2 at its widest and pinching to nothing at the cusp. The wedge picture explains it — area accumulates as r², so the fat side dominates far more than its width suggests. Switch the curve to cos(3θ) and the panel reports 1.5708 — which is π/2, exactly twice the three-petal rose's real area of π/4. The integral is not wrong; the range is. An odd-petal rose is traced completely as θ runs to π and is drawn over again by 2π, so the wedges get counted twice. Set the upper limit to 3.14159 and 0.7854 appears. Polar area is the one integral where the limits are a modelling decision rather than a formality.

$A = \frac{3 π}{2} = 4.7124 = 1.5 \times the unit circle; r (π) = 0$

References (2)

Who first used these coordinates, and for what: J. L. Coolidge, "The Origin of Polar Coordinates." The American Mathematical Monthly 59(2), 78–85, 1952.
The logarithmic spiral in shells and horns — the reason polar form is worth the change of variable: D’Arcy W. Thompson, On Growth and Form, Chapter XI: The Equiangular Spiral. Cambridge University Press, 1917.

By the 42Math team · Updated 2026-07-27

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Cardioid - r = 1+cos(θ) over a full turn encloses A = 4.7124, which is 3π/2 — one and a half times the unit circle
Rose (3) - cos(3θ) needs only θ up to π to close all three petals, enclosing A = 0.7854 = π/4
Rose (8) - cos(4θ) draws eight petals, not four — an even k doubles them — and needs the full 2π for A = 1.5708 = π/2
Limacon - r = 1+2cos(θ) goes negative: at θ = π the readout shows r = −1.0000 at the point (1.000, 0.000), and the inner loop is the result