§ 01Why Use Polar Coordinates?
Some curves are almost impossible to describe with y as a function of x, yet they have a devastatingly simple polar equation. A circle of radius 3 centred at the origin is x² + y² = 9 in Cartesian coordinates, but just r = 3 in polar. A four-petalled rose is a tangle in Cartesian form; in polar, it is r = cos(2θ).
The key insight is that Cartesian coordinates are built around rectangular grids, they naturally describe objects aligned with horizontal and vertical directions. Polar coordinates are built around rotation: they naturally describe objects with circular or angular symmetry. For curves that "spin outward" from the origin (spirals, roses, cardioids, lemniscates), polar coordinates give equations that are not just simpler to write but easier to integrate, differentiate, and visualise.
In multivariable calculus, polar coordinates become indispensable when evaluating double integrals over circular regions. The change-of-variables formula introduces the Jacobian factor r, transforming complicated Cartesian integrals into straightforward iterated integrals. The famous Gaussian integral, the foundation of probability theory, is essentially impossible without switching to polar coordinates.
§ 02The Polar Coordinate System
A point P in the plane is described in polar coordinates by an ordered pair (r, θ), where:
Non-Uniqueness of Polar Representations
Unlike Cartesian coordinates, every point has infinitely many polar representations. The point (2, π/3) is the same as (2, π/3 + 2π), (2, π/3 − 2π), and also (−2, π/3 + π). This non-uniqueness must be handled carefully when finding intersections of polar curves, you must check all representations, not just the ones your algebra produces.
§ 03Converting Between Polar and Cartesian
The conversion formulas follow directly from right-triangle trigonometry applied to the diagram above. They are among the most-used formulas in all of calculus:
Note: when computing θ = arctan(y/x), always check the quadrant of (x, y), arctan only returns values in (−π/2, π/2), so you may need to add π for points in quadrants II or III.
| Goal | Formula | Notes |
|---|---|---|
| Polar → Cartesian | x = r cosθ, y = r sinθ | Direct substitution. Always works. |
| Cartesian → Polar (r) | r = √(x² + y²) | Always non-negative if you want the principal value. |
| Cartesian → Polar (θ) | θ = atan2(y, x) | Use atan2 (or check quadrant manually). Range: (−π, π]. |
| r² in expressions | r² = x² + y² | Useful when converting x² + y² in Cartesian equations. |
Convert the point (−√3, 1) from Cartesian to polar coordinates, giving θ ∈ [0, 2π).
tanθ = 1/(−√3), so θ = arctan(−1/√3). The basic angle is π/6 (30°), but (−√3, 1) is in Quadrant II, so:
θ = π − π/6 = 5π/6Verify: x = 2cos(5π/6) = 2·(−√3/2) = −√3 ✓
y = 2sin(5π/6) = 2·(1/2) = 1 ✓
Convert r = 4sinθ to Cartesian form and identify the curve.
Use r² = x² + y² and r sinθ = y.
The polar equation r = 4sinθ describes a circle of radius 2 passing through the origin, centred at (0, 2).
§ 04Symmetry Tests for Polar Curves
Exploiting symmetry can cut your sketching work in half (or more). The three standard symmetry tests for r = f(θ) are:
§ 05Common Polar Curves — A Visual Gallery
The following six curve families appear constantly in calculus. Recognising them by their equation shape saves enormous time when sketching or setting up integrals.
§ 06Area Enclosed by a Polar Curve
The area formula in polar coordinates is derived by approximating the region with thin circular sectors rather than rectangles. A sector of radius r and angle dθ has area ½r²dθ. Summing these sectors gives the definite integral:
where the curve is traced from θ = α to θ = β. The limits must be chosen so the curve is traced exactly once: doubling up causes the area to be counted twice.
Area Between Two Polar Curves
When one curve r = f(θ) lies outside another r = g(θ) on [α, β], the area between them is:
Finding the limits α, β requires solving f(θ) = g(θ): and checking all polar representations of intersection points.
Find the total area enclosed by the cardioid r = 2(1 + cosθ).
The cardioid is traced once for θ ∈ [0, 2π]. Using symmetry (x-axis), compute the upper half (θ ∈ [0, π]) and double:
A = 2 · ½ ∫₀π [2(1 + cosθ)]² dθ = ∫₀π 4(1 + cosθ)² dθFind the area of one petal of the four-petalled rose r = 3cos(2θ).
The right-most petal is traced when r ≥ 0 and θ ∈ [−π/4, π/4]. At θ = ±π/4: cos(2·π/4) = cos(π/2) = 0 → r = 0 (petal tip returns to origin).
Total area of all four petals = 4 × 9π/8 = 9π/2.
Find the area inside r = 3sinθ but outside r = 1 + sinθ.
On [π/6, 5π/6], r = 3sinθ is the outer curve and r = 1 + sinθ is the inner curve.
A = ½ ∫π/65π/6 [(3sinθ)² − (1 + sinθ)²] dθ§ 07Arc Length of a Polar Curve
To find the arc length of a polar curve r = f(θ) from θ = α to θ = β, we use the parametric arc length formula with x(θ) = r cosθ and y(θ) = r sinθ:
Derivation: dx/dθ = (dr/dθ)cosθ − r sinθ and dy/dθ = (dr/dθ)sinθ + r cosθ. Then (dx/dθ)² + (dy/dθ)² = r² + (dr/dθ)², giving the formula above after simplification using sin² + cos² = 1.
Find the total arc length of the cardioid r = a(1 − cosθ) (a > 0).
Use the identity 1 − cosθ = 2sin²(θ/2):
√(2a²·2sin²(θ/2)) = 2a|sin(θ/2)|For θ ∈ [0, 2π], sin(θ/2) ≥ 0, so |sin(θ/2)| = sin(θ/2).
Find the length of the spiral r = θ for 0 ≤ θ ≤ 2π.
dr/dθ = 1, so r² + (dr/dθ)² = θ² + 1.
L = ∫₀2π √(θ² + 1) dθUsing the standard result ∫√(θ²+1)dθ = ½[θ√(θ²+1) + ln|θ + √(θ²+1)|] + C:
L = ½[2π√(4π²+1) + ln(2π + √(4π²+1))] ≈ ½[2π·6.361 + ln(6.361 + 2π)] ≈ 21.26§ 08Further Worked Examples
Convert x² + y² − 6x = 0 to polar form and describe the curve.
Replace x² + y² with r² and x with r cosθ:
r² − 6r cosθ = 0So r = 0 (the origin) or r = 6cosθ.
This is a circle of radius 3 centred at the Cartesian point (3, 0). The original equation x² + y² − 6x = 0 completes the square to (x−3)² + y² = 9, confirming this.
Test the three-petalled rose r = sin(3θ) for x-axis, y-axis, and origin symmetry.
Replace θ → −θ: r = sin(−3θ) = −sin(3θ) ≠ sin(3θ). Test fails, but check (−r, −θ): −r = sin(−3θ) = −sin(3θ), so r = sin(3θ). The curve IS symmetric about the x-axis via the second form.
Replace θ → π − θ: sin(3(π−θ)) = sin(3π − 3θ) = sin(3π)cos(3θ) − cos(3π)sin(3θ) = 0 + sin(3θ) = sin(3θ). Test passes → symmetric about y-axis.
Replace r → −r: −r = sin(3θ) → r = −sin(3θ) ≠ sin(3θ). Not origin-symmetric (the three petals are not reflected through the origin to give the same curve).
r = sin(3θ) is symmetric about both coordinate axes. It has 3 petals, all traced as θ goes from 0 to π. Total area = 3 × (½ ∫₀π/3 sin²(3θ) dθ) = 3π/4 (for a = 1).
Find the total area enclosed by the lemniscate r² = 4cos(2θ).
r² ≥ 0 requires cos(2θ) ≥ 0. This holds for θ ∈ [−π/4, π/4] ∪ [3π/4, 5π/4]. By symmetry, compute the right lobe (θ ∈ [−π/4, π/4]) and double.
For the general lemniscate r² = a²cos(2θ), the total area is a², a strikingly clean result.
§ 09Common Mistakes
§ 10Test Your Understanding
Ten questions spanning conversions, curve identification, symmetry, area calculations, and arc length. Work each one through before revealing the explanation.