Polar Curves and Area
Polar Curves and Area
Polar coordinates describe points by their distance from the origin (r) and angle from the positive x-axis (θ), rather than by (x, y). Many beautiful curves have simple polar equations, and finding enclosed areas uses a dedicated integral formula.
Polar Coordinates Basics
A point P has polar coordinates (r, θ) where:
- r = distance from the pole (origin) to P
- θ = angle measured anticlockwise from the initial line (positive x-axis)
The relationship to Cartesian coordinates:
- x = r cos θ, y = r sin θ
- r² = x² + y², tan θ = y/x
Standard Polar Curves
| Equation | Curve type | Notes |
|---|---|---|
| r = a | Circle | Centre origin, radius a |
| r = a cos θ | Circle | Centre (a/2, 0), radius a/2 |
| r = a sin θ | Circle | Centre (0, a/2), radius a/2 |
| r = a(1 + cos θ) | Cardioid | Heart-shaped, passes through origin |
| r = a(1 + 2cos θ) | Limaçon with loop | Inner loop when coefficient of cos > 1 |
| r = a cos 2θ | Rose (4 petals) | Petals at 0, π/2, π, 3π/2 |
| r = a cos 3θ | Rose (3 petals) | r = a cos nθ has n petals if n odd, 2n if n even |
| r² = a² cos 2θ | Lemniscate | Figure-of-eight shape |
| r = aθ | Spiral of Archimedes | Spiral outward |
| r = ae^(bθ) | Logarithmic spiral | Equiangular spiral |
Sketching Polar Curves
Method:
1. Find where r = 0 (the curve passes through the origin at these θ values)
2. Find the maximum value of r and the θ where it occurs
3. Check for symmetry:
- r(−θ) = r(θ) → symmetric about the initial line (x-axis)
- r(π − θ) = r(θ) → symmetric about θ = π/2 (y-axis)
4. Plot key points and join smoothly
Example: r = 2(1 + cos θ) (cardioid)
- r = 0 when cos θ = −1, so θ = π
- Maximum r = 4 when θ = 0
- Symmetric about θ = 0 (since cos(−θ) = cos θ)
- At θ = π/2: r = 2
Area Enclosed by a Polar Curve
The area enclosed between a polar curve r = f(θ) and the pole, from θ = α to θ = β, is:
A = (1/2) ∫_α^β r² dθ
This formula comes from summing thin triangular sectors: each sector has area ≈ (1/2)r²δθ.
Example: Find the area enclosed by the cardioid r = a(1 + cos θ).
The full curve is traced for 0 ≤ θ ≤ 2π, but by symmetry we can compute 0 to π and double:
A = 2 × (1/2) ∫₀^π a²(1 + cos θ)² dθ
= a² ∫₀^π (1 + 2cos θ + cos²θ) dθ
= a² ∫₀^π (1 + 2cos θ + (1 + cos 2θ)/2) dθ
= a² [3θ/2 + 2sin θ + sin 2θ/4]₀^π
= a² × 3π/2 = 3πa²/2
Area Between Two Polar Curves
If two curves r₁ = f(θ) and r₂ = g(θ) with r₁ > r₂ enclose a region between angles α and β:
A = (1/2) ∫_α^β (r₁² − r₂²) dθ
Finding intersection points: Set f(θ) = g(θ) and solve. Also check r = 0 for both curves — the origin may be a shared point at different θ values.
Tangents to Polar Curves
To find the gradient dy/dx at a point on r = f(θ):
dy/dx = (dr/dθ sin θ + r cos θ) / (dr/dθ cos θ − r sin θ)
This uses x = r cos θ, y = r sin θ with the chain rule.
Tangent parallel to the initial line: dy/dθ = 0 (numerator = 0)
Tangent perpendicular to the initial line: dx/dθ = 0 (denominator = 0)
Tangent at the pole (r = 0): If r = 0 at θ = α, the tangent at the pole is the line θ = α.
Worked Example: Rose Curve Area
Find the area of one petal of r = 3 cos 2θ.
One petal lies where r ≥ 0. For cos 2θ ≥ 0: −π/4 ≤ θ ≤ π/4 (the first petal).
A = (1/2) ∫_{−π/4}^{π/4} 9 cos²2θ dθ
By symmetry = ∫₀^{π/4} 9 cos²2θ dθ = (9/2) ∫₀^{π/4} (1 + cos 4θ) dθ
= (9/2)[θ + sin 4θ/4]₀^{π/4} = (9/2)(π/4 + 0) = 9π/8
The total area of all 4 petals is 4 × 9π/8 = 9π/2.
Exam Tips
- The area formula uses r², not r — don't forget to square
- Always determine the correct limits of integration by finding where r = 0 or where the curve completes
- Exploit symmetry to simplify calculations (most standard curves have at least one axis of symmetry)
- For rose curves r = a cos nθ, one petal spans an angle of π/n — the limits are ±π/(2n)
- When a curve has r < 0 for some θ, those portions are plotted in the opposite direction — be careful when finding areas
- To check your limits, count how many times the curve passes through the origin in [0, 2π]