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Areas enclosed by polar curves

Edexcel IAL Further Mathematics · WFM02 Further Pure Mathematics 2

The area swept out by a polar curve between two angles is found with a single integral. Most questions need a sketch for the limits, a double angle formula, and an exact answer.

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The formula

The area bounded by the curve r = f(θ) and the half-lines θ = α and θ = β is

A = ½ ∫ from α to β of r2 dθ.

It comes from adding thin sectors, each with area ½r2 δθ. Do not forget the ½.

Integrating r2

Expand r2, then use cos2 θ = ½(1 + cos 2θ) and sin2 θ = ½(1 - cos 2θ).

For a cardioid: (1 + cos θ)2 = 1 + 2cos θ + cos2 θ = 3/2 + 2cos θ + ½cos 2θ.

Finding the limits

Sketch the curve. For a closed curve traced once, integrate over one full turn. For one loop of a rose, use consecutive values where r = 0, such as 0 and π/3 for r = a sin 3θ. Symmetry can halve the work: integrate over half and double.

Area between two curves

Find where the curves meet by solving r1 = r2. Then split the region and use the correct curve and limits for each part, integrating each separately; the June 2025 examiners' report says the most common error was giving both integrals the same limits. Give exact answers when asked: keep π and surds.

Worked example

Find the area enclosed by the cardioid r = a(1 + cos θ).

  1. A = ½ ∫ from 0 to 2π of a2(1 + cos θ)2 dθ.
  2. (1 + cos θ)2 = 3/2 + 2cos θ + ½cos 2θ.
  3. ∫ from 0 to 2π = [3θ/2 + 2sin θ + ¼sin 2θ] = 3π.
  4. A = ½a2 × 3π.

Answer: A = 3πa2/2

Key idea: A = ½∫r2 dθ. Sketch for the limits, use double angle formulae, and keep separate integrals for separate pieces.

Check yourself

  1. The area enclosed by r = 2(1 + cos θ) is kπ. Find k.
    Show answer

    6. Using 3πa2/2 with a = 2: 3π × 4/2 = 6π.

  2. The area of one loop of r = 4sin 3θ is kπ. Find k. Give your answer as a fraction in its simplest form.
    Show answer

    4/3. ½∫ from 0 to π/3 of 16sin2 3θ dθ = 4∫(1 - cos 6θ) dθ = 4[θ - sin 6θ/6] from 0 to π/3 = 4π/3.

  3. Find the area swept out by the spiral r = θ from θ = 0 to θ = π, to 3 decimal places.
    Show answer

    5.168. ½∫ from 0 to π of θ2 dθ = ½ × π3/3 = π3/6 = 5.168.

  4. A student finds the area inside r = 3 + cos θ as ∫ from 0 to 2π of (3 + cos θ)2 dθ. What is wrong?
    • The limits should be 0 to π
    • The factor ½ is missing
    • r should not be squared
    • Nothing: this is correct
    Show answer

    The factor ½ is missing. The area formula is ½∫r2 dθ.

  5. Find the area of the region inside r = 2 + 2cos θ and outside r = 3, to 3 decimal places.
    Show answer

    4.653. The curves meet where 2 + 2cos θ = 3, so θ = ±π/3. Area = ½∫ from -π/3 to π/3 of [(2 + 2cos θ)2 - 9] dθ = (9√3 - 2π)/2 = 4.653.

  6. The curves r = 2 + 2cos θ and r = 3 meet at θ = kπ with 0 < k < 1. Find k. Give your answer as a fraction in its simplest form.
    Show answer

    1/3. 2 + 2cos θ = 3 gives cos θ = 1/2, so θ = π/3.

  7. Find the area of one loop of the lemniscate r2 = 9cos 2θ.
    Show answer

    4.5. One loop: -π/4 ≤ θ ≤ π/4. ½∫9cos 2θ dθ = ½ × 9 × [sin 2θ/2] from -π/4 to π/4 = ½ × 9 × 1 = 9/2.

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Last updated 8 November 2026. Spotted a mistake? Tell us.