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AP® Calculus BC · Unit 9 · Topic 9.8

Area of a Polar Region or the Area Bounded by a Single Polar Curve: Revision Notes (Calculus BC 9.8)

One-page recap of polar area: where ½∫r² dθ comes from, choosing limits that trace a region once, petals and loops, the identities you need, and common errors.

Short on time? This page is the recap. For the sector diagram, explanations and worked examples, use the full study guide. BC-only material. Prerequisites (polar curves 9.7, Riemann sums and definite integrals 6.2–6.4) are on the Calculus BC roadmap.

Recap

  • A polar region is bounded by rays from the pole and a curve r = f(θ), so slice it into thin sectors, not rectangles.
  • A sector of radius r and angle Δθ has area ½ r² Δθ. Adding sectors gives a Riemann sum; its limit is the integral.
  • The rate at which area is swept out is dA/dθ = ½ r².
  • Limits matter more than the integration. Sketch first.

Key relationships

Situation Area Note
Curve and two rays ½ ∫ from α to β of r² dθ Region swept once from α to β
Whole closed curve, r never 0 ½ ∫ over one full trace Often 0 to 2π, but check
One petal or loop ½ ∫ between consecutive zeros of r r = 0 at the pole
Symmetric region 2 × (area of one half) Only if the halves match
By hand sin²kθ = ½(1 − cos 2kθ), cos²kθ = ½(1 + cos 2kθ) Use before integrating

Traced twice? The circle r = 6 cos θ is traced once on [0, π] (area 9π). On [0, 2π] the integral gives 18π, double the true area.

Mistakes to avoid

  1. Forgetting the ½ or forgetting to square r.
  2. Expanding (a + b cos θ)² wrongly: the middle term 2ab cos θ is easy to lose.
  3. 0 to 2π by habit, which double-counts curves that repeat.
  4. Wrong zeros for a petal: use consecutive zeros with the petal between them.
  5. Integrating sin²θ without the identity.
  6. Degree mode on the calculator, or giving a number with no integral written down.

Quick self-check

  1. Find the area of the region bounded by r = 5 and the rays θ = 0 and θ = π/2. (25π/4, a quarter disc)
  2. Find the area of one petal of r = sin 4θ. (Petal on 0 ≤ θ ≤ π/4: ½ ∫ sin²4θ dθ = π/16)
  3. Find the area swept out by r = e^θ for 0 ≤ θ ≤ 1. ((e² − 1)/4 ≈ 1.597)
  4. Why does ½ ∫ from 0 to 2π of (2 cos 3θ)² dθ not give the area of the rose? (The rose is traced twice on [0, 2π]; the integral gives 2π but the area is π.)

Next: practice questions.

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