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9.8 Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve

Syllabus
2020
Topic
9.8
Level
—

Accumulate Polar Area as Thin Sectors

A small change dθd\theta sweeps out a thin sector of radius r=f(θ)r=f(\theta). Because a sector with angle dθd\theta has area approximately 12r2dθ\tfrac12r^2d\theta, adding all sectors gives the area bounded by one polar curve.

A=\frac12\int_{\alpha}^{\beta}[r(\theta)]^2,d\theta

  1. Identify the curve r=f(θ)r=f(\theta) and the region inside it.
  2. Choose [α,β][\alpha,\beta] so the desired region is traced exactly once; use symmetry only when the chosen interval covers a known fraction.
  3. Square the entire expression for rr, multiply the integral by 1/21/2, and evaluate.
  4. Check that the result is nonnegative and that no lobe or sector was counted twice.

Example: the cardioid r=1+cos⁡θr=1+\cos\theta is traced once for 0≤θ≤2π0\le\theta\le2\pi. Its enclosed area is
A=12∫02π(1+cos⁡θ)2 dθ=12∫02π(1+2cos⁡θ+cos⁡2θ) dθ.A=\frac12\int_0^{2\pi}(1+\cos\theta)^2\,d\theta=\frac12\int_0^{2\pi}(1+2\cos\theta+\cos^2\theta)\,d\theta.
Over a full period, ∫02π1 dθ=2π\int_0^{2\pi}1\,d\theta=2\pi, ∫02π2cos⁡θ dθ=0\int_0^{2\pi}2\cos\theta\,d\theta=0, and ∫02πcos⁡2θ dθ=π\int_0^{2\pi}\cos^2\theta\,d\theta=\pi. Therefore A=12(3π)=3π2A=\tfrac12(3\pi)=\tfrac{3\pi}{2} square units.

Do not compute ∫r dθ\int r\,d\theta: polar area depends on r2r^2 and includes the factor 1/21/2. Also, bounds describe a traversal, not merely the visible width of a sketch; if the interval traces the same region twice, the integral double-counts its area.