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C54 · Unit C9 Parametric, polar and vector calculus

Intersecting polar curves and shared regions

BC extensionReview editionNot yet human-reviewed
Time
About 45 minutes, untimed
Course framework reference
AP 9.9 · BC
Builds on
Part of

Progress on this lesson counts in every pathway listed.

Understand the idea

For an area common to two polar regions with nonnegative radii, use the smaller radius on each angular interval. Find angles where the radii agree, then test the ordering between crossings. Symmetry can reduce work, but confirm its axis and multiplier. Also check origin intersections that do not arise from equating radii at the same angle.

Worked example

In the upper half-plane, the common region inside r equals 1 and r equals 2sin theta changes boundary at sin theta equals 1 over 2, so theta equals pi /6 and 5 pi /6. Area is one half [∫ sub 0 ^( pi /6)4sin squared theta d theta +∫_( pi /6)^(5 pi /6)1d theta +∫_(5 pi /6)^ pi 4sin squared theta d theta ] equals 2 pi /3 minus square root of 3 over 2.

Explore the mathematics

Intersecting polar curves and shared regions

Shade a shared polar region using the smaller radius and a boundary switch.

Setting 3 of 5

Scroll sideways to see the full diagram.

Which radius bounds the shared region at each angle?Shared region: r=2sinθ and r=2cosθSplit at θ=π/4

Predict before continuing

Which radius bounds the shared region at each angle?

Read the complete explanation
  1. Which radius bounds the shared region at each angle?
  2. In the first quadrant, the smaller radius is the shared-region boundary.
  3. The boundary switches where sinθ=cosθ, at θ=π/4.
  4. Integrate ½min(r₁²,r₂²) from 0 to π/2, splitting at π/4. The common area is π/2−1.
Intersecting polar curves and shared regions · visual worked method
Intersecting polar curves and shared regions · visual worked methodIn the upper half-plane, the common region inside r=1 and r=2sinθ changes boundary at sinθ=1/2, so θ=π/6 and 5π/6. Area is ½[∫₀^(π/6)4sin²θ dθ+∫_(π/6)^(5π/6)1dθ+∫_(5π/6)^π4sin²θ dθ]=2π/3−√3/2.Worked example · 1/2In the upper half-plane, thecommon region inside r=1 andr=2sinθ changes boundary atsinθ=1/2, so θ=π/6 and 5π/6. AreaisWorked example · 2/2½[∫₀^(π/6)4sin²θdθ+∫_(π/6)^(5π/6)1dθ+∫_(5π/6)^π4sin²θdθ]=2π/3−√3/2.

In the upper half-plane, the common region inside r=1 and r=2sinθ changes boundary at sinθ=1/2, so θ=π/6 and 5π/6. Area is ½[∫₀^(π/6)4sin²θ dθ+∫_(π/6)^(5π/6)1dθ+∫_(5π/6)^π4sin²θ dθ]=2π/3−√3/2.

Discuss: explain what each label represents, then draw a related example with different values.

A common misconception

For an overlap use the smaller radial bound, not the larger. One difference-of-squares integral across a boundary switch is generally wrong.

Investigate and explain

Shade common interior ray by ray. Compare it with the union and with the region inside one curve but outside the other.
Identify one step in the worked example, name the mathematical fact that permits it, and state its domain conditions.