Polar coordinates describe points using a distance from the origin and an angle from a reference ray. This system is especially useful for curves with circular, spiral, or petal-like shapes that are awkward in rectangular coordinates. To find area in polar coordinates, calculus adds up many tiny wedge-shaped sectors instead of vertical rectangles.
The central formula is A = 1/2 ∫_α^β r(θ)^2 dθ, which comes from the area of a circular sector.
Key Facts
- Area enclosed by one polar curve: A = 1/2 ∫_α^β r(θ)^2 dθ.
- Area between two polar curves: A = 1/2 ∫_α^β (r_outer(θ)^2 - r_inner(θ)^2) dθ.
- A small polar sector has approximate area dA = 1/2 r^2 dθ.
- Use radians for θ in polar area integrals.
- Intersection angles are found by solving r_1(θ) = r_2(θ), but also check the origin when r = 0.
- For symmetric polar curves, find one matching region and multiply by the number of identical regions.
Vocabulary
- Polar coordinates
- A coordinate system that locates a point by its distance r from the origin and its angle θ from a reference direction.
- Polar curve
- A graph described by an equation r = f(θ), where the radius changes as the angle changes.
- Sector
- A wedge-shaped region bounded by two rays from the origin and an arc.
- Bounds of integration
- The starting and ending angle values α and β that define the part of the polar curve being measured.
- Area between curves
- The area of a region found by subtracting the inner polar radius squared from the outer polar radius squared before integrating.
Common Mistakes to Avoid
- Forgetting the factor 1/2 is wrong because polar area comes from sector area, not rectangle area.
- Using r instead of r^2 is wrong because the area of a sector depends on the square of the radius.
- Choosing angle bounds from the picture without solving intersections can give the wrong region because polar curves may cross at unexpected angles.
- Assuming the outer curve is the same for the whole interval can be wrong because two polar curves may switch which one is farther from the origin.
Practice Questions
- 1 Find the area enclosed by one petal of r = 2 sin(3θ). Use the interval 0 ≤ θ ≤ π/3.
- 2 Find the area inside r = 4 cos θ for -π/2 ≤ θ ≤ π/2.
- 3 For the region between r = 3 and r = 1 + 2 cos θ, explain how you would decide which curve is outer on the interval before setting up the area integral.