Area in Polar Coordinates

You’ll be able to

Compute the area enclosed by a polar curve
Find the area between two polar curves
Determine the correct limits of integration in θ

Polar coordinates and sweeping sectors

In polar coordinates a point is (r, θ): a distance r from the origin at angle θ. A polar curve r = f(θ) sweeps out a region as θ increases. To find its enclosed area, you do not use rectangles — you use thin circular sectors. A sector of radius r and tiny angle dθ has area ½r² dθ (a slice of a circle). Summing these wedge-shaped slivers as θ runs over the region gives the polar area integral.

Area enclosed by a polar curve

A = ½ ∫_α^β [ r(θ) ]² dθ

The ½r² comes from the area of a circular sector. Integrate over the θ-interval that traces the region exactly once.

Area between two polar curves

For the region between an outer curve r = R(θ) and an inner curve r = r(θ), subtract the sector areas: A = ½ ∫_α^β ([R(θ)]² − [r(θ)]²) dθ. Note you square each radius before subtracting — (R² − r²), never (R − r)². The limits α and β are the angles where the curves intersect or where the desired region begins and ends, found by setting R(θ) = r(θ) or by tracing the petal.

Worked example

Find the area enclosed by one petal of the rose r = 2·sin(2θ), which is traced as θ goes from 0 to π/2.

1.Use the polar area formula: A = ½ ∫₀^{π/2} [2·sin(2θ)]² dθ.
2.Square the radius: [2·sin(2θ)]² = 4·sin²(2θ), so A = ½ ∫₀^{π/2} 4·sin²(2θ) dθ = 2∫₀^{π/2} sin²(2θ) dθ.
3.Use sin²(u) = (1 − cos(2u))/2: sin²(2θ) = (1 − cos(4θ))/2.
4.Integrate: 2·∫₀^{π/2} (1 − cos 4θ)/2 dθ = ∫₀^{π/2}(1 − cos 4θ) dθ = [θ − (sin 4θ)/4]₀^{π/2} = π/2.

Answer: One petal has area π/2. The key steps are squaring the radius and choosing θ-bounds that trace exactly one petal.

Watch out

Getting the θ-limits right is half the battle. Limits that trace the curve twice double your area, and limits that stop short cut it off. For a rose petal, integrate over the θ-interval where r goes from 0 back to 0 exactly once.

Checkpoint

What is the area enclosed by one full loop of a polar curve r = f(θ) from θ = α to θ = β?

Checkpoint

For the region between an outer polar curve R(θ) and an inner curve r(θ), the area is:

On the exam

The number-one polar-area error is writing (R − r)² instead of (R² − r²). Square each radius separately, then subtract. When in doubt, compute the outer area and inner area as two integrals and subtract the results.

Answer the 2 checkpoints as you read.