Polar Slopes and Regions Between Two Curves

You’ll be able to

Find dy/dx for a polar curve by treating it as a parametric curve in θ
Choose correct limits of integration for a polar region, including a single petal
Set up the area bounded by two polar curves, using their intersection angles

A polar curve is a parametric curve in disguise

Given r = f(θ), write x = r cos θ = f(θ)cos θ and y = r sin θ = f(θ)sin θ. Now the parametric machinery applies unchanged: dy/dx = (dy/dθ)/(dx/dθ), with each of those computed by the product rule. The slope of a polar curve is not dr/dθ — that is how the radius changes, a different quantity entirely, and confusing the two is the standard polar error.

Choosing the limits is most of the work

The area swept by r = f(θ) from α to β is (1/2)∫ᵅᵝ r² dθ. Getting α and β right is where the marks are. For a single petal of r = a·sin(kθ) or a·cos(kθ), integrate between two consecutive angles where r = 0. For a full closed curve, integrate over one complete traversal — and note that r = a·cos(2θ) traces all four petals over [0, 2π] while r = a·cos(3θ) traces all three over [0, π], so "0 to 2π" is not a safe default. Sweeping a curve twice doubles the area.

Regions between two polar curves

For the region inside r₁ and outside r₂, the area is (1/2)∫ᵅᵝ (r₁² − r₂²) dθ — the squares are subtracted, not the radii. The limits α and β are the angles where the curves meet, found by setting f₁(θ) = f₂(θ). Polar intersections need care: two curves can cross at a point that the two equations reach at different θ values, and the origin is a common such case, so sketching before integrating is worth the thirty seconds.

Polar area and slope

A = (1/2)∫ᵅᵝ r² dθ · between curves: (1/2)∫ᵅᵝ (r_outer² − r_inner²) dθ · dy/dx = (dy/dθ)/(dx/dθ)

Square first, then subtract. (r₁ − r₂)² is not r₁² − r₂².

Worked example

Find the area of one petal of the four-petaled rose r = 3cos(2θ).

1.A petal begins and ends where r = 0, so solve 3cos(2θ) = 0: cos(2θ) = 0 at 2θ = ±π/2, giving θ = −π/4 and θ = π/4.
2.Those are consecutive zeros, so one petal is swept as θ runs from −π/4 to π/4.
3.Set up the area: A = (1/2)∫ from −π/4 to π/4 of (3cos 2θ)² dθ = (9/2)∫ cos²(2θ) dθ.
4.Use the identity cos²(u) = (1 + cos 2u)/2: (9/2)∫ (1 + cos 4θ)/2 dθ = (9/4)[θ + sin(4θ)/4] from −π/4 to π/4.
5.Evaluate: at π/4, sin(π) = 0 so the bracket is π/4; at −π/4 it is −π/4. The difference is π/2, giving A = (9/4)(π/2) = 9π/8 ≈ 3.53.

Answer: One petal has area 9π/8. Integrating from 0 to 2π instead would give the area of all four petals, and integrating from 0 to π/4 would give half a petal.

Watch out

The 1/2 in front of ∫r² dθ is part of the formula, not an option. Dropping it doubles every polar area, and it is the most frequent single-symbol loss on this topic.

Checkpoint

Which integral gives the area inside r = 4 and outside r = 2 + 2cos θ, between the angles α and β where they intersect?

Checkpoint

For the polar curve r = f(θ), the slope dy/dx of the tangent line is:

On the exam

Polar free-response questions almost always award a point for correct limits separate from the integrand. Even if the algebra stalls, write the fully set-up integral with its bounds.

Answer the 2 checkpoints as you read.