Unit 9: Parametric Equations, Polar Coordinates & Vector-Valued Functions

Area in Polar Coordinates

Using a definite integral to find the area of a region bounded by a polar curve.

Advanced25 min lesson3 min readUpdated August 12, 2026Author not yet attributed

Prerequisites

Motion with Vector-Valued Functions

Area and Slope in a Coordinate System Built on Angle

The area-under-a-curve formulas so far all summed thin rectangles. A polar curve r(θ) isn't naturally sliced into rectangles — but it can be sliced into thin pie-shaped sectors instead. Before reading on: if a full circle of radius r has area πr², what fraction of that area would a thin sector spanning just Δθ radians (out of the full 2π) have?

Definition — Area in Polar Coordinates

A thin sector spanning angle Δθ, at radius r(θ), has area approximately (Δθ/2π)·πr² = (1/2)r²Δθ — the fraction of a full circle's area that Δθ represents. Summing these sectors from θ = α to θ = β and taking the limit as Δθ → 0 gives A = (1/2)∫_α^β [r(θ)]² dθ.

A = \frac{1}{2}\int_{\alpha}^{\beta} \big[r(\theta)\big]^2\, d\theta

Worked Example — Finding the Area Enclosed by a Polar Circle

Find the area enclosed by r = 3sin(θ) (a circle of diameter 3, traced once as θ goes from 0 to π, since r = 0 at both ends and reaches its max, 3, at θ = π/2). A = (1/2)∫₀^π (3sinθ)² dθ = (9/2)∫₀^π sin²θ dθ. Using sin²θ = (1−cos2θ)/2: ∫₀^π sin²θ dθ = (1/2)[θ − sin(2θ)/2]₀^π = (1/2)(π) = π/2. So A = (9/2)(π/2) = 9π/4. Check independently: this circle has radius 3/2 (diameter 3), so its area should be π(3/2)² = 9π/4 — matching exactly.

A polar curve's slope also comes from a parametric derivative — since x = r cosθ and y = r sinθ are both functions of θ (with r itself a function of θ), dy/dx = (dy/dθ)/(dx/dθ), the same idea from the parametric-derivative lesson with θ playing the role of t. Using the product rule on x and y: dy/dθ = r'sinθ + r cosθ, and dx/dθ = r'cosθ − r sinθ.

Worked Example — Finding the Slope of a Polar Curve

Find dy/dx for r = 3sinθ at θ = π/6. r = 3(1/2) = 3/2. r' = 3cosθ, so r' = 3(√3/2) = 3√3/2 at θ = π/6. Numerator: r'sinθ + r cosθ = (3√3/2)(1/2) + (3/2)(√3/2) = 3√3/4 + 3√3/4 = 3√3/2. Denominator: r'cosθ − r sinθ = (3√3/2)(√3/2) − (3/2)(1/2) = 9/4 − 3/4 = 3/2. dy/dx = (3√3/2)/(3/2) = √3.

Tip

A polar curve's area formula uses r² dθ, not r dθ — a sector's area scales with the square of its radius, the same reason a circle's area is πr² rather than a linear function of r.

Common Mistakes

Using dy/dx = dr/dθ for a polar curve's slope.
dr/dθ measures how the radius changes with angle, not the slope of the tangent line in the xy-plane — the actual slope requires converting through x = r cosθ and y = r sinθ first, as the product-rule derivation above shows.

Key Takeaways

Polar area sums thin circular sectors: A = (1/2)∫[r(θ)]²dθ, derived from a sector's fraction of a full circle.
A polar curve's slope, dy/dx, is a parametric derivative in disguise, with θ as the shared parameter.

Summary

This closes Unit 9: parametric, vector-valued, and polar representations all extend calculus beyond a single y = f(x). Unit 10 turns to a different kind of extension — sums with infinitely many terms.

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