MathsPolar coordinates › Areas with polar coordinates

Areas with polar coordinates

Sweep a radius round a polar curve and it brushes out area in thin sectors. Half r squared, integrated over the angle, measures any region a polar equation encloses.

Builds on Polar curves and Compound angles and the harmonic form.

IN THIS TOPIC

  • Apply A = ½∫r² dθ with limits that trace the region exactly once.
  • Square and simplify r with double angle identities before integrating.
  • Find the area of a region bounded by two polar curves.

COMMON MISCONCEPTION

Area in polar coordinates is the integral of r dθ, by analogy with the area under y = f(x).

Sectors, not strips

Between θ and θ + dθ the radius sweeps a thin sector, near enough a triangle of area ½r × r dθ. Integrating the sectors gives the polar area formula:

A=12αβr2dθA = \frac{1}{2} \text{∫}_{α}^{β} r^{2} \, dθIN THE FORMULAE BOOKLET

It is printed in the booklet under Area of a sector, so the marks here go for choosing limits that trace the region once and for squaring r cleanly, not for recall. The r is squared, and the square is the point. A strip under a cartesian graph has area y dx, but a sector's area grows with the square of its radius. Sanity check on r = a over a full turn: ½ × a² × 2π = πa², the circle's area.

Polar area swept in thin sectors: each wedge holds ½r² dθ, integrated from α round to βθ = αθ = βone wedge: ½r² dθA = ½∫r² dθ
FIG. 1A polar region cut into thin sectors: each contributes ½r² dθ, and the integral sweeps them from α round to β.

WORKED EXAMPLE

A quarter of a circle

Find the area swept by r = 2 as θ runs from 0 to π/2.

A = ½∫4 dθ from 0 to π/2 = 2 × π/2 = π.

The full circle has area 4π and a quarter of it is π, so the formula agrees with geometry before any hard curve is attempted.

Cardioids and petals

Real questions square a trig expression, so the double angle identity cos²θ = (1 + cos 2θ)/2 does the heavy lifting. Keep limits that trace the region once, and use symmetry to halve the work wherever the curve allows.

WORKED EXAMPLE

The area inside a cardioid

Find the area enclosed by r = 1 + cos θ.

A = ½∫(1 + cos θ)² dθ over 0 to 2π = ½∫(1 + 2 cos θ + cos²θ) dθ.

The cos θ term integrates to zero over a full turn, and cos²θ contributes π via the double angle.

A = ½(2π + 0 + π) = 3π/2.

The inside of the cardioid r = 1 + cos θ: ½∫r² dθ over a full turn gives 3π/2area = 3π/2r = 1 + cos θ½∫(1 + cos θ)² dθ from 0 to 2π
FIG. 2The cardioid r = 1 + cos θ with its interior shaded: ½∫r² dθ over a full turn gives exactly 3π/2.

INDEPENDENT PRACTICE

One petal of a rose

Find the area of one loop of r = cos 2θ, using the loop traced for −π/4 ≤ θ ≤ π/4.

Show the working

A = ½∫cos²2θ dθ = ½∫(1 + cos 4θ)/2 dθ over the loop.

The cos 4θ part integrates to zero across the symmetric limits, leaving ½ × ½ × π/2.

A = π/8. All four petals together cover π/2, half the unit circle's area.

Two curves, one region

When a region is trapped between two polar curves, subtract the sector integrals over the same range of θ, so A = ½∫(router² − rinner²) dθ. Everything then hangs on the limits, which come from solving r₁ = r₂. Get those wrong and no amount of correct integration will save the answer.

WORKED EXAMPLE

Inside the cardioid, outside the circle

Find the area of the region inside r = 1 + cos θ and outside r = 1.

The curves meet where 1 + cos θ = 1, so cos θ = 0 and θ = ±π/2.

Between those angles the cardioid is the outer curve, so A = ½∫[(1 + cos θ)² − 1] dθ from −π/2 to π/2 = ½∫(2 cos θ + cos²θ) dθ.

The first part gives ½ × 4 = 2 and the second gives ½ × π/2.

A = 2 + π/4, a shade under 2.79.

One warning. Two polar curves can pass through the pole at different values of θ and still intersect there, so solving r₁ = r₂ will miss that crossing. Check the pole separately whenever both curves reach it.

ASSESSMENT FOCUS

  • Square r before integrating, then reach for the double angle identities immediately.
  • Choose limits by finding where r = 0. A loop runs between consecutive zeros.
  • Over a full turn, plain cos θ and sin θ terms vanish. Say so instead of integrating them longhand.
  • For a region between curves, state the intersection angles as a separate line of working.

CHECK YOURSELF

Find the area swept by r = 3 as θ runs from 0 to 2π/3.

Show a hint

½∫9 dθ over the sector.

Show the answer

A = ½ × 9 × 2π/3 = 3π. As a check, that is one third of the full circle's 9π, matching the one-third turn.

Polar area is ½∫r² dθ, so the radius comes in squared.

Square out, deploy cos²θ = (1 + cos 2θ)/2, and set limits between zeros of r or at intersections.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

6 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

Or read them with their worked answers on the areas with polar coordinates questions page.

CHECK YOUR PROGRESS

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  • Apply A = ½∫r² dθ with limits that trace the region exactly once.
  • Square and simplify r with double angle identities before integrating.
  • Find the area of a region bounded by two polar curves.

Open the full revision checklist to see every objective in the course in one place.