9.8 Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve

BC only. The area swept out by a polar curve r = f(θ) from θ = α to θ = β is ½∫ r² dθ, taken from α to β. The formula adds up thin pie-shaped wedges, and choosing the right θ-limits is most of the work.

Key terms

polar area
sector
limits in θ
r²

Where the formula comes from

This whole unit is BC only. A sector (pie slice) of a circle with radius r and angle Δθ, in radians, has area ½r²Δθ. Slice the region swept out by r = f(θ) into thin wedges. Each wedge is nearly a sector with radius f(θ). Add them up and let the wedges get thin:

A = ½ ∫ from α to β of [f(θ)]² dθ.

This requires r to trace the region's boundary exactly once as θ runs from α to β. The formula uses r², so a negative r still gives positive area.

Choosing the limits

Sketch the curve, or graph it on a calculator in polar mode, to see the region.
Find where r = 0. Those angles are where the curve passes through the origin, and they're often the edges of loops or petals.
Choose α and β so the region is traced exactly once.
Use symmetry when it helps: compute one petal or half the region, then multiply.

Shapes you'll see

Curve	Shape	One full trace
r = a (constant)	circle of radius a	0 to 2π
r = a cos θ or a sin θ	circle through the origin with diameter a (ignoring sign)	0 to π
r = a(1 + cos θ)	cardioid (heart shape)	0 to 2π
r = a + b cos θ, with b bigger than a in size	limaçon with an inner loop	0 to 2π
r = a sin(nθ) or a cos(nθ)	rose with n petals if n is odd, 2n if n is even	0 to π for n odd, 0 to 2π for n even

Double counting

Some curves retrace themselves. The circle r = 2 cos θ is traced once from 0 to π; integrating from 0 to 2π counts its area twice. A rose with an odd number of petals is traced once from 0 to π. If an answer comes out twice as large as you expected, check whether your interval goes around twice.

You'll almost always need ∫ sin² or ∫ cos² by hand. Use the identities sin² θ = (1 − cos 2θ)/2 and cos² θ = (1 + cos 2θ)/2. With a calculator, just enter the integral.

Worked examples

Try each one yourself first, then open the solution.

Example 1
Area of a cardioid
Find the area enclosed by r = 1 + cos θ.
Show the solutionHide the solution
1. Step 1: The cardioid is traced once for 0 ≤ θ ≤ 2π.
2. Step 2: A = ½ ∫ from 0 to 2π of (1 + cos θ)² dθ = ½ ∫ from 0 to 2π of (1 + 2 cos θ + cos² θ) dθ.
3. Step 3: Over a full period, ∫ cos θ dθ = 0 and ∫ cos² θ dθ = π (using cos² θ = (1 + cos 2θ)/2).
4. Step 4: A = ½(2π + 0 + π) = 3π/2.
Answer: 3π/2 ≈ 4.712
Example 2
One petal of a rose
Find the area of one petal of r = 2 sin(3θ).
Show the solutionHide the solution
1. Step 1: r = 0 when sin(3θ) = 0, at θ = 0 and θ = π/3. Between them r > 0, so that's one petal.
2. Step 2: A = ½ ∫ from 0 to π/3 of 4 sin²(3θ) dθ = 2 ∫ from 0 to π/3 of (1 − cos 6θ)/2 dθ.
3. Step 3: = ∫ from 0 to π/3 of (1 − cos 6θ) dθ = [θ − sin(6θ)/6] from 0 to π/3 = π/3.
4. Step 4: Trap: the whole rose (3 petals) has area π. Integrating ½∫ r² dθ from 0 to 2π gives 2π, because the curve traces every petal twice.
Answer: π/3 ≈ 1.047
Example 3
Inner loop of a limaçon
Find the area inside the inner loop of r = 1 + 2 cos θ.
Show the solutionHide the solution
1. Step 1: r = 0 when cos θ = −½, at θ = 2π/3 and 4π/3. Between them, r < 0, and that's the inner loop.
2. Step 2: A = ½ ∫ from 2π/3 to 4π/3 of (1 + 2 cos θ)² dθ.
3. Step 3: Expand: 1 + 4 cos θ + 4 cos² θ = 3 + 4 cos θ + 2 cos 2θ.
4. Step 4: Antiderivative: 3θ + 4 sin θ + sin 2θ. Evaluated from 2π/3 to 4π/3, this gives 2π − 3√3.
5. Step 5: A = ½(2π − 3√3) = π − (3√3)/2 ≈ 0.544.
Answer: π − (3√3)/2 ≈ 0.544

Common mistakes

Forgetting the ½ in front.
Using r instead of r² in the integrand.
Using limits that trace the region twice, or picking limits from the x-axis instead of from where r = 0.
Integrating sin² θ as if it were sin θ. Use the half-angle identity.

On the exam

Polar area appears on BC free response often, usually with a calculator. Write ½ ∫ r² dθ with the correct limits, then the value.
Without a calculator, expect a curve where the half-angle identity finishes the job.

Connected topics

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