Finding the Area of the Region Bounded by Two Polar Curves: Study Guide (Calculus BC 9.9)

BC-only material. Polar area is part of Calculus BC only. Calculus AB students do not need this topic.

Before you start: prerequisites

This topic joins two ideas you already have: the single-curve polar area formula from Topic 9.8 and the “outer minus inner” thinking from areas between curves. If any row is shaky, revisit it from the Calculus BC roadmap first.

Prerequisite	Topic	What you use it for here
Polar coordinates and sketching r = f(θ)	9.7	Seeing the region and which curve is outer
Area bounded by one polar curve	9.8	Each piece of a region is ½ ∫ r² dθ
Area between two curves (rectangular)	8.4	The same “outer minus inner” idea
Trig equations and double-angle identities	Precalculus	Finding intersection angles; integrating cos²θ and sin²θ

Notation on this page: ∫ from a to b of f(θ) dθ is a definite integral with lower limit a and upper limit b. Angles are in radians. R is the curve farther from the pole and r is the curve nearer to it, on the interval being used.

From one curve to two

In Topic 9.8 you cut a polar region into thin sectors (pizza slices) with their point at the pole. A sector of radius r and angle dθ has area ½ r² dθ, so a region swept by one curve from θ = α to θ = β has area ½ ∫ from α to β of r² dθ.

Now suppose the region lies between two curves. On each ray from the pole, the region starts at the inner curve r = g(θ) and stops at the outer curve R = f(θ). A thin slice of the region is a big sector with a small sector removed:

slice area ≈ ½ R² dθ − ½ r² dθ = ½ (R² − r²) dθ

Add up the slices from α to β:

Area = ½ ∫ from α to β of ( [f(θ)]² − [g(θ)]² ) dθ, where f(θ) ≥ g(θ) ≥ 0 on [α, β]

This is the polar version of “top minus bottom”. The important difference is that you square each radius before subtracting. For a slice with R = 3, r = 2 and dθ = 0.1, the true area is ½(9 − 4)(0.1) = 0.25. The wrong form ½(R − r)² dθ gives ½(1)(0.1) = 0.05, far too small.

The formula has the same three ingredients every time: limits (where the region starts and stops), an outer curve and an inner curve. The rest of this guide is about finding those three ingredients reliably.

Step 1: find where the curves meet

The limits of integration are usually angles where the curves cross. To find them:

1. Solve f(θ) = g(θ) over one full trace of each curve (often 0 ≤ θ < 2π, or 0 ≤ θ ≤ π for a circle such as r = a sin θ).
2. Check the pole separately. A curve passes through the pole when r = 0. Two curves can both reach the pole at different angles, so the pole is a shared point that the equation f(θ) = g(θ) never finds.

Example of the pole trap. Take r = 2 cos θ and r = 1 − cos θ. Solving 2 cos θ = 1 − cos θ gives cos θ = ⅓, so two points with r = ⅔. But r = 2 cos θ is 0 at θ = π/2, and r = 1 − cos θ is 0 at θ = 0. Both curves pass through the pole, at different angles, so the curves actually share three points. A sketch shows this at once.

Why does this happen? A point in polar form has many names: the pole is (0, θ) for every θ. Solving f(θ) = g(θ) only finds places where both curves are at the same point at the same angle.

Step 2: decide which curve is outer

Between two consecutive intersection angles, one curve stays farther from the pole than the other (they cannot swap without crossing). So pick any test angle inside the interval and compare the two r-values. The larger one is the outer curve R on that whole interval.

Write this test down in a free-response answer. It is your justification for the order of subtraction.

Step 3: identify the type of region

You are asked for the area…	Set-up
inside f and outside g	½ ∫ (f² − g²) dθ over the angles where f ≥ g
inside both curves	Split at each intersection angle. On each piece use the curve nearer the pole: ½ ∫ (nearer curve)² dθ
inside f only (no second curve)	½ ∫ f² dθ, as in Topic 9.8

A region inside both curves has the pole as one corner of each slice, so each piece is a single-curve area. That is why it needs no subtraction, only splitting.

Worked example 1: inside a cardioid, outside a circle (no calculator)

Question. Find the exact area of the region that lies inside the cardioid r = 2 + 2 cos θ and outside the circle r = 3.

1. **Sketch.** The circle has radius 3. The cardioid has r = 4 at θ = 0 and r = 0 at θ = π (its cusp is at the pole). So the cardioid pokes out beyond the circle only on the right, near θ = 0 (Figure 1). 2. **Find where they meet.** 2 + 2 cos θ = 3 gives cos θ = ½, so θ = −π/3 and θ = π/3, where r = 3. Pole check: the circle never reaches the pole, so there are no other shared points. 3. **Decide which is outer.** Test θ = 0: cardioid r = 4, circle r = 3. So on −π/3 < θ < π/3 the cardioid is outer. 4. **Write the integral.** **A = ½ ∫ from −π/3 to π/3 of [ (2 + 2 cos θ)² − 3² ] dθ** 5. **Expand and use cos²θ = ½(1 + cos 2θ).** (2 + 2 cos θ)² = 4 + 8 cos θ + 4 cos²θ = 6 + 8 cos θ + 2 cos 2θ. Subtract 9: the integrand is **−3 + 8 cos θ + 2 cos 2θ**. 6. **Integrate.** An antiderivative is −3θ + 8 sin θ + sin 2θ. - At θ = π/3: −π + 8(√3/2) + √3/2 = −π + 4√3 + √3/2. - At θ = −π/3: π − 4√3 − √3/2. - Difference: −2π + 8√3 + √3 = **9√3 − 2π**. 7. **Multiply by ½.** A = ½(9√3 − 2π) = **9√3/2 − π ≈ 4.653** square units.

Checks.

• Symmetry: the region is symmetric about the x-axis, so A = 2 × ½ ∫ from 0 to π/3 of (…) dθ = ∫ from 0 to π/3 of (−3 + 8 cos θ + 2 cos 2θ) dθ. This gives 9√3/2 − π again. ✓
• Size: the region is a thin crescent. The whole circle has area 9π ≈ 28.3, so 4.653 is sensible.
• Wrong form: ½ ∫ from −π/3 to π/3 of (2 + 2 cos θ − 3)² dθ ≈ 0.544. If you get this, you squared the gap instead of each radius.

Worked example 2: the region inside both curves (calculator)

Question. Find the area of the region that lies inside both the circle r = 4 sin θ and the cardioid r = 1 + cos θ. Give the answer to 3 decimal places.

1. Sketch. r = 4 sin θ, for 0 ≤ θ ≤ π, is a circle of diameter 4 sitting on the pole, above the x-axis. r = 1 + cos θ is a cardioid with r = 2 at θ = 0 and its cusp at the pole (θ = π). The shared region is the small lens-shaped area in Figure 2.
2. Find where they meet. Solve 4 sin θ = 1 + cos θ with a calculator: θ = α ≈ 0.48996 (store it unrounded). There r ≈ 1.882. θ = π also solves the equation: both curves are at the pole there. The circle also reaches the pole at θ = 0, so the pole is a corner of the region.
3. Decide which curve is nearer the pole on each interval.
   • Test θ = 0.2 (between 0 and α): circle r ≈ 0.795, cardioid r ≈ 1.980. The circle is nearer.
   • Test θ = π/2 (between α and π): circle r = 4, cardioid r = 1. The cardioid is nearer.
4. Write the integral as two pieces, each a single-curve area: A = ½ ∫ from 0 to α of (4 sin θ)² dθ + ½ ∫ from α to π of (1 + cos θ)² dθ Stop at θ = π: for π < θ < 2π the circle has r < 0 and retraces itself, and the lower half of the cardioid is outside the circle.
5. Evaluate with the calculator. First piece ≈ 0.2989. Second piece ≈ 1.4143.
6. Add. A ≈ 1.713 square units.

Checks.

• Size: the region must be smaller than each whole curve. The circle has area 4π ≈ 12.566 and the cardioid 3π/2 ≈ 4.712. 1.713 is smaller than both. ✓
• Intersection: in fact α = 2 arctan(¼), with sin α = 8/17 and cos α = 15/17. Then 4 sin α = 32/17 and 1 + cos α = 32/17, both ≈ 1.882. ✓ (You do not need this exact angle; the calculator value is enough.)
• Wrong choices: using the farther curve on each piece (the cardioid on [0, α] and the circle on [α, π]) gives about 13.209, which is bigger than the whole circle. That is impossible for a region inside both curves.

Answer. About 1.713 square units.

Using symmetry and the calculator well

• Symmetry. If both curves use only cos θ, the region is symmetric about the x-axis: integrate over the top half and double. If both use only sin θ, it is symmetric about the y-axis: integrate over the right-hand half and double. Say which symmetry you are using, and check the halves really match on your sketch.
• Store, do not retype. Store each intersection angle in the calculator and use the stored value as a limit. Rounding α to 0.49 before integrating can change the third decimal place of the answer.
• Write the integral first. On a free-response question the set-up (limits, ½, outer² − inner²) usually earns a point on its own. A decimal with no integral shown does not show your reasoning.
• Exact or decimal? Without a calculator, intersection angles are usually standard angles such as π/6 or π/3, and you finish by hand as in Worked example 1. With a calculator, give 3 decimal places.

A note on negative r

The formula ½ ∫ r² dθ squares r, so a negative r never makes an area negative. But a negative r plots the point on the opposite side of the pole. This can make a curve retrace part of itself (as r = 4 sin θ does for π < θ < 2π), so integrating over too wide an interval counts some area twice. Choose limits so that each part of the region is swept exactly once, and let the sketch guide you.

Common misconceptions

• “Area = ½ ∫ (R − r)² dθ.” You must square each radius, then subtract: ½ ∫ (R² − r²) dθ.
• Dropping the ½. The ½ comes from the sector area ½ r² dθ. Without it every answer doubles.
• Missing the pole. Solving f(θ) = g(θ) finds meetings at the same angle only. Check where each curve has r = 0.
• Using the wrong curve for “inside both”. On each piece use the curve nearer the pole, not the farther one. A test angle settles it.
• Not splitting. A single integral cannot describe “inside both” when the nearer curve changes at the intersection angle.
• Integrating over 0 to 2π by habit. Use the angles where the region actually is. Over a full turn, a curve such as r = 4 sin θ is traced twice.
• Subtracting in the wrong order. If an “area” comes out negative, the inner and outer curves are swapped on that interval.
• Rounding the limits early. Store the calculator’s intersection angles and round only the final area.

Where this leads

This is the last topic of Unit 9. Polar questions often combine an area with the derivative work of Topic 9.7 (dr/dθ and dy/dx), so review Topics 9.7 to 9.9 together. Next, Unit 10 (BC only) starts infinite series with Topic 10.1, Defining Convergent and Divergent Infinite Series. Use the Calculus BC roadmap to see the order.

Try the practice questions now, then use the revision notes and the checklist to consolidate.