AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/9/9-9)
Unit 9 · Topic 9.9
BC only9.9 Finding the Area of the Region Bounded by Two Polar Curves
BC only. To find the area between two polar curves, find the angles where they meet, then integrate ½(R² − r²) dθ, where R is the outer curve and r the inner one. For the region inside both curves, split the integral where the inner curve switches.
Key terms
- intersection points
- outer curve
- inner curve
- area between polar curves
Inside one curve, outside another
This whole unit is BC only. Picture each wedge as a pie slice with a bite taken out: the outer curve sets the slice, and the inner curve sets the bite. The area is
A = ½ ∫ from α to β of (R² − r²) dθ,
where R is the outer curve's r-value (farther from the origin) and r is the inner curve's. Note that it's R² − r², not (R − r)², just like washers in 8.11.
The limits α and β are usually the angles where the curves intersect.
Finding intersection angles
- Set the two r-expressions equal and solve for θ: for 3 cos θ = 1 + cos θ, cos θ = ½, so θ = ±π/3.
- Check whether the curves also meet at the origin. Both can pass through the pole at different angles, which setting r-values equal won't find.
- With a calculator, graph both curves in polar mode and solve for the intersection angles, storing them.
The region inside both curves
For the overlap of two regions, the boundary switches from one curve to the other at the intersection points. On each θ-interval, use whichever curve is closer to the origin (the inner one) as the boundary, and add the pieces:
A = ½ ∫ (inner curve)² dθ over the first interval + ½ ∫ (inner curve)² dθ over the next interval, and so on.
Symmetry often halves the work.
A sketch is essential
These problems are hard to set up without a picture. Sketch both curves, mark the intersection points, shade the region and draw one ray from the origin through the region. Wherever that ray enters the region is the inner boundary, and wherever it leaves is the outer boundary. If the entry or exit curve changes as the ray rotates, you need to split the integral there.
A useful check: the area inside one curve but outside another plus the overlap should equal the total area inside the first curve.
On a calculator question, you can often avoid deciding which curve is outer by integrating ½|R² − r²|, but only over an interval where the region really is bounded by those two curves.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Inside a circle, outside a cardioid
Find the area of the region inside r = 3 cos θ and outside r = 1 + cos θ.
Show the solutionHide the solution
- Step 1: Intersections: 3 cos θ = 1 + cos θ → cos θ = ½ → θ = −π/3 and π/3.
- Step 2: For −π/3 < θ < π/3, test θ = 0: 3 > 2, so the circle is outside. R = 3 cos θ, r = 1 + cos θ.
- Step 3: A = ½ ∫ from −π/3 to π/3 of [9 cos² θ − (1 + cos θ)²] dθ.
- Step 4: By symmetry, A = ∫ from 0 to π/3 of [8 cos² θ − 2 cos θ − 1] dθ = ∫ from 0 to π/3 of [4 + 4 cos 2θ − 2 cos θ − 1] dθ.
- Step 5: = [3θ + 2 sin 2θ − 2 sin θ] from 0 to π/3 = π + 2(√3/2) − 2(√3/2) = π.
Answer: π ≈ 3.142
- Example 2
Trap: the overlap needs a split
Find the area of the region inside both r = 3 cos θ and r = 1 + cos θ.
Show the solutionHide the solution
- Step 1: Using a single integral of ½(R² − r²) doesn't work here: you want the region inside both, so on each interval you use only the inner curve.
- Step 2: For 0 ≤ θ ≤ π/3, the cardioid is inside (closer to the origin). For π/3 ≤ θ ≤ π/2, the circle is inside (it reaches the origin at θ = π/2).
- Step 3: Top half: ½ ∫ from 0 to π/3 of (1 + cos θ)² dθ + ½ ∫ from π/3 to π/2 of 9 cos² θ dθ.
- Step 4: The region is symmetric about the x-axis, so double it: A = ∫ from 0 to π/3 of (1 + cos θ)² dθ + ∫ from π/3 to π/2 of 9 cos² θ dθ = 5π/4.
- Step 5: Check: the circle's total area is π(3/2)² = 9π/4, and 9π/4 − π (from the first example) = 5π/4. ✓
Answer: 5π/4 ≈ 3.927
Common mistakes
- Writing ½ ∫ (R − r)² dθ instead of ½ ∫ (R² − r²) dθ.
- Swapping inner and outer curves, which gives a negative area.
- Missing an intersection at the origin.
- Using one integral for an overlap region whose boundary switches curves.
On the exam
- Free-response polar questions often ask for the area inside one curve and outside another, sometimes with a calculator. Give the intersection angles, the setup and the value.
- A quick sketch with the intersection angles labeled helps you and the reader see which curve is outer on each interval.
Connected topics
- Unit 99.8 Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve
- Unit 99.7 Defining Polar Coordinates and Differentiating in Polar Form
- Unit 88.11 Volume with Washer Method: Revolving Around the x- or y-Axis
- Unit 88.6 Finding the Area Between Curves That Intersect at More Than Two Points
Videos
Check yourself
4 questions on 9.9 Finding the Area of the Region Bounded by Two Polar Curves. Pick an answer to see if you got it, and why.
Which of the following gives the area of the region inside the polar curve r = 3cos θ and outside the polar curve r = 1 + cos θ?
What is the area of the region that lies inside the polar curve r = 4sin θ and outside the polar curve r = 1 + sin θ?
What is the area of the region that lies inside both of the polar curves r = 2sin θ and r = 2cos θ?
Which of the following gives the area of the region inside the polar curve r = 2 and outside the polar curve r = 2 − 2cos θ?
0 of 4 answered