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Area Bounded by
Two Polar Curves

Learn how to compute the area between two polar curves using definite integrals, a key skill for the AP Calculus BC exam.

AP Calculus BC

What Are Polar Curves?

In polar coordinates, every point in the plane is described by a distance \( r \) from the origin and an angle \( \theta \) measured from the positive \( x \)-axis. A polar curve is defined by a function \( r = f(\theta) \).

\( r \)
The radial distance from the pole (origin) to the point on the curve.
\( \theta \)
The angle swept counterclockwise from the positive \( x \)-axis.

Area Under a Single Polar Curve

Before tackling two curves, recall how to find the area enclosed by one polar curve \( r = f(\theta) \) between angles \( \alpha \) and \( \beta \):

$$ A = \frac{1}{2}\int_{\alpha}^{\beta} \bigl[f(\theta)\bigr]^2 \, d\theta $$

Visualizing the Region Between Two Curves

The Core Formula

If \( f(\theta) \geq g(\theta) \geq 0 \) on the interval \( [\alpha, \beta] \), the area of the region between the two curves is:

$$ A = \frac{1}{2}\int_{\alpha}^{\beta}\Bigl(\bigl[f(\theta)\bigr]^2 - \bigl[g(\theta)\bigr]^2\Bigr)\,d\theta $$

Why Does It Work?

1
Thin sector: At angle \( \theta \), a tiny sector of width \( d\theta \) has area \( dA = \frac{1}{2}r^2\,d\theta \).
2
Outer minus inner: The sliver between the curves has area \( \frac{1}{2}\bigl[f(\theta)^2 - g(\theta)^2\bigr]\,d\theta \).
3
Integrate: Sum all slivers from \( \alpha \) to \( \beta \) to get \( A = \frac{1}{2}\int_{\alpha}^{\beta}\bigl[f(\theta)^2 - g(\theta)^2\bigr]\,d\theta \).

Rectangular vs. Polar Area

Rectangular Polar
Element of area \( [f(x) - g(x)]\,dx \) \( \tfrac{1}{2}[f(\theta)^2 - g(\theta)^2]\,d\theta \)
Shape of slice Thin rectangle Thin circular sector
Integration variable \( x \) (horizontal) \( \theta \) (angle)
Squaring? Not squared \( r \) is squared
Factor None \( \frac{1}{2} \)

Step-by-Step Approach

1
Sketch both curves on the same polar grid to identify the enclosed region.
2
Find intersection angles by setting \( f(\theta) = g(\theta) \) and solving for \( \theta \). These give \( \alpha \) and \( \beta \).
3
Determine outer vs. inner: On \( [\alpha, \beta] \), identify which curve is farther from the origin: \( f(\theta) \geq g(\theta) \).
4
Set up the integral: \( A = \frac{1}{2}\int_{\alpha}^{\beta}\bigl[f(\theta)^2 - g(\theta)^2\bigr]\,d\theta \).
5
Evaluate the integral using identities, substitution, or a calculator.

Example 1: Two Circles

Find the area of the region inside \( r = 4\cos\theta \) and outside \( r = 2 \).

Example 1: Solution

Outer: \( f(\theta) = 4\cos\theta \). Inner: \( g(\theta) = 2 \). Limits: \( \alpha = -\frac{\pi}{3} \), \( \beta = \frac{\pi}{3} \).

$$ A = \frac{1}{2}\int_{-\pi/3}^{\pi/3}\bigl[(4\cos\theta)^2 - (2)^2\bigr]\,d\theta $$
$$ = \frac{1}{2}\int_{-\pi/3}^{\pi/3}\bigl[16\cos^2\theta - 4\bigr]\,d\theta $$

Example 2: Cardioid and Circle

Find the area inside \( r = 1 + \cos\theta \) and outside \( r = 1 \).

Example 2: Solution

Outer: \( f(\theta) = 1 + \cos\theta \). Inner: \( g(\theta) = 1 \). Limits: \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \).

$$ A = \frac{1}{2}\int_{-\pi/2}^{\pi/2}\bigl[(1+\cos\theta)^2 - 1\bigr]\,d\theta $$
$$ = \frac{1}{2}\int_{-\pi/2}^{\pi/2}\bigl[2\cos\theta + \cos^2\theta\bigr]\,d\theta $$

Example 3: Rose and Circle

Find the area inside \( r = \sin 2\theta \) and outside \( r = \frac{1}{\sqrt{2}} \) in the first quadrant.

Using Symmetry to Simplify

Many polar regions have symmetry about the polar axis (\( \theta = 0 \)) or the line \( \theta = \frac{\pi}{2} \). When this happens, you can integrate over half the region and multiply by 2.

Symmetric about \( \theta = 0 \)
If \( f(-\theta) = f(\theta) \) and \( g(-\theta) = g(\theta) \), integrate from \( 0 \) to \( \beta \) and double.
Split integrals
If one curve is outer on part of the interval and inner on another, split into separate integrals.

Common Mistakes to Avoid

✕ Forgetting to square
Writing \( f(\theta) - g(\theta) \) instead of \( f(\theta)^2 - g(\theta)^2 \). Polar area uses squared radial functions.
✕ Omitting the \( \frac{1}{2} \)
The sector element is \( \frac{1}{2}r^2\,d\theta \), not \( r^2\,d\theta \). Always include the factor of \( \frac{1}{2} \).
✕ Wrong limits
Using \( x \)-values or failing to solve \( f(\theta) = g(\theta) \) correctly. Always solve for \( \theta \) at intersections.

Formula Summary

Scenario Formula
Single polar curve \( \frac{1}{2}\int_\alpha^\beta [f(\theta)]^2\,d\theta \)
Between two curves \( \frac{1}{2}\int_\alpha^\beta \bigl[f(\theta)^2 - g(\theta)^2\bigr]\,d\theta \)
With symmetry \( 2 \cdot \frac{1}{2}\int_0^\beta \bigl[f(\theta)^2 - g(\theta)^2\bigr]\,d\theta \)
Full enclosed region May require splitting: sum integrals on subintervals

Worked practice: Practice Problems, Problem 4 — applied

Work through this original lesson problem, then reveal the solution and illustrations.

A satellite's radar sweeps a region bounded by the cardioid r = 2 + 2 cos θ and the circle r = 3. (a) Find all angles θ in [0, 2π] where the two curves intersect. (b) Determine which curve is farther from the origin for θ ∈ (0, θ₁), where θ₁ is the smaller intersection angle. (c) Set up, but do not evaluate, an integral expression for the total area of the region inside the cardioid and outside the circle. (d) Now evaluate the integral from part (c) to find the exact area.

Show answer and explanation

Rubric (5 points): Point 1 (Intersection points): Set 2 + 2 cos θ = 3, so cos θ = ½, giving θ = π/3 and θ = 5π/3 (equivalently θ = −π/3). Award 1 point for both correct angles. Point 2 (Identifying outer/inner): Test θ = 0: cardioid gives r = 4, circle gives r = 3. Since 4 > 3, the cardioid is outer on (−π/3, π/3). Award 1 point. Point 3 (Integral setup): By symmetry about the polar axis, A = 2 × ½ ∫ from 0 to π/3 [(2 + 2 cos θ)² − 9] dθ = ∫ from 0 to π/3 [(2 + 2 cos θ)² − 9] dθ. Award 1 point for a correct integral with correct limits and the factor of 2 from symmetry. Point 4 (Expansion and simplification): (2 + 2 cos θ)² = 4 + 8 cos θ + 4 cos²θ. Subtracting 9: −5 + 8 cos θ + 4 cos²θ. Using cos²θ = (1 + cos 2θ)/2: −5 + 8 cos θ + 2 + 2 cos 2θ = −3 + 8 cos θ + 2 cos 2θ. Award 1 point for correct simplification. Point 5 (Evaluation): ∫ from 0 to π/3 (−3 + 8 cos θ + 2 cos 2θ) dθ = [−3θ + 8 sin θ + sin 2θ] from 0 to π/3 = (−π + 8(√3/2) + √3/2) − 0 = −π + 4√3 + √3/2 = −π + 9√3/2. So A = −π + 9√3/2 = (9√3 − 2π)/2. Award 1 point for the correct final answer.

Region outside r=3 and inside r=2+2cosθ over −π/3≤θ≤π/3, with area 9√3/2−π.
The shaded region is inside r=2+2cosθ and outside r=3 for −π/3≤θ≤π/3. Its area is 9√3/2−π square model units; the diagram does not establish real orbital or geographic coverage.

Question 1 of 5

Which integral represents the area between \( r = f(\theta) \) (outer) and \( r = g(\theta) \) (inner) from \( \alpha \) to \( \beta \)?

Question 2 of 5

To find the area inside \( r = 3 \) and outside \( r = 2 + \cos\theta \), the first step is to find the limits of integration. What equation do you solve?

Question 3 of 5

What is the area inside \( r = 2 \) and outside \( r = 2(1 - \cos\theta) \) for \( 0 \leq \theta \leq \frac{\pi}{2} \)?

Question 4 of 5

A student writes the area between two polar curves as \( \frac{1}{2}\int_\alpha^\beta\bigl[f(\theta) - g(\theta)\bigr]^2\,d\theta \). What is wrong with this setup?

Question 5 of 5

The curves \( r = 4\sin\theta \) and \( r = 2 \) intersect at \( \theta = \frac{\pi}{6} \) and \( \theta = \frac{5\pi}{6} \). On \( \bigl[\frac{\pi}{6}, \frac{5\pi}{6}\bigr] \), which is the outer curve?