Finding the area of regions bounded by two polar curves

AP Calculus BC· AP Calculus BC CED — Parametric Equations, Polar Coordinates, and Vector-Valued Functions· 14 min read

1. Finding Intersection Points of Two Polar Curves★★☆☆☆⏱ 3 min

Before you can calculate the area of a bounded region between two polar curves, you must first find all intersection points to identify the $θ$ -values that bound your region of interest. Unlike Cartesian curves, polar curves can intersect at two distinct types of points: points that satisfy both equations for the same values of $(r, θ)$ , and the pole (origin), which can be reached by both curves at different values of $θ$ , so it is often missed.

Set $r_{1} (θ) = r_{2} (θ)$ and solve for $θ$ over the interval $0 \leq θ < 2 π$ .
Check if each curve passes through the pole separately: if $r_{1} (θ_{1}) = 0$ for some $θ_{1}$ and $r_{2} (θ_{2}) = 0$ for some $θ_{2}$ , the pole is an intersection even if it does not appear when solving $r_{1} = r_{2}$ .

📐 Worked Example

Find all intersection points of the polar curves $r = 3 sin θ$ and $r = 1 + sin θ$ .

1
Set $r_{1} = r_{2}$ and solve for $θ$ :
2
$3\sin\theta = 1 + \sin\theta \implies 2\sin\theta = 1 \implies \sin\theta = \frac{1}{2}$
3
Over $0 \leq θ < 2 π$ , solutions are $θ = \frac{π}{6}$ and $θ = \frac{5 π}{6}$ .
4
Check for the pole as an intersection:
5
For $r = 3 sin θ$ , $r = 0$ when $θ = 0$ and $θ = π$ . For $r = 1 + sin θ$ , $r = 0$ only when $sin θ = - 1$ , so it never equals 0. The pole is not an intersection.
6
Final intersections: $(\frac{3}{2}, \frac{π}{6})$ and $(\frac{3}{2}, \frac{5 π}{6})$ .

Exam tip:

Always check the pole as an intersection, even if solving $r_{1} = r_{2}$ gives no solutions. Roughly 1 in 3 AP polar area problems require this check to get correct integral bounds.

2. Area When One Curve Is Entirely Inside Another★★☆☆☆⏱ 4 min

The most common problem type on the AP exam is finding the area of a region inside a larger outer curve and entirely outside a smaller inner curve, where the entire inner curve lies inside the outer curve for all values of $θ$ . This case is simpler than overlapping intersecting curves because you do not need to split the integral into multiple intervals. The formula derives from subtracting the area of the inner curve from the area of the outer curve:

A = \frac{1}{2}\int_{0}^{2\pi} \left( \left[r_{\text{out}}(\theta)\right]^2 - \left[r_{\text{in}}(\theta)\right]^2 \right) d\theta

The $\frac{1}{2}$ factor comes from the area of an infinitesimal polar sector, which is $\frac{1}{2} r^{2} d θ$ . Because the inner curve is smaller than the outer curve for all $θ$ , the integrand $r_{out}^{2} - r_{in}^{2}$ is always positive, so no absolute value or splitting is needed.

📐 Worked Example

Find the area of the region inside the circle $r = 5$ and outside the cardioid $r = 2 (1 + sin θ)$ . Confirm the cardioid is entirely inside the circle first.

1
Confirm the cardioid is entirely inside the circle: the maximum value of $r$ for the cardioid is $2 (1 + 1) = 4 < 5$ , so $r_{in} < r_{out}$ for all $θ$ , so we can use the single-integral formula.
2
Set up the area integral:
3
$A = \frac{1}{2}\int_{0}^{2\pi} \left[ 5^2 - \left(2(1+\sin\theta)\right)^2 \right] d\theta$
4
Simplify the integrand using the power-reduction identity $sin^{2} θ = \frac{1 - c o s 2 θ}{2}$ :
5
$25 - 4(1 + 2\sin\theta + \sin^2\theta) = 21 - 8\sin\theta - 4\sin^2\theta = 19 - 8\sin\theta + 2\cos 2\theta$
6
Integrate and evaluate from $0$ to $2 π$ :
7
$A = \frac{1}{2} \left[ 19\theta + 8\cos\theta + \sin 2\theta \right]_0^{2\pi} = \frac{1}{2} \left( 38\pi \right) = 19\pi$

Exam tip:

Always confirm that one curve is entirely inside another before using this single-integral formula. If any part of the inner curve extends outside the outer curve, you will need to split the integral at intersection points.

3. Area of Overlapping Intersecting Polar Curves★★★☆☆⏱ 5 min

When two polar curves cross each other (neither is entirely inside the other), the bounded region between them is split into intervals where one curve is the outer radius and the other is the inner radius. For these cases, you must split the integral at each intersection point, and use the outer radius for each interval. The general formula for the area of a region bounded by two intersecting curves is:

A = \sum_{i=0}^{n-1} \frac{1}{2}\int_{\theta_i}^{\theta_{i+1}} \left( \left[r_{\text{out}}(\theta)\right]^2 - \left[r_{\text{in}}(\theta)\right]^2 \right) d\theta

Symmetry is often used to simplify calculations by cutting the interval in half and doubling the result, which reduces the chance of arithmetic error.

📐 Worked Example

Find the area of the region that lies inside both $r = 2 cos θ$ and $r = 1$ .

1
Intersections are at $θ = \frac{π}{3}$ and $θ = \frac{5 π}{3}$ , with no intersection at the pole. By symmetry across the x-axis, we calculate twice the area for $0 \leq θ \leq \frac{π}{2}$ : for $0 \leq θ \leq \frac{π}{3}$ , the outer curve is $r = 2 cos θ$ ; for $\frac{π}{3} \leq θ \leq \frac{π}{2}$ , the outer curve is $r = 1$ .
2
Set up the symmetric integral:
3
$A = 2\left( \frac{1}{2}\int_{0}^{\frac{\pi}{3}} (2\cos\theta)^2 d\theta + \frac{1}{2}\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} 1^2 d\theta \right) = \int_{0}^{\frac{\pi}{3}} 4\cos^2\theta d\theta + \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} 1 d\theta$
4
Simplify and integrate using power-reduction $cos^{2} θ = \frac{1 + c o s 2 θ}{2}$ :
5
$4\cos^2\theta = 2(1 + \cos 2\theta) \\ \int_{0}^{\frac{\pi}{3}} (2 + 2\cos 2\theta) d\theta = \left[ 2\theta + \sin 2\theta \right]_0^{\frac{\pi}{3}} = \frac{2\pi}{3} + \frac{\sqrt{3}}{2} \\ \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} 1 d\theta = \frac{\pi}{6}$
6
Add the results to get the final area:
7
$A = \frac{2\pi}{3} + \frac{\sqrt{3}}{2} + \frac{\pi}{6} = \frac{5\pi}{6} + \frac{\sqrt{3}}{2}$

✓ Quick check

Test your understanding with this AP-style multiple choice question:

What is the area of the region inside $r = 3 sin θ$ and outside $r = 1 + sin θ$ ?
- $π$
- $2 π$
- $3$
- $π + 3$
Reveal answer
A —
Correct! The intersections are at $θ = \frac{π}{6}$ and $\frac{5 π}{6}$ , setting up the integral and simplifying gives a final area of $π$ .

Exam tip:

Use symmetry to cut your work in half on the AP exam. Most polar curves are symmetric across the x-axis or y-axis, so you can avoid integrating over a full $0$ to $2 π$ interval.

4. Common Pitfalls

Wrong move:

Forgetting to check the pole as an intersection point, leading to missing bounds when two curves cross at the origin but reach it at different $θ$ .

Why:

Students assume all intersections come from solving $r_{1} = r_{2}$ , but the pole can be an intersection even if it does not appear in that solution set.

Correct move:

After solving $r_{1} = r_{2}$ , always check if each curve passes through the pole separately; if both do, add the pole as an intersection and adjust your bounds.

Wrong move:

Using $r_{1} - r_{2}$ instead of $\frac{1}{2} (r_{1}^{2} - r_{2}^{2})$ in the area formula.

Why:

Students confuse polar area with Cartesian area, where you subtract the functions directly. Polar area relies on the area of a sector, which depends on the square of the radius.

Correct move:

Always start any polar area calculation by writing the $\frac{1}{2}$ factor and squaring both radii before subtracting.

Wrong move:

Not splitting the integral at intersection points when curves cross, leading to using the wrong inner/outer radius over an interval.

Why:

Students assume one curve is larger everywhere just because it is larger at some angles.

Correct move:

After finding all intersection points, test the radius of each curve at a sample angle in each interval between intersections to confirm which is outer and which is inner.

Wrong move:

Integrating over $0$ to $2 π$ for a rose curve $r = sin n θ$ with even $n$ , leading to doubling the area incorrectly.

Why:

Students default to $0$ to $2 π$ for all closed polar curves, but even $n$ roses trace all petals over $0$ to $π$ .

Correct move:

Remember that rose curves $r = sin n θ$ or $r = cos n θ$ are traced exactly once over $0$ to $π$ for even $n$ , and $0$ to $2 π$ for odd $n$ .

Wrong move:

Forgetting to apply the power-reduction identity to $sin^{2} θ$ or $cos^{2} θ$ , leading to an integrand that cannot be evaluated correctly.

Why:

Students memorize the area formula but forget that squaring the trigonometric polar function requires simplification before integration.

Correct move:

Whenever you have $sin^{2} θ$ or $cos^{2} θ$ in the integrand, apply the power-reduction identity immediately before integrating.

5. Quick Reference Cheatsheet

Category	Formula/Rule	Notes
Area of single polar region	$A = \frac{1}{2} \int_{α}^{β} r (θ)^{2} d θ$	For region bounded by one polar curve from $α$ to $β$
Polar intersection check	Solve $r_{1} (θ) = r_{2} (θ)$ ; 2. Check pole	Pole is intersection even if not found by $r_{1} = r_{2}$
Area inside $r_{o u t}$ , outside $r_{in}$	$A = \frac{1}{2} \int_{0}^{2 π} (r_{o u t}^{2} - r_{in}^{2}) d θ$	Use only if $r_{in} < r_{o u t}$ for all $θ$
Area of intersecting curves	$A = \sum \frac{1}{2} \int_{θ_{i}}^{θ_{i + 1}} (r_{o u t}^{2} - r_{in}^{2}) d θ$	Split integral at each intersection, confirm inner/outer per interval
Power-reduction: $sin^{2} θ$	$sin^{2} θ = \frac{1 - c o s 2 θ}{2}$	Required for integrating squared trigonometric terms
Power-reduction: $cos^{2} θ$	$cos^{2} θ = \frac{1 + c o s 2 θ}{2}$	Required for integrating squared trigonometric terms
X-axis symmetry for polar area	$A = 2 \cdot A (θ \in [0, π])$	Cuts work in half, reduces arithmetic error
Full trace interval for rose curves	$[0, π]$ for even $n$ , $[0, 2 π]$ for odd $n$	Prevents double-counting area for even $n$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · BC
FRQ polar area between two curves
2022 · BC MCQ
Set up polar area integral

What's Next

Mastering the area of regions bounded by two polar curves is a critical prerequisite for finding arc length of polar curves, the next major topic in Unit 9 of the AP Calculus BC CED. Arc length calculations for polar curves also rely on integrating squared trigonometric terms, so the power-reduction and integration techniques you practiced here transfer directly. Beyond Unit 9, this topic builds core integration fluency that is tested across all multiple-choice and free-response sections of the AP exam, and it lays the groundwork for double integrals in polar coordinates if you continue to multivariable calculus after AP. Correctly identifying intersection bounds and setting up area integrals correctly is foundational for all advanced polar applications.