To find the area between two polar curves, use $A=\frac{1}{2}\int_a^b (r_{\text{outer}}^2-r_{\text{inner}}^2)\, d\theta$ . The hard part is not the integral; it is sketching the curves, identifying which radius is outer and which is inner, and setting the correct $\theta$ bounds by solving $r_1(\theta)=r_2(\theta)$ .

Why This Matters for the AP Calculus Exam

This is a BC-only topic in Unit 9, which carries a meaningful share of the BC exam. Polar area shows up in both multiple-choice and free-response style problems, so you may be asked to set up an integral by hand on the no-calculator section or evaluate one on the calculator section. The skills that earn credit are choosing the right setup, identifying outer versus inner curves, and picking accurate bounds. Clear notation and a correct integral setup matter for showing your reasoning, even when a calculator handles the arithmetic.

This topic also reinforces earlier ideas. The area formula comes from the same Riemann sum thinking you used for area in rectangular coordinates, and it leans on trig identities, radian measure, and the area of a sector.

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Key Takeaways

The single-curve formula is $A=\frac{1}{2}\int_a^b r^2\, d\theta$ . For two curves, subtract the squared radii: $A=\frac{1}{2}\int_a^b (r_{outer}^2 - r_{inner}^2)\, d\theta$ .
You subtract the squares of the radii, not the difference of the radii squared. $(r_{outer}^2 - r_{inner}^2)$ is correct, $(r_{outer}-r_{inner})^2$ is wrong.
Find bounds by solving $r_1(\theta)=r_2(\theta)$ , then use the graph to keep only the angles that border your region.
Always work in radians, not degrees.
Sketch first. The picture tells you which curve is outer (farther from the origin) and which is inner for the interval you care about.
Expect to use trig identities like the double angle formulas to evaluate $r^2$ integrals on no-calculator questions.

Equation for Area: One Curve vs. Two Curves

The area bounded by a single polar curve (from Topic 9.8) is:

A=\frac{1}{2}\int_a^b r^2\, d\theta

Here $a$ and $b$ are the angle bounds and $r$ is the radius of the curve.

To handle the region between two curves, subtract the squared radii:

A=\frac{1}{2}\int_a^b (r_2^2 - r_1^2)\, d\theta

In this setup $r_2$ is the outer radius (farther from the origin) and $r_1$ is the inner radius. Your bounds are the angles where $r_1 = r_2$ .

For a review of the single-curve case, see Topic 9.8.

Worked Example: Area Between Two Polar Curves

Let $R$ be the region bounded between $\textcolor{red}{r=3}$ and $\textcolor{blue}{r = 3-2\sin(2\theta)}$ in the second quadrant. Find the area between the two functions.

1. Define your inner and outer radius

Decide which curve is inner and which is outer for this region. A quick trick: picture yourself standing at the origin and looking outward. The first curve you reach is the inner curve, and the next one is the outer curve.

\textcolor{red}{r_1 = 3} \\ \textcolor{blue}{r_2 = 3-2\sin(2\theta)}

In the second quadrant, the circle $r=3$ is reached first, so it is the inner curve. The other curve is the outer curve.

2. Find the bounds

In polar coordinates, the radius is the distance from the origin and $\theta$ is the angle measured from the positive $x$ -axis. Use radians, not degrees.

The problem asks for the second quadrant, so the angles run from $\frac{\pi}{2}$ to $\pi$ .

To confirm this systematically, set the radii equal and solve for $\theta$ :

3-2\sin(2\theta) = 3 \\ -2\sin(2\theta)=0 \\ \sin(2\theta) = 0 \\ 2\sin(\theta)\cos(\theta) = 0 \\ \theta = 0,\frac{\pi}{2},\pi,\frac{3\pi}{2}

Since you want the second quadrant, keep the bounds $\frac{\pi}{2}$ to $\pi$ .

3. Set up the integral and evaluate

With your bounds and radii chosen, plug into the formula:

A = \frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} \left[(3-2\sin(2\theta))^2 - (3)^2\right] d\theta

On a no-calculator question you would expand the squared term and use a trig identity to integrate. On a calculator question you can evaluate the definite integral directly. If your setup gives a negative value, that is a sign your outer and inner curves are swapped for this interval, so recheck which radius is larger here.

How to Use This on the AP Calculus Exam

Free Response

Sketch or read the graph carefully before writing anything. The sketch decides outer versus inner.
Write the integral with the correct bounds and the correct order, $r_{outer}^2 - r_{inner}^2$ . A clean setup shows your reasoning even if a calculator does the final number.
Keep the factor of $\frac{1}{2}$ out front. Dropping it is one of the most common point losses.

Problem Solving

Solve $r_1(\theta)=r_2(\theta)$ to find candidate bounds, then use the picture to pick the angles that actually border your region. Not every solution is a bound you use.
Use symmetry when you can. If a region is symmetric, integrate over half and double, which keeps the algebra cleaner.

Common Trap

If the outer and inner curves switch which one is farther from the origin partway through, split the integral at the angle where they cross and set up each piece separately.

Common Misconceptions

Subtracting radii instead of squared radii. The formula uses $r_{outer}^2 - r_{inner}^2$ , not $(r_{outer}-r_{inner})^2$ . These are not the same thing.
Forgetting the $\frac{1}{2}$ . Every polar area integral carries a factor of $\frac{1}{2}$ . Leaving it out doubles your answer.
Using degrees. Polar area integrals require radian measure. Plugging in degrees gives a wrong result.
Treating every intersection angle as a bound. Setting $r_1=r_2$ gives several solutions. Only the angles that border your specific region are the correct limits, so check against the graph.
Assuming the outer curve stays outer. Over a full interval, which curve is farther from the origin can change. When it does, split the integral so each piece uses the correct outer and inner radius.
Confusing this with rectangular area between curves. In rectangular coordinates you integrate $top - bottom$. In polar coordinates you integrate the difference of squared radii with a $\frac{1}{2}$ factor, because area is built from sectors, not vertical strips.

Vocabulary

The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.

Term	Definition
areas of regions	The measure of the two-dimensional space enclosed by one or more curves.
definite integral	The integral of a function over a specific interval [a, b], representing the net signed area between the curve and the x-axis.
polar curve	Curves defined by equations in polar coordinates, where points are located by a distance r from the origin and an angle θ from the positive x-axis.

Frequently Asked Questions

What formula do I use for the area between two polar curves?

Use A = (1/2)∫[r_outer(θ)^2 - r_inner(θ)^2] dθ over the angle interval that traces the region exactly once. The key is subtracting the squared radii, not subtracting the radii first.

How do I find the theta bounds for area between polar curves?

Set the two polar equations equal, solve r1(θ) = r2(θ), and then use the graph or the stated region to keep only the intersection angles that actually border the region. Some solutions may not be part of the region you need.

How do I know which polar curve is outer and which is inner?

For the interval you are using, the outer curve is the radius farther from the origin and the inner curve is closer to the origin. Check the graph or test a theta value in the interval. If the curves switch, split the integral.

Why is there a one-half in the polar area formula?

Polar area comes from sector area, not rectangular strips. A small sector has area about (1/2)r^2 dθ, so definite integrals for polar area include the factor of one-half.

What mistakes should I avoid on AP Calculus BC polar area problems?

Common mistakes include forgetting the one-half, using degrees instead of radians, using (r_outer - r_inner)^2, choosing every intersection as a bound, or failing to split the integral when outer and inner curves change.

Is finding area between two polar curves on AP Calculus AB or BC?

This is a BC-only Topic 9.9 skill. You should be able to calculate areas of regions defined by polar curves using definite integrals and clearly justify your setup on AP Calculus BC questions.