9.9 - Finding the Area of the Region Bounded by Two Polar Curves

Review

The area bounded by a single polar curve

r = f(\theta)

A = \frac{1}{2}\int_{\alpha}^{\beta} r^2 \, d\theta

Same reminders as before: keep the

\frac{1}{2}

, square the whole function, get the bounds right.

Area Common to Two Polar Curves

When two polar curves overlap, you have to think carefully about which curve actually forms the boundary at each angle.

The Rotating Line Method

Put a line at the origin pointing along the positive x-axis ( $\theta = 0$ )

Rotate it counterclockwise

At each angle, ask which curve the line hits first on its way out

Here’s the key rule: for the region common to both curves (inside both), the boundary at each angle is the curve closer to the origin, meaning the one with the smaller

r

. Whenever the two curves swap which one is closer, you split the integral there.

(This is the opposite of what your intuition might say. If you’re inside both regions, you can only go out as far as the nearer boundary lets you.)

Example 1: Two Overlapping Circles

Find the area common to

r = 3\sin\theta

and

r = 3\cos\theta

9.9 - Finding the Area of the Region Bounded by Two Polar Curves

Step 1: Find the intersection.

3\sin\theta = 3\cos\theta \quad \Rightarrow \quad \tan\theta = 1 \quad \Rightarrow \quad \theta = \frac{\pi}{4}

Both curves also pass through the origin:

r = 3\sin\theta

\theta = 0

, and

r = 3\cos\theta

\theta = \frac{\pi}{2}

Step 2: Which curve is closer to the origin?

From $\theta = 0$ to $\theta = \frac{\pi}{4}$ : $\sin\theta < \cos\theta$ , so $r = 3\sin\theta$ is the inner curve

From $\theta = \frac{\pi}{4}$ to $\theta = \frac{\pi}{2}$ : $r = 3\cos\theta$ is the inner curve

Step 3: Set up. Use the inner curve on each piece.

A = \frac{1}{2}\int_{0}^{\pi/4} 9\sin^2\theta \, d\theta + \frac{1}{2}\int_{\pi/4}^{\pi/2} 9\cos^2\theta \, d\theta

By symmetry the two pieces are equal, so

A = 2 \cdot \frac{9}{2}\left(\frac{\pi}{8} - \frac{1}{4}\right) = 9\left(\frac{\pi}{8} - \frac{1}{4}\right) \approx 1.284

Sanity check: each circle has radius

1.5

and area

\approx 7.07

. An overlap of

1.284

is a reasonable fraction of that.

Using Symmetry to Simplify

If the region is symmetric, compute one piece and multiply.

Example 2: Circle and Cardioid

Find the area inside both

r = 4

and

r = 4(1 + \cos\theta)

Step 1: Find the intersections.

4 = 4(1 + \cos\theta) \quad \Rightarrow \quad \cos\theta = 0 \quad \Rightarrow \quad \theta = \frac{\pi}{2}, \frac{3\pi}{2}

Step 2: Which curve is inner? The cardioid

4(1 + \cos\theta)

is bigger than 4 exactly when

\cos\theta > 0

On $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ : the cardioid is farther out, so the circle $r = 4$ is the boundary

On $\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$ : the cardioid is the boundary

Step 3: Set up.

A = \frac{1}{2}\int_{-\pi/2}^{\pi/2} 4^2 \, d\theta + \frac{1}{2}\int_{\pi/2}^{3\pi/2} \left[4(1 + \cos\theta)\right]^2 \, d\theta

First piece:

\frac{1}{2}(16)(\pi) = 8\pi \approx 25.133

Second piece:

\approx 5.699

A \approx 30.83 \text{ square units}

Subtracting a Gap: Inside One, Outside Another

To find the area inside curve

r_1

but outside curve

r_2

(where

r_1 > r_2

on the interval):

A = \frac{1}{2}\int_{\alpha}^{\beta} \left[r_1^2 - r_2^2\right] \, d\theta

Equivalently:

A = \frac{1}{2}\int_{\alpha}^{\beta} r_1^2 \, d\theta - \frac{1}{2}\int_{\alpha}^{\beta} r_2^2 \, d\theta

Critical: each radius gets squared individually. Never write

(r_1 - r_2)^2

Example 3: Limaçon Outside a Circle

Find the area inside

r = 4 + 2\cos\theta

and outside

r = 3

Step 1: Intersections.

4 + 2\cos\theta = 3 \quad \Rightarrow \quad \cos\theta = -\frac{1}{2} \quad \Rightarrow \quad \theta = \frac{2\pi}{3}, \frac{4\pi}{3}

Step 2: Where is the limaçon outside the circle? Wherever

\cos\theta > -\frac{1}{2}

, meaning

-\frac{2\pi}{3} < \theta < \frac{2\pi}{3}

Step 3: Set up, using symmetry about the x-axis.

A = \frac{1}{2}\int_{-2\pi/3}^{2\pi/3} \left[(4 + 2\cos\theta)^2 - 3^2\right] d\theta = \int_{0}^{2\pi/3} \left[(4 + 2\cos\theta)^2 - 9\right] d\theta

Expand the bracket:

(4 + 2\cos\theta)^2 - 9 = 9 + 16\cos\theta + 2\cos(2\theta)

A = \left[9\theta + 16\sin\theta + \sin(2\theta)\right]_0^{2\pi/3} \approx 31.84 \text{ square units}

Example 4: Circle Outside a Cardioid

Find the area inside

r = 2

and outside

r = 2(1 - \cos\theta)

Step 1: Intersections.

2 = 2(1 - \cos\theta) \quad \Rightarrow \quad \cos\theta = 0 \quad \Rightarrow \quad \theta = \pm\frac{\pi}{2}

Step 2: Where is the circle outside the cardioid? Wherever

\cos\theta > 0

, meaning

-\frac{\pi}{2} < \theta < \frac{\pi}{2}

Step 3: Set up with symmetry.

A = \frac{1}{2}\int_{-\pi/2}^{\pi/2} \left[2^2 - \left(2(1-\cos\theta)\right)^2\right] d\theta = \int_{0}^{\pi/2} \left[4 - 4(1-\cos\theta)^2\right] d\theta

\approx 4.858 \text{ square units}

Example 5: Cardioid Outside a Circle

Find the area inside

r = 2 + 2\sin\theta

and outside

r = 2

Step 1: Intersections.

2 + 2\sin\theta = 2 \quad \Rightarrow \quad \sin\theta = 0 \quad \Rightarrow \quad \theta = 0, \pi

Step 2: Where is the cardioid outside the circle? Wherever

\sin\theta > 0

, meaning

0 < \theta < \pi

Step 3: Set up.

A = \frac{1}{2}\int_{0}^{\pi} \left[(2 + 2\sin\theta)^2 - 2^2\right] d\theta = \frac{1}{2}\int_{0}^{\pi} \left[8\sin\theta + 4\sin^2\theta\right] d\theta

= \frac{1}{2}\left[16 + 2\pi\right] = 8 + \pi \approx 11.14 \text{ square units}

Common Mistakes

Squaring the difference instead of differencing the squares. Wrong:

\frac{1}{2}\int (r_1 - r_2)^2 d\theta

. Right:

\frac{1}{2}\int (r_1^2 - r_2^2) d\theta

Using the outer curve for a “common to both” region. For the region inside both curves, the boundary is the nearer curve, not the farther one.

Wrong bounds. Always verify with a graph, or trace which curve is the boundary at each angle.

Forgetting symmetry. If you computed half, remember to double it. Or cancel the

\frac{1}{2}

against the factor of 2 right away.

Not finding all the intersections. Set

r_1 = r_2

, but also check where each curve hits the origin (

r = 0

Step-by-Step Process

Graph both curves and look at the region

Find the intersections by setting $r_1 = r_2$

Check for symmetry, since it can halve your work

Use the rotating line method: sweep counterclockwise and figure out which curve forms the boundary in each angular interval, splitting the integral where it changes

Set up the integrals:

For a region bounded by one curve: $\frac{1}{2}\int r^2\,d\theta$

For “inside one, outside another”: $\frac{1}{2}\int (r_{\text{outer}}^2 - r_{\text{inner}}^2)\,d\theta$

Evaluate

Apply the symmetry factor if you used one

Multiple Choice Strategy

On the AP exam you often just have to pick the correct integral setup. Check:

Is the $\frac{1}{2}$ there?

Are the radii squared individually, not as a difference?

Are $\alpha$ and $\beta$ right for the shaded region?

Does the integral switch at the intersection points?

Is symmetry handled correctly?

Key Takeaways

Area between polar curves often needs multiple integrals with different $r$ functions

Rotating line method: sweep from the origin and see which curve bounds the region

For a region common to both curves, the boundary is the curve with the smaller $r$

Find intersections by setting $r_1 = r_2$ , and also check where each $r = 0$

Symmetry saves work: compute one piece and multiply

Gap formula: $A = \frac{1}{2}\int_{\alpha}^{\beta} (r_1^2 - r_2^2)\,d\theta$

Square individually, never $(r_1 - r_2)^2$

Verify bounds with technology

Don’t forget the $\frac{1}{2}$

Some curves don’t start at $\theta = 0$ where you’d expect

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