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AP® Calculus AB-BC

Area Between Two Polar Curves: AP® Calculus AB–BC Review

area between two polar curves

Polar area problems appear again and again on the AP® Calculus exams because they mix three tested skills: graph reading, integration, and limit setting. Unlike Cartesian area questions, these items use circles and spirals instead of rectangles. Therefore, students must think in angles, not x-values. This post shows how to use the formula for area between polar curves, set correct bounds, and sidestep common mistakes.

Quick Polar Coordinates Refresher

Key Ideas

  • Pole = origin
  • Polar axis = positive x-axis
  • Radius r = directed distance from the pole
  • Angle \theta = counter-clockwise measure from the polar axis

Sketching Polar Graphs in Seconds

Symmetry checks save valuable time:

TestHowResult if symmetric
Polar axisReplace \theta by -\thetaMirror across x-axis
y-axisReplace \theta by \pi-\thetaMirror across y-axis
OriginReplace r by -r or \theta by \theta+\pi180° rotation

Choose a few key angles, plot points, and connect smoothly.

Mini-Example

Graph r = 2 + 2\cos\theta (a cardioid).

1. Test symmetry: replacing \theta by -\theta leaves r unchanged; therefore the graph is symmetric about the polar axis.

2. Table of values:

\theta0\pi/2\pi
r420

3. Plot and draw a heart-shaped curve opening right.

Area of a Single Polar Region

Why the Formula Works

Think of carving a pizza. Each tiny slice has area approximately \tfrac12 r^{2}\Delta\theta. Summing and taking the limit forms the definite integral:

\text{Area} = \tfrac12\int_{\alpha}^{\beta} r^{2}d\theta

Single-Curve Example

Find the area of one petal of r = 3\sin 2\theta.

1. A petal ends where r=0. Solve 3\sin 2\theta = 0 to get \theta = 0 and \theta = \tfrac{\pi}{2}.

2. Because the graph is four-petaled and symmetric, one petal spans 0 to \tfrac{\pi}{2}.

3. Apply the formula:

\text{Area} = \tfrac12\int_{0}^{\pi/2} (3\sin 2\theta)^{2}d\theta = \tfrac12\int_{0}^{\pi/2} 9\sin^{2} 2\theta d\theta

Using the identity \sin^{2}x = \tfrac12(1 - \cos 2x) and evaluating produces \tfrac{9\pi}{8}.

Area Between Two Polar Curves

The Main Formula

\text{Area} = \tfrac12\int_{\alpha}^{\beta}\left(r_{\text{outer}}^{2} - r_{\text{inner}}^{2}\right)d\theta

Visualizing “Outer” vs. “Inner”

At a chosen angle \theta, two radii are drawn. The longer one is outer; the shorter one is inner. The integral subtracts inner area from outer area slice by slice.

Four-Step Checklist

  1. Sketch or trace both curves.
  2. Locate intersection angles by solving r_{1}=r_{2}.
  3. Choose correct \theta bounds; sometimes separate integrals are needed.
  4. Integrate r_{\text{outer}}^{2}-r_{\text{inner}}^{2} and multiply by \tfrac12.

Worked Example #1 – Cardioid vs. Circle

Find the area between r = 2(1+\cos\theta) and the circle r=2.

Step 1: Graph

The cardioid envelopes the circle on the right side, yet the circle sticks out on the left. A quick sketch shows that from -\tfrac{\pi}{2} to \tfrac{\pi}{2} the cardioid is outside.

Step 2: Intersection Angles

Solve 2(1+\cos\theta) = 21+\cos\theta=1\cos\theta=0\theta=\pm\tfrac{\pi}{2}.

Step 3: Set Up Integral

Outer radius: r_{\text{outer}} = 2(1+\cos\theta).

Inner radius: r_{\text{inner}} = 2.

\text{Area} = \tfrac12\int_{-\pi/2}^{\pi/2}\left([2(1+\cos\theta)]^{2} - (2)^{2}\right)d\theta

Step 4: Integrate

Expand: [2(1+\cos\theta)]^{2} = 4(1+\cos\theta)^{2} = 4(1 + 2\cos\theta + \cos^{2}\theta).

So

\text{Area} = \tfrac12\int_{-\pi/2}^{\pi/2}\bigl(4 + 8\cos\theta + 4\cos^{2}\theta - 4\bigr)d\theta= \tfrac12\int_{-\pi/2}^{\pi/2}\bigl(8\cos\theta + 4\cos^{2}\theta\bigr)d\theta

Because the interval is symmetric and \cos\theta is even, double the 0→π/2 integral:

\text{Area} = \int_{0}^{\pi/2}\bigl(8\cos\theta + 4\cos^{2}\theta\bigr)d\theta

Now integrate:

  • \int 8\cos\theta d\theta = 8\sin\theta
  • Rewrite \cos^{2}\theta = \tfrac12(1+\cos 2\theta).

Therefore,

\text{Area} = \Bigl[8\sin\theta + 4\cdot\tfrac12(\theta+\tfrac12\sin 2\theta)\Bigr]_{0}^{\pi/2} = \Bigl[8\sin\theta + 2\theta + \sin 2\theta\Bigr]_{0}^{\pi/2}

Plugging in gives 8(1) + 2(\pi/2) + 0 - (0 + 0 + 0) = 8 + \pi.

So the area between the two curves equals 8 + \pi square units.

Using Symmetry & Graphing Tricks

Recognizing Symmetry

If both curves share symmetry, halve or quarter the integral, then multiply back. This move cuts time and algebra errors.

Leveraging Technology

Graphing calculators show where curves overlap. However, always list the intersection algebraically because free-response rubrics award setup points.

Example #2 – Half-Petal Region

Find the shaded half-petal area between r = 4\sin\theta (outer) and r = 4\sin 2\theta (inner) above the polar axis.

1. Intersections: solve 4\sin\theta = 4\sin 2\theta\sin\theta = 2\sin\theta\cos\theta\sin\theta (1-2\cos\theta)=0. So \theta=0 or \cos\theta=\tfrac12\theta=\tfrac{\pi}{3}.

2. Bounds: 0 to \tfrac{\pi}{3}.

3. Integral:

\text{Area} = \tfrac12\int_{0}^{\pi/3}\left((4\sin2\theta)^{2} - (4\sin \theta)^{2}\right)d\theta

4. Simplify and integrate.

Common Pitfalls and Pro Tips

  • Forgetting to square radii—always integrate r^{2}.
  • Mixing up outer and inner when curves cross several times; therefore sketch first.
  • Overlapping regions sometimes need piecewise integrals; check if one curve switches roles.

Quick Reference Vocabulary Chart

TermConcise Definition
Polar axisReference ray (positive x-axis)
PoleOrigin in polar coordinates
Radius rDistance from pole to point
Angle \thetaRotation from polar axis
Polar curveEquation relating r and \theta
Intersection angle\theta where two curves meet
Outer radiusLarger r at a given \theta
Inner radiusSmaller r at a given \theta
Definite integralSigned area under a curve
Area between two curve polarRegion bounded by two polar graphs
Formula for area between polar curves\tfrac12\int(r_{\text{outer}}^{2}-r_{\text{inner}}^{2})d\theta

Conclusion

Successfully finding the area between two polar curves hinges on one clean strategy: sketch, label outer and inner radii, set the right intersection bounds, then apply the formula for area between polar curves. Consequently, always square radii and watch for symmetry to shorten computations. Practicing a variety of polar area problems will build confidence and secure valuable AP® points.

Sharpen Your Skills for AP® Calculus AB-BC

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