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AP Calculus BC — Unit 9.9

Finding the area of the region bounded by two polar curves

Key Concepts Packet + Worked Examples + Interactive MCQ Practice (36 Questions)

Unit 9.9 Key Concepts Packet
1) Area Between Two Polar Curves: Core Formula
Must-know setup
If the region is bounded by two polar curves \(r_{\text{outer}}(\theta)\) and \(r_{\text{inner}}(\theta)\) on \([\alpha,\beta]\) (and the region is traced once), then:
\( \displaystyle A=\frac12\int_{\alpha}^{\beta}\Big(r_{\text{outer}}(\theta)^2-r_{\text{inner}}(\theta)^2\Big)\,d\theta \).
  • It is the polar version of “washer method”: outer radius squared minus inner radius squared.
  • The bounds must match the region once (avoid double-counting).
2) Finding Intersection Angles
Intersection rule (most common)
To find where two polar curves intersect, you typically solve \(r_1(\theta)=r_2(\theta)\) using the same angle \(\theta\).
  • After solving, confirm the intersection actually lies on the region you want (due to possible retracing or negative radius representations).
  • Those intersection angles usually become your integration bounds.
3) Choosing the Outer vs. Inner Curve (and Switching)
Outer/inner can change
Sometimes one curve is outer on part of the interval, but the other curve becomes outer on another part. In that case, you must split the integral where the “outer” curve switches.
  • Use a quick test angle in each sub-interval (plug in \(\theta\) to compare \(r_1\) and \(r_2\)).
  • When switching happens, you do piecewise area: \( \frac12\int (r_{\text{outer}}^2-r_{\text{inner}}^2)\,d\theta \) on each piece.
4) Using Symmetry Correctly
  • If the region is symmetric (about the \(x\)-axis, \(y\)-axis, or the pole), you can compute a partial area and multiply.
  • Symmetry is only valid if your chosen interval traces a non-overlapping portion of the region.
5) Negative Radius and Tracing Issues
Key identity
\((r,\theta)\equiv(-r,\theta+\pi)\).
  • Even though the integrand uses squares, negative radius can change how the curve is traced and where intersections “really” occur.
  • Always use bounds that match the region you intend (not just a random full period).
6) Calculator Expectations + Common AP Traps
  • AP may require a numeric value for \( \frac12\int (r_{\text{outer}}^2-r_{\text{inner}}^2)\,d\theta \).
  • Trap: forgetting the \(\frac12\).
  • Trap: mixing up which curve is outer (or failing to split when it switches).
  • Trap: using \( \int (r_{\text{outer}}-r_{\text{inner}})\,d\theta \) (wrong—must square).
  • Important: If you ever see raw backslashes instead of formatted math, it means MathJax did not typeset that section—this page forces typesetting on load and when solutions open.
Worked Examples + Notes

Example 1 — Between a circle and a line in polar

Find the area of the region bounded by \(r=2\cos\theta\) and \(r=1\).

Step 1: Find intersections: \(2\cos\theta=1\Rightarrow \cos\theta=\frac12\Rightarrow \theta=\pm \frac{\pi}{3}\).
Step 2: On \(\left[-\frac{\pi}{3},\frac{\pi}{3}\right]\), \(2\cos\theta\ge 1\), so outer is \(r_{\text{outer}}=2\cos\theta\), inner is \(r_{\text{inner}}=1\).
Step 3: Area setup: \( \displaystyle A=\frac12\int_{-\pi/3}^{\pi/3}\Big((2\cos\theta)^2-1^2\Big)\,d\theta \).
Conclusion: Correct area between two curves uses “outer squared minus inner squared,” with intersection angles as bounds.

Example 2 — One curve is always outside

Find the area between \(r=2\sin\theta\) and \(r=\sin\theta\) on \([0,\pi]\).

Step 1: On \([0,\pi]\), \(\sin\theta\ge 0\) and \(2\sin\theta\ge \sin\theta\), so outer is \(2\sin\theta\).
Step 2: Area setup: \( \displaystyle A=\frac12\int_0^{\pi}\Big((2\sin\theta)^2-(\sin\theta)^2\Big)\,d\theta \).
Conclusion: If outer/inner never switch, you do one integral with the given bounds.

Example 3 — When the outer curve switches

Consider \(r_1=1+\cos\theta\) and \(r_2=1-\cos\theta\). Describe how to set up the area between them over \([0,2\pi]\).

Step 1: Intersections: \(1+\cos\theta=1-\cos\theta\Rightarrow \cos\theta=0\Rightarrow \theta=\frac{\pi}{2},\frac{3\pi}{2}\).
Step 2: On \((-\frac{\pi}{2},\frac{\pi}{2})\), \(\cos\theta>0\) so \(1+\cos\theta\) is outer; on \((\frac{\pi}{2},\frac{3\pi}{2})\), \(\cos\theta<0\) so \(1-\cos\theta\) is outer.
Step 3: Piecewise setup: \( \displaystyle A=\frac12\int_0^{\pi/2}\big((1+\cos\theta)^2-(1-\cos\theta)^2\big)\,d\theta+\frac12\int_{\pi/2}^{3\pi/2}\big((1-\cos\theta)^2-(1+\cos\theta)^2\big)\,d\theta+\frac12\int_{3\pi/2}^{2\pi}\big((1+\cos\theta)^2-(1-\cos\theta)^2\big)\,d\theta \).
Conclusion: Switching means you split the integral at intersection angles.

Example 4 — Common mistake check

A student sets up \( \int_{\alpha}^{\beta}(r_{\text{outer}}-r_{\text{inner}})\,d\theta \) for polar area between two curves. What is wrong?

Step 1: Polar area comes from sector area \( \frac12 r^2\,d\theta \), so radii must be squared.
Conclusion: Correct is \( \frac12\int (r_{\text{outer}}^2-r_{\text{inner}}^2)\,d\theta \).
Note: This page forces MathJax typesetting on load and when solution panels open.
Unit 9.9 Multiple-Choice Practice (36 Questions)
Answer Key