Unit 9.8 – Area of Polar Regions BC ONLY

AP® Calculus BC | Finding Areas in Polar Coordinates

Why This Matters: Area in polar coordinates uses a completely different approach than rectangular coordinates! Instead of rectangles, we use circular sectors (pie slices). This topic is essential for finding areas enclosed by polar curves like circles, cardioids, and rose petals. This appears on virtually every BC exam!

📐 The Polar Area Formula

Area of a Polar Region

THE FUNDAMENTAL FORMULA:

For a curve \(r = f(\theta)\) from \(\theta = \alpha\) to \(\theta = \beta\):

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

\[ A = \frac{1}{2}\int_\alpha^\beta [f(\theta)]^2 \, d\theta \]

Why This Formula?

The area is found by summing infinitesimal sectors (pie slices). Each sector has area:

\[ dA = \frac{1}{2}r^2 \, d\theta \]

This is half the sector formula from geometry!

📝 Critical: Don't forget the \(\frac{1}{2}\) factor! This is THE most common error in polar area problems.

🔄 Area Between Two Polar Curves

Area Between Two Curves

THE FORMULA:

For curves \(r = f(\theta)\) (outer) and \(r = g(\theta)\) (inner):

\[ A = \frac{1}{2}\int_\alpha^\beta \left([r_{\text{outer}}]^2 - [r_{\text{inner}}]^2\right) \, d\theta \]

\[ A = \frac{1}{2}\int_\alpha^\beta \left([f(\theta)]^2 - [g(\theta)]^2\right) \, d\theta \]

Key Points:

Square EACH function before subtracting
NOT \((r_{\text{outer}} - r_{\text{inner}})^2\)!
Outer curve is the one farther from the origin
Find intersection points to determine bounds

🔁 Using Symmetry to Simplify

Exploiting Symmetry:

If the curve is symmetric:

About the x-axis: Calculate for \(\theta \in [0, \pi]\) and multiply by 2
About the y-axis: Calculate for \(\theta \in [0, \pi/2]\) and multiply by 2
About the origin: Calculate for one petal/loop and multiply accordingly

💡 Pro Tip: For rose curves and cardioids, use symmetry to avoid integration over negative \(r\) values!

📖 Comprehensive Worked Examples

Example 1: Area of a Circle

Problem: Find the area enclosed by \(r = 2\cos\theta\).

Solution:

Step 1: Identify the curve

This is a circle through the origin. It traces out from \(\theta = -\pi/2\) to \(\theta = \pi/2\).

Using symmetry, integrate from 0 to \(\pi/2\) and multiply by 2.

Step 2: Set up integral

\[ A = 2 \cdot \frac{1}{2}\int_0^{\pi/2} (2\cos\theta)^2 \, d\theta = \int_0^{\pi/2} 4\cos^2\theta \, d\theta \]

Step 3: Use identity \(\cos^2\theta = \frac{1 + \cos 2\theta}{2}\)

\[ A = \int_0^{\pi/2} 4 \cdot \frac{1 + \cos 2\theta}{2} \, d\theta = \int_0^{\pi/2} (2 + 2\cos 2\theta) \, d\theta \]

\[ = \left[2\theta + \sin 2\theta\right]_0^{\pi/2} = \pi + 0 - 0 - 0 = \pi \]

ANSWER: \(A = \pi\) (This is a circle of radius 1!)

Example 2: Area of a Cardioid

Problem: Find the area enclosed by \(r = 1 + \cos\theta\).

Setup:

A cardioid traces from \(\theta = 0\) to \(\theta = 2\pi\).

\[ A = \frac{1}{2}\int_0^{2\pi} (1 + \cos\theta)^2 \, d\theta \]

Expand:

\[ A = \frac{1}{2}\int_0^{2\pi} (1 + 2\cos\theta + \cos^2\theta) \, d\theta \]

\[ = \frac{1}{2}\int_0^{2\pi} \left(1 + 2\cos\theta + \frac{1 + \cos 2\theta}{2}\right) \, d\theta \]

\[ = \frac{1}{2}\int_0^{2\pi} \left(\frac{3}{2} + 2\cos\theta + \frac{\cos 2\theta}{2}\right) \, d\theta \]

Integrate:

\[ A = \frac{1}{2}\left[\frac{3\theta}{2} + 2\sin\theta + \frac{\sin 2\theta}{4}\right]_0^{2\pi} = \frac{1}{2} \cdot 3\pi = \frac{3\pi}{2} \]

Example 3: Area of One Petal of a Rose

Problem: Find the area of one petal of \(r = 3\sin(2\theta)\).

Find one petal:

One petal is traced when \(2\theta\) goes from 0 to \(\pi\), so \(\theta\) goes from 0 to \(\pi/2\).

\[ A = \frac{1}{2}\int_0^{\pi/2} [3\sin(2\theta)]^2 \, d\theta = \frac{9}{2}\int_0^{\pi/2} \sin^2(2\theta) \, d\theta \]

Use identity:

\[ \sin^2(2\theta) = \frac{1 - \cos(4\theta)}{2} \]

\[ A = \frac{9}{2}\int_0^{\pi/2} \frac{1 - \cos(4\theta)}{2} \, d\theta = \frac{9}{4}\left[\theta - \frac{\sin(4\theta)}{4}\right]_0^{\pi/2} = \frac{9\pi}{8} \]

Example 4: Area Between Two Curves

Problem: Find the area inside \(r = 2 + 2\cos\theta\) and outside \(r = 2\).

Find intersections:

\[ 2 + 2\cos\theta = 2 \quad \Rightarrow \quad \cos\theta = 0 \quad \Rightarrow \quad \theta = \pm\frac{\pi}{2} \]

Set up integral (using symmetry):

\[ A = 2 \cdot \frac{1}{2}\int_0^{\pi/2} [(2+2\cos\theta)^2 - 2^2] \, d\theta \]

\[ = \int_0^{\pi/2} [(4 + 8\cos\theta + 4\cos^2\theta) - 4] \, d\theta \]

\[ = \int_0^{\pi/2} (8\cos\theta + 4\cos^2\theta) \, d\theta \]

(Evaluate using \(\cos^2\theta\) identity)

📊 Common Polar Curve Areas

Famous Polar Curve Areas
Curve	Equation	Total Area
Circle	\(r = a\)	\(\pi a^2\)
Cardioid	\(r = a(1 \pm \cos\theta)\)	\(\frac{3\pi a^2}{2}\)
Rose (n odd)	\(r = a\cos(n\theta)\)	\(\frac{n\pi a^2}{4}\) (all petals)
Rose (n even)	\(r = a\cos(n\theta)\)	\(\frac{n\pi a^2}{2}\) (all petals)
One Rose Petal	\(r = a\sin(n\theta)\)	\(\frac{\pi a^2}{4n}\)

💡 Essential Tips & Strategies

✅ Success Strategies:

NEVER forget the \(\frac{1}{2}\): This is the #1 mistake!
Square the function: \(r^2\), not \(r\)
Use symmetry: Reduces calculation and errors
Find intersection points: For bounds and between curves
Sketch the curve: Helps visualize the region
For \(\cos^2\) or \(\sin^2\): Use power-reduction formulas
Check which curve is outer: Compare \(r\) values
Calculator allowed: Often for these integrals

🔥 Power-Reduction Formulas (MEMORIZE!):

\(\cos^2\theta = \frac{1 + \cos(2\theta)}{2}\)
\(\sin^2\theta = \frac{1 - \cos(2\theta)}{2}\)
\(\sin\theta\cos\theta = \frac{\sin(2\theta)}{2}\)

❌ Common Mistakes to Avoid

Mistake 1: Forgetting the \(\frac{1}{2}\) factor (MOST COMMON!)
Mistake 2: Not squaring \(r\) before integrating
Mistake 3: Using \((r_{\text{outer}} - r_{\text{inner}})^2\) instead of \(r_{\text{outer}}^2 - r_{\text{inner}}^2\)
Mistake 4: Wrong bounds (not finding intersection points)
Mistake 5: Not using power-reduction formulas for \(\sin^2\) or \(\cos^2\)
Mistake 6: Forgetting to expand \((a + b\cos\theta)^2\)
Mistake 7: Calculator in degree mode instead of radians
Mistake 8: Counting area twice (not recognizing symmetry)
Mistake 9: Integration errors (especially with trig)
Mistake 10: Not checking which curve is outer

📝 Practice Problems

Find the area:

Inside \(r = 3\sin\theta\)
Inside \(r = 2 + 2\sin\theta\) (cardioid)
One petal of \(r = 4\cos(3\theta)\)
Inside \(r = 4\) and outside \(r = 2\)

Answers:

\(\frac{9\pi}{4}\)
\(6\pi\)
\(\frac{4\pi}{9}\)
\(12\pi\)

✏️ AP® Exam Success Tips

What AP® Graders Look For:

Write the formula: \(A = \frac{1}{2}\int r^2\,d\theta\) with the \(\frac{1}{2}\)!
Show squaring: Write \([f(\theta)]^2\) explicitly
State bounds clearly: \(\int_\alpha^\beta\)
For between curves: Show both squared terms
Show expansion: When squaring binomials
Use symmetry justification: If doubling an integral
Show integration work: Or state using calculator
Simplify answer: Leave in exact form with \(\pi\)

💯 Exam Strategy:

Identify the curve type
Determine bounds (use symmetry if possible)
Write \(A = \frac{1}{2}\int_\alpha^\beta r^2\,d\theta\) with THE \(\frac{1}{2}\)
Square the function: \([f(\theta)]^2\)
Expand if needed
Use power-reduction formulas
Integrate
Simplify and include \(\pi\)

⚡ Quick Reference Guide

POLAR AREA ESSENTIALS

Single Curve:

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

Between Two Curves:

\[ A = \frac{1}{2}\int_\alpha^\beta (r_{\text{outer}}^2 - r_{\text{inner}}^2) \, d\theta \]

Power-Reduction:

\(\cos^2\theta = \frac{1 + \cos(2\theta)}{2}\)
\(\sin^2\theta = \frac{1 - \cos(2\theta)}{2}\)

Remember:

NEVER forget the \(\frac{1}{2}\)!
Square \(r\), not just \(r\)
Use symmetry to simplify

Master Polar Area! The fundamental formula: \(A = \frac{1}{2}\int_\alpha^\beta r^2\,d\theta\) where the \(\frac{1}{2}\) comes from the sector area formula. For curve \(r = f(\theta)\), substitute and square: \(\frac{1}{2}\int[f(\theta)]^2\,d\theta\). Between two curves: \(A = \frac{1}{2}\int(r_{\text{outer}}^2 - r_{\text{inner}}^2)\,d\theta\)—square EACH function then subtract. Power-reduction formulas essential: \(\cos^2\theta = \frac{1+\cos(2\theta)}{2}\), \(\sin^2\theta = \frac{1-\cos(2\theta)}{2}\). Use symmetry to simplify: integrate over symmetric portion and multiply. Common curves: cardioid area = \(\frac{3\pi a^2}{2}\), rose petal = \(\frac{\pi a^2}{4n}\). Most common error: FORGETTING THE \(\frac{1}{2}\)! Second most common: not squaring \(r\). Third: using \((r_1-r_2)^2\) instead of \(r_1^2-r_2^2\). This is guaranteed BC content—appears every year! Practice until the \(\frac{1}{2}\) becomes automatic! 🎯✨