Slicing the Sector Slides SLICING THE SECTOR
Introduction to Polar Area
Unit: Polar Integration
The "Pizza Slice" Problem
In Rectangular Calculus, we accumulate area using rectangles of width \( dx \).
In Polar Calculus, the radius sweeps out sectors (wedges) of angle \( d\theta \).
How do we sum an infinite number of these wedges to find the total area?
RECTANGULAR \( (dx) \) POLAR \( (d\theta) \)
The Geometry of a Sector
The area of a circle is \( A = \pi r^2 \).
A sector is a fraction of a circle: \[ \text{Area} = \left( \frac{\theta}{2\pi} \right) \pi r^2 \]
Simplifying the formula: \[ \text{Area} = \frac{1}{2} r^2 \theta \]
\( r \)
\( \theta \)
One slice = \(\frac{1}{2} r^2 \theta\)
Summing to Infinity
1. Define the tiny angle
\( d\theta \)
2. Define the tiny area
\( dA = \frac{1}{2} r^2 d\theta \)
3. Accumulate!
\( \int dA \)
The Polar Area Formula
\[ A = \frac{1}{2} \int_{\alpha}^{\beta} [r(\theta)]^2 \, d\theta \]
Crucial Step: Finding \(\alpha\) and \(\beta\)
1
Visual Setup
Trace the curve from start to finish. In which direction is \(\theta\) moving?
2
The "Sweep"
The area is always "swept out" from the pole (origin) out to the curve.
3
Algebraic Check
Set \( r = 0 \) to find where the loop begins and ends at the origin.
Think of \(\theta\) as a radar arm sweeping through a region.
Slicing the Sector Worksheet Slicing the Sector
Lesson 1: Introduction to Polar Integration
Name: ____________________
Date: ____________________
Part 1: The Pizza Slice Inquiry
Imagine a large pizza with a radius of \( r = 8 \) inches. You want to calculate the area of exactly one slice that spans an angle of \( \pi/4 \) radians.
Geometric Approach:
Area of a full circle = \( \pi r^2 \)
1. What fraction of a full circle (\( 2\pi \) radians) is a slice of \( \pi/4 \) radians?
2. Calculate the exact area of this slice using basic geometry.
The "Wedge" Concept:
In calculus, we think of this slice as a "sector" with area \( A = \frac{1}{2} r^2 \theta \).
3. Use the formula \( A = \frac{1}{2} r^2 \theta \) with \( r=8 \) and \( \theta = \pi/4 \). Do you get the same answer as above?
Part 2: From Sum to Integral
If a polar curve's radius changes as the angle \( \theta \) changes (\( r = f(\theta) \)), we can't just use one big sector. We must sum up an infinite number of infinitesimally small sectors with angle \( d\theta \).
The Polar Area Formula
\[ A = \frac{1}{2} \int_{\alpha}^{\beta} [r(\theta)]^2 \, d\theta \]
Where \( \alpha \) and \( \beta \) are the start and end angles of the region.
Check for Understanding:
4. Why is there a \( \frac{1}{2} \) in front of the integral? Base your answer on the geometry of a sector.
5. If we were calculating the area of a circle \( r = 4 \) from \( \theta = 0 \) to \( \theta = 2\pi \), set up the integral. Then evaluate it to verify it equals \( 16\pi \).
Setup:
Evaluation:
Part 3: Basic Practice
Set up (but do not solve) the integrals for the area of the following regions:
Region A: A semi-circle with radius \( r = 3 + \cos(\theta) \) for \( 0 \leq \theta \leq \pi \)
Region B: The region enclosed by \( r = \theta \) for \( 0 \leq \theta \leq \pi \)
Challenge Question:
Suppose a curve is defined as \( r = 2\sin(\theta) \). This curve forms a circle. What interval of \( \theta \) should you use to calculate the area of the entire circle exactly once? Explain your reasoning.
Polar Integration Answer Key Answer Key & Teacher Reference
Sequence: Polar Integration Expedition
Teacher Resource
Lesson 1: Slicing the Sector
1. Fraction: \( \frac{\pi/4}{2\pi} = \frac{1}{8} \)
2. Area (Geometry): \( \frac{1}{8} \pi (8^2) = 8\pi \approx 25.13 \text{ in}^2 \)
3. Formula: \( A = \frac{1}{2}(8^2)(\pi/4) = 8\pi \). Yes, they match.
4. Why 1/2? The sector area formula is derived from the circle area \( \pi r^2 \). Since a sector is \( \frac{\theta}{2\pi} \) of a circle, the area is \( \frac{\theta}{2\pi} \cdot \pi r^2 = \frac{1}{2} r^2 \theta \).
5. Setup/Eval: \( \frac{1}{2} \int_0^{2\pi} 4^2 d\theta = \frac{1}{2} [16\theta]_0^{2\pi} = 16\pi \).
Region A: \( A = \frac{1}{2} \int_0^{\pi} (3 + \cos\theta)^2 d\theta \)
Region B: \( A = \frac{1}{2} \int_0^{\pi} \theta^2 d\theta \)
Challenge: \( [0, \pi] \). Since \( \sin\theta \) is positive for \( 0 \to \pi \), the circle is completed once. If you go to \( 2\pi \), you trace the circle twice.
Lesson 2: The One Petal Problem
1a. Zeros: \( 4\sin(3\theta) = 0 \implies 3\theta = 0, \pi \implies \theta = 0, \pi/3 \).
1b. Interval: \( [0, \pi/3] \)
2a. Cardioid Pole: Yes, at \( \theta = \pi \) where \( \cos(\pi) = -1 \).
2b. Interval: \( [0, 2\pi] \)
3. Rose Area: Limits \( [-\pi/6, \pi/6] \) or \( 2 \times [0, \pi/6] \).
4. Symmetry: Integrating from 0 avoids negative signs and often makes evaluating trig functions simpler as \( \sin(0)=0 \) and \( \cos(0)=1 \).
5. Cardioid Area: \( A = \frac{1}{2} \int_0^{2\pi} (1-\sin\theta)^2 d\theta = \frac{1}{2} \int (1 - 2\sin\theta + \sin^2\theta) d\theta = \frac{1}{2} [ \theta + 2\cos\theta + \frac{1}{2}\theta - \frac{1}{4}\sin 2\theta ]_0^{2\pi} = \frac{3\pi}{2} \).
Lesson 3: Boundary Battles
1. Intersections: \( 3 = 2+2\cos\theta \implies \cos\theta = 1/2 \implies \theta = \pi/3, 5\pi/3 \text{ or } \pm \pi/3 \).
2. Intersections: \( \sqrt{2} = 2\sin\theta \implies \sin\theta = \sqrt{2}/2 \implies \theta = \pi/4, 3\pi/4 \).
The One Petal Problem Slides The One Petal Problem
Mastering Single-Curve Boundaries
Lesson 2: Angular Limits
The "Petal" Paradox
Suppose we want the area of one petal of the rose curve \( r = 3\cos(2\theta) \).
The formula is ready: \( \frac{1}{2} \int_{\alpha}^{\beta} (3\cos 2\theta)^2 d\theta \).
The Big Question:
What are the limits \(\alpha\) and \(\beta\)?
Where does this start and stop?
Strategy: Hunting for Zeros
A "petal" starts and ends at the Pole (Origin).
\( r = 0 \)
Solve for \( \theta \)
For \( r = \cos(2\theta) \):
\( \cos(2\theta) = 0 \)
\( 2\theta = \pm \pi/2 \)
\( \theta = \pm \pi/4 \)
The Interval:
\( [-\pi/4, \pi/4] \)
This captures exactly one petal center-right.
The Power of Symmetry
Integrating trigonometric functions can be messy. Avoid limits with negative values when possible.
Instead of:
\( \frac{1}{2} \int_{-\pi/4}^{\pi/4} r^2 d\theta \)
Use Symmetry:
\( 2 \cdot \left[ \frac{1}{2} \int_{0}^{\pi/4} r^2 d\theta \right] \)
Why double it?
\( 0 \to \pi/4 \)
Calculate half, double for the whole petal.
The Integration "Cheat Code"
Polar area integrals almost always contain squared trig terms. Use these identities or perish!
Cosine
\[ \cos^2(u) = \frac{1 + \cos(2u)}{2} \]
Sine
\[ \sin^2(u) = \frac{1 - \cos(2u)}{2} \]
The One Petal Problem Worksheet Petal Hunting
Lesson 2: Calculating Area of Single Polar Regions
Name: ____________________
Date: ____________________
Part 1: Hunting for the Pole
To find the area of a single "petal" or "loop," you must first identify when the curve passes through the origin (the pole). This happens when \( r = 0 \).
1. For the rose curve \( r = 4\sin(3\theta) \):
a) Set \( r = 0 \) and solve for the first two positive values of \( \theta \).
b) What is the interval \( [\alpha, \beta] \) that traces out exactly one petal?
2. For the cardioid \( r = 2 + 2\cos(\theta) \):
a) Does this curve ever pass through the pole? If so, at what angle \( \theta \)?
b) Determine the interval of \( \theta \) required to trace the entire cardioid exactly once.
Part 2: The Area Calculation
Problem 3: The Three-Leaf Rose
Calculate the exact area of one petal of \( r = 2\cos(3\theta) \).
Step 1: Determine Limits (\( r=0 \))
Step 2: Set up the Integral
Step 3: Evaluate (Use Power Reduction Identities!)
Part 3: Symmetry & Strategy
4. Strategic Limits:
Explain why using the integral \( 2 \cdot \frac{1}{2} \int_{0}^{\pi/6} (2\cos 3\theta)^2 d\theta \) is often easier than using \( \frac{1}{2} \int_{-\pi/6}^{\pi/6} (2\cos 3\theta)^2 d\theta \).
5. Area of a Cardioid:
Find the total area enclosed by the cardioid \( r = 1 - \sin(\theta) \). Show all work including the use of trig identities.
Boundary Battles Slides Boundary Battles
Integrating Between Multiple Polar Curves
Lesson 3: Complex Regions
The Shaded Challenge
How do we find the area of the region inside the circle \( r = 3 \) but outside the cardioid \( r = 2 + 2\cos\theta \)?
The Strategy:
Find intersection points
Determine the "Outer" and "Inner" curve
Subtract!
Conflict Zone
Step 1: Locate the Battle Lines
Where do the curves meet?
\( r_1 = r_2 \)
Solve for \( \theta \)
Example
\( 3 = 2 + 2\cos\theta \)
Isolate
\( \cos\theta = 1/2 \)
Target
\( \theta = \pm \pi/3 \)
Scenario A: The "Donut" Method
When one curve is always further from the pole than the other...
Subtraction Formula
\[ A = \frac{1}{2} \int_{\alpha}^{\beta} (R^2 - r^2) \, d\theta \]
\( R \) = Outer Curve, \( r \) = Inner Curve
Always measure radius from the pole out to the curve.
Visualize the "sweep" between your intersection angles.
Scenario B: The "Switch" Method
The "Inside Both" Problem
If the region is bounded by different curves at different angles...
YOU MUST SPLIT!
\( \int r_1^2 + \int r_2^2 \)
!
Crucial Realization: Area "Inside Both" is not subtraction. It is the accumulation of two separate sweeps.
Ask yourself: "As I rotate the radar arm, which curve is providing the ceiling?"
Boundary Battles Worksheet Shaded Region Challenge
Lesson 3: Area Between Two Polar Curves
Name: ____________________
Date: ____________________
Part 1: The Intersection Duel
Before setting up any integral, you must find exactly where the curves clash. Solve the following equations for \( \theta \) on the interval \( [0, 2\pi] \).
1. \( r = 3 \) and \( r = 2 + 2\cos(\theta) \)
Angles of intersection:
2. \( r = \sqrt{2} \) and \( r = 2\sin(\theta) \)
Angles of intersection:
Part 2: Strategic Setup
Scenario: The "Moon" Region
You want the area of the region inside the circle \( r = 3 \) but outside the cardioid \( r = 2 + 2\cos(\theta) \).
3. Should you subtract two integrals or add two separate integrals for this specific region? Justify your choice based on the distance of each curve from the pole.
4. Set up the integral(s) for this region. Use symmetry to simplify if possible.
Scenario: The "Eye" Region
You want the area of the region shared by (inside both) \( r = 2 \) and \( r = 4\sin(\theta) \).
5. Why is this a "Switch" problem? Identify the angle \( \theta \) at which the boundary curve changes from the circle to the cardioid (or vice versa).
6. Set up the compound integral for this shared region.
Part 3: Final Battle
Problem 7: The Inner Loop Challenge
The limacon \( r = 1 + 2\cos(\theta) \) has an "inner loop." To find its area, you must integrate between the angles where the curve crosses itself at the pole.
A) Find the limits (\( r=0 \)):
B) Setup the Integral:
C) Solve for the Area (Show all calculus steps):
Perimeter in Polar Slides Perimeter in Polar
Measuring the Boundaries of Curves
Lesson 4: Arc Length
The "String" Experiment
If you laid a string along the boundary of a cardioid \( r = 1 + \cos\theta \), how long would it be?
We already know Area (\( \frac{1}{2} \int r^2 d\theta \)). Now we need Arc Length.
Calculus Connection:
Just like in rectangular form, we sum up infinite tiny segments \( ds \).
The Foundation
Recall the Parametric Arc Length formula:
\[ L = \int \sqrt{ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 } \, dt \]
By substituting \( x = r\cos\theta \) and \( y = r\sin\theta \), we get the Polar version...
The Polar Arc Length Formula
\[ L = \int_{\alpha}^{\beta} \sqrt{ r^2 + \left(\frac{dr}{d\theta}\right)^2 } \, d\theta \]
Radial Distance
\( r^2 \)
Rate of Change
\( (r')^2 \)
Applying the Formula
Find the length of the circle \( r = 4\sin\theta \) from \( 0 \leq \theta \leq \pi \).
1. Differentiate
\( r = 4\sin\theta \)
\( r' = 4\cos\theta \)
2. Substitute
\[ \sqrt{(4\sin\theta)^2 + (4\cos\theta)^2} \] \[ \sqrt{16(\sin^2\theta + \cos^2\theta)} \]
3. Solve
\[ \sqrt{16} = 4 \] \[ L = \int_{0}^{\pi} 4 \, d\theta = 4\pi \]
Makes sense! A circle with diameter 4 has circumference \( 4\pi \).
Watch the Limits!
Many polar curves (like rose petals or circles) complete their path in \( \pi \) radians rather than \( 2\pi \).
Pro-Tip:
Always graph the function or check the period to ensure you aren't measuring the perimeter twice!
Avoid Overlap
Perimeter in Polar Worksheet The Perimeter Path
Lesson 4: Arc Length of Polar Curves
Name: ____________________
Date: ____________________
Part 1: The Radial Derivative
The arc length formula requires the square of the derivative of the radius with respect to theta. Find \( \frac{dr}{d\theta} \) for each of the following:
1. \( r = 2 - 2\sin\theta \)
\( \frac{dr}{d\theta} = \)
2. \( r = \cos(3\theta) \)
\( \frac{dr}{d\theta} = \)
3. \( r = e^{\theta} \) (Logarithmic Spiral)
\( \frac{dr}{d\theta} = \)
4. \( r = \theta^2 \)
\( \frac{dr}{d\theta} = \)
Part 2: Arc Length Setups
Recall: \( L = \int_{\alpha}^{\beta} \sqrt{ r^2 + \left(\frac{dr}{d\theta}\right)^2 } \, d\theta \)
5. The Full Cardioid: \( r = 1 + \cos\theta \)
Set up the integral to find the total perimeter of the cardioid. Do not evaluate.
6. One Petal: \( r = \sin(2\theta) \)
Set up the integral for the perimeter of exactly ONE petal. Be careful with your limits!
7. Concept Check: Circle Comparison
Consider the circle \( r = 6 \). Using the polar arc length formula, show how it simplifies to the standard circumference formula \( C = 2\pi r \).
Part 3: Evaluation Challenge
Problem 8: The Logarithmic Spiral
Find the total length of the spiral \( r = e^{2\theta} \) for \( 0 \leq \theta \leq \ln(3) \). This curve has a very clean arc length evaluation!
Step 1: Compute \( (r')^2 \)
Step 2: Simplify the Integrand \( \sqrt{r^2 + (r')^2} \)
Step 3: Evaluate the Integral
The Polar Estate Project Guide The Polar Estate
Surveying & Design Specification
Project 05-P
The Brief
You have been commissioned by the Curious Geometries Land Trust to design a new public park. The catch? The donors have mandated that the park's boundaries must be defined strictly by polar equations.
Your job is to produce a formal "Estate Blueprint" that includes a visual design and precise mathematical calculations for material costs.
Required Elements
2+ Intersecting Curves
Exact Area Calculation
Exact Perimeter Calculation
Material Cost Analysis
Phase 1: Design
Sketch your proposed park layout below. You must use at least two different polar equations (e.g., a circle and a cardioid, or two rose curves) to define your region.
Blueprints: Sketch Layout & Equations Here
Equation \( r_1 \):
Equation \( r_2 \):
Phase 2: The Survey
1. Area of Sod (Land Area)
Calculate the total area of your shaded region. Identify intersection points and set up the integral(s) precisely.
Show full integration work here
2. Boundary Length (Fencing)
Calculate the total perimeter of the park's outer boundary. Use the polar arc length formula.
Show full integration work here
Cost Summary Report
Sodding Cost ($5.00 / unit area)
$ ____________________
Fencing Cost ($12.00 / unit length)
$ ____________________
Total Project Budget
$ _______________
