Polar Area Slides Polar Area Formula
Shifting from rectangular slices to radial sectors in the pursuit of area.
Calculus BC
Unit: Radial Integration
The Hook
Slicing for Accumulation
Cartesian Slicing
Think of a loaf of bread. We find the area by summing thin rectangular vertical slices.
Area = \(\int_a^b f(x) \, dx\)
Polar Slicing
Think of a pizza. We find the area by summing thin wedges (sectors) originating from the center.
Area = ???
The Sector Foundation
To build our integral, we need the geometry of a single "slice." In Polar, that slice is a Circular Sector .
Area = \(\frac{1}{2}r^2\theta\)
Circle Area
\(A = \pi r^2\)
Fraction of Circle
\(\text{Fraction} = \frac{\theta}{2\pi}\)
Sector Area
\(A = \pi r^2 \cdot \left(\frac{\theta}{2\pi}\right) = \frac{1}{2}r^2\theta\)
The Formula
Summing infinitely thin sectors from \(\alpha\) to \(\beta\):
\[A = \int_{\alpha}^{\beta} \frac{1}{2} [r(\theta)]^2 \, d\theta\]
\(r(\theta)\) Distance from Pole
\(d\theta\) Infinitesimal Angle
\([\alpha, \beta]\) Angular Bounds
Guided Example: Circle Area
Find the area of the circle defined by r = 3.
1. Setup
Radius function: \(r = 3\)
Bounds: One full revolution (\(0\) to \(2\pi\))
2. Integrate
\(A = \int_{0}^{2\pi} \frac{1}{2} (3)^2 \, d\theta\)
\(A = \int_{0}^{2\pi} \frac{9}{2} \, d\theta\)
\(A = \left[ \frac{9}{2}\theta \right]_0^{2\pi}\)
\(A = \frac{9}{2}(2\pi) - 0\)
\(A = 9\pi\)
Matches \(A = \pi r^2\)! Our formula works.
Think & Discuss
Variable Radius
What happens to our "slices" when \(r\) is not constant, like in \(r = \cos(2\theta)\)? How do the wedges change length?
Angular Bounds
Why is finding the correct \(\alpha\) and \(\beta\) often the hardest part of polar integration?
Sector Secrets Worksheet Sector Secrets Worksheet
Mastering the Geometry of Polar Area
Name:
Date:
Concept Foundations
1. Write the integral formula for the area of a polar region bounded by \(r = f(\theta)\) from \(\theta = \alpha\) to \(\theta = \beta\).
2. Briefly explain why we use the term \(\frac{1}{2}r^2\) in the polar area integral instead of just \(r\), as we do with heights in Cartesian integrals (\(\int f(x) \, dx\)).
Computational Practice
3. Find the area of the region bounded by the graph of r = 2 + \cos(\theta) for \(0 \leq \theta \leq 2\pi\).
Lima\u00e7on
4. Calculate the area of one petal of the rose curve r = 4\sin(3\theta).
Rose Curve
5. Consider the polar curve r = \theta for \(0 \leq \theta \leq \pi\).
Spiral
a) Sketch the curve below:
b) Set up and evaluate the integral:
The Radar Challenge
A radar sweeps an area defined by \(r = 6\sqrt{\sin(\theta)}\) from \(\theta = 0\) to \(\theta = \pi\). Find the total area scanned by the radar. Explain why the result is an integer.
Polar Area Teacher Key Teacher Guide
Lesson 1: The Polar Area Formula
Radial Integration Sequence
Calculus BC / 12th Grade
Key Objective
Students will derive and apply the formula \(A = \int \frac{1}{2}r^2 d\theta\) to find areas of polar regions.
Common Myth
Students often forget the \(1/2\) or forget to square \(r\), treating it like \(y\) in rectangular integrals.
Prerequisites
Evaluating basic integrals, trigonometric identities, polar coordinate plotting.
Worksheet Answer Key
1 & 2. Concept Foundations
Formula: \(A = \int_{\alpha}^{\beta} \frac{1}{2} [r(\theta)]^2 \, d\theta\)
Explanation: In Cartesian, we sum rectangles with area \(f(x) \Delta x\). In Polar, we sum circular sectors with area \(\frac{1}{2}r^2 \Delta \theta\). The \(\frac{1}{2}\) and the square come from the geometry of a circle slice.
3. Lima\u00e7on Area (\(r = 2 + \cos\theta\))
\(A = \frac{1}{2} \int_0^{2\pi} (2 + \cos\theta)^2 \, d\theta = \frac{1}{2} \int_0^{2\pi} (4 + 4\cos\theta + \cos^2\theta) \, d\theta\)
Use identity: \(\cos^2\theta = \frac{1 + \cos(2\theta)}{2}\)
\(A = \frac{1}{2} \int_0^{2\pi} (4.5 + 4\cos\theta + 0.5\cos(2\theta)) \, d\theta\)
\(A = \frac{1}{2} [4.5\theta + 4\sin\theta + 0.25\sin(2\theta)]_0^{2\pi} = \frac{1}{2} (9\pi) = 4.5\pi\)
4. Rose Petal (\(r = 4\sin(3\theta)\))
Bounds: \(r=0\) when \(3\theta = 0, \pi \Rightarrow \theta = 0, \pi/3\).
\(A = \frac{1}{2} \int_0^{\pi/3} (4\sin(3\theta))^2 \, d\theta = 8 \int_0^{\pi/3} \sin^2(3\theta) \, d\theta\)
\(A = 8 \int_0^{\pi/3} \frac{1 - \cos(6\theta)}{2} \, d\theta = 4 [\theta - \frac{1}{6}\sin(6\theta)]_0^{\pi/3} = 4\pi/3\)
The Radar Challenge
\(A = \frac{1}{2} \int_0^{\pi} (6\sqrt{\sin\theta})^2 \, d\theta = \frac{1}{2} \int_0^{\pi} 36\sin\theta \, d\theta\)
\(A = 18 \int_0^{\pi} \sin\theta \, d\theta = 18 [-\cos\theta]_0^{\pi}\)
\(A = 18 ( -(-1) - (-1) ) = 18(2) = 36\)
The square root and the square "cancel," leaving a simple sine integral, resulting in the integer 36.
Inner Loop Slides Inner Loops
Finding the hidden area within lima\u00e7ons and finding the exact boundaries of radial loops.
Lesson 02
Solving for \(r = 0\)
The Lima\u00e7on Dilemma
Why \(0\) to \(2\pi\) is not always the answer.
When a polar curve crosses the pole (\(r=0\)), it creates loops.
Consider \(r = 1 + 2\cos\theta\)
If we integrate from \(0 \to 2\pi\), we get the total area (outer + inner).
To find JUST the inner loop, we need the exact angles where the curve enters and leaves the pole.
INNER
The Hunt for Zero
Finding the loop boundaries means solving the equation:
\(r(\theta) = 0\)
Step 1: Isolate Trig
Set the function to zero and solve for the trigonometric part.
Ex: \(\cos\theta = -1/2\)
Step 2: Reference Angles
Identify the two specific \(\theta\) values within one period.
Ex: \(\theta = 2\pi/3, 4\pi/3\)
Worked Example: The Small Loop
Function
\(r = 1 + 2\cos\theta\)
We want the area of the inner loop only.
Calculations
1. \(0 = 1 + 2\cos\theta \rightarrow \cos\theta = -1/2\)
2. \(\theta = \frac{2\pi}{3}, \frac{4\pi}{3}\)
3. \(A = \frac{1}{2} \int_{2\pi/3}^{4\pi/3} (1+2\cos\theta)^2 \, d\theta\)
Pro-Tip: Use symmetry! You can integrate from \(2\pi/3\) to \(\pi\) and multiply by 2.
\(A = \int_{2\pi/3}^{\pi} (1+2\cos\theta)^2 \, d\theta\)
Visualizing Negative r
Why does the loop happen?
For certain angles, the equation produces a negative value for \(r\). In polar coordinates, this plots the point in the opposite quadrant .
Area is Always Positive
Even when \(r\) is negative, \([r(\theta)]^2\) is positive. Integration still sums area correctly, provided your bounds are perfect.
Q1
Plotting \(r < 0\) here
Lima\u00e7on Loop Lab The Lima\u00e7on Loop Lab
Exploring Areas of Inner Regions
Student Name
The Rule of r=0
To find the area of a loop, you must first find the "Pole Crossing" angles. These are the \(\theta\) values where the graph enters and exits the origin.
Task 1: The Single Inner Loop
Consider the polar equation r = 1 - 2\sin(\theta).
a) Find the values of \(\theta\in[0, 2\pi]\) where \(r = 0\).
b) Sketch the graph (focus on the loops).
c) Set up the integral for the area of the inner loop only .
SYMMETRY SHORTCUT
Task 2: The Cardioid vs. Lima\u00e7on
Integrate the area of one loop of the rose curve \(r = 3\sin(2\theta)\). Then, use your result to predict the total area of all four loops.
Setup and Calculation:
Individual Petal Area:
Predicted Total Area:
The "Outside But Inside" Puzzle
Calculate the area of the region inside the outer loop but outside the inner loop for the curve \(r = 2 + 4\cos\theta\).
1. Outer Area Strategy
Total area from \(0 \to 2\pi\):
2. Inner Loop Correction
Area to subtract:
Final Evaluation
Show your work here
Inner Loop Teacher Key Teacher Guide
Lesson 2: Inner Loop Calculations
Task 1: \(r = 1 - 2\sin\theta\)
r=0 Intersection:
\(0 = 1 - 2\sin\theta \Rightarrow \sin\theta = 1/2\)
\(\theta = \pi/6, 5\pi/6\)
Area Integral:
\(A = \frac{1}{2} \int_{\pi/6}^{5\pi/6} (1 - 2\sin\theta)^2 \, d\theta\)
Evaluating this yields: \(\pi - \frac{3\sqrt{3}}{2} \approx 0.544\)
Task 2: Rose Curve \(r = 3\sin(2\theta)\)
Single Petal: \(r=0\) at \(\theta=0\) and \(2\theta=\pi \Rightarrow \theta=\pi/2\).
\(A = \frac{1}{2} \int_0^{\pi/2} (3\sin(2\theta))^2 \, d\theta = \frac{9}{2} \int_0^{\pi/2} \sin^2(2\theta) \, d\theta = \frac{9}{4} \int_0^{\pi/2} (1 - \cos(4\theta)) \, d\theta = \frac{9\pi}{8}\)
Total Area: \(4 \times \frac{9\pi}{8} = \frac{9\pi}{2}\).
The "Outside But Inside" Puzzle
Curve: \(r = 2 + 4\cos\theta\)
Total Area (\(0 \to 2\pi\)):
\(A_{tot} = \frac{1}{2} \int_0^{2\pi} (2+4\cos\theta)^2 \, d\theta = 12\pi\)
Note: This counts the inner loop TWICE because the integral covers the full rotation.
Strategy:
To find the area between the loops, compute the "Outer" area (subtracting the inner loop once) and then subtract the inner loop AGAIN.
Wait! Actually:
Correct Method: Area_{Outer} - Area_{Inner}. Bounds for Inner: \(\cos\theta = -1/2 \Rightarrow \theta \in [2\pi/3, 4\pi/3]\).
Misconception Alert
Students often struggle with the fact that \(r\) can be negative. Remind them that for the purpose of area , the \([r(\theta)]^2\) term ensures we are summing positive radial "squares," so we don't get "negative area" the way we do when a function drops below the x-axis in Cartesian.
Bounded Regions Slides Bounded Regions
Calculating the shared space and differences between multiple polar curves.
Lesson 03
Intersections & Compound Integrals
The Radar Overlap
Acting as Land Surveyors
Imagine two radar stations. Each station detects objects within a specific polar boundary.
To find the shared detection zone , we need to know:
Where do the signals meet? (Intersection)
Which curve is "inner" for which interval?
SHARED AREA
Strategies for Interaction
Between Curves
Find the area outside one curve and inside another.
Area = \(\int \frac{1}{2}(r_{out}^2 - r_{in}^2) \, d\theta\)
Shared / Inner Region
Find the area belonging to both curves.
Requires Splitting the integral at intersection points.
Finding Intersections
To find where r₁(\theta) meets r₂(\theta), set them equal:
\(r_1(\theta) = r_2(\theta)\)
Warning!
Points may also intersect at the pole (\(r=0\)) at different \(\theta\) values. Check for \(r=0\) separately!
Example Setup
\(r = 3\sin\theta\)
\(r = 1 + \sin\theta\)
\(3\sin\theta = 1 + \sin\theta\)
\(2\sin\theta = 1\)
\(\sin\theta = 1/2\)
\(\theta = \pi/6, 5\pi/6\)
The Art of the Split
For shared areas, we often change which function we follow.
\[\text{Total Area} = \int_{0}^{\alpha} \frac{1}{2}r_1^2 \, d\theta + \int_{\alpha}^{\beta} \frac{1}{2}r_2^2 \, d\theta\]
"Follow the curve that is closest to the origin for the shared region."
Shared Territory Worksheet Shared Territory
Complex Polar Regions
Sector Grade A+
Navigator:
Coordinate:
Analysis 01: The Overlap
Compound Integral
Find the area of the region shared by the circles r = 1 and r = 2\sin(\theta).
1. Intersection (\(\theta\))
2. Sketch Intersection
3. Integral Setup
Analysis 02: Outside/Inside
Find the area of the region that is inside the cardioid r = 3 + 3\sin(\theta) but outside the circle r = 4.5.
Determine the angular bounds by solving for intersection, then apply the difference of squares formula.
The Surveyor's Challenge
The boundaries of three radio towers are given by:
Identify the area that is contained within all three circles .
Visual Proof
Draw the triple overlap
Calculations Bounded Regions Teacher Key Teacher Guide
Lesson 3: Bounded Polar Regions
Analysis 01: Circle Shared Area
Equations: \(r = 1\) and \(r = 2\sin\theta\)
1. Intersections:
\(1 = 2\sin\theta \Rightarrow \sin\theta = 1/2\)
\(\theta = \pi/6, 5\pi/6\)
2. Integral Setup:
Shared area follows \(r = 2\sin\theta\) from \(0 \to \pi/6\), then \(r = 1\) from \(\pi/6 \to 5\pi/6\), then \(r = 2\sin\theta\) again. By symmetry:
\(A = 2 \times [ \frac{1}{2}\int_0^{\pi/6} (2\sin\theta)^2 \, d\theta + \frac{1}{2}\int_{\pi/6}^{\pi/2} (1)^2 \, d\theta ]\)
Analysis 02: Outside/Inside
Equations: \(r = 3+3\sin\theta\) (Cardioid) and \(r = 4.5\) (Circle)
Intersections: \(3+3\sin\theta = 4.5 \Rightarrow 3\sin\theta = 1.5 \Rightarrow \sin\theta = 1/2\).
\(\theta = \pi/6, 5\pi/6\).
Integral: Cardioid is outer, Circle is inner.
\(A = \frac{1}{2} \int_{\pi/6}^{5\pi/6} [ (3+3\sin\theta)^2 - (4.5)^2 ] \, d\theta\)
The Surveyor's Challenge
The three circles are centered at (2,0), (0,2), and (0,0). They all have radius 2 (or 4 for the polar diameter curves).
Intersection Analysis:
Tower A & B meet at \(\theta = \pi/4\).
Tower A & C meet where \(4\cos\theta = 2 \Rightarrow \theta = \pi/3\).
Tower B & C meet where \(4\sin\theta = 2 \Rightarrow \theta = \pi/6\).
Integral Strategy:
The shared area in the first quadrant follows Tower B (\(4\sin\theta\)) from \(0\) to \(\pi/6\), then Tower C (\(2\)) from \(\pi/6\) to \(\pi/3\), then Tower A (\(4\cos\theta\)) from \(\pi/3\) to \(\pi/2\).
\(A = \int_0^{\pi/6} \frac{1}{2}(4\sin\theta)^2 \, d\theta + \int_{\pi/6}^{\pi/3} \frac{1}{2}(2)^2 \, d\theta + \dots\)
Teaching Tip: Choosing the Formula
If students are confused whether to subtract or split/add , use the "Ray Method." Draw a ray from the pole through the region. If the ray passes through two curves to get into and out of the region, use subtraction. If the region is bounded only by the pole and different curves at different angles, use splitting/addition.
Polar Arc Length Slides Polar Arc Length
Measuring the winding paths of spirals, cardioids, and rose curves.
Lesson 04
The String Challenge
Measuring the Wind
String vs. Calculus
How do we measure the length of a nautilus shell?
In Cartesian, we use the Pythagorean distance formula summed over a path. In Polar, the distance is driven by both radial change and angular rotation .
We need a way to combine \(dr\) and \(d\theta\) into a single length element \(ds\).
From Parametric to Polar
Parametric Base
\(L = \int \sqrt{ (x')^2 + (y')^2 } \, dt\)
Polar Form
\(L = \int_{\alpha}^{\beta} \sqrt{ r^2 + (\frac{dr}{d\theta})^2 } \, d\theta\)
Why \(r^2 + (r')^2\)?
Think of a tiny triangle. One side is the change in distance from the pole (\(dr\)). The other side is the arc created by rotating (\(r d\theta\)). The hypotenuse \(ds\) follows:
Length of a Cardioid
\(r = 1 + \cos\theta\)
Preparation:
Find \(\frac{dr}{d\theta}\)
Square both \(r\) and \(r'\)
Sum and simplify using trig identities
Calculations
\(r' = -\sin\theta\)
\(r^2 + (r')^2 = (1+\cos\theta)^2 + (-\sin\theta)^2\)
\(= 1 + 2\cos\theta + \cos^2\theta + \sin^2\theta\)
\(= 2 + 2\cos\theta\)
\(L = \int_0^{2\pi} \sqrt{2 + 2\cos\theta} \, d\theta\)
The Archimedean Spiral
How long is the spiral r = \theta as it makes one full rotation?
\(L = \int_0^{2\pi} \sqrt{\theta^2 + 1} \, d\theta\)
"This integral requires a Trig Substitution or a Formula Table. Calculus is the only string that can measure this perfectly."
Spiral Shell Worksheet The Spiral Shell Challenge
Calculating Distance in a Radial World
STUDENT:
LENGTH:
Master Formula
L = \int_{\alpha}^{\beta} \sqrt{r^2 + (r')^2} \, d\theta
1 The Circle Check
Use the arc length formula to find the circumference of the circle r = 5 from \(0 \leq \theta \leq 2\pi\). Does it match \(C = 2\pi r\)?
Show derivatives and setup...
2 The Heart's Edge
Calculate the total arc length of the cardioid r = 2 + 2\cos(\theta).
Focus on the integration of the trig identity...
3 The Winding Spiral
An Archimedean spiral is given by r = 4\theta. Find the length of the spiral from \(\theta = 0\) to \(\theta = \pi\).
Space for complex spiral calculation
Reflection Question
Explain in your own words why we must include the \(r^2\) term inside the square root of the polar arc length formula, even if the radius isn't changing. What geometric part of the movement does it represent?
Spiral Shell Teacher Key Teacher Guide
Lesson 4: Polar Arc Length
1. Circle Check (\(r = 5\))
\(r = 5, r' = 0\)
\(L = \int_0^{2\pi} \sqrt{5^2 + 0^2} \, d\theta = \int_0^{2\pi} 5 \, d\theta = [5\theta]_0^{2\pi} = 10\pi\)
Matches \(C = 2\pi(5) = 10\pi\)!
2. The Cardioid (\(r = 2+2\cos\theta\))
\(r' = -2\sin\theta\)
\(r^2 + (r')^2 = (4+8\cos\theta+4\cos^2\theta) + 4\sin^2\theta = 8 + 8\cos\theta\)
Using identity: \(1+\cos\theta = 2\cos^2(\theta/2)\)
\(L = \int_0^{2\pi} \sqrt{16\cos^2(\theta/2)} \, d\theta = \int_0^{2\pi} 4|\cos(\theta/2)| \, d\theta\)
Due to absolute value/symmetry: \(2 \times \int_0^{\pi} 4\cos(\theta/2) \, d\theta = 8 [2\sin(\theta/2)]_0^{\pi} = 16\)
Total Length = 16 units.
3. The Spiral (\(r = 4\theta\))
\(r' = 4\)
\(L = \int_0^{\pi} \sqrt{(4\theta)^2 + 4^2} \, d\theta = 4 \int_0^{\pi} \sqrt{\theta^2 + 1} \, d\theta\)
Apply provided integral formula with \(a=1, x=\theta\):
\(L = 4 [ \frac{1}{2}\theta\sqrt{\theta^2+1} + \frac{1}{2}\ln|\theta+\sqrt{\theta^2+1}| ]_0^{\pi}\)
\(L = 2\pi\sqrt{\pi^2+1} + 2\ln|\pi+\sqrt{\pi^2+1}|\)
Evaluation: \(\approx 24.36\) units.
Reflection Guide
The \(r^2\) term represents the angular component of the arc length. Even if the distance from the pole is constant (\(dr = 0\)), the rotation itself covers distance. This distance is the arc length of a circle, \(s = r\theta\). In the infinitesimal limit, this is \((r \, d\theta)\), which squared is \(r^2 d\theta^2\). Without it, you'd only be measuring radial "stretching," not the curve itself.
Surface Solids Slides Surface Solids
Synthesizing arc length and coordinate rotation to find the skin of a polar solid.
Lesson 05
Final Synthesis
Spinning the Curve
Creating 3D Surfaces from 2D Graphs
In Cartesian, surface area is the sum of circumferences of circles generated by the function:
In Polar, we rotate around:
The Polar Axis (rotation around horizontal)
The line \(\theta = \pi/2\) (rotation around vertical)
Visual Thought
Imagine spinning a single petal of a rose curve. What shape does the skin form?
Surface Area Formulas
About Polar Axis
Radius of rotation is \(y = r\sin\theta\).
\(S = \int_{\alpha}^{\beta} 2\pi (r\sin\theta) \sqrt{r^2 + (r')^2} \, d\theta\)
About \(\theta = \pi/2\)
Radius of rotation is \(x = r\cos\theta\).
\(S = \int_{\alpha}^{\beta} 2\pi (r\cos\theta) \sqrt{r^2 + (r')^2} \, d\theta\)
Synthesis: Sphere Check
Rotate the semicircle r = 2\cos\theta from \(0 \leq \theta \leq \pi/2\) around the polar axis.
Arc Length Element:
\(\sqrt{ (2\cos\theta)^2 + (-2\sin\theta)^2 } = 2\)
Integration
\(S = \int_0^{\pi/2} 2\pi (2\cos\theta\sin\theta)(2) \, d\theta\)
\(S = 8\pi \int_0^{\pi/2} \sin(2\theta) \, d\theta\)
\(S = 8\pi [-\frac{1}{2}\cos(2\theta)]_0^{\pi/2}\)
\(S = 4\pi ( -(-1) - (-1) ) = 8\pi\)
Matches \(4\pi r^2\) for \(r=1.41\)? No, wait! Check dimensions...
Closing the Loop
Complexity
Why is surface area the ultimate test of polar mastery? What previously learned skills must you combine to solve a single problem?
Real-World
Where would a radial surveyor or engineer need surface area over simple area or arc length? (e.g., Antenna design, optics)
Radial Revolution Lab Radial Revolution Lab
Surface Area Synthesis
SOLVER:
Around Polar Axis
\(S = 2\pi \int r\sin\theta \sqrt{r^2+(r')^2} \, d\theta\)
Around \(\theta = \pi/2\)
\(S = 2\pi \int r\cos\theta \sqrt{r^2+(r')^2} \, d\theta\)
1. Rotating the Cardioid
Find the surface area of the solid generated by revolving the top half of the cardioid r = 1 + \cos\theta (from \(0 \leq \theta \leq \pi\)) about the polar axis.
Setup the integral and use the \(\sqrt{r^2+(r')^2} = 2\cos(\theta/2)\) simplification from the previous lesson.
2. The Vertical Spin
Consider the circle r = 4\sin\theta. Revolve this circle about the line \(\theta = \pi/2\). What shape is formed, and what is its surface area?
Hint: This forms a torus (donut) or a sphere depending on the bounds. Perform the calculus to verify.
The Petal Rotation Challenge
Rotate one petal of the rose curve r = \cos(2\theta) (where \(-\pi/4 \leq \theta \leq \pi/4\)) about the polar axis.
1. Set up the complete integral.
Integral Setup
Analysis of Difficulty
Radial Revolution Teacher Key Teacher Guide
Lesson 5: Surface Area Revolution
1. Rotating the Cardioid
\(r = 1+\cos\theta\)
\(ds = \sqrt{r^2+(r')^2} = \sqrt{2+2\cos\theta} = 2\cos(\theta/2)\)
\(S = \int_0^{\pi} 2\pi (1+\cos\theta)\sin\theta \cdot 2\cos(\theta/2) \, d\theta\)
Using \(1+\cos\theta = 2\cos^2(\theta/2)\) and \(\sin\theta = 2\sin(\theta/2)\cos(\theta/2)\):
\(S = \int_0^{\pi} 2\pi (2\cos^2(\theta/2)) (2\sin(\theta/2)\cos(\theta/2)) (2\cos(\theta/2)) \, d\theta\)
\(S = 16\pi \int_0^{\pi} \cos^4(\theta/2)\sin(\theta/2) \, d\theta\)
Let \(u = \cos(\theta/2), du = -1/2\sin(\theta/2)d\theta\)
\(S = 16\pi [-2/5 \cos^5(\theta/2)]_0^{\pi} = 16\pi (0 - (-2/5)) = 32\pi/5\)
2. The Vertical Spin
The circle \(r = 4\sin\theta\) is a circle of radius 2 centered at (0, 2). Rotating it around the vertical line \(\theta = \pi/2\) (which is the y-axis) rotates the circle on its own diameter.
This forms a SPHERE with radius 2.
\(S = \int_0^{\pi} 2\pi (4\sin\theta\cos\theta)(4) \, d\theta = 16\pi [-\cos^2\theta]_0^{\pi} = 0?\)
The Petal Rotation Challenge
Setup for \(r = \cos(2\theta)\):
\(S = 2\pi \int_{-\pi/4}^{\pi/4} \cos(2\theta)\sin\theta \sqrt{\cos^2(2\theta) + 4\sin^2(2\theta)} \, d\theta\)
Why is it harder? Unlike the cardioid, the square root term does not simplify nicely with trig identities. It requires numerical integration or very complex substitutions (\(\sqrt{1+3\sin^2(2\theta)}\)).
Closing Strategy
Encourage students to always check for geometric symmetry first. Many of these integrals can be simplified by integrating from 0 to \(\pi/2\) or using known volume/surface area formulas to verify their answers. This builds confidence in the heavy algebraic lifting required for these problems.
