Curve Stretching Slides Curve Stretching
The Calculus of Arc Length
The Mapmaker's Problem
Imagine you have a string and a map. To measure the distance of a winding mountain road, you lay the string along the curve, then pull it straight and measure it with a ruler.
"How can we do this mathematically without the string?"
road_geometry.v1
The Strategy: Zoom In
1
Break the curve into \( n \) small segments.
2
Treat each segment as a straight line (a chord).
3
Use the distance formula for each segment:
\( d = \sqrt{(\Delta x)^2 + (\Delta y)^2} \)
The Algebra Magic
We can rewrite the distance formula to involve the derivative:
\[ \Delta L \approx \sqrt{(\Delta x)^2 + (\Delta y)^2} \]
\[ \Delta L \approx \sqrt{1 + \left(\frac{\Delta y}{\Delta x}\right)^2} \cdot \Delta x \]
As \( \Delta x \to 0 \), the sum becomes an integral and \( \frac{\Delta y}{\Delta x} \to f'(x) \).
The Arc Length Formula
If \( f' \) is continuous on \( [a, b] \), the length of the curve \( y = f(x) \) from \( x = a \) to \( x = b \) is:
\[ L = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} \, dx \]
Guided Practice
Find the length of the arc of the curve \( y = \frac{4\sqrt{2}}{3}x^{3/2} - 1 \) on the interval \( [0, 1] \).
Step 1: Find the derivative
\( y' = \frac{4\sqrt{2}}{3} \cdot \frac{3}{2}x^{1/2} = 2\sqrt{2}x^{1/2} \)
Step 2: Square the derivative
\( (y')^2 = (2\sqrt{2}x^{1/2})^2 = 8x \)
Step 3: Set up the integral
\( L = \int_{0}^{1} \sqrt{1 + 8x} \, dx \)
What could go wrong?
Wait! Is the integral of \( \sqrt{1 + [f'(x)]^2} \) always easy to solve?
"Most arc length integrals are notoriously difficult or impossible to solve analytically. We often need numerical methods or very specific functions to get a clean answer!"
Arc Length Mastery Worksheet Arc Length Mastery
Applications of Integration • Lesson 1
Student Name
Date
Formula Recall
\( L = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} \, dx \)
Ensure \( f' \) is continuous on \( [a, b] \).
01
The Warming Up Curve
Find the exact length of the curve \( y = \frac{1}{3}(x^2 + 2)^{3/2} \) from \( x = 0 \) to \( x = 3 \).
Step A: Find and simplify \( f'(x) \)
Step B: Simplify \( \sqrt{1 + [f'(x)]^2} \)
Step C: Integrate to find \( L \)
02
The Logarithmic Path
Determine the length of the arc of the function \( y = \ln(\cos x) \) on the interval \( [0, \pi/4] \).
Work Area
Final Answer:
03
Horizontal Perspective
Sometimes it is easier to integrate with respect to \( y \). Find the length of the curve \( x = \frac{1}{6}y^3 + \frac{1}{2y} \) from \( y = 1 \) to \( y = 2 \).
Hint: Use the formula \( L = \int_{c}^{d} \sqrt{1 + [g'(y)]^2} \, dy \).
Work Area
Final Answer:
Conceptual Challenge
Explain why the function \( f(x) = \sqrt[3]{x} \) on the interval \( [-1, 1] \) might pose a problem for the standard arc length formula. What happens to the derivative at \( x=0 \)? How might you resolve this?
Arc Length Mastery Key Answer Key
Arc Length Mastery Worksheet
Teacher Resource
01
The Warming Up Curve
Step A: Derivative
\( y = \frac{1}{3}(x^2+2)^{3/2} \implies y' = \frac{1}{3} \cdot \frac{3}{2}(x^2+2)^{1/2} \cdot (2x) = x\sqrt{x^2+2} \)
Step B: Simplify Integral Term
\( \sqrt{1 + (y')^2} = \sqrt{1 + (x\sqrt{x^2+2})^2} = \sqrt{1 + x^2(x^2+2)} = \sqrt{x^4 + 2x^2 + 1} = \sqrt{(x^2+1)^2} = x^2 + 1 \)
Step C: Integrate
\( L = \int_{0}^{3} (x^2+1) \, dx = \left[ \frac{1}{3}x^3 + x \right]_{0}^{3} = (9 + 3) - 0 = 12 \)
02
The Logarithmic Path
\( y = \ln(\cos x) \implies y' = \frac{-\sin x}{\cos x} = -\tan x \)
\( \sqrt{1 + (y')^2} = \sqrt{1 + (-\tan x)^2} = \sqrt{1 + \tan^2 x} = \sqrt{\sec^2 x} = \sec x \)
\( L = \int_{0}^{\pi/4} \sec x \, dx = \left[ \ln|\sec x + \tan x| \right]_{0}^{\pi/4} \)
\( L = \ln(\sqrt{2} + 1) - \ln(1 + 0) = \ln(\sqrt{2} + 1) \approx 0.881 \)
03
Horizontal Perspective
\( x = \frac{1}{6}y^3 + \frac{1}{2}y^{-1} \implies x' = \frac{1}{2}y^2 - \frac{1}{2}y^{-2} = \frac{1}{2}(y^2 - y^{-2}) \)
\( 1 + (x')^2 = 1 + \frac{1}{4}(y^4 - 2 + y^{-4}) = \frac{1}{4}(y^4 + 2 + y^{-4}) = \left(\frac{1}{2}(y^2 + y^{-2})\right)^2 \)
\( L = \int_{1}^{2} \frac{1}{2}(y^2 + y^{-2}) \, dy = \frac{1}{2} \left[ \frac{1}{3}y^3 - y^{-1} \right]_{1}^{2} \)
\( L = \frac{1}{2} \left( (\frac{8}{3} - \frac{1}{2}) - (\frac{1}{3} - 1) \right) = \frac{1}{2} (\frac{13}{6} + \frac{2}{3}) = \frac{1}{2} (\frac{17}{6}) = \frac{17}{12} \)
Conceptual Challenge Answer
For \( f(x) = \sqrt[3]{x} \), the derivative \( f'(x) = \frac{1}{3}x^{-2/3} \) is undefined (infinite) at \( x = 0 \). This means the curve has a vertical tangent line at the origin, violating the requirement for a continuous derivative on the interval.
Resolution: You can solve this by switching variables. Let \( x = y^3 \). Then \( dx/dy = 3y^2 \), which is continuous everywhere. Integrate with respect to \( y \) from \( y = -1 \) to \( y = 1 \).
The Outer Shell Slides The Outer Shell
Surface Area of Revolution
Wrapping Problems
Peeling a label off a soup can is easy: it's just a rectangle.
"But what about a lightbulb? A football? A vase?"
We need a way to measure the area of curved 3D surfaces created by spinning a line.
Axis of Revolution
The Frustum
Just as arc length uses chords , surface area uses frustums (cones with the tops cut off).
Area of a Frustum:
\( S = 2\pi r L \)
where \( r \) is the average radius and \( L \) is the slant height.
L
The Integral Formula
By summing infinite tiny frustums, we get the total Surface Area of Revolution about the x-axis:
\[ S = \int_{a}^{b} 2\pi f(x) \sqrt{1 + [f'(x)]^2} \, dx \]
Remember: \( 2\pi \cdot (\text{Radius}) \cdot (\text{Arc Length}) \)
Which Axis?
About the X-axis
Radius is the height of the curve:
\( r = y = f(x) \)
About the Y-axis
Radius is the distance from the y-axis:
\( r = x \)
Choose your variable carefully! Sometimes \( dy \) is easier than \( dx \).
Practice: The Parabola
Revolve \( y = x^3 \) from \( x = 0 \) to \( x = 1 \) about the x-axis .
1. Radius
\( r = x^3 \)
2. Arc Segment
\( dL = \sqrt{1 + (3x^2)^2} \, dx \)
3. Setup
\( S = \int_{0}^{1} 2\pi x^3 \sqrt{1 + 9x^4} \, dx \)
Solution Tip: Use u-substitution with \( u = 1+9x^4 \).
Shell Surface Workshop Worksheet Shell Surface Workshop
Applications of Integration • Lesson 2
Student Name
About X-Axis:
About Y-Axis:
1. The Parabolic Bowl
Find the exact surface area of the solid generated by revolving the curve \( y = \sqrt{x} \) from \( x = 0 \) to \( x = 4 \) about the x-axis .
2. The Vertical Revolution
Find the surface area generated by revolving the line segment \( y = 2x \) from \( x = 0 \) to \( x = 2 \) about the y-axis .
3. The Tricky Arc
Consider the curve \( y = \frac{x^3}{6} + \frac{1}{2x} \). If this arc is revolved about the x-axis on the interval \( [1, 2] \), determine the resulting surface area.
Result:
Extension
Compare the surface area of a cylinder (radius \( r \), height \( h \)) to the surface area generated by revolving the line segment \( y = r \) from \( x = 0 \) to \( x = h \) about the x-axis. Why do the formulas match?
Shell Surface Workshop Key Solution Guide
Shell Surface Workshop
Answer Key
01
The Parabolic Bowl
\( y = \sqrt{x} \implies y' = \frac{1}{2\sqrt{x}} \)
\( \sqrt{1 + (y')^2} = \sqrt{1 + \frac{1}{4x}} = \sqrt{\frac{4x+1}{4x}} = \frac{\sqrt{4x+1}}{2\sqrt{x}} \)
\( S = \int_{0}^{4} 2\pi \sqrt{x} \left( \frac{\sqrt{4x+1}}{2\sqrt{x}} \right) \, dx = \int_{0}^{4} \pi \sqrt{4x+1} \, dx \)
Use u-sub: \( u = 4x+1, du = 4dx \). Bounds: \( 0 \to 1, 4 \to 17 \).
\( S = \frac{\pi}{4} \int_{1}^{17} u^{1/2} \, du = \frac{\pi}{4} \left[ \frac{2}{3}u^{3/2} \right]_{1}^{17} = \frac{\pi}{6}(17\sqrt{17} - 1) \approx 36.177 \)
02
The Vertical Revolution
Revolve \( y = 2x \) about y-axis. Rewrite as \( x = \frac{1}{2}y \). Bounds for \( y \) are \( [0, 4] \).
\( x' = \frac{1}{2} \implies \sqrt{1 + (x')^2} = \sqrt{1 + 1/4} = \frac{\sqrt{5}}{2} \)
\( S = \int_{0}^{4} 2\pi (\frac{1}{2}y) \left( \frac{\sqrt{5}}{2} \right) \, dy = \int_{0}^{4} \frac{\pi\sqrt{5}}{2} y \, dy \)
\( S = \frac{\pi\sqrt{5}}{2} \left[ \frac{1}{2}y^2 \right]_{0}^{4} = \frac{\pi\sqrt{5}}{2} \cdot 8 = 4\pi\sqrt{5} \approx 28.123 \)
03
The Tricky Arc
Note: This derivative setup is similar to the arc length worksheet.
\( y = \frac{1}{6}x^3 + \frac{1}{2}x^{-1} \implies y' = \frac{1}{2}(x^2 - x^{-2}) \implies \sqrt{1 + (y')^2} = \frac{1}{2}(x^2 + x^{-2}) \)
\( S = \int_{1}^{2} 2\pi (\frac{x^3}{6} + \frac{1}{2x}) \left( \frac{1}{2}(x^2 + x^{-2}) \right) \, dx \)
\( S = \pi \int_{1}^{2} (\frac{x^3}{6} + \frac{1}{2x})(x^2 + \frac{1}{x^2}) \, dx = \pi \int_{1}^{2} (\frac{x^5}{6} + \frac{x}{6} + \frac{x}{2} + \frac{1}{2x^3}) \, dx \)
\( S = \pi \int_{1}^{2} (\frac{1}{6}x^5 + \frac{2}{3}x + \frac{1}{2}x^{-3}) \, dx \)
\( S = \pi [ \frac{1}{36}x^6 + \frac{1}{3}x^2 - \frac{1}{4}x^{-2} ]_{1}^{2} = \pi [ (\frac{64}{36} + \frac{4}{3} - \frac{1}{16}) - (\frac{1}{36} + \frac{1}{3} - \frac{1}{4}) ] = \frac{47\pi}{16} \)
The Balance Point Slides The Balance Point
Moments & Center of Mass in 1D
The Seesaw Physics
Why does a 50kg child balance a 100kg adult if they sit further from the center?
Torque = Force × Distance
In calculus, we call this a Moment .
100
50
Variable Density
Not every rod has the same weight throughout. Imagine a rod made of lead on one end and wood on the other.
Let \( \rho(x) \) be the density function (mass per unit length).
Mass \( M = \int_{a}^{b} \rho(x) \, dx \)
Finding the Pivot
The Moment about the Origin is the sum of (mass × position):
\( M_0 = \int_{a}^{b} x \cdot \rho(x) \, dx \)
The Center of Mass \( \bar{x} \) is the point where the total mass balances:
\[ \bar{x} = \frac{M_0}{M} \]
Guided Problem
A rod lies on the x-axis from \( x=0 \) to \( x=4 \). Its density is \( \rho(x) = 2x + 1 \). Find \( \bar{x} \).
Step 1: Total Mass
\( M = \int_{0}^{4} (2x+1) \, dx \)
\( M = [x^2 + x]_0^4 = 16 + 4 = 20 \)
Step 2: Moment \( M_0 \)
\( M_0 = \int_{0}^{4} x(2x+1) \, dx \)
\( M_0 = \int_{0}^{4} (2x^2 + x) \, dx \)
\( M_0 = [\frac{2}{3}x^3 + \frac{1}{2}x^2]_0^4 = \frac{128}{3} + 8 = \frac{152}{3} \)
\( \bar{x} = \frac{152/3}{20} = \frac{38}{15} \approx 2.53 \)
Sanity Check
Does our answer make sense? The density \( \rho(x) = 2x + 1 \) increases as \( x \) increases. The rod is heavier on the right side.
Conclusion: \( \bar{x} \) should be greater than the midpoint (2).
Result \( 2.53 > 2 \). Logic holds!
Linear Mass Lab Worksheet Linear Mass Lab
Applications of Integration • Lesson 3
Student Name
Key Formulas
\( M = \int \rho(x) \, dx \)
\( M_0 = \int x\rho(x) \, dx \)
\( \bar{x} = M_0 / M \)
1
Uniform Rod
A rod of length 6 cm has a constant density \( \rho(x) = 5 \) g/cm. Prove using integration that the center of mass is at its geometric center (\( x = 3 \)).
2
Tapered Density
A thin wire is located on the interval \( [1, 3] \) with a linear density function \( \rho(x) = 2x - 1 \). Find the total mass and the center of mass.
Total Mass calculation
Moment calculation
\( M = \)
\( \bar{x} = \)
3
Heavy Endings
A rod on the interval \( [0, \pi] \) has a density of \( \rho(x) = \sin(x) \). Without calculating, where do you expect the center of mass to be? Then, calculate the exact value of \( \bar{x} \) to verify.
Prediction: ____________________________________________________________________________________
Calculus Verification
The Physics Connection
If you cut the rod at \( x = \bar{x} \), will the two pieces have the same mass? Why or why not?
Linear Mass Lab Key Grading Guide
Linear Mass Lab
Answer Key
1
Uniform Rod
\( M = \int_{0}^{6} 5 \, dx = [5x]_0^6 = 30 \)
\( M_0 = \int_{0}^{6} x(5) \, dx = [\frac{5}{2}x^2]_0^6 = \frac{5}{2}(36) = 90 \)
\( \bar{x} = 90/30 = 3 \). (Verified!)
2
Tapered Density
Mass
\( M = \int_{1}^{3} (2x-1) \, dx \)
\( M = [x^2 - x]_1^3 = (9-3) - (1-1) \)
\( M = 6 \)
Moment
\( M_0 = \int_{1}^{3} (2x^2 - x) \, dx \)
\( M_0 = [\frac{2}{3}x^3 - \frac{1}{2}x^2]_1^3 \)
\( M_0 = (18 - 4.5) - (\frac{2}{3} - 0.5) = 13.5 - \frac{1}{6} = \frac{40}{3} \)
\( \bar{x} = \frac{40/3}{6} = \frac{20}{9} \approx 2.22 \)
3
Heavy Endings
Prediction: Because \( \sin(x) \) is symmetric about \( x = \pi/2 \), we expect the center of mass to be at \( \pi/2 \).
\( M = \int_{0}^{\pi} \sin x \, dx = [-\cos x]_0^\pi = -(-1) - (-1) = 2 \)
\( M_0 = \int_{0}^{\pi} x \sin x \, dx \) (Use Integration by Parts: \( u=x, dv=\sin x dx \))
\( M_0 = [-x \cos x]_0^\pi + \int_{0}^{\pi} \cos x \, dx = \pi + [\sin x]_0^\pi = \pi \)
\( \bar{x} = \pi/2 \approx 1.57 \)
Reflection Answer
No , the masses of the two pieces are not necessarily equal. The center of mass is the balance point where the total torque (moment) is zero. One side might be shorter but denser, while the other is longer but lighter. They balance each other out in terms of rotation, but their absolute masses could differ.
The Geometric Heart Slides The Geometric Heart
Centroids of 2D regions
The Balancing Act
If you cut a shape out of cardboard, where could you place your finger so it balances perfectly?
This point \( (\bar{x}, \bar{y}) \) is called the Centroid .
\( (\bar{x}, \bar{y}) \)
Shortcut: Symmetry
If a region has an axis of symmetry , the centroid must lie on that axis.
If there are two axes , the centroid is at their intersection.
Uniform density (a lamina ) means we only care about geometry, not weight!
The 2D Integral Formulas
Horizontal Balance (\( \bar{x} \))
\( \bar{x} = \frac{1}{A} \int_{a}^{b} x [f(x) - g(x)] \, dx \)
Total Moment about Y / Area
Vertical Balance (\( \bar{y} \))
\( \bar{y} = \frac{1}{A} \int_{a}^{b} \frac{1}{2} [f(x)^2 - g(x)^2] \, dx \)
Total Moment about X / Area
Conceptual Breakdown
Think of a vertical rectangle at position \( x \):
Its area is \( [f(x) - g(x)] \, dx \).
Its x-position is simply \( x \).
Its y-center is the average: \( \frac{f(x) + g(x)}{2} \).
Moment of the strip about X-axis:
\( dM_x = \text{Area} \times \text{Center}_y \)
\( dM_x = [f(x)-g(x)] \cdot \frac{f(x)+g(x)}{2} \, dx \)
Center of Strip
Centroid Workshop Worksheet Centroid Workshop
Applications of Integration • Lesson 4
Student Name
A
Simple Regions
Find the centroid of the region bounded by \( y = 4 - x^2 \) and the x-axis.
Observation: Is this region symmetric? If so, what is the immediate value of \( \bar{x} \)?
Calculation of Area (A)
Calculation of \( \bar{y} \)
Centroid: ( _______ , _______ )
B
Between Two Curves
Find the centroid of the region bounded by \( y = x \) and \( y = x^2 \).
Step 1: Area and Limits of Integration
Step 2: Moment about Y-axis (\( M_y \))
Step 3: Moment about X-axis (\( M_x \))
Final Centroid Coordinates
x =
y =
Physics Check
Is it possible for the centroid of a region to lie outside the boundaries of the region itself? Provide a sketch or example of a shape where this might occur (think of a "C" or "U" shape).
Centroid Workshop Key Instructor Key
Centroid Workshop
Reference Only
A
Simple Regions
Region: \( y = 4 - x^2 \) and x-axis. Limits: \( x = -2 \) to \( x = 2 \).
Symmetry: The region is symmetric about the y-axis, so \( \bar{x} = 0 \).
Area
\( A = \int_{-2}^{2} (4 - x^2) \, dx = [4x - \frac{1}{3}x^3]_{-2}^{2} = (8 - 8/3) - (-8 + 8/3) = 16 - 16/3 = 32/3 \)
Moment about X-axis (\( M_x \))
\( M_x = \int_{-2}^{2} \frac{1}{2}(4-x^2)^2 \, dx = \frac{1}{2} \int_{-2}^{2} (16 - 8x^2 + x^4) \, dx \)
\( M_x = \frac{1}{2} [16x - \frac{8}{3}x^3 + \frac{1}{5}x^5]_{-2}^{2} = \frac{1}{2} [ (32 - 64/3 + 32/5) - (-32 + 64/3 - 32/5) ] = 32 - 64/3 + 32/5 = \frac{256}{15} \)
\( \bar{y} = \frac{256/15}{32/3} = \frac{256}{15} \cdot \frac{3}{32} = \frac{8}{5} = 1.6 \). Centroid: (0, 1.6)
B
Between Two Curves
Bounds: \( y = x \) (top) and \( y = x^2 \) (bottom). Intersection at \( x = 0, 1 \).
\( A = \int_{0}^{1} (x - x^2) \, dx = [\frac{1}{2}x^2 - \frac{1}{3}x^3]_0^1 = 1/6 \)
\( M_y = \int_{0}^{1} x(x - x^2) \, dx = \int_{0}^{1} (x^2 - x^3) \, dx = [\frac{1}{3}x^3 - \frac{1}{4}x^4]_0^1 = 1/12 \)
\( M_x = \int_{0}^{1} \frac{1}{2}(x^2 - (x^2)^2) \, dx = \frac{1}{2} \int_{0}^{1} (x^2 - x^4) \, dx = \frac{1}{2} [\frac{1}{3}x^3 - \frac{1}{5}x^5]_0^1 = \frac{1}{2} (2/15) = 1/15 \)
\( \bar{x} = \frac{1/12}{1/6} = 1/2 = 0.5 \)
\( \bar{y} = \frac{1/15}{1/6} = 6/15 = 2/5 = 0.4 \)
Centroid: (0.5, 0.4)
Physics Check Answer
Yes . The centroid can lie outside the region. This is common for non-convex shapes like a horseshoe, a ring (annulus), or a U-shaped pipe. The "average" position of the geometry is in the empty space between the arms of the shape.
The Pappus Shortcut Slides The Pappus Shortcut
Unifying Centroids, Surface Area, and Volume
The Doughnut Challenge
How quickly can you calculate the volume of a torus (doughnut)?
Standard calculus (Washer Method) requires a complex integral.
Pappus's Theorem lets you do it in seconds with simple geometry.
Centroid travels in a circle
Theorem 1: Volume
The volume \( V \) of a solid of revolution is equal to the area \( A \) of the generating region multiplied by the distance \( d \) traveled by its centroid.
\[ V = A \cdot d \]
\[ V = A \cdot (2\pi \bar{r}) \]
Theorem 2: Surface Area
The surface area \( S \) is equal to the arc length \( L \) of the generating curve multiplied by the distance \( d \) traveled by its centroid.
\[ S = L \cdot d \]
\[ S = L \cdot (2\pi \bar{r}) \]
Application: The Torus
A circle of radius \( r \) is revolved around an axis at distance \( R \) (where \( R > r \)).
Step 1: Area of circle \( A = \pi r^2 \)
Step 2: Radius of centroid \( \bar{r} = R \)
Step 3: Distance traveled \( d = 2\pi R \)
Volume \( V = (\pi r^2)(2\pi R) = 2\pi^2 r^2 R \)
Why use Pappus?
Simplicity
Avoids messy integrals of radicals and squares if you already know the area and centroid.
Unified Logic
Links centroids (Lesson 4), Arc Length (Lesson 1), and Surface Area (Lesson 2) into one theorem.
"Geometry is the skeleton of calculus."
The Pappus Challenge Worksheet The Pappus Challenge
Applications of Integration • Lesson 5
Student Name
Theorem Recall
Theorem for Volume
\( V = 2\pi \bar{r} A \)
Theorem for Surface Area
\( S = 2\pi \bar{r} L \)
\( \bar{r} \) is the perpendicular distance from the axis of revolution to the centroid of the generating region (or curve).
01
The Instant Torus
Find the volume and surface area of a torus generated by revolving a circle of radius 2 centered at (5, 0) about the y-axis .
Volume Analysis
\( V = \)
Surface Area Analysis
\( S = \)
02
Triangle Revolution
Consider a triangle with vertices at (1, 1), (3, 1), and (2, 4). If this triangle is revolved about the x-axis , what is the volume of the resulting solid?
Part A: Calculate the Area of the Triangle
Part B: Find the Centroid (\( \bar{y} \)) of the Triangle
Part C: Apply Pappus's Theorem
03
The Reverse Logic
A solid of revolution has a volume of \( 100\pi \) and was generated by revolving a region with area 10 about the line \( x = -2 \). How far from the y-axis is the centroid of the generating region located?
Distance from y-axis:
The Pappus Challenge Key Master Key
The Pappus Challenge
Teacher Solution
01
The Instant Torus
Volume
\( A = \pi (2^2) = 4\pi \)
\( \bar{r} = 5 \) (distance from y-axis)
\( d = 2\pi(5) = 10\pi \)
\( V = 4\pi \cdot 10\pi = 40\pi^2 \)
Surface Area
\( L = 2\pi(2) = 4\pi \)
\( \bar{r} = 5 \)
\( d = 2\pi(5) = 10\pi \)
\( S = 4\pi \cdot 10\pi = 40\pi^2 \)
02
Triangle Revolution
Vertices: (1, 1), (3, 1), (2, 4).
Area: Base = 2, Height = 3. \( A = \frac{1}{2}(2)(3) = 3 \).
Centroid \( \bar{y} \): Average of y-coordinates: \( (1+1+4)/3 = 6/3 = 2 \).
Revolves about x-axis: \( \bar{r} = \bar{y} = 2 \).
\( V = A \cdot (2\pi \bar{r}) = 3 \cdot (2\pi \cdot 2) = 12\pi \approx 37.7 \)
03
The Reverse Logic
\( V = A \cdot d \implies 100\pi = 10 \cdot d \implies d = 10\pi \)
\( d = 2\pi \bar{r} \implies 10\pi = 2\pi \bar{r} \implies \bar{r} = 5 \)
The radius of revolution is 5. The axis is at \( x = -2 \).
If the axis is \( x = -2 \), and the radius is 5, the x-coordinate of the centroid is \( -2 + 5 = 3 \).
Distance from y-axis = 3 units.
Teaching Tip: Highlight how Problem 2 is much simpler with Pappus than with the disk/washer method. To use washers, students would have to split the integral into two parts (from \( x=1 \) to \( x=2 \) and \( x=2 \) to \( x=3 \)). Pappus bypasses this entirely!
