Vertical Slicing Slides VERTICAL SLICING
Area Between Two Curves
The Logo Challenge
Imagine you're designing a new tech logo. The shaded region is bounded by two distinct parabolas:
f(x) = -x² + 4 (Upper)
How can we calculate the exact area of this shape to determine ink costs?
The Calculus Connection
Vertical Slicing
We slice the region into infinite rectangles of width \(dx\).
Height = f(x) - g(x)
"Top function minus Bottom function"
The Formula
\[ A = \int_{a}^{b} [f(x) - g(x)] \, dx \]
Where \(a\) and \(b\) are the x-coordinates of the intersection points.
The Blueprint
1
Find Intersections
Set \(f(x) = g(x)\) to find your limits \(a\) and \(b\).
2
Identify Top vs. Bottom
Check which function is greater on the interval \([a, b]\).
3
Set Up & Integrate
Integrate the difference: \( \int_{a}^{b} (Top - Bottom) \, dx \)
Warning: Collision Ahead!
Curves Crossing?
If the curves cross within your interval, the "Top" and "Bottom" swap places!
The Fix: Split the integral at the intersection point and use absolute values or separate setups.
[Visual: S-curve crossing a line]
Don't let negative area cancel your hard work.
Curve Capture Worksheet CURVE CAPTURE
Area Between Curves: Vertical Slicing
Name:
Date:
The Mission: Use your integration skills to determine the exact area of the regions captured between two functions. For each problem, sketch the region, identify the upper and lower functions, and set up the integral before evaluating.
1
Simple Capture
Find the area of the region bounded by:
f(x) = x² + 2
Integral Setup:
Sketch Area
Show Your Work:
2
Intersecting Boundaries
Find the area of the region bounded by the curves:
f(x) = 2 - x²
Note: You must find the intersection points first!
Intersections:
Integral Setup:
Sketch Area
Evaluation & Solution:
3
The S-Curve Crossing
Calculate the total area bounded by \( f(x) = x^3 \) and \( g(x) = x \).
Think about the symmetry and where the functions cross!
Curve Capture Answer Key Answer Key
Curve Capture: Vertical Slicing
Teacher Reference
Calculus Applications
1
Simple Capture Solution
Setup
\[ \int_{0}^{3} (x^2 + 2 - x) \, dx \]
Antiderivative
\[ \left[ \frac{1}{3}x^3 + 2x - \frac{1}{2}x^2 \right]_0^3 \]
Final Answer
\( 13.5 \) or \( \frac{27}{2} \) units²
Teaching Tip:
Ensure students correctly identify \(x^2+2\) as the upper function. Since parabolas with positive leading coefficients open upwards and the vertex is at (0,2), it stays above \(y=x\) on the interval [0,3].
2
Intersecting Boundaries Solution
1. Find Intersections
\( 2 - x^2 = x \implies x^2 + x - 2 = 0 \)
2. Setup & Solve
\[ \int_{-2}^{1} (2 - x^2 - x) \, dx = \left[ 2x - \frac{x^3}{3} - \frac{x^2}{2} \right]_{-2}^{1} \]
Final Answer
\( 4.5 \) or \( \frac{9}{2} \) units²
Grading Note:
Look for the identification of intersection points. Many students forget to find limits and assume they are given. Point out the parabola is "on top" because its vertex is higher and it opens down.
3
The S-Curve Crossing Solution
The functions intersect at \( x = -1, 0, 1 \). Due to symmetry, students can compute the area on [0,1] and double it.
Split Setup
\[ \int_{-1}^{0} (x^3 - x) dx + \int_{0}^{1} (x - x^3) dx \] OR \[ 2 \int_{0}^{1} (x - x^3) dx \]
Final Result
\( 0.5 \) or \( \frac{1}{2} \) units²
Horizontal Slicing Slides HORIZONTAL SHIFT
Integrating with Respect to Y
The Vertical Nightmare
Consider the region bounded by:
x = y² - 2
If we try vertical slicing (\(dx\)), we must split the region and solve for \(y\) in terms of \(x\)—which involves ugly square roots!
There has to be a better way...
The Horizontal Pivot
New Perspective
Instead of vertical rectangles, we use horizontal strips with width \(dy\).
Length = Right - Left
The dy Formula
\[ A = \int_{c}^{d} [f(y) - g(y)] \, dy \]
"Integrate from bottom boundary \(c\) to top boundary \(d\)."
Which Slice is Right?
Vertical (dx)
Functions are \(y = f(x)\)
Upper/Lower stay consistent
Multi-part intervals (need 2+ integrals)
Horizontal (dy)
Functions are \(x = f(y)\)
Right/Left stay consistent
Simplifies complex x-boundaries
Horizontal Shift Worksheet Horizontal Shift
Integrating With Respect To Y
Student Name:
Technique Spotlight
When curves are defined as functions of \(y\) (e.g., \(x = f(y)\)), or when vertical slicing requires splitting an integral, horizontal slicing is your most efficient tool. Remember the mantra: Right Curve minus Left Curve from \(y=c\) to \(y=d\).
01 Sideways Parabola
Standard Application
Find the area of the region bounded by:
x = y² - 4
Intersection Points (find y):
Definite Integral Setup:
Region Sketch
Detailed Evaluation:
02 Strategic Selection
Analyze First
Given the region bounded by \( y = \sqrt{x} \), \( y = x - 2 \), and the x-axis:
The Problem
Sketch the region. Why is vertical slicing inefficient here? How does horizontal slicing simplify the setup?
Sketch here
1. Convert functions to x = f(y)
2. Setup & Evaluate Area
Final Answer (Numerical):
Horizontal Shift Answer Key Answer Key
Horizontal Shift: Integrating With Respect To Y
Teacher Reference
Calculus Applications
01
Sideways Parabola Solution
Find Intersections
\( y^2 - 4 = 2 - y \)
Setup (Right - Left)
\[ \int_{-3}^{2} [(2 - y) - (y^2 - 4)] \, dy = \int_{-3}^{2} (6 - y - y^2) \, dy \]
Final Answer
\( 20.833... \) or \( \frac{125}{6} \) units²
Evaluation Check:
\( [6y - \frac{y^2}{2} - \frac{y^3}{3}]_{-3}^{2} \)
\( (12 - 2 - \frac{8}{3}) - (-18 - \frac{9}{2} + 9) \)
Pedagogy Note:
Remind students that \(x=2-y\) is the "Right" function because for any \(y\), its x-value is greater on the interval of interest.
02
Strategic Selection Solution
1. Conversion
\( y = \sqrt{x} \implies x = y^2 \)
2. Limits & Setup
Intersects at \( y^2 = y + 2 \implies y = 2, -1 \). But the region is also bounded by the x-axis (\(y=0\)). Thus limits are \(0\) to \(2\).
\[ \int_{0}^{2} [(y + 2) - y^2] \, dy \]
The "Efficiency" Argument:
Vertical slicing would require two integrals: \( \int_{0}^{2} \sqrt{x} \, dx + \int_{2}^{4} (\sqrt{x} - (x-2)) \, dx \).
Result
\( \frac{10}{3} \)
Disk Method Slides THE SPIN CYCLE
Volumes via The Disk Method
From 2D to 3D
Think of a potter's wheel .
A flat, 2D wooden tool (the profile) is held against spinning clay. As it rotates 360°, a perfectly symmetrical 3D vase emerges.
Calculus lets us measure the exact volume of that vase by summing infinite "disks" of clay.
Anatomy of a Slice
The Representative Disk
Shape: Cylinder (very thin)Radius (R): Distance from axis to functionThickness: \(dx\) (or \(dy\))
Disk Volume Formula
\[ V = \pi \int_{a}^{b} [R(x)]^2 \, dx \]
Because Volume of a cylinder is \( \pi r^2 h \)
Choose Your Axis
Around X-Axis
Slices are vertical. Integrate with respect to \(x\).
Radius = f(x)
Around Y-Axis
Slices are horizontal. Integrate with respect to \(y\).
Radius = f(y)
Visualizing Rotation
[Place 3D Volume Animation Video Here]
Disk Drive Worksheet DISK DRIVE
Solid of Revolution Analysis
Operator:
Mission Date:
Core Formula
V = π ∫ [R(x)]² dx
The Objective: Generate solids by rotating the shaded regions about the specified axis. Set up the integral and calculate the total volume. Watch your units and remember your π!
01
Parabolic Spinner
Profile Function
y = √x
Bounded by: x = 4, y = 0
Rotate around X-Axis
Radius Function R(x):
Volume Integral:
Evaluation & Total Volume:
02
Linear Cone
Profile Function
y = 2x
Bounded by: y = 4, x = 0
Rotate around Y-Axis
Strategic Note
Since we rotate around the y-axis, you must convert the function to x = f(y).
Converted Function:
Radius Function R(y):
Volume Setup & Calculation:
Conceptual Check:
Explain why rotating \(y = 2x\) around the y-axis creates a cone, while rotating the same function around the x-axis would create a different solid.
Disk Drive Answer Key Answer Key
Disk Drive: Volumes of Revolution
Instructor Key
01
Parabolic Spinner Solution
Integral Setup
\( V = \pi \int_{0}^{4} (\sqrt{x})^2 \, dx \)
Evaluation
\( V = \pi [ \frac{1}{2}x^2 ]_{0}^{4} \)
Total Volume
8π
≈ 25.133 units³
02
Linear Cone Solution
Conversion & Setup
\( y = 2x \implies x = \frac{1}{2}y \)
Evaluation
\( V = \pi [ \frac{1}{12}y^3 ]_{0}^{4} \)
Total Volume
16π/3
≈ 16.755 units³
Teaching Tip: Help students visualize the cone's base at \(y=4\) and its tip at the origin. Remind them to always square the radius before integrating.
Washer Method Slides THE HOLE STORY
Rotational Volumes: The Washer Method
What if there's a gap?
The Disk Method works when a region is flush against the axis of rotation.
But what happens when the region is "floating"?
We get a hole.
We create a solid with a cavity—like a bead, a pipe, or a stack of CDs.
The Washer Slice
Subtracting the Void
The Outer & Inner
R Outer Radius: Axis to the far curve
r Inner Radius: Axis to the near curve
The Washer Formula
\[ V = \pi \int_{a}^{b} ([R(x)]^2 - [r(x)]^2) \, dx \]
"Volume = Total Solid minus the Empty Core"
The Shifted Axis
What if we rotate around a line like y = -2 or y = 10?
Rule of Thumb:
Rotating around y = K (below region)
Radius = f(x) - K
Rotating around y = K (above region)
Radius = K - f(x)
Washer Wisdom Worksheet WASHER WISDOM
Complex Rotational Volumes
Student:
Washer Strategy
Identify your Outer Radius (R) and Inner Radius (r) . Subtract the squares, multiply by \(\pi\), and integrate over your bounds.
V = π ∫ (R² - r²) dx
1
The Hollow Bead
Region Definitions:
Upper: y = x² + 1
Outer Radius R(x):
Inner Radius r(x):
Integral Setup:
Evaluation Work:
2
Shifted Axis Challenge
Rotation about y = 5
Find the volume of the region bounded by \( y = \sqrt{x} \) and \( y = 0 \) for \( 0 \leq x \leq 4 \), rotated around the horizontal line \( y = 5 \).
Visual Aid
y = 5 (Axis)
Distance Axis to Furthest Point (R):
Distance Axis to Nearest Point (r):
Setup & Calculation:
Final Volume:
Washer Wisdom Answer Key Answer Key
Washer Wisdom: Instructor Solutions
Level: Advanced Integration
01
The Hollow Bead Solution
Radii Identification
R(x) = x² + 1 (Outer)
Setup
\[ V = \pi \int_{0}^{2} ((x^2 + 1)^2 - x^2) \, dx \] \[ V = \pi \int_{0}^{2} (x^4 + 2x^2 + 1 - x^2) \, dx \] \[ V = \pi \int_{0}^{2} (x^4 + x^2 + 1) \, dx \]
Final Calculation
136π/15
≈ 28.484 units³
Eval Check: \( \pi [ \frac{1}{5}x^5 + \frac{1}{3}x^3 + x ]_0^2 = \pi (\frac{32}{5} + \frac{8}{3} + 2) \)
02
Shifted Axis Challenge Solution
Radii (Axis at y=5)
Since the region (y=0 to y=√x) is below y=5:
R(x) = 5 - 0 = 5 (Outer)
Setup
\[ V = \pi \int_{0}^{4} [5^2 - (5 - \sqrt{x})^2] \, dx \] \[ V = \pi \int_{0}^{4} [25 - (25 - 10\sqrt{x} + x)] \, dx \] \[ V = \pi \int_{0}^{4} (10\sqrt{x} - x) \, dx \]
Final Result
136π/3
≈ 142.419 units³
Teaching Point: Emphasize that \(R\) is the distance to the furthest point on the region, which in this case is the x-axis (\(y=0\)).
Strategy Synthesis Slides STRATEGY SYNTHESIS
Mastering the Choice
Integration Flowchart
1
Area or Volume?
Are we looking for 2D coverage or 3D rotation?
2
dx or dy?
Which slice prevents splitting the integral? Right/Left vs Top/Bottom?
3
Disk or Washer?
For volumes: Is the region flush against the axis or "floating"?
Speed Setup Round
Ready?
Region: \( y = x^2 \) and \( y = x + 2 \)
Identify Method
Washer Method (dx)
The Bounds
\( x \in [-1, 2] \)
"First one to write the full integral on their whiteboard wins!"
Why Mastery Matters
Manufacturing Efficiency
Engineers use these methods to calculate material costs, fluid displacement, and structural weight of symmetrical parts.
3D Printing Paths
Slicing algorithms in 3D printers literally compute these integrals to determine layer paths.
Integration Ironman Challenge Worksheet INTEGRATION IRONMAN
The Final Synthesis Challenge
Participant:
Score:
Mission Briefing
In this mastery workshop, the goal is not just computation—it is strategic selection . For each scenario, analyze the region and rotation axis, then select the most efficient integration path. No points are awarded for multi-integral solutions if a single integral solution exists!
I
The Strategy Draft
Region Alpha
y = x²
1. Best Method for Area:
dx
dy
Explain your choice based on function orientation.
Region Beta
y = x - 1
1. Best Method for Area:
dx
dy
Why is the other method inefficient here?
II
The Rotational Gauntlet
The Setup
Rotate the region bounded by \( y = \frac{1}{x} \), \( y = 0 \), \( x = 1 \), and \( x = e \) about the y-axis .
HINT: Consider if Disk/Washer is viable vs the upcoming Shell Method (not yet learned). Can we use dy?
Method Selected:
Bounds (Lower/Upper):
Integral Expression (Set Up Only):
The Setup
Rotate the region bounded by \( y = \sqrt{x} \) and \( y = x^2 \) about the line y = -1 .
Method Selected:
Radii (R and r):
Integral Expression (Set Up Only):
Integration Ironman Answer Key Answer Key
Integration Ironman: Mastery Solutions
Final Synthesis Lesson
I
Strategy Draft Solutions
Region Alpha (y=x², x=y²)
Choice: Either (Equally Efficient)
Both functions are easily invertible. dx: \( \int_0^1 (\sqrt{x} - x^2) dx \)
Region Beta (y=x-1, y²=2x+6)
Choice: Horizontal (dy)
Reasoning: The left boundary is a single sideways parabola (\(x = \frac{1}{2}y^2 - 3\)). Using dx would require splitting the integral because the "bottom" function changes from the lower half of the parabola to the line.
II
The Rotational Gauntlet Solutions
Problem A (y-axis rotation)
Method & Setup
Method: Washer Method (dy)
Intersects at (1,1) and (e, 1/e). Bound by y=0. However, this region is tricky! From y=0 to y=1/e, it's a solid disk from x=1 to x=e. From y=1/e to y=1, it's a washer.
Setup (dy):
Problem B (Axis y = -1)
Radii Identification
Method: Washer (dx)
Axis is BELOW the region.
R(x) = (Upper Function) - (-1) = √x + 1
r(x) = (Lower Function) - (-1) = x² + 1
Integral Setup:
Instructor Note: These problems are designed to highlight edge cases. Problem A shows that "simple" looking regions can lead to complex multi-part integrals, setting the stage for the Shell Method in future units. Problem B tests radius shifting logic.
