Calculus II ep03: Area between curves (Sep 14, 2026)

Overview

Prof Staecker develops the area-between-curves method from Riemann-sum rectangles: integrate the upper function minus the lower function, splitting at intersections when their order changes. He shows when to switch to horizontal slices and integrate right minus left, then introduces volumes of revolution using thin cylindrical slices to derive the disk formula.

Key takeaways

For vertical slices, area between curves is ∫ₐᵇ(top function − bottom function) dx; the bounds and which curve is on top must be determined from the graph.
When the upper and lower curves switch at an intersection, split the region into separate integrals; for y = 2x + 1 and y = −x, the switch occurs at x = −1/3.
Horizontal slices solve regions that are awkward to describe as single-valued y-functions: rewrite boundaries as x(y), integrate right minus left, and use y-values as bounds.
For y = ½x and x = y² − 3, horizontal slicing gives bounds −1 to 3 and integrand 2y − (y² − 3), for an area of 32/3.
Rotating y = f(x) about the x-axis creates circular cross-sections of radius f(x); integrating their areas gives V = π∫ₐᵇ[f(x)]² dx.
Prof Staecker advises using AI for assistance rather than submitting generated homework solutions, because quiz performance requires independently producing the reasoning.

Chapters

0:00 Course Website, Homework, and Class Resources

Prof Staecker directs students to the class website for homework assignments, accessible through the syllabus QR code or a Blackboard link.
Recordings and notes from earlier class meetings are linked on the website for students who miss class.

1:53 Prof Staecker’s Guidance on AI and Homework

Prof Staecker says copying AI-generated solutions into assignments counts as cheating, even though AI tools can solve many homework problems.
AI may be used for help, but reading an answer is not the same as being able to produce a solution independently on a quiz.
He recommends asking him for help when possible and cautions that AI can create a false sense of understanding.

4:31 Mastery Checklist and Quiz Practice Problems

The class website includes a replacement copy of the mastery checklist and a seven-page set of practice problems.
The practice set provides two representative problems for each section on the mastery list.
Prof Staecker recommends trying the relevant problems before the weekly Thursday quizzes, which do not allow calculators.

5:39 Deriving the Biggie-Minus-Smalls Area Formula

A region between curves can be approximated by narrow rectangles of width Δx and height f(xᵢ) − g(xᵢ).
Taking the Riemann-sum limit gives area = ∫ₐᵇ [f(x) − g(x)] dx when f is the upper curve and g is the lower curve.
Prof Staecker’s “biggie minus smalls” rule means subtracting the lower y-value from the higher y-value.

12:43 Finding the Area Between y = x and y = x²

The curves intersect where x = x², giving x = 0 and x = 1 as the integration bounds.
On 0 ≤ x ≤ 1, the line y = x lies above the parabola y = x², despite x² being larger for some other x-values.
The area is ∫₀¹(x − x²) dx = 1/6.

18:20 Splitting a Bow-Tie Region at the Line Intersection

For y = 2x + 1 and y = −x, the curves intersect at x = −1/3; the pictured region extends from x = −3 to x = 2.
The upper curve changes at x = −1/3, so one integral cannot use a single top-minus-bottom expression across the entire region.
Prof Staecker sets up separate integrals on [−3, −1/3] and [−1/3, 2], reversing which line is subtracted from which.

28:04 Why a Sideways Parabola Complicates Vertical Slices

The example combines y = ½x with the sideways parabola x = y² − 3.
The parabola has two y-values for many x-values, so representing it as a single top or bottom function of x can fail across the shaded region.
Solving for y with a square root captures only one branch unless both positive and negative roots are handled.

35:54 Switching to Horizontal Slices: Right Minus Left

Prof Staecker avoids the multiple-branch problem by integrating horizontally, expressing both curves as x in terms of y.
For horizontal slices, the area integrand is the rightmost x-value minus the leftmost x-value.
In the example, the line becomes x = 2y while the parabola remains x = y² − 3.

38:43 Solving Sideways Intersections and Setting Bounds

Equating x = 2y and x = y² − 3 gives y² − 2y − 3 = 0, which factors as (y − 3)(y + 1) = 0.
The intersection heights are y = −1 and y = 3, so the integral uses those y-bounds.
The area setup is ∫₋₁³ [2y − (y² − 3)] dy, using the line as the right boundary and the parabola as the left.

45:13 Student Practice: Intersections and Sideways Regions

Prof Staecker gives students two shaded-area exercises and time to work collaboratively before reviewing the setups.
The first exercise requires finding curve intersections and splitting the region where the upper and lower functions switch.
The second exercise practices horizontal integration for a region bounded by a sideways parabola and a line.

56:08 Solving the Two-Lobed Region Between Quadratics and a Line

For y = 3x − x² and y = 3 − x, intersections occur at x = 1 and x = 3; the shaded region also begins at x = 0.
The top curve changes at x = 1, so the setup uses integrals over [0, 1] and [1, 3] with the order of subtraction reversed.
Each lobe has area 4/3, giving total shaded area 8/3.

1:00:45 Horizontal-Slice Practice with a Parabola and a Line

The second practice region is bounded by x = y² and the line x = y − 1, with y-values from 0 to 1.
The parabola is the right boundary and the line is the left boundary, so the integrand is y² − (y − 1).
Evaluating ∫₀¹(y² − y + 1) dy gives an area of 5/6.

1:04:24 Introducing Volumes of Revolution Around the x-Axis

Prof Staecker introduces volumes formed by rotating a curve segment from x = a to x = b around the x-axis.
The resulting three-dimensional shape can resemble a rounded, symmetrical object, and its volume can be found by adding thin slices.
The section shifts from two-dimensional area applications to volume calculations using integrals.

1:07:54 Modeling Revolution Slices as Thin Cylinders

A rotation slice has thickness Δx and a circular cross-section, making each piece a short cylinder.
The cylinder radius is the function value f(xᵢ), the distance from the x-axis to the curve.
Using cylinder volume πr²h, each slice has approximate volume π[f(xᵢ)]²Δx.

1:13:35 The Disk Formula for Volume of Revolution

Adding the cylindrical slices and taking the Riemann-sum limit yields V = ∫ₐᵇ π[f(x)]² dx.
The radius is squared because each cross-section’s area is πr²; the slice thickness becomes dx.
Prof Staecker recommends memorizing the formula and remembering its origin in circular cross-sections.

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Prof Staecker.