Calculus II ep06: Volume with cylindrical shells (Sep 21, 2026)

Overview

Prof Staecker develops the cylindrical-shell method for volumes of revolution, emphasizing that each setup combines 2π, the distance from the rotation axis, and the vertical height of a shell; examples cover axes on either side of a region and heights bounded by two curves. The class then begins the textbook’s logarithm-first treatment of Section 6.2*: defining ln x as the integral of 1/t from 1 to x, deriving its 1/x derivative with the Fundamental Theorem of Calculus, and practicing chain, product, and quotient rules.

Key takeaways

A cylindrical-shell volume integral has the structure 2π∫(distance to the axis)(vertical height) dx; the radius is a geometric distance and may be x − a or a − x.
For a region between curves, shell height is upper function minus lower function; for y = x − x² and y = x² on [0, 1], it simplifies to 2x − 2x².
When an axis lies to the right of the region, as x = π/2 does for the region on [0, π/4], the shell radius is π/2 − x rather than x − π/2.
The textbook’s starred logarithm section defines ln x as ∫₁ˣ (1/t) dt, making ln 2 the area under y = 1/t and approximately 0.693.
The integral definition of ln immediately yields its derivative: by the Fundamental Theorem of Calculus, d(ln x)/dx = 1/x; for ln u, the chain rule gives u′/u.
Prof Staecker distinguishes quiz coverage: washers and area between curves are on the current week’s quiz, while cylindrical shells are deferred to the next week.

Chapters

0:00 Cylindrical Shells: Radius, Height, and the 2π Formula

Rotating a thin rectangle around the y-axis creates a cylindrical shell with thickness Δx.
The shell’s circumference is 2πx and its height is f(x), giving volume element 2πx f(x) dx.
Summing shells leads to the volume integral 2π∫ x f(x) dx for rotation about the y-axis.

6:00 Generalizing Shell Radius and Height Beyond the y-Axis

Prof Staecker reframes x in the shell formula as distance from the rotation axis, not necessarily the coordinate x.
For an axis other than the y-axis, the radius is a horizontal distance such as x − a or a − x.
Shell height may require subtracting a lower curve from an upper boundary rather than using f(x) alone.

9:00 Finding Shell Height Between y = 4 and y = x²

For the region between y = 4 and y = x² on x = 1 to x = 2, rotation is about the y-axis.
The shell radius is x, while the vertical height is top minus bottom: 4 − x².
The setup is 2π∫₁² x(4 − x²) dx; multiply out before integrating.

12:30 Rotating y = 2x − 1 Around the Vertical Line x = 1

The region under y = 2x − 1 from x = 2 to x = 3 is revolved around x = 1.
Its shell height is 2x − 1, measured from the curve down to y = 0.
Its radius is x − 1, so the volume setup is 2π∫₂³ (x − 1)(2x − 1) dx.

18:00 Combining Nonstandard Radius and Height for √x and x

The region between y = √x and y = x runs from their intersections at x = 0 and x = 1.
Rotating around x = −2 gives a shell radius of x − (−2) = x + 2.
Since √x is above x on [0, 1], the shell height is √x − x, producing 2π∫₀¹ (x + 2)(√x − x) dx.

21:00 When the Rotation Axis Lies to the Right of the Region

The region between sin x and cos x extends from x = 0 to their intersection at x = π/4.
For rotation around x = π/2, the radius is π/2 − x because the axis is to the right.
The height is cos x − sin x, so the setup is 2π∫₀^(π/4) (π/2 − x)(cos x − sin x) dx.
Expanding produces terms such as x cos x and x sin x, which require integration by parts.

25:30 Shell-Method Practice with Two Parabolas and Multiple Axes

Students set up rotations of the region between y = x − x² and y = x², whose intersections are x = 0 and x = 1.
The first exercise uses the y-axis; two more versions change the rotation axis to x = −3 and x = 2.
A separate exercise initially has a copied equation error; Prof Staecker corrects it to y = x² + 2x rather than y = 2x + 2.

41:30 Checking the Parabola-Region Shell Setups

For the region between y = x − x² and y = x² on [0, 1], the height is (x − x²) − x² = 2x − 2x².
The y-axis radius is x, giving 2π∫₀¹ x(2x − 2x²) dx; evaluating gives volume π/3.
For rotation around x = −3, replace the radius with x + 3; for rotation around x = 2, use 2 − x.
The height and integration interval stay the same across all three axis choices.

45:30 Sketching and Setting Up the Corrected x² + 2x Region

The corrected second region is bounded by y = x² + 2x and y = 3.
Solving x² + 2x = 3 gives (x + 3)(x − 1) = 0, so the limits are x = −3 and x = 1.
On that interval, the height is 3 − 2x − x²; rotation around x = −4 gives radius x + 4.
The setup is 2π∫₋₃¹ (x + 4)(3 − 2x − x²) dx; expanding produces 12 − 5x − 6x² − x³.

52:00 Shells Wrap-Up and Quiz Coverage

Prof Staecker concludes the volume-of-revolution section after covering both washers and cylindrical shells.
The current week’s quiz covers area between curves and washers, but not shells.
Cylindrical-shell problems are scheduled for the following week’s quiz.

53:30 Section 6.2*: Defining the Natural Logarithm by an Integral

The next three sections cover exponential and logarithmic functions using the textbook’s starred sections 6.2*, 6.3*, and 6.4*.
Unlike the usual approach of defining logarithms as inverses of exponentials, this presentation introduces logarithms first.
The definition is ln x = ∫₁ˣ (1/t) dt, the signed area under y = 1/t from 1 to x.
The area interpretation makes ln 2 approximately 0.693 and allows numerical estimation with Riemann sums or Simpson’s rule.

1:00:00 Deriving d(ln x)/dx = 1/x from the Fundamental Theorem

Differentiating ln x = ∫₁ˣ (1/t) dt uses the Fundamental Theorem of Calculus to give d(ln x)/dx = 1/x.
The integration variable t is distinct from the upper bound x, which becomes the input to the integrand.
For 4 ln(5x² + 1), the chain rule gives 4 · 10x/(5x² + 1).
Prof Staecker notes that logarithm derivatives usually require applying the chain rule to the expression inside ln.

1:05:00 Product Rule with Logarithms and In-Class Derivative Practice

For x⁵ ln(4x − x²), use the product rule and differentiate the logarithm’s argument by the chain rule.
The derivative is x⁵(4 − 2x)/(4x − x²) + 5x⁴ ln(4x − x²).
Students then practice derivatives combining logarithms with products, quotients, and nested functions.
The exercises review product, quotient, and chain rules alongside d(ln u)/dx = u′/u.

1:08:00 Solutions: Logarithmic Quotients and Nested Chain Rules

For a logarithmic quotient, Prof Staecker applies the quotient rule, differentiating the numerator and denominator separately.
For a nested expression involving ln and a square root, he rewrites the root as a 1/2 power to make the chain-rule layers explicit.
The nested example applies successive derivatives to the outer ln, the square-root power, and the inner sine function.
The class ends with reminders to arrange quiz makeups at least one day in advance.

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.