Calculus II ep04: Volume of revolution (Sep 16, 2026)

Overview

Prof Staecker develops the disk and washer methods for volumes of revolution, starting from the cylinder formula and showing how to set up integrals with radii measured from the axis of rotation. Examples derive the volume formulas for a radius-2 sphere and a cone, handle rotation around the y-axis, and use washers to find volumes between two curves.

Key takeaways

For rotation around the x-axis, a solid with no central hole has volume π∫ₐᵇ[f(x)]² dx; the function value is the disk radius.
The standard sphere and cone volume formulas follow from disk integration: a radius-2 sphere has volume 32π/3, and a cone has volume (1/3)πr²h.
For rotation around the y-axis, rewrite the boundary as x in terms of y and integrate with dy; revolving y = x² from y = 0 to 1 produces volume π/2.
Washer cross-sections require subtracting squared radii, not radii: the integrand is π(R_outer² − R_inner²).
For curves y = x and y = x² on [0, 1], the outer radius is x and the inner radius is x², yielding volume 2π/15 around the x-axis.

Chapters

0:00 Homework, GradeScope, and Quiz Logistics

Homework is due tonight; Prof Staecker asks students to report any GradeScope access or submission problems.
Tomorrow's quiz begins class, lasts about 15 minutes, and must be completed on paper and submitted as phone photos through GradeScope.
The two quiz questions cover basic derivatives and integrals, then u-substitution; the second requires substitution and the first does not.

1:41 Disk Method: Volume from Cylinder Slices

Revolving a graph y = f(x) from x = a to x = b around the x-axis creates a solid whose volume is π∫ₐᵇ[f(x)]² dx.
The formula comes from summing thin cylindrical slices: each slice has circular area πr² and thickness dx.
Prof Staecker recommends memorizing the formula for an upcoming quiz.

3:54 A Parabolic Horn from Revolving y = x²

Revolving y = x² over 0 ≤ x ≤ 1 around the x-axis forms a horn-shaped solid.
Substitution into the disk formula gives V = π∫₀¹(x²)² dx = π∫₀¹x⁴ dx.
Integrating x⁴ yields x⁵/5, so the solid's volume is π/5.

7:42 Deriving the Radius-2 Sphere Volume

A sphere of radius 2 is formed by revolving the upper semicircle x² + y² = 4, or y = √(4 − x²), over −2 ≤ x ≤ 2.
The disk integral is π∫₋₂²(√(4 − x²))² dx; squaring the radius cancels the square root.
Integrating 4 − x² and evaluating at −2 and 2 gives 32π/3, matching the sphere formula (4/3)πr³ for r = 2.

17:39 Deriving the Cone Formula with a Linear Radius

A cone of radius r and height h comes from revolving the line y = (r/h)x over 0 ≤ x ≤ h around the x-axis.
The disk integral is π∫₀ʰ((r/h)x)² dx, with the constants r²/h² factored outside.
Integrating x² gives h³/3 at the upper bound; canceling h² leaves V = (1/3)πr²h.

27:15 Rotating Around the y-Axis: A Paraboloid Basin

For y = x² revolved around the y-axis from y = 0 to y = 1, rewrite the curve as x = √y.
Use y-bounds and dy because the slices are taken perpendicular to the y-axis.
The disk setup π∫₀¹(√y)² dy simplifies to π∫₀¹y dy, giving a basin volume of π/2.
Only the positive square root is needed: revolving that half of the graph around the y-axis generates the full symmetric solid.

32:36 Washer Method: Subtracting the Central Hole

When a revolved region leaves a hollow center, each cross-section is a washer, also called an annulus.
A washer's area is π(R_outer² − R_inner²), found by subtracting the inner circle's area from the outer circle's area.
For rotation around the x-axis, integrate the squared outer radius minus the squared inner radius, multiplying by π.

40:41 Worked Washer Example: y = 2x Outside y = x

On 0 ≤ x ≤ 1, revolving the region between y = 2x and y = x around the x-axis gives outer radius 2x and inner radius x.
The setup is π∫₀¹[(2x)² − x²] dx = π∫₀¹3x² dx.
Evaluating the integral gives a volume of π.

44:28 Practice Washer Setup: Between y = x and y = x²

For 0 ≤ x ≤ 1, the region between y = x and y = x² is revolved around the x-axis.
Since x is greater than x² on this interval, the outer radius is x and the inner radius is x².
The volume is π∫₀¹(x² − x⁴) dx = π(1/3 − 1/5) = 2π/15.

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.