Difficulty: Intermediate to Advanced | Prerequisites: Areas between curves (Sections 6.1-6.2), basic integration techniques, comfort with visualising 3D shapes from 2D cross-sections.
TL;DR: You can find the volume of a 3D solid by slicing it into thin pieces (disks, washers, or shells) and integrating. The disk/washer method slices perpendicular to the axis of revolution; the shell method wraps cylindrical layers around it. Choosing the right method depends on which one gives you a cleaner integral.
Volumes of solids of revolution are the centrepiece of the "applications of integration" block in Calculus II. The idea is straightforward: take a flat region (the kind you found the area of in Sections 6.1-6.2), spin it around a line, and ask how much space the resulting 3D shape occupies. This is how engineers calculate volumes of tanks, nozzles, and any rotationally symmetric object. You should already be comfortable finding areas between curves and setting up definite integrals. The new skill here is geometric imagination: turning a 2D picture into a 3D solid and picking the slicing method that makes the integral simplest.
Solid of revolution
A 3D solid formed by rotating a 2D region around a fixed line (the axis of revolution). Think of it as spinning a flat shape on a lathe.
Axis of revolution
The line you rotate around. Common choices: the x-axis, y-axis, or a line like x = 2, y = -1. The axis does not have to touch the region.
Disk method
Used when the region has no hole after rotation. Each cross-section perpendicular to the axis is a solid circle (disk). Volume = π ∫ [R(x)]² dx, where R is the radius from the axis to the curve.
Washer method
Used when the region, once rotated, has a hole in the middle (like a donut cross-section). Each cross-section is a ring. Volume = π ∫ [R_outer² - R_inner²] dx. In simple terms, you subtract the inner hole from the outer disk.
Shell method (cylindrical shells)
Instead of slicing perpendicular to the axis, you wrap concentric cylinders around it. Volume = 2π ∫ (radius)(height) dx. This is often easier when the curves are functions of x but you are rotating around a vertical line, or vice versa.
Cross-sectional area method
For solids that are not surfaces of revolution. You know the base shape and the cross-section type (squares, equilateral triangles, semicircles, etc.) and integrate A(x) dx, where A(x) is the area of each cross-section.
Radius (in disk/washer problems)
The distance from the axis of revolution to the curve. When the axis is not the x- or y-axis itself, the radius is the curve's value minus the axis's position (or the reverse), not just the curve's value.
Disk/Washer Method (slicing perpendicular to the axis)
Identify the axis of revolution and the bounded region.
Determine the outer radius R(x) and, if there is a gap, the inner radius r(x). Each is the distance from the axis to the respective curve.
When the axis is not y = 0 or x = 0, adjust: if rotating around y = -1, and the curve is y = f(x), the radius is f(x) - (-1) = f(x) + 1.
Integrate:
V = \pi \int_a^b \bigl[R(x)^2 - r(x)^2\bigr]\, dxIf there is no inner radius (a solid disk), drop the r(x) term.
Example approach (Problem 3): y = 2√(x-1) and y = x-1, rotated about x = -1
The axis is a vertical line, so you integrate with respect to y (or use shells in x). Using washers in y: solve for x in terms of y. From y = 2√(x-1), get x = (y/2)² + 1 = y²/4 + 1. From y = x - 1, get x = y + 1.
The outer radius (distance from x = -1 to the farther curve) and inner radius must be identified for each y.
Integrate over the y-interval where the curves bound the region. Answer: V = 96π/5.
Example approach (Problem 5): y = x - x², y = 0, about x = 2
The axis x = 2 is vertical, the curves are in terms of x. Shells are natural here: radius = 2 - x, height = x - x².
V = 2π ∫ from 0 to 1 of (2 - x)(x - x²) dx.
Shell Method (wrapping cylinders around the axis)
When the axis of revolution is parallel to the variable you are integrating over (e.g., rotating around a vertical axis but your curves are y = f(x)), shells avoid having to re-express curves.
The radius of each shell is the distance from the axis to the strip. The height is the length of the strip.
Integrate:
V = 2\pi \int_a^b (\text{radius})(\text{height})\, dxExample approach (Problem 11): y = x² - 6x + 9 and y = -x² + 6x - 1, about x = 8
Using shells: radius = 8 - x, height = (-x² + 6x - 1) - (x² - 6x + 9) = -2x² + 12x - 10.
Find where the curves intersect: x² - 6x + 9 = -x² + 6x - 1 gives 2x² - 12x + 10 = 0, so x = 1 and x = 5.
V = 2π ∫ from 1 to 5 of (8 - x)(-2x² + 12x - 10) dx.
Cross-Sectional Area Method (non-revolution solids)
When cross-sections are known geometric shapes (equilateral triangles, squares, semicircles), express the side length in terms of the base's width at position x.
Find A(x) for that shape and integrate.
Example approach (Problem 4): base is a disk of radius r, cross-sections are equilateral triangles
At position x, the width of the disk is 2√(r² - x²).
The side of the equilateral triangle is s = 2√(r² - x²).
Area of an equilateral triangle: A = (√3/4)s².
V = ∫ from -r to r of (√3/4)(2√(r² - x²))² dx = √3 ∫ from -r to r of (r² - x²) dx.
Choosing your method
If the axis is horizontal and curves are y = f(x): disks/washers in x is natural.
If the axis is vertical and curves are y = f(x): shells in x, or convert to x = g(y) and use washers in y.
If the axis is offset (not x = 0 or y = 0): adjust every radius by the offset distance.
The guiding question: "Which method lets me write one integral with functions I already have?"
Disk method
V = \pi \int_a^b [R(x)]^2\, dxWasher method
V = \pi \int_a^b \bigl[R_{\text{outer}}(x)^2 - R_{\text{inner}}(x)^2\bigr]\, dxShell method
V = 2\pi \int_a^b (\text{radius})(\text{height})\, dxCross-sectional area
V = \int_a^b A(x)\, dxEquilateral triangle area (for cross-section problems)
A = \frac{\sqrt{3}}{4}\, s^2Using the curve's y-value as the radius when the axis is offset. If the axis of revolution is y = -2, the radius is not f(x) but f(x) - (-2) = f(x) + 2. This is the single most common error in this topic.
Confusing when to use shells vs. washers. Students sometimes force the washer method when shells would be simpler, or vice versa. Neither is universally better. The choice depends on the axis orientation relative to your variable.
Squaring before subtracting (washer method). The formula is π∫(R² - r²), not π∫(R - r)². These are different expressions. Always square each radius first, then subtract.
Forgetting the 2π in the shell method. The shell formula has a factor of 2π, not just π. This comes from the circumference of each cylindrical shell.
Setting up cross-section problems with the wrong side length. For the equilateral-triangle cross-section problem, the side length equals the full width of the base at that slice (the chord of the disk), not the radius.
⚠️ The exam heavily tests volumes. Problems 3 through 7 and 10 through 11 on the review are all volume problems, making this the largest single topic by question count.
⚠️ Expect at least one problem with an offset axis (rotation about x = -1, y = -2, x = 8, etc.). Practise adjusting the radius formula.
⚠️ The exam may specify which method to use ("use disks/rings" or "use cylindrical shells"). Read the instructions carefully.
⚠️ Cross-section problems (like the equilateral triangle one) are less common but straightforward marks if you know the area formulas for basic shapes.
⚠️ Problems 6, 7, and 10 all involve non-standard axes. This is clearly a focus of the course.
True or false: The washer method formula is π∫(R - r)² dx. False. It is π∫(R² - r²) dx.
Fill in the blank: In the shell method, the volume element is 2π × ______ × ______ × dx. Radius, height.
True or false: When rotating about x = 8, the shell radius for a vertical strip at position x is 8 - x. True.
Fill in the blank: For a solid with known cross-sections, V = ∫ ______ dx. A(x), the cross-sectional area.
True or false: You can always use either shells or washers for any volume of revolution problem. True in principle, though one is often far simpler than the other.
Q: (Problem 3) Find the volume of the solid obtained by rotating the region bounded by y = 2√(x-1) and y = x - 1 about the line x = -1. (Answer: 96π/5)
A: Find intersections: set 2√(x-1) = x - 1. Let u = x - 1, so 2√u = u, giving u = 0 or u = 4, hence x = 1, x = 5. Using shells (integrating in x): radius = x - (-1) = x + 1, height = 2√(x-1) - (x-1). V = 2π ∫ from 1 to 5 of (x + 1)[2√(x-1) - (x-1)] dx. Evaluate to get 96π/5.
Q: (Problem 4) Find the volume of the solid whose base is a disk of radius r and whose cross-sections are equilateral triangles.
A: At position x, the chord length is 2√(r² - x²). Each cross-section is an equilateral triangle with side s = 2√(r² - x²). Area = (√3/4)s² = √3(r² - x²). V = √3 ∫ from -r to r of (r² - x²) dx = (4√3/3)r³.
Q: (Problem 5) Find the volume of the solid obtained by rotating the region bounded by y = x - x² and y = 0 about the line x = 2.
A: Use shells. The region sits between x = 0 and x = 1 (where x - x² = 0). Radius = 2 - x, height = x - x². V = 2π ∫ from 0 to 1 of (2 - x)(x - x²) dx. Expand, integrate, and simplify.
Q: (Problem 6) Find the volume of the solid obtained by rotating the region bounded by x = (y-2)² and y = x about y = -1.
A: The axis is horizontal, and one curve is naturally in terms of y. Find intersections of x = (y-2)² and y = x by substituting: y = (y-2)², giving y² - 5y + 4 = 0, so y = 1 and y = 4. Use washers in y: outer radius = y - (-1) = y + 1, inner radius = (y-2)² - (-1) = (y-2)² + 1. V = π ∫ from 1 to 4 of [(y+1)² - ((y-2)²+1)²] dy.
Q: (Problem 7) Given the region bounded by x = |y| and x = 2 - y², find the volume generated by revolving about (a) the y-axis, (b) x = 4.
A: (a) Shells in x or washers in y. The region is symmetric about the x-axis. Using washers in y: outer radius = 2 - y², inner radius = |y|. V = π ∫ of [(2-y²)² - y²] dy over the relevant bounds. (b) Using washers in y with the axis at x = 4: outer radius = 4 - |y|, inner radius = 4 - (2 - y²) = 2 + y². Integrate accordingly.
Q: (Problem 10) Use disks/rings to find the volume of the solid from rotating the region bounded by y = 6e^(-2x) and y = 6 + 4x - 2x² between x = 0 and x = 1 about y = -2.
A: The axis is y = -2. Outer radius = (6 + 4x - 2x²) - (-2) = 8 + 4x - 2x². Inner radius = 6e^(-2x) - (-2) = 6e^(-2x) + 2. V = π ∫ from 0 to 1 of [(8 + 4x - 2x²)² - (6e^(-2x) + 2)²] dx.
Volumes of revolution build directly on areas between curves (the region you rotate is the region whose area you computed earlier). The geometric reasoning here, slicing a shape into infinitesimally thin pieces and summing, is the same principle behind arc length (Section 8.2) and surface area of revolution.
The shell method reappears in physics (moments of inertia for cylindrical objects), and the disk/washer method underpins fluid-pressure calculations where you integrate cross-sectional areas.
Tags: volume of revolution, disk method, washer method, shell method, cylindrical shells, cross-sectional area, equilateral triangle cross-section, axis of revolution, offset axis, Calculus II, Calc 2, NYU, sections 6.1 6.2, solids of revolution, volume integral