Disk vs washer vs shell comes down to one question, not three: is your slice perpendicular or parallel to the axis of rotation? Perpendicular slices give you circles — disks if solid, washers if there is a hole. Parallel slices give you cylindrical shells. That is the entire decision, and it is answerable before you write a single integral sign.
What follows is not three separate recipes. It is one region, rotated once, set up all three ways, arriving at the same number — because seeing 8π come out of both a washer integral and a shell integral is what finally makes the choice feel obvious. After that: rotating about lines that are not the x- or y-axis (where most marks get lost), a formula card for both orientations, and an independent way to check your answer that does not involve redoing the integral.
Disk vs Washer vs Shell: The One Question That Picks It
Draw the region. Draw the axis. Now draw a representative rectangle inside the region and ask what happens when it spins.
A rectangle perpendicular to the axis sweeps out a flat circular plate. If the region touches the axis, that plate is solid — a disk, area πr². If there is a gap between the axis and the region, the plate has a bite out of the middle — a washer, area π(R² − r²). A rectangle parallel to the axis sweeps out a hollow tube — a cylindrical shell, unrolled area 2πrh.
The practical version of the rule is about which variable you integrate in. Rotating about a horizontal line means perpendicular slices are vertical, so disks and washers integrate dx. Rotating about a vertical line means perpendicular slices are horizontal, so disks and washers integrate dy. Shells always run the other way round.
| Axis of rotation | Disk / Washer | Shell |
|---|---|---|
| Horizontal (x-axis, y = c) | Integrate dx, functions of x | Integrate dy, functions of y |
| Vertical (y-axis, x = c) | Integrate dy, functions of y | Integrate dx, functions of x |
| Which to prefer | When the curve is already solved for the slicing variable | When solving for the other variable is ugly or impossible |
That last row is the real tiebreaker, and it is worth stating bluntly: pick the method that avoids inverting the function. Rotate y = x³ + x about the y-axis and washers demand x as a function of y — good luck. Shells need only x³ + x, exactly as given. Meanwhile a region bounded by y = √x rotated about the x-axis is trivial with disks and needlessly fiddly with shells. Neither method is better; one is usually less work, and the algebra tells you which.
If you want the setup confirmed before committing to it, the volume of revolution calculator accepts either orientation and reports the integral it built, which makes it a fast way to sanity-check a bound or a radius you are unsure about.
One Region, Three Methods, One Answer
Take the region bounded by y = x², the x-axis, and the line x = 2. Rotate it about the y-axis. Here is that same solid computed three ways.
Method 1 — Shells (integrate dx)
Vertical rectangles run parallel to the y-axis, so each sweeps a shell. Radius = x, height = x², thickness = dx.
V = ∫02 2πx · x² dx = 2π ∫02 x³ dx = 2π [x⁴/4]02 = 2π(4) = 8π
Method 2 — Washers (integrate dy)
Horizontal rectangles are perpendicular to the y-axis. At height y the solid runs from the parabola x = √y out to the line x = 2, so R = 2 and r = √y, with y from 0 to 4.
V = ∫04 π(2² − (√y)²) dy = π ∫04 (4 − y) dy = π [4y − y²/2]04 = π(16 − 8) = 8π
Method 3 — Disks, by subtraction
The washer integral is really two disk integrals. The full cylinder of radius 2 and height 4 has volume π(2²)(4) = 16π. The solid carved out inside the parabola is ∫04 π(√y)² dy = π[y²/2]04 = 8π.
16π − 8π = 8π
Three setups, three different variables of integration, one answer. Notice the effort gap though: the shell version needed no inverse function and no subtraction, and it was a single power rule. That is not luck — the region was described in terms of x, and shells about a vertical axis integrate in x. Match the method to how the region is already written and the algebra mostly disappears.
Worth trying yourself: rotate the same region about the x-axis instead. Now disks integrate in x (V = ∫02 π(x²)² dx = 32π/5) and it is the shell version that needs the inverse. Same region, opposite recommendation. Run both through the disk method calculator and the shell method calculator to watch the setups swap roles.
Rotating About a Line That Isn’t an Axis
This is where the marks go. Rotate about y = −1 or x = 3 and the formulas are unchanged — but every radius must be measured from that line, not from zero. Students who have only ever rotated about the axes tend to plug the function straight in and lose the shift.
The rule is mechanical. For a horizontal axis y = c, the radius to a curve y = f(x) is |f(x) − c|. For a vertical axis x = c, the radius to a point at horizontal position x is |x − c|. Absolute values matter when the axis cuts through or sits above the region.
| Rotate about | Washer radii | Shell radius |
|---|---|---|
| x-axis (y = 0) | R = f(x), r = g(x) | y |
| y = c, region above it | R = f(x) − c, r = g(x) − c | y − c |
| y = c, region below it | R = c − g(x), r = c − f(x) | c − y |
| y-axis (x = 0) | R, r as functions of y | x |
| x = c, region to the left | radii measured from x = c | c − x |
Worked shift: y = x², y = 0, x = 2, rotated about x = 3
Shells, integrating dx. The axis sits to the right of the region, so the radius is 3 − x, not x. Height is still x².
V = ∫02 2π(3 − x)(x²) dx = 2π ∫02 (3x² − x³) dx = 2π [x³ − x⁴/4]02 = 2π(8 − 4) = 8π
It coincidentally matches the y-axis answer here. Use the radius x instead of 3 − x and you get 8π as well — which is exactly why this shift is so easy to get away with on this particular region and so costly on any other. Test the rule on a region that is not symmetric about x = 1.5 before trusting it.
One more trap specific to washers: when the axis passes through the region, you cannot use a single washer integral at all. Split the region at the axis and treat each piece separately, because the outer radius is measured in opposite directions on either side. Shells handle this more gracefully, which is another reason to reach for them when the geometry looks awkward.
A Formula Card for Both Orientations
Everything above, compressed. The variable in the differential is the one you are slicing along; the functions must be written in that same variable.
| Method | Horizontal axis | Vertical axis | Use when |
|---|---|---|---|
| Disk | π∫ [f(x)]² dx | π∫ [f(y)]² dy | Region touches the axis, no gap |
| Washer | π∫ ([R(x)]² − [r(x)]²) dx | π∫ ([R(y)]² − [r(y)]²) dy | Gap between axis and region |
| Shell | 2π∫ y · w(y) dy | 2π∫ x · h(x) dx | Inverting the function is painful |
| Surface area | 2π∫ f(x)√(1 + [f′(x)]²) dx | 2π∫ x√(1 + [dx/dy]²) dy | You need the skin, not the filling |
That last row catches people out constantly. “Surface area of revolution” is a genuinely different integral — it carries the arc length factor √(1 + [f′]²), because you are wrapping a curve rather than stacking circles. Squaring the radius there is a guaranteed wrong answer. The surface area of revolution calculator keeps the two apart, and the arc length calculator handles the √(1 + [f′]²) piece on its own if you want to build it up in stages.
Note also that the shell formulas swap radius and height relative to the disk formulas. In 2π∫ x · h(x) dx about a vertical axis, x is the radius and h(x) is the vertical extent of the region. Writing 2π∫ h(x) · x dx is the same thing; writing 2π∫ x² dx because you half-remembered a π r² is not.
Check the Answer Without Redoing the Integral
Recomputing a volume with the same method usually reproduces the same mistake. Two independent checks are worth more than a second attempt.
Pappus’s centroid theorem. If a plane region of area A is rotated about an external axis, the volume is V = 2π d̄ A, where d̄ is the distance from the axis to the region’s centroid. For the parabola region above: A = ∫02 x² dx = 8/3, and the centroid’s x-coordinate is (∫02 x·x² dx)/A = 4/(8/3) = 3/2. So V = 2π(3/2)(8/3) = 8π. Independent route, same answer — that is a real check, not a re-run. It only works when the axis does not cut the region, which is precisely the washer/shell case anyway.
Bracket it with a cylinder. Before integrating, bound the answer. The parabola solid sits inside a cylinder of radius 2 and height 4, so V must land between 0 and 16π — and since the parabola splits that cylinder into two visually comparable halves, somewhere near 8π. An answer of 40π fails the upper bound outright; an answer of 0.8π is an order of magnitude too small. Both are caught in two seconds without touching the algebra.
Where these setups actually break
Squaring a difference instead of differencing the squares. π∫(R − r)² dx is not π∫(R² − r²) dx. On R = 2, r = 1 that is 1π versus 3π — a threefold error that still looks plausible.
Using x-bounds with a dy integral. Switching to washers about the y-axis means the limits become y-values. The parabola region runs x from 0 to 2 but y from 0 to 4.
Forgetting the 2π in shells. Shells carry 2πrh, disks carry πr². Dropping the 2 halves the volume.
Wrong radius after a shift. Rotating about x = 3 with radius x rather than 3 − x. Always re-derive the radius from a sketch, never from memory.
Ignoring where the curves cross. If f and g swap places inside the interval, the outer radius changes identity and the integral must be split at the crossing point.
For the messier integrands that come out of shifted axes, evaluating the definite integral itself is often the slowest part. The definite integral calculator takes that step off your hands so the effort stays on the setup, which is where the understanding lives — and where examiners award most of the marks.
Four Regions Worth Setting Up
Set up the integral only — do not evaluate. Choosing correctly is the skill; the antiderivative is bookkeeping.
- y = √x, y = 0, x = 4, about the x-axis. Region touches the axis and is already solved for y, so disks in dx: π∫04 x dx. Answer 8π.
- y = x², y = x, about the y-axis. Curves cross at (0,0) and (1,1), with y = x outside. Shells avoid two inverse functions: 2π∫01 x(x − x²) dx = π/6.
- y = eˣ, y = 0, x = 0, x = 1, about the y-axis. Inverting gives x = ln y and an awkward split at y = 1. Shells: 2π∫01 x eˣ dx, one integration by parts, answer 2π.
- y = x², y = 4, about y = 4. The axis is the top boundary, so the radius is 4 − x² and disks work in dx: π∫−22 (4 − x²)² dx = 512π/15. Note the radius, not the function.
Three of those four favour a method that is not the one most courses teach first, which is the point. The disk method gets introduced first because it is simplest, not because it is most often the right tool. Once a region is bounded by two curves or the axis is shifted, shells usually win on setup cost — and setup cost is what determines whether you finish the question.
To check any of these, the solid of revolution calculator handles all three methods from one region description, while the washer method calculator and cylindrical shell calculator show the individual radii they built, which is where a mis-set bound shows up first.
For the underlying theory and the standard statements of these formulas, the Wikipedia entry on solids of revolution covers the disk, washer, and shell integrals alongside Pappus’s theorem in a single reference.
