What the disk and washer methods do
Spin a plane region about a line and you sweep out a solid. The disk and washer methods find its volume by slicing the solid perpendicular to the axis of revolution. Each slice is a thin circular plate: a full disk if the region touches the axis, an annulus — a washer — if it does not. The area of an annulus with outer radius R and inner radius r is π(R² − r²), so a slice of thickness dx has volume π(R² − r²)dx, and the integral adds them all up.
The whole method is that one sentence. Everything students find hard about it is bookkeeping: which function is the outer radius, which is the inner, whether to integrate in x or in y, and what the radii become when the axis is not a coordinate axis. This calculator handles the last of those explicitly by asking where the axis is, so R = |outer − k| and r = |inner − k| rather than R = outer.
Solids of revolution are not just an exam topic. A pressure vessel head, a turned table leg, a wine glass, a machined pulley, a rocket nozzle and a lens blank are all solids of revolution, and their volumes — and therefore their masses and material costs — come from exactly this integral. CAD software computes them the same way, with the profile curve supplied by the sketch.
Setting up the integral correctly
Start by drawing the axis. The slices must be perpendicular to it, and that single fact decides the variable of integration: a horizontal axis takes vertical slices and you integrate in x; a vertical axis takes horizontal slices and you integrate in y. Choosing the wrong one is the most common setup error, and it usually announces itself by making the radii impossible to express.
Next, measure the radii from the axis, not from the origin. If you revolve the region between y = 3 and y = 1 about the line y = 1, the outer radius is 3 − 1 = 2 and the inner radius is 1 − 1 = 0, so you get a solid cylinder of radius 2 rather than a washer between 3 and 1. Getting this wrong is the second most common error and it usually produces an answer that is too large. Taking the absolute value, as the calculator does, keeps the arithmetic right even when a boundary sits on the far side of the axis from the one you had in mind — though if the region genuinely straddles the axis, the solid overlaps itself and the integral no longer describes it.
Then identify which boundary is outer. "Outer" means further from the axis, which is not always the larger function value. Revolving about y = 4 the curve y = 1 is further away than y = 3, so y = 1 is the outer boundary. The calculator warns you when the inner radius exceeds the outer one anywhere on the range, because then the integrand goes negative and the answer is not a volume.
The π(R² − r²) form is not the same as π(R − r)², and confusing them is the third classic error. Squaring is not linear: the annulus between radii 2 and 1 has area π(4 − 1) = 3π, not π(2 − 1)² = π. Square each radius separately and subtract afterwards.
Worked example: y = x from 0 to 3, revolved about the x-axis
The region under the line y = x, above the x-axis, between x = 0 and x = 3, is a right triangle. Spinning it about the x-axis makes a cone of base radius 3 and height 3.
- Choose the variable. The axis is horizontal, so slice vertically and integrate in x.
- Write the radii. The axis is y = 0, so R(x) = |x − 0| = x and r(x) = 0. No hole, so this is the disk method.
- Set up. V = π∫₀³ x² dx.
- Integrate. ∫₀³x²dx = [x³/3]₀³ = 27/3 = 9.
- Multiply by π. V = 9π = 28.274334.
- Check against geometry. A cone's volume is πr²h/3 = π(3²)(3)/3 = 9π. Identical, as it must be.
Now add a hole. Suppose the region is between y = x and y = x/2 over the same interval, still revolved about the x-axis. Then R = x, r = x/2, and R² − r² = x² − x²/4 = 3x²/4. So V = π∫₀³(3x²/4)dx = π(3/4)(9) = 6.75π = 21.205750. Note that this is 3/4 of the solid cone, not 1/2, even though the inner curve is half the height of the outer one — because area goes as the square of the radius, and 1 − (1/2)² = 3/4. That is the same trap as π(R−r)², seen from the other side.
Finally move the axis. Revolve the original triangle about y = −1 instead. Now R(x) = |x − (−1)| = x + 1 and r(x) = |0 − (−1)| = 1, so R² − r² = (x+1)² − 1 = x² + 2x. Then V = π∫₀³(x² + 2x)dx = π[x³/3 + x²]₀³ = π(9 + 9) = 18π = 56.548668, exactly twice the volume of the cone. Nothing about the region changed; only the distance to the axis did.
Reading the output and checking it
The single best check is a geometric bound. Your solid sits inside a cylinder of radius R_max and length L, so the volume cannot exceed πR_max²L. The calculator reports both the largest outer radius and the length along the axis, so that bound takes five seconds to apply. For the default cone, π(3²)(3) = 27π and the answer is 9π — one third, exactly as a cone should be.
The mean cross-sectional area is the second check. It must lie between the smallest and largest slice areas in the station table. If it does not, either a boundary is undefined somewhere between the stations or the inner and outer functions cross. The station table also makes the shape of the solid readable: an area column that rises monotonically describes a solid that flares steadily, while one that peaks in the middle describes a barrel.
Watch for a negative or suspiciously small integrand. If the calculator warns that the inner boundary is further from the axis than the outer boundary somewhere, the integral is subtracting rather than adding over that stretch, and the reported figure is not the volume of anything. Fix it by swapping the two functions, or by splitting the range at the crossing point and running the calculator twice.
When the region touches the axis, the inner radius is zero and this is the disk method; the calculator says so. That is the normal case for a solid turned part. A non-zero inner radius means a through hole along the whole length — a tube, a ring, a bushing.
Reference: standard solids and their disk or washer setups
| Solid | Region and axis | Integral | Volume |
|---|---|---|---|
| Cylinder, r = 2, h = 5 | y = 2, about y = 0, x from 0 to 5 | π∫₀⁵ 4 dx | 20π ≈ 62.831853 |
| Cone, r = 3, h = 3 | y = x, about y = 0, x from 0 to 3 | π∫₀³ x² dx | 9π ≈ 28.274334 |
| Sphere, r = 1 | y = √(1−x²), about y = 0, x from −1 to 1 | π∫₋₁¹ (1−x²) dx | 4π/3 ≈ 4.188790 |
| Paraboloid | x = y², about x = 0, y from 0 to 1 | π∫₀¹ y⁴ dy | π/5 ≈ 0.628319 |
| Tube, R = 2, r = 1, L = 5 | y = 2 and y = 1, about y = 0 | π∫₀⁵ (4−1) dx | 15π ≈ 47.123890 |
| Truncated cone (frustum) | y = x, about y = 0, x from 1 to 3 | π∫₁³ x² dx | 26π/3 ≈ 27.227136 |
The frustum row is a useful check of the standard formula πh(R² + Rr + r²)/3: with h = 2, R = 3, r = 1 that gives 2π(9 + 3 + 1)/3 = 26π/3, matching the integral exactly.
Common errors and the limits of this method
- Using π(R − r)² instead of π(R² − r²). The area of an annulus is the difference of two circle areas, not the area of a circle of the difference.
- Measuring radii from the origin when the axis is elsewhere. Every radius is a distance to the axis of revolution. Subtract k first, then take the magnitude.
- Slicing parallel to the axis. Disks and washers require perpendicular slices. If perpendicular slicing gives radii you cannot write down, that is the signal to use cylindrical shells instead.
- Revolving a region that straddles the axis. The two halves sweep the same solid, so the integral double-counts. Revolve only the part on one side.
- Forgetting to re-solve for the other variable. Rotating about a vertical axis needs x as a function of y. The curve y = x² for x ≥ 0 becomes x = √y.
- Assuming an answer that lacks π is wrong. The integral column and the volume column differ by exactly a factor of π; both are shown so you can quote either form.
- Expecting this to handle solids with non-circular cross-sections. Those are general cross-section problems, V = ∫A(x)dx, and only reduce to this when A is an annulus.
When to reach for cylindrical shells instead
Use shells when the perpendicular slice would need two different formulas or when solving for the other variable is ugly. Revolving the region under y = x sin x about the y-axis is a shell problem: V = 2π∫x·f(x)dx is trivial, while inverting the function to get a radius in terms of y is impossible in closed form. Shells and washers always agree when both apply; pick whichever gives an integrand you can evaluate. The quadrature underneath is the same machinery used by the Riemann sum calculator.
Where this fits with the other integral applications
Volumes of revolution belong to a family of applications that all replace a geometric quantity with an integral of a local element. The area under the curve is the Riemann sum limit; the length along the curve is the arc-length integral; the surface area of the same solid you have just measured is ∫2πR ds, which combines both — a radius from this page and a length element from that one. Noticing that the surface integral uses ds and the volume integral uses dx is the difference between a correct surface area and a common wrong answer.
To set the limits you often first need the intersection of two curves, which is a root-finding problem: solve outer(x) − inner(x) = 0 with the bisection method calculator and use the roots as your limits. Where the profile has a turning point that changes which boundary is outer, locate it with the critical points calculator and split the integral there.
The calculator evaluates the integral with composite Simpson's rule on the panel count you choose, and the boundary functions are evaluated directly rather than differentiated, so the numerical error here is smaller than in the arc-length case. For any integrand that is a polynomial of degree three or less — which covers the cone, the cylinder and the sphere — Simpson's rule is exact to rounding, which is why the reference table above reproduces to every digit shown.