Volumes by Slicing: Disks and Washers
Slice a solid into thin slabs of cross-sectional area \(A(x)\) and thickness \(dx\), then add them up: \(V = \int_a^b A(x)\,dx\). For a solid of revolution every slice is a disk or a washer, so \(A(x) = \pi\big(R^2 - r^2\big)\).
By the end you'll be able to set up and evaluate \(V = \int_a^b A(x)\,dx\) for any sliceable solid, choose the right outer radius \(R\) and inner radius \(r\) for disk/washer problems — including a shifted axis of revolution — and avoid the \((R - r)^2\) trap.
Predict: the gap between \(y = x\) and \(y = x^2\) is widest at \(x = 0.5\) — does the washer's area peak there too? Slide x to check.
The region between y = x (outer) and y = x² (inner) on \([0, 1]\) is revolved about the x-axis. The slider positions one slicing plane: the left panel shows the slab in the side view, the middle panel shows the annulus face-on with radii R = x and r = x², and the right panel shows A(x) = π(R² − r²) accumulating into the volume. Hover the washer or the area curve for exact values.
R = 0.500 · r = 0.250 · A(x) = π(R² − r²) = 0.589 · volume so far V(x) = 0.111 of total 2π/15 ≈ 0.419
The volume of a solid is \(\int A(x)\,dx\), the integral of its cross-sectional area; for solids of revolution this gives the disk/washer method, \(A(x) = \pi(R^2 - r^2)\).
One principle: stack the slices
In the previous lesson you built area by stacking thin (top − bottom) strips. Volume works the same way, one dimension up: cut the solid into slabs perpendicular to the x-axis, each of thickness \(dx\). A slab at position \(x\) with cross-sectional area \(A(x)\) has volume \(A(x)\,dx\), and summing infinitely many infinitely thin slabs is exactly an integral:
$$V = \int_a^b A(x)\,dx.$$
This works for any solid — pyramids, wedges, weird sculptures — as long as you can write a formula for \(A(x)\). The cross-sections don't have to be circles.
Picture a loaf of bread. Its volume is the sum of the volumes of its slices, and each slice's volume is (face area) × (thickness). Now revolve a region about the x-axis, as in the demo above: every slice's face becomes a circle (a disk, radius \(R(x) = f(x)\)) or a ring (a washer, outer \(R\), inner \(r\)) — that's why the middle panel of the demo shows the slice face-on. The washer's area is the big disk minus the hole: \(A(x) = \pi R^2 - \pi r^2 = \pi(R^2 - r^2)\).
This is how you'd compute the material in a lathe-turned chair leg, the capacity of a vase, or the mass of a machined bushing (a cylinder with a bore — literally a stack of washers). Engineers set \(R(x)\) to the outer profile, \(r(x)\) to the bore profile, and integrate \(\pi(R^2 - r^2)\) along the axis.
Disks, washers, and the classic trap
Revolving the region under \(y = f(x)\) about the x-axis (no gap) gives disks: \(V = \pi\int_a^b [f(x)]^2\,dx\). Revolving the region between an outer curve and an inner curve gives washers: \(V = \pi\int_a^b \big(R^2 - r^2\big)\,dx\). The trap: \(R^2 - r^2\) is not \((R - r)^2\). You square each radius separately and then subtract — a washer is one disk's area minus another's, not the square of a height difference.
If the axis of revolution is shifted — say the line \(y = -1\) instead of the x-axis — measure every radius as a distance to that axis: a curve at height \(y\) is \(y + 1\) away from \(y = -1\). And always check with a test point which curve is farther from the axis: that one is \(R\).
Revolve \(y = \sqrt{x}\) on \([0, 4]\) about the x-axis. No gap, so disks with \(R(x) = \sqrt{x}\): \(V = \pi\int_0^4 (\sqrt{x})^2\,dx = \pi\int_0^4 x\,dx = \pi\left[\tfrac{x^2}{2}\right]_0^4 = \pi \cdot 8 = \) \(8\pi\).
Revolve \(y = x^2\) on \([0, 1]\) about the x-axis. Again no gap, so disks with \(R(x) = x^2\): \(A(x) = \pi (x^2)^2 = \pi x^4\). Now finish it: \(V = \pi\int_0^1 x^4\,dx = \) ____
Reveal the answer
\(V = \pi\left[\tfrac{x^5}{5}\right]_0^1 = \) \(\tfrac{\pi}{5}\). Note the order of operations: square the radius first (\(x^2 \to x^4\)), then integrate. Squaring after integrating gives nonsense.
More info — why the washer area peaks where it does (not where the region is widest)
In the demo, the vertical gap \(x - x^2\) is largest at \(x = 0.5\), but \(A(x) = \pi(x^2 - x^4)\) peaks at \(x = \tfrac{1}{\sqrt{2}} \approx 0.707\). That's because revolving weights each strip by its distance from the axis: the same vertical gap sweeps out far more area when it sits farther from the axis (bigger circles). This is exactly why you can't reuse the (top − bottom) integrand from the area-between-curves setup — squaring the radii separately, \(R^2 - r^2\), is what encodes "distance from the axis matters." Paul's Notes (Dive deeper below) works several examples where choosing and measuring the radii is the whole game.
Check your understanding
Revolve the region under \(y = x\) on \([0, 3]\) about the x-axis. What is the volume of the resulting cone?
A region between an outer curve \(R(x)\) and an inner curve \(r(x)\) (both measured from the x-axis) is revolved about the x-axis. Which integral gives the volume?
The region between \(y = x\) (top) and \(y = x^2\) (bottom) on \([0, 1]\) is revolved about the line \(y = -1\) instead of the x-axis. What are the correct outer and inner radii?
On \([0, 1]\), the area between \(y = x\) and \(y = x^2\) is \(\int_0^1 (x - x^2)\,dx = \tfrac16\). If the same region is revolved about the x-axis, which integral gives the volume?
Recap
- Slicing principle: \(V = \int_a^b A(x)\,dx\) — works for any solid whose cross-sectional area \(A(x)\) you can write down.
- Disks (no gap): \(V = \pi\int_a^b [f(x)]^2\,dx\). Washers (gap): \(V = \pi\int_a^b (R^2 - r^2)\,dx\).
- Square each radius separately: \(R^2 - r^2 \ne (R - r)^2\) unless \(r = 0\).
- Shifted axis: measure \(R\) and \(r\) as distances to the axis of revolution (e.g. axis \(y = -1\) turns height \(y\) into radius \(y + 1\)); the curve farther from the axis is the outer radius.
Dive deeper
- Paul's Online Math Notes — Volumes: Method of Rings See disk and washer examples with different axes of revolution.
Sources
- Volumes by Slicing — Disks and Washers