Volumes by Cylindrical Shells
Instead of slicing a solid of revolution perpendicular to the axis, the shell method wraps it in thin nested cylinders parallel to the axis: each shell contributes \(2\pi(\text{radius})(\text{height})\,dx\), and the integral adds them all up.
By the end you'll be able to set up and evaluate \(V=\int 2\pi(\text{radius}) (\text{height})\,dx\) for a region revolved about a vertical axis — including shifted axes — and recognize when shells are less work than washers.
Predict: as the shell radius grows, does its lateral area \(2\pi r\,h\) keep growing even while the height shrinks? Slide the radius out toward the edge to check.
The region under \(y = 4 - x^2\) on \([0,2]\) is revolved about the y-axis. In shell view the slider picks a vertical strip at radius \(r\); the top-down panel shows the nested shells it sweeps out. Toggle to washer view to slice the same solid the other way — horizontally, at a height \(y\) — and compare the setups. Hover or focus the strip or the ring for exact values.
Shell at r = 1.00: height h = 3.00, lateral area 2πrh ≈ 18.85 · volume of shells so far ≈ 11.00 of 8π ≈ 25.13
The cylindrical-shells method computes a volume of revolution as \(\int 2\pi(\text{radius})(\text{height})\,dx\), summing thin nested cylindrical shells — often simpler than washers when revolving around an axis parallel to the slicing direction.
Take a vertical strip of the region at position \(x\): width \(dx\), height \(f(x)\). Spin it about the \(y\)-axis and it sweeps out a thin cylindrical shell — a soup-can wall — of radius \(x\), height \(f(x)\), and thickness \(dx\). Unroll that wall flat and you get a rectangular slab whose length is the circumference \(2\pi x\), so its volume is \(dV = 2\pi x\, f(x)\,dx\). Nest all the shells from \(x=a\) out to \(x=b\), as in the top-down panel above, and the total volume is \[V = \int_a^b 2\pi\,x\,f(x)\,dx.\] More generally, \(V = \int 2\pi(\text{radius})(\text{height})\,dx\), where the radius is the distance from the strip to the axis of revolution and the height is the strip's vertical extent (\(f(x)-g(x)\) between two curves).
Why keep two methods? Reach for shells when the axis of revolution runs parallel to the strips you'd naturally draw. Revolving \(y=f(x)\) about a vertical line with washers means solving \(y=f(x)\) for \(x\) — inverting \(f\), which can be messy or split into branches. Shells skip the inversion entirely: you integrate in \(x\) using \(f\) exactly as given. That's what the toggle above shows — the washer view needs \(x=\sqrt{4-y}\), the shell view just reads off \(h = 4 - x^2\). Same solid, same \(8\pi\), less algebra.
Revolve the region under \(y=x^2\) on \([0,2]\) about the \(y\)-axis. Radius \(=x\), height \(=x^2\): \[V=\int_0^2 2\pi\,x\cdot x^2\,dx = 2\pi\int_0^2 x^3\,dx = 2\pi\left[\frac{x^4}{4}\right]_0^2 = 2\pi\cdot 4 = \textbf{8}\boldsymbol{\pi}.\] Now shift the axis: revolve the same region about the line \(x=3\) instead. The radius from the strip at \(x\) to that axis is \(3-x\), so \(V=\int_0^2 2\pi(3-x)\,x^2\,dx = 2\pi\left[x^3 - \frac{x^4}{4}\right]_0^2 = 2\pi(8-4) = 8\pi\). The radius is always the distance to the axis of revolution — not automatically \(x\).
Revolve the region under \(y=x\) on \([0,3]\) about the \(y\)-axis. Radius \(=x\), height \(=x\), so \(V=\int_0^3 2\pi\,x\cdot x\,dx = 2\pi\int_0^3 x^2\,dx = 2\pi\left[\;\underline{\qquad}\;\right]_0^3 = \) ____
Reveal the answer
The antiderivative is \(\frac{x^3}{3}\), so \(V = 2\pi\cdot\frac{27}{3} = \textbf{18}\boldsymbol{\pi}\). Sanity check without calculus: this solid is a cylinder of radius 3 and height 3 (volume \(27\pi\)) minus the cone \(y=x\) carves out (\(\frac{1}{3}\cdot 27\pi = 9\pi\)) — and \(27\pi - 9\pi = 18\pi\). It checks out.
More info — why the factor is 2π, and where the peak shell lives
The \(2\pi\) is the shell's circumference, not an area formula: unrolling a cylinder of radius \(r\) gives a rectangle of length \(2\pi r\). That's why a far-out shell can contribute more volume than a taller one near the axis — in the demo, the lateral area \(2\pi r(4-r^2)\) keeps rising until \(r=\sqrt{4/3}\approx 1.15\), even though the height falls the whole way, because the growing circumference outweighs the shrinking height for a while. This is the same area-times-thickness accumulation you used for disks in the previous lesson — only the slicing direction changed. For side-by-side setups of both methods on the same solids, see the Paul's Online Math Notes link under Dive deeper.
Check your understanding
Using shells, find the volume when the region under \(y = x^3\) on \([0,1]\) is revolved about the \(y\)-axis.
A region occupying \(0 \le x \le 2\) is revolved about the vertical line \(x = 5\). What is the radius of the shell swept out by the strip at position \(x\)?
You revolve the region under \(y=f(x)\), \(a \le x \le b\), about the \(y\)-axis. Why are shells often easier than washers here?
The region under \(y=\sqrt{x}\) on \([0,4]\) is revolved about the \(y\)-axis. Shells give \(V=\int_0^4 2\pi x\sqrt{x}\,dx = 128\pi/5\). Which washer setup computes the same volume?
Recap
- Shell method: \(V = \int 2\pi(\text{radius})(\text{height})\,dx\) — for the region under \(y=f(x)\) revolved about the \(y\)-axis, \(V=\int_a^b 2\pi\,x\,f(x)\,dx\).
- The radius is the distance from the strip to the axis of revolution: it's \(x\) for the \(y\)-axis but \(3-x\) for the line \(x=3\), \(x+1\) for \(x=-1\), and so on.
- Height is the strip's vertical extent — \(f(x)\), or \(f(x)-g(x)\) between two curves. Don't swap it with the radius.
- The \(2\pi\) comes from the shell's circumference (unroll the cylinder into a slab of length \(2\pi r\)).
- Prefer shells over washers when the axis of revolution is parallel to your strips — it avoids inverting \(f\); both methods agree when both apply.
Dive deeper
- Paul's Online Math Notes — Volumes: Method of Cylinders Compare shell setups against the disk/washer setups for the same solids.
Sources
- Volumes by Cylindrical Shells