Volumes by Cylindrical Shells

The disk method cut a solid into circular slabs perpendicular to the axis of rotation. When you rotate about the y-axis but your region is described as a function of x, those slabs become a nuisance — you would have to invert f. The shell method dodges that entirely: instead of slicing across, it builds the solid out of thin nested cylindrical shells, like the rings of a tree.

Each shell is a thin-walled tube of radius x, height f(x), and wall-thickness dx. Summing them across [a, b] gives

V = \int_a^{b} 2\pi x\, f(x)\,dx.

Unroll one shell — the derivation

The factor 2\pi x looks mysterious until you slit a shell open and flatten it.

Step 1 — picture one shell. Rotate a thin vertical strip of the region, at distance x from the y-axis, about that axis. It sweeps out a hollow tube of radius x and height f(x), with wall thickness equal to the strip's width dx.

Step 2 — slit it and unroll. Cut the tube straight down one side and flatten it. It becomes a thin rectangular sheet: its width is the circumference of the circle it used to be, 2\pi x; its height is f(x); and its thickness is still dx.

Step 3 — the shell's volume is that sheet's. Volume = (circumference) × (height) × (thickness):

\Delta V = \underbrace{2\pi x}_{\text{circumference}} \cdot \underbrace{f(x)}_{\text{height}} \cdot \underbrace{\Delta x}_{\text{thickness}}.

Step 4 — nest the shells and refine. Sum over all radii from a to b and let the walls grow infinitely thin:

V = \lim_{\Delta x \to 0} \sum 2\pi x_i\, f(x_i)\,\Delta x = \int_a^{b} 2\pi x\, f(x)\,dx.

Worked example: rotate y = x^2 on [0, 1] about the y-axis

Step 1 — identify radius and height. A shell at position x has radius x and height f(x) = x^2.

Step 2 — build the integrand 2\pi x\, f(x):

V = \int_0^{1} 2\pi x \cdot x^2\,dx = 2\pi \int_0^{1} x^3\,dx.

Step 3 — integrate and evaluate.

V = 2\pi \left[\frac{x^4}{4}\right]_0^{1} = 2\pi \cdot \frac{1}{4} = \frac{\pi}{2}.

The bowl-shaped solid swept out by y = x^2 around the y-axis has volume \pi/2. (Notice how the extra factor of x from the radius nudged the power from x^2 up to x^3 — shells integrate a power higher than disks would.)

Rotate the region under y = f(x) \ge 0 on [a, b] (with 0 \le a) about the y-axis. Its volume is

V = \int_a^{b} 2\pi x\, f(x)\,dx,

Shells and disks compute the same volume; they differ only in which variable you must integrate. The rule of thumb: slice perpendicular to the axis of rotation for disks, parallel to it for shells — and pick whichever spares you from inverting a function.

Rotate the region under y = x^2, 0 \le x \le 1, about the y-axis. With shells you integrate the given f(x) = x^2 directly — one clean line, as above. With disks (slabs perpendicular to the y-axis) you must first solve x = \sqrt{y}, integrate in y, and subtract from the enclosing cylinder:

V = \pi(1)^2(1) - \pi \int_0^{1} (\sqrt{y})^2\,dy = \pi - \pi \cdot \frac{1}{2} = \frac{\pi}{2}.

Same \pi/2 — but you had to invert the curve and reason about a hole. The shell setup needed neither. Whenever rotating a "y in terms of x" region about the y-axis, reach for shells first.

Unroll the shell yourself

On the left, the curve y = x^2 with a shell drawn as a vertical strip at radius x. On the right, that same shell unrolled into a flat rectangle of width 2\pi x (its circumference) and height f(x) = x^2. Slide the radius and watch the rectangle stretch: its area is the shell's lateral area 2\pi x\, f(x), and the readout shows it.