Volumes by Slicing

Look at a loaf of sliced bread. Nobody is surprised that the whole loaf is exactly the sum of its slices — put the slices back together and you get the loaf. Now make the slices thinner. And thinner. In the limit of wafer-thin slices, "add up the slices" becomes an integral, and suddenly you can compute the volume of shapes far stranger than bread: cones, vases, doughnuts, trumpet bells — anything you can describe with a function.

This is not just a maths trick. A CT scanner genuinely sees your body this way: it images one thin cross-section at a time, then stacks hundreds of them into a 3-D reconstruction. Radiologists literally scroll through your slices. The mathematics of "a solid is the stack of its cross-sections" runs modern medicine.

Here is the whole idea in one line. Cut a solid into thin slabs perpendicular to the x-axis. If the slab at position x has cross-sectional area A(x) and thickness dx, its volume is A(x)\,dx — area times thickness, just like any thin slab. Stack the slabs from a to b:

V = \int_a^{b} A(x)\,dx.

Compare it with what you already know: area was a sum of thin strips of height f(x); volume is a sum of thin slabs of area A(x). One dimension up, same machine. The whole art of this topic is finding A(x) — everything after that is an ordinary integral.

Yes — and you can see why with a stack of coins. Pile 50 coins into a neat cylinder. Now push the stack sideways so it leans like the Tower of Pisa, or twist it into a spiral. The shape has changed completely, but the volume hasn't budged: it is the same 50 coins. Every horizontal level still contains exactly one coin of exactly the same area.

That is Cavalieri's principle (Bonaventura Cavalieri, 1598–1647, a student of Galileo): if two solids have equal cross-sectional areas at every height, they have equal volumes. Cavalieri used it two generations before calculus existed — his "indivisibles" were slices in all but name, and they let him find volumes that had defeated geometers since the Greeks. Our formula V = \int A(x)\,dx is Cavalieri's principle grown up: the volume depends only on the areas A(x), not on how the slices are stacked.

First payoff: the cone formula, finally explained

You have probably used V = \tfrac{1}{3}\pi r^2 h for years without anyone telling you where the mysterious \tfrac13 comes from. Slicing answers it in four lines.

Stand a cone on its tip at the origin, opening along the x-axis, with height h and base radius r. Slice it perpendicular to its axis: every slice is a disc. How big is the disc at position x?

Similar triangles. The radius grows in proportion to the distance from the tip: at the tip (x = 0) the radius is 0; at the base (x = h) it is r. So at position x the radius is \frac{r}{h}x, and the disc's area is

A(x) = \pi \left(\frac{r}{h}x\right)^{2} = \frac{\pi r^2}{h^2}\,x^2.

Stack the discs.

V = \int_0^{h} \frac{\pi r^2}{h^2}\,x^2\,dx = \frac{\pi r^2}{h^2}\left[\frac{x^3}{3}\right]_0^{h} = \frac{\pi r^2}{h^2}\cdot\frac{h^3}{3} = \frac{1}{3}\pi r^2 h.

There it is. The \tfrac13 is the fingerprint of \int x^2\,dx = \tfrac{x^3}{3} — the cone's radius grows linearly, so its slice area grows like x^2, and integrating x^2 is what divides by three. A cylinder's slices are all the same size (constant A, no shrinking towards a tip), which is exactly why a cone holds precisely one third of the cylinder that encloses it.

Solids of revolution: the disc method, derived slab by slab

The cone was easy because we could see each slice was a disc. There is a whole factory of solids like that: take the region under y = f(x) on [a, b] and spin it about the x-axis. It sweeps out a solid of revolution — a shape with rotational symmetry, like anything turned on a potter's wheel or a lathe — and every cross-section is a filled circle: a disc.

Step 1 — find the radius of one slice. At position x, the curve sits at height f(x). When that point whirls around the axis it traces a circle of radius exactly r = f(x). The function value is the radius.

Step 2 — the cross-sectional area is a circle's. A disc of radius f(x) has area

A(x) = \pi r^2 = \pi\, f(x)^2.

Step 3 — one thin disc's volume. Give the slab thickness \Delta x; it is a squat cylinder of volume

\Delta V = \pi\, f(x)^2\,\Delta x.

Step 4 — stack the discs and refine. Sum over a partition and let the slices get infinitely thin; the Riemann sum becomes an integral:

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

Notice this is just the general slicing formula with A(x) = \pi f(x)^2 plugged in — the disc method is not a new idea, only the most popular costume the slicing idea wears.

Watch one disc sweep out

Below is the curve y = \sqrt{x} with a single representative slice drawn as a vertical bar of radius f(x) (above and below the axis — the full diameter of the disc it generates), plus a thin ellipse suggesting the circle that bar traces as it spins. Slide the position x: the readout reports the disc's radius f(x) and its cross-sectional area \pi f(x)^2.

Two things to notice as you slide. First, the area readout is \pi x exactly — because f(x)^2 = (\sqrt{x})^2 = x, squaring kills the square root. Second, the volume we are about to compute is nothing but the accumulation of that readout across [0, 1].

Worked example: rotate y = \sqrt{x} on [0, 1]

Step 1 — square the radius. Here f(x) = \sqrt{x}, so f(x)^2 = x — the messy square root vanishes the moment it is squared. (This is common: functions with square roots are deliberately popular in disc problems precisely because the squaring cleans them up.)

Step 2 — assemble the integral.

V = \pi \int_0^{1} \big(\sqrt{x}\big)^2 dx = \pi \int_0^{1} x\,dx.

Step 3 — integrate and evaluate.

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

The trumpet-shaped solid swept out by y = \sqrt{x} holds exactly \pi/2 cubic units. Sanity check: it fits inside the cylinder of radius 1 and length 1 (volume \pi), and it is fatter than the cone with those dimensions (volume \pi/3). Sure enough, \pi/3 < \pi/2 < \pi. Always run a bracket check like this — it catches most slips for free.

Washers: solids with a hole

Spin the region between two curves and the solid comes out hollow — think of a napkin ring or a pipe. Each slice is now a disc with a smaller disc punched out of its middle: an annulus, or washer (named after the little metal rings used with bolts). If the outer edge is at radius R(x) and the hole has radius r(x), the slice's area is the big disc minus the hole:

A(x) = \pi R(x)^2 - \pi\, r(x)^2 = \pi\big(R(x)^2 - r(x)^2\big).

Worked example. Rotate the region between y = x (on top) and y = x^2 (underneath) on [0, 1] about the x-axis.

Step 1 — identify outer and inner radii. Between 0 and 1 the line sits above the parabola, so the outer radius is R(x) = x and the inner radius (the hole) is r(x) = x^2.

Step 2 — square each radius separately, then subtract.

A(x) = \pi\big(x^2 - x^4\big).

Step 3 — integrate.

V = \pi \int_0^{1} \big(x^2 - x^4\big)\,dx = \pi\left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^{1} = \pi\left(\frac{1}{3} - \frac{1}{5}\right) = \frac{2\pi}{15}.

Equivalently: the solid from spinning y = x alone is a cone of volume \pi/3; drilling out the solid from y = x^2 removes \pi/5; what's left is \pi/3 - \pi/5 = 2\pi/15. A washer problem is always two disc problems subtracted.

A solid of revolution about the x-axis, with cross-sections perpendicular to that axis, has volume V = \int_a^{b} A(x)\,dx, where:

Two squaring slips cause almost every wrong answer in this topic:

The formula V = \int A(x)\,dx never assumed a circle — that was a special case. Some of the prettiest solids have cross-sections that are squares, triangles, or semicircles, with no rotation in sight.

Take a solid whose base is the disk x^2 + y^2 \le 1 and whose slices perpendicular to the x-axis are squares. At position x the base runs from y = -\sqrt{1 - x^2} to +\sqrt{1 - x^2}, a side of length s = 2\sqrt{1 - x^2}. The square slice has area A(x) = s^2 = 4(1 - x^2), so

V = \int_{-1}^{1} 4(1 - x^2)\,dx = 4\left[x - \frac{x^3}{3}\right]_{-1}^{1} = 4 \cdot \frac{4}{3} = \frac{16}{3}.

No \pi anywhere — the moment the cross-section stops being a circle, the \pi simply doesn't appear. Find A(x) honestly and the integral does the rest.

A CT (computed tomography) scanner fires thin X-ray beams through your body from hundreds of angles and measures how much each beam is absorbed. Clever mathematics — the Radon transform, worked out by Johann Radon in 1917, decades before any scanner existed — turns those absorption readings back into a picture of one cross-sectional slice. The scanner then steps along your body a few millimetres at a time, imaging slice after slice, and software stacks them into a full 3-D model: V = \int A(x)\,dx made of flesh and bone. Radiologists even measure tumour volumes exactly the way this page does — area of the tumour in each slice, times slice thickness, summed. Godfrey Hounsfield and Allan Cormack shared the 1979 Nobel Prize in Medicine for making it practical: a mathematician's slicing idea, saving lives daily.

See it explained