In short

Rotate a region about an axis and you get a solid. Its volume is an integral, and there are three standard ways to set it up. The disk method slices the solid into thin circles, giving V = \pi \int [f(x)]^2\,dx. The washer method handles solids with a hole, giving V = \pi \int ([f(x)]^2 - [g(x)]^2)\,dx. The shell method slices the solid into thin cylindrical shells, giving V = 2\pi \int x\,f(x)\,dx. Pick the method that matches the geometry of the region and the axis of rotation.

Take the region under the curve y = \sqrt{x} from x = 0 to x = 4. It sits in the first quadrant, looking like a curved triangle. Now spin that region around the x-axis like a pizza dough being tossed — full revolution, 360°. What you end up with is a three-dimensional solid that looks like a horn: it starts at a point at the origin and flares out to a circular face of radius 2 at x = 4. What is the volume of this horn?

You already know how to compute areas by slicing a region into thin strips. The same idea works for volumes — you slice a solid into thin pieces whose volume you can compute, then add them up with an integral. The trick is choosing the right kind of slice. For a solid of revolution there are three natural choices, each suited to a different situation.

This article builds all three methods from scratch, shows you when each one is the right tool, and ends with two full worked examples.

The disk method

Go back to the horn. Slice the solid perpendicular to the x-axis at position x. What you see in the cross-section is a circle — because the region was spun all the way around the x-axis, every point at height y on the original curve sweeps out a full circle of radius y.

At x, the curve has height y = \sqrt{x}. The cross-sectional circle has radius \sqrt{x}, so its area is \pi (\sqrt{x})^2 = \pi x.

A thin slice of the solid at x is almost a cylinder with circular base of area \pi x and thickness dx. Its volume is approximately \pi x\, dx. Adding up all the slices from x = 0 to x = 4:

V = \int_0^4 \pi x\, dx = \pi \left[\frac{x^2}{2}\right]_0^4 = 8\pi

The horn has volume 8\pi cubic units. You never needed to picture the whole 3D shape — slicing reduced it to a 1D integral.

The region under $y = \sqrt{x}$ from $0$ to $4$ and its reflection below the x-axis — a vertical cross-section of the horn obtained by revolving the region. At each $x$, the solid has a circular cross-section of radius $\sqrt{x}$ (shown at $x = 2$ as the red vertical segment). Each disk has area $\pi x$ and thickness $dx$.

This is the disk method, and the recipe is short.

Disk Method

If a region bounded by y = f(x) \geq 0 and the x-axis between x = a and x = b is rotated about the x-axis, the volume of the resulting solid is

V = \pi \int_a^b \bigl[f(x)\bigr]^2\,dx

If the rotation is about the y-axis instead, and the region is bounded by x = g(y) \geq 0 between y = c and y = d, the volume is

V = \pi \int_c^d \bigl[g(y)\bigr]^2\,dy

Where the formula comes from. At position x (or y), the cross-section is a disk of radius f(x) (or g(y)). The disk's area is \pi r^2 and its thickness is dx (or dy). Integrating gives the total volume. That is the entire derivation — one line of geometry and one integral sign.

The washer method

Disks work when the cross-section is a full circle. But sometimes you rotate a region that does not touch the axis of rotation — there is a gap between the region and the axis — and the cross-section becomes a ring with a hole in the middle. A ring with inner radius r and outer radius R is called a washer. Its area is \pi R^2 - \pi r^2.

Consider the region between y = x^2 and y = 2x from x = 0 to x = 2. The line y = 2x is on top (at x = 1 it gives 2; the parabola gives 1). Rotate this region about the x-axis.

At position x, the outer boundary of the rotated region is y = 2x and the inner boundary is y = x^2. The cross-section is a washer with outer radius 2x and inner radius x^2. Its area is

\pi (2x)^2 - \pi (x^2)^2 = \pi(4x^2 - x^4)

Volume of the solid:

V = \int_0^2 \pi (4x^2 - x^4)\,dx = \pi \left[\frac{4x^3}{3} - \frac{x^5}{5}\right]_0^2 = \pi\left(\frac{32}{3} - \frac{32}{5}\right) = \frac{64\pi}{15}

Washer Method

If a region bounded above by y = f(x) and below by y = g(x), with f(x) \geq g(x) \geq 0 on [a, b], is rotated about the x-axis, the resulting solid has volume

V = \pi \int_a^b \bigl[(f(x))^2 - (g(x))^2\bigr]\,dx

The formula subtracts the inner volume (generated by g) from the outer volume (generated by f).

Critical point about the washer formula. You subtract the squares, not the squares of the difference. The volume generated by the lower curve is \pi \int g^2\,dx and the volume generated by the upper curve is \pi \int f^2\,dx, and the region between them produces the difference. The expression \pi \int (f - g)^2\,dx is wrong — it does not correspond to any geometric volume.

The shell method

Now a different picture. Instead of slicing perpendicular to the axis of rotation, slice the region parallel to the axis — into thin vertical strips. When you rotate one of these strips about the axis, it sweeps out a thin cylindrical shell, like a soup-can wall with almost no thickness.

Take a vertical strip at position x of width dx and height f(x) in the region under y = f(x). Rotate this strip about the y-axis. It sweeps out a cylindrical shell with:

If you unroll the shell, it becomes a flat rectangular sheet. Its length is the circumference 2\pi x, its height is f(x), and its thickness is dx. So its volume is

dV = 2\pi x \cdot f(x)\,dx

Adding up all the shells from x = a to x = b:

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

This is the shell method, and it is the right tool when the disk method would force you to express things as x = g(y) — in particular, when the axis of rotation is the y-axis but the curve is naturally written as y = f(x).

Shell Method

If a region bounded above by y = f(x) \geq 0, below by the x-axis, between x = a and x = b (with 0 \leq a < b) is rotated about the y-axis, the volume of the resulting solid is

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

For rotation about a vertical line x = x_0 (where x_0 \leq a), the shell radius is x - x_0, and the formula becomes

V = 2\pi \int_a^b (x - x_0)\,f(x)\,dx

Why this is sometimes essential. Take y = \sin x on [0, \pi] rotated about the y-axis. Using disks, you would have to invert: x = \arcsin y on one branch, x = \pi - \arcsin y on the other, and the integral becomes messy. Using shells:

V = 2\pi \int_0^\pi x \sin x\,dx

One integration by parts and you are done: V = 2\pi \int x\sin x\,dx = 2\pi[-x\cos x + \sin x]_0^\pi = 2\pi[\pi + 0 - 0 - 0] = 2\pi^2.

Shells turned a two-branch inversion problem into a single clean integral.

The region under $y = \sin x$ on $[0, \pi]$. A thin vertical strip at position $x$ (red) rotates about the y-axis to form a cylindrical shell of radius $x$, height $\sin x$, and thickness $dx$. Integrating all the shells from $0$ to $\pi$ gives the volume $2\pi^2$.

Which method to use

With three methods, how do you pick?

Match the slicing direction to the axis of rotation. Disks and washers slice perpendicular to the axis; shells slice parallel to the axis.

Disks/washers are natural when the curve is y = f(x) and you rotate about a horizontal axis (typically the x-axis). The cross-sections are easy to visualise: circles (or rings) with radii read directly off the curves.

Shells are natural when the curve is y = f(x) and you rotate about a vertical axis (typically the y-axis). Instead of inverting the function, you keep it as-is and let the shell radius be the distance from the axis.

The rule of thumb: if using one method would force you to invert the function, try the other method first. Most problems are doable with either, but one will usually be much cleaner.

Worked examples

Example 1: Sphere from rotating a semicircle

Derive the formula for the volume of a sphere of radius R by rotating the upper half of the circle x^2 + y^2 = R^2 about the x-axis.

Step 1. Set up the function.

The upper half of the circle is y = \sqrt{R^2 - x^2}, valid for x \in [-R, R].

Why: solving x^2 + y^2 = R^2 for y and taking the positive root gives the upper semicircle. Rotating this region about the x-axis sweeps out a full ball.

Step 2. Set up the disk integral.

The region sits between the curve and the x-axis. Rotating about the x-axis gives disks, so use the disk method:

V = \pi \int_{-R}^{R} \bigl[\sqrt{R^2 - x^2}\bigr]^2\,dx = \pi \int_{-R}^{R}(R^2 - x^2)\,dx

Why: the square of \sqrt{R^2 - x^2} is just R^2 - x^2 — the square root vanishes nicely, which is why the sphere is a clean target for this method.

Step 3. Evaluate the integral.

\pi \int_{-R}^{R}(R^2 - x^2)\,dx = \pi\left[R^2 x - \frac{x^3}{3}\right]_{-R}^{R}
= \pi\left[\left(R^3 - \frac{R^3}{3}\right) - \left(-R^3 + \frac{R^3}{3}\right)\right]
= \pi\left[\frac{2R^3}{3} + \frac{2R^3}{3}\right] = \frac{4\pi R^3}{3}

Why: the integrand is an even function, so the two halves of the integral are equal. The algebra reduces to twice the value at R.

Step 4. Interpret.

V = \frac{4}{3}\pi R^3 — the formula you have seen in geometry since middle school. Calculus has rederived it from first principles: you built the ball out of thin disks and added them up.

Why: the formula is not just a fact from a textbook — it follows from one integral. Every other sphere-volume property (density, moment of inertia, surface area) can be derived from the same disk picture.

Result: V = \dfrac{4}{3}\pi R^3.

A cross-section of the sphere of radius $R = 2$ rotating about the x-axis. The red vertical segment at $x = 1$ is the diameter of one disk slice, of length $2\sqrt{R^2 - x^2}$. Stacking all such disks from $x = -R$ to $x = R$ reconstructs the full ball of volume $\frac{4}{3}\pi R^3$.

This is one of those results where calculus rederives a piece of middle-school geometry, and that is the whole point. The volume formula for the sphere is not magical — it is one disk integral.

Example 2: A torus by the shell method

Find the volume of the torus (doughnut) formed by rotating the disk (x - 3)^2 + y^2 \leq 1 about the y-axis.

Step 1. Understand the geometry.

The disk (x - 3)^2 + y^2 \leq 1 is a disk of radius 1 centred at (3, 0). It sits entirely to the right of the y-axis (in fact, x ranges from 2 to 4). Rotating this disk about the y-axis sweeps out a torus — a doughnut shape with major radius 3 (distance from y-axis to centre of disk) and minor radius 1 (radius of the disk itself).

Why: the disk does not cross or touch the y-axis, so every point in the disk is at some positive distance from the axis. Rotating traces out a ring-shaped solid.

Step 2. Set up shells.

Use horizontal strips of the disk — shells of radius x. At each x \in [2, 4], the vertical strip inside the disk has height 2\sqrt{1 - (x-3)^2} (the full vertical extent of the disk at that x, from -\sqrt{1-(x-3)^2} up to +\sqrt{1-(x-3)^2}).

Shell contribution at x:

dV = 2\pi x \cdot 2\sqrt{1 - (x-3)^2}\,dx = 4\pi x\sqrt{1 - (x-3)^2}\,dx

Why: the shell has radius x, height equal to the vertical extent of the disk at that x, and thickness dx. The factor of 2 in the height appears because the strip extends above and below the x-axis.

Step 3. Substitute to clean the integral.

Let u = x - 3, so x = u + 3 and dx = du. When x = 2, u = -1; when x = 4, u = 1.

V = 4\pi \int_{-1}^{1}(u + 3)\sqrt{1 - u^2}\,du
= 4\pi \int_{-1}^{1}u\sqrt{1 - u^2}\,du + 12\pi \int_{-1}^{1}\sqrt{1 - u^2}\,du

Why: the substitution centres the disk at the origin in u-coordinates, which makes symmetry visible. The integrand splits into an odd part (u\sqrt{1 - u^2}) and an even part (\sqrt{1 - u^2}).

Step 4. Evaluate each piece.

The first integral: u\sqrt{1 - u^2} is an odd function on a symmetric interval, so \int_{-1}^{1} u\sqrt{1-u^2}\,du = 0.

The second integral: \int_{-1}^{1}\sqrt{1 - u^2}\,du is the area of a half-disk of radius 1, which is \pi/2.

Putting it together:

V = 4\pi \cdot 0 + 12\pi \cdot \frac{\pi}{2} = 6\pi^2

Why: odd-function symmetry kills one integral for free. The other integral is recognisable as a known area — faster than computing with trigonometric substitution.

Result: The torus has volume V = 6\pi^2.

The cross-section of the torus: a disk of radius $1$ centred at $(3, 0)$, rotated about the y-axis (dashed red). The shell at $x = 2.5$ (red segment) has radius $2.5$ and height $2\sqrt{1 - (2.5 - 3)^2} = \sqrt 3$. The total volume is $6\pi^2$.

You can also check this torus volume against Pappus's theorem: the volume of a solid of revolution equals the area of the cross-section times the distance travelled by its centroid. Here the cross-section is a disk of area \pi, its centroid is at (3, 0), and as it rotates about the y-axis the centroid travels a circle of circumference 2\pi \cdot 3 = 6\pi. Product: \pi \cdot 6\pi = 6\pi^2. Same answer, and Pappus's theorem is really just the shell method abstracted to a single line.

Common confusions

Going deeper

If you can set up a volume integral for any reasonable rotation problem, you have it. The rest of this section zooms out to see where the three methods come from and how they connect.

Everything is one integral

Disks, washers, and shells are not three different techniques. They are three choices of how to slice a solid before integrating. The underlying statement is much simpler:

The volume of a solid is V = \int A(s)\,ds, where s is a coordinate that picks out a cross-section and A(s) is the cross-sectional area at that value of s.

For disks: s = x, cross-section perpendicular to the x-axis, area = \pi[f(x)]^2. For washers: s = x, cross-section perpendicular to the x-axis, area = \pi[f^2 - g^2]. For shells: the slicing is different. A thin vertical strip is rotated about the axis to trace out a surface, not a cross-section. The shell's volume is computed as the product of its surface area 2\pi x \cdot f(x) and its thickness dx.

All three methods are instances of the general principle: slice the solid into pieces whose volumes you know, then integrate.

Pappus's theorem

Pappus's theorem (also called the Pappus-Guldinus theorem) states: if a planar region of area A is rotated about an axis that does not cross it, the volume of the resulting solid equals A times the distance travelled by the centroid of the region.

V = 2\pi \cdot \bar{x}_c \cdot A

where \bar{x}_c is the distance from the axis to the centroid of the region.

This is an enormous shortcut. For the torus, A = \pi(1)^2 = \pi, \bar{x}_c = 3, so V = 2\pi \cdot 3 \cdot \pi = 6\pi^2. One line, no integration.

The proof is just the shell method:

V = 2\pi \int x\,f(x)\,dx = 2\pi \cdot \bar{x}_c \cdot A

because \int x\,f(x)\,dx = \bar{x}_c \cdot A by definition of the centroid. Pappus's theorem and the shell method are equivalent.

Generalising beyond solids of revolution

The slice-and-integrate principle works for any solid, not just solids of revolution. If you can describe the cross-section of a solid at every height (or at every position along some axis), you can compute its volume.

For example, a pyramid with a square base of side a and height h has cross-sections that are smaller squares. At height z above the base, the cross-section is a square with side a(1 - z/h), so area a^2(1 - z/h)^2. Volume:

V = \int_0^h a^2(1 - z/h)^2\,dz = a^2 \cdot \frac{h}{3} = \frac{1}{3}a^2 h

The volume of any cone-like solid with base area A and height h is \frac{1}{3}Ah. The sphere, the pyramid, the cone, and every solid of revolution all come from the same one-integral idea.

Simpson's paradox for volumes

Not really a paradox, but a surprise: two solids can have the same cross-sectional area at every height and therefore the same volume, even though they look nothing alike. This is Cavalieri's principle, and it predates calculus by a century — the idea survived because it is true even for solids whose shapes cannot be described with simple functions.

Cavalieri's principle in one sentence: if every horizontal slice of solid A has the same area as the corresponding slice of solid B, then A and B have the same volume. This is what justifies using the shear of a pyramid (cutting it into slabs and sliding them sideways) without changing its volume. The \int A(s)\,ds formulation of volume is, at bottom, a continuous version of Cavalieri.

Where this leads next

Volumes of revolution are the first case of "integrate to get a 3D answer." The same ideas extend naturally to lengths, surface areas, and higher-dimensional measurements.