Annulus area:
formula and worked example

Where ring-shaped areas turn up

An annulus is the flat region between two circles that share the same centre. You meet it every time something round has a round hole: a flat washer, a CD, a gasket, the cross-section of a hollow pipe wall, a circular running track lane, a paved path around a circular pond or a round tree guard.

In each case the question is the same: how much material, paint or surface lies between the inner and outer edges, ignoring the hole. The area also underlies volumes, because multiplying it by a length gives the volume of a hollow tube.

Annuli also sit behind many volume calculations. A hollow tube, a ring-shaped seal or a drilled disc is just an annulus extended through a thickness, so getting the ring area right is the first half of any such volume.

The formula, one circle minus another

Area equals pi times the outer radius squared minus the inner radius squared. Think of it as the big disc minus the disc you cut out. The outer radius R is the distance from the centre to the outside edge; the inner radius r is the distance to the edge of the hole. R must be larger than r, or the ring does not exist.

Pulling pi out as a common factor avoids two separate multiplications and, more importantly, makes it easy to see the structure: R^2 - r^2 equals (R + r)(R - r). That second form says the area is the average circumference times the ring width, which is a useful sense check.

In many drawings you are given outer and inner diameters instead of radii. Halve both before squaring, or use the equivalent form pi / 4 x (D^2 - d^2). Picking one convention for a whole job and sticking to it prevents most arithmetic errors.

A ring from radius 4 out to radius 6

Let the outer radius be 6 units and the inner radius 4 units, so the ring is 2 units wide.

1. 1Square the outer radius: 6 x 6 = 36.
2. 2Square the inner radius: 4 x 4 = 16.
3. 3Subtract: 36 - 16 = 20.
4. 4Multiply by pi: 20 x 3.14159 = 62.832 square units.

Cross-check with the average-circumference idea: the mean radius is 5, the mean circumference is 2 x pi x 5 = 31.416, and times the width of 2 gives 62.832. Two routes, one answer.

Interpreting the ring in practice

Once you have the area, the rest depends on what the ring is made of or covers. For a steel washer, multiply by thickness to get volume and then by density to get mass. For a paved path, multiply by the cost per square metre and add an allowance for cuts along the curve. For a gasket, the area tells you the sheet size required, though nesting waste can be significant when rings are cut from rectangular stock.

Notice also that a thin ring far from the centre can have a larger area than a thick ring close in. A 1-unit-wide ring from 9 to 10 covers about 59.7 square units; a 3-unit-wide ring from 0 to 3 covers about 28.3. Distance from the centre matters as much as width.

Growing only the outer radius

Raising R from 6 to 6.6 pushes the area from 62.832 to 86.582, a jump of about 38%. The ring width went from 2 to 2.6, a 30% change, yet the area rose faster because the added strip lies at the outside, where each extra unit of width covers more length. Compare that with a plain disc of the same growth, which gains 21%.

The lesson for estimating is that ring area is quite sensitive to the outer edge. If the hole is fixed, an extra few millimetres on the outer radius of a washer or a few centimetres on a path width add up faster than intuition suggests.

Subtraction slips specific to rings

• Squaring the difference: pi x (6 - 4)^2 = 12.57, only a fifth of the true 62.83.
• Subtracting the radii first and then multiplying by pi x 2, which gives the circumference of the width, not an area.
• Entering diameters in a formula that wants radii, which inflates the answer by a factor of four.
• Swapping inner and outer, which gives a negative area and is a clear sign that the inputs are the wrong way round.

Another failure shows up when the hole is not centred. If the hole is offset, the area is still the full disc minus the hole, so the formula still holds, but the ring is no longer uniform in width. The formula remains correct as long as the hole lies completely inside the outer edge.

Scenario: a paved border around a round pond

A garden pond has a radius of 2.5 m and you want a 0.8 m paved border, so the outer radius is 3.3 m. The border area is pi x (3.3^2 - 2.5^2) = pi x (10.89 - 6.25) = pi x 4.64 = 14.58 square metres. Paving slabs are rectangular, so plan for extra for cuts around the curve, and consider whether the pond lining needs its own overlap under the first course. The same annulus calculation, applied twice, would give the area of the water surface (pi x 6.25) versus the dry paving.