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Annulus

Annulus
Washer

The region between two concentric circles which have different radii.

 

Area of annulus \( = \pi \left( {{R^2} - {r^2}} \right)\)

This is an annulus. It is the region between two concentric circles. The larger circle has radius uppercase R and the smaller has radius lowercase r. 

Key Formula

A=π(R2−r2)A = \pi(R^2 - r^2)
Where:
  • AA = Area of the annulus
  • RR = Radius of the outer (larger) circle
  • rr = Radius of the inner (smaller) circle

Worked Example

Problem: A circular garden path surrounds a fountain. The outer edge of the path has a radius of 10 m, and the inner edge (the fountain boundary) has a radius of 6 m. Find the area of the path.
Step 1: Identify the two radii. The outer radius is R = 10 m and the inner radius is r = 6 m.
R=10,r=6R = 10, \quad r = 6
Step 2: Apply the annulus area formula.
A=π(R2−r2)=π(102−62)A = \pi(R^2 - r^2) = \pi(10^2 - 6^2)
Step 3: Compute the squares and subtract.
A=π(100−36)=64πA = \pi(100 - 36) = 64\pi
Step 4: Evaluate numerically if needed.
A≈201.06 m2A \approx 201.06 \text{ m}^2
Answer: The area of the garden path is 64π≈201.0664\pi \approx 201.06 square metres.

Another Example

Problem: A washer (flat ring) has an outer diameter of 8 cm and an inner diameter of 3 cm. What is its area?
Step 1: Convert diameters to radii by dividing by 2.
R=82=4 cm,r=32=1.5 cmR = \frac{8}{2} = 4 \text{ cm}, \quad r = \frac{3}{2} = 1.5 \text{ cm}
Step 2: Substitute into the annulus area formula.
A=π(42−1.52)=π(16−2.25)=13.75πA = \pi(4^2 - 1.5^2) = \pi(16 - 2.25) = 13.75\pi
Step 3: Evaluate numerically.
A≈43.20 cm2A \approx 43.20 \text{ cm}^2
Answer: The area of the washer is 13.75π≈43.2013.75\pi \approx 43.20 square centimetres.

Frequently Asked Questions

Can you factor the annulus area formula?
Yes. Because R2−r2R^2 - r^2 is a difference of two squares, it factors as (R+r)(R−r)(R + r)(R - r). So the area can also be written as A=π(R+r)(R−r)A = \pi(R + r)(R - r). This form is sometimes easier to use when you know the sum and difference of the radii.
What is the difference between an annulus and a circle?
A circle encloses a single round region with one radius. An annulus is the region between two concentric circles, so it has a hole in the middle. You can think of an annulus as a full disk with a smaller disk removed from its centre.

Annulus vs. Disk (Circle region)

A disk is the full interior of a single circle, with area πR2\pi R^2. An annulus removes a concentric inner disk, leaving only the ring-shaped region with area π(R2−r2)\pi(R^2 - r^2). If the inner radius r=0r = 0, the annulus reduces to a full disk.

Why It Matters

Annuli appear whenever you need the area of a ring-shaped object—pipes, washers, circular tracks, tree trunk cross-sections, or the surface of a CD. In calculus, thin annuli are the building blocks of the washer method used to compute volumes of solids of revolution. Understanding annuli also reinforces the key skill of subtracting areas to find regions between curves.

Common Mistakes

Mistake: Subtracting the radii first, then squaring: writing π(R−r)2\pi(R - r)^2 instead of π(R2−r2)\pi(R^2 - r^2).
Correction: You must square each radius separately before subtracting. (R−r)2≠R2−r2(R - r)^2 \neq R^2 - r^2 because the cross term 2Rr2Rr is missing. Always compute R2R^2 and r2r^2 first.
Mistake: Confusing diameter with radius when given diameter measurements.
Correction: Remember to halve each diameter before substituting into the formula. Using diameters directly will give an area four times too large.

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