Case 1
- Outer radius
- 5
- Inner radius
- 3
50.265482
Open with these valuesAnnulus Area Calculator
Change any value to update the result.
50.265482units²
Ring area: 50.265482 units²Examples
How the result moves
Inner radius → Ring area (units²)Your current valid inputs, with one value varied across the range.
How the result moves
Held fixed: Outer radius 5.000.
| Inner radius | Ring area |
|---|---|
| 2.000 | 65.973446 |
| 3.000Your value | 50.265482 |
| 4.000 | 28.274334 |
A washer, a pipe cross-section, a running track: the big circle minus the hole, π(R² − r²). Note the squares — a ring from 5.9 to 6 is only a tenth wide but still covers 3.74 square units, because the circumference it runs around is long.
Worked examples
Case 2
- Outer radius
- 10
- Inner radius
- 5
235.619449
Open with these valuesCase 3
- Outer radius
- 6
- Inner radius
- 5.9
3.738495
Open with these valuesHow it's calculated
A = π × (R² − r²)
- StepMeasure the outer radius of the ring.Outer radius=5
- StepMeasure the inner radius — the hole.Inner radius=3
- ResultRead the area in the square of the unit you entered.50.265482units²
What this number means
Square first, then subtract
The ring is the difference of two circle areas, π(R² − r²) — it is not π(R − r)². For 5 and 3 that is 16π, about 50.27 square units, where the wrong order would leave 4π.
A narrow ring is not a small ring
From 5.9 to 6 the ring is only a tenth wide, yet it covers 3.738495 square units. The circumference it runs around is long, and the squares carry that.
The width alone does not fix the area
A ring 2 wide covers 16π when it runs from 3 to 5, but only 12π from 2 to 4. The same width around a smaller circle covers less.
Pipes: use the radius of the bore
Enter the outer radius and the inner radius of the bore. If only the wall thickness is known, subtract it from the outer radius to get the inner one.
Commonly misread
MisconceptionA ring from 5.9 to 6 is a tenth wide, so its area is π × 0.1².
CorrectionThat is the area of a small circle of radius 0.1, about 0.03. The ring itself covers 3.738495 square units, because R² − r² is not (R − r)².
MisconceptionTwo rings of the same width have the same area.
CorrectionOnly if they sit at the same radius. From 3 to 5 the area is 16π, from 2 to 4 only 12π.
MisconceptionWith no hole the formula does not apply.
CorrectionAn inner radius of zero is allowed and simply removes the hole. The expression reduces to πR², the full circle.
Reference table
| Outer, inner | Exact | Area |
|---|---|---|
| 6, 5.9 | 1.19π | 3.738495 |
| 2, 1 | 3π | 9.424778 |
| 4, 2 | 12π | 37.699112 |
| 5, 3 | 16π | 50.265482 |
| 10, 5 | 75π | 235.619449 |
Questions
How do you calculate the area of an annulus?
Square both radii, subtract the inner from the outer, and multiply by π. From 5 down to 3 gives 16π, about 50.27 square units.
Can I just use the width of the ring?
Not on its own. A ring 2 wide covers 16π when it runs from 3 to 5, but 12π when it runs from 2 to 4 — the same width around a smaller circle covers less.
How do I get the cross-section of a pipe?
Use the outer radius and the inner radius of the bore. If you have the wall thickness, subtract it from the outer radius to get the inner one.
What if the inner radius is zero?
Then there is no hole and the formula reduces to the full circle, πR².