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Theorem — what does that mean?
The circumference of the tube, 2πr, carried once round a circle of radius R, 2πR: four pi squared R r.
In Swedish: Torusens area
A
The circumference of the tube, 2πr, carried once round a circle of radius R, 2πR: four pi squared R r.
A≈394.7842cm²
A equals (4 times π squared) times (5 times 2), is approximately (4 times 9.8696) times 10, is approximately 39.4784 times 10, is approximately 394.7842 cm²
You are given
Round to 1 decimal.
Pick the unit your lengths are in. Areas are then labelled in square units and volumes in cubic units; nothing is converted.
R is the major radius, from the centre of the hole to the centre of the tube, and r the minor radius, the radius of the tube. Unlike the volume, the area has both radii to the first power.
R = 5 and r = 2 cm: A = 4π² · 5 · 2 = 40π² ≈ 394.8 cm². The number equals the volume’s for this pair, 2π² · 5 · 4, only because r² = 2r when r = 2; at r = 3 the area is 60π² and the volume 90π².
The tube’s cross-section is a circle with circumference 2πr. Its centre travels round a circle of radius R, a distance of 2πR, and the surface swept out is the circumference times the distance: 2πr · 2πR = 4π²Rr. That is Pappus’s second theorem, and it is the same reasoning that gives a cylinder’s curved surface as circumference times length.
The formula needs R ≥ r. At R = r the hole closes and the inner surface meets itself at a point; below it the inner surface is inside the solid and the product counts area that is not on the outside.
Why does the inner side count the same as the outer?
The inner side is shorter round but the outer side longer, and they average out to exactly the middle
circle, 2πR. That is what Pappus’s theorem says.
How much paint for a lifebuoy?
R = 30 cm and r = 10 cm: 4π² · 300 ≈ 11,800 cm², about 1.2 m², both sides included since a torus has
only one.
What is the area of a doughnut’s glaze?
The top half: 2π²Rr, if the glaze stops at the equator.
| Solve for | Rearranged | With the numbers | Comes to |
|---|---|---|---|
| R | A4π²r | 394.84π²·2 | 5 cm |
| A | 4π²Rr | 4π²·5·2 | 394.8 cm² |
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