In 30 seconds
This 3Blue1Brown excerpt sketches a projection from a sphere onto an enclosing cylinder. The local area changes cancel, and unwrapping the cylinder yields the familiar surface-area formula.
- 1The formulaA sphere’s surface area is four times the area of a great circle.
- 2Project to a cylinderCompare sphere patches with their projections on the enclosing cylinder.
- 3Unwrap the cylinderCircumference times height gives 2πr × 2r = 4πr².
Takeaways
- A sphere and its tightly enclosing cylinder have equal surface and lateral areas, respectively.
- The projection’s local stretching and compression cancel in the infinitesimal limit.
- The cylinder’s circumference times its height gives 4πr².
Read the edited text by chapter
A sphere and its cylinder
A projection argument leads to the surface-area formula.
A sphere of radius r has surface area 4πr², four times the area of a circle with the same radius. This short 3Blue1Brown excerpt sketches a geometric explanation associated with Archimedes.
Imagine a cylinder tightly enclosing the sphere, ignoring the cylinder’s circular caps. Project points outward from the cylinder’s axis onto its side; the argument compares small patches of the sphere with their projected patches on the cylinder.
Near a point, projection compresses the patch in one direction and stretches it in the other. In the infinitesimal limit, the area factors cancel, so the sphere’s area equals the lateral area of the enclosing cylinder.
Unwrap the cylinder into a rectangle: its width is the circumference 2πr and its height is 2r. Multiplying gives 4πr²; the narrator also compares this rectangle with four unwrapped circles.
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