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Volume and surface area of a sphere (Archimedes)

Statement

A sphere of radius rr has volume V=43πr3V = \dfrac{4}{3}\pi r^3 and surface area A=4πr2A = 4\pi r^2; both are equal to 23\dfrac{2}{3} of the volume 2πr32\pi r^3 and total surface area 6πr26\pi r^2 of its circumscribing right circular cylinder of radius rr and height 2r2r.

Why is it true?

Slice a hemisphere of radius rr and a right circular cylinder of radius rr and height rr with an inverted cone of radius rr and height rr hollowed out of it by a horizontal plane at height zz above the base. By the Pythagorean theorem, the disk cross-section of the hemisphere has radius r2−z2\sqrt{r^2 - z^2} and area π(r2−z2)\pi(r^2 - z^2), which is identical to the area πr2−πz2\pi r^2 - \pi z^2 of the ring cross-section of the hollowed cylinder. Because every horizontal slice has the same area, the hemisphere has volume πr3−13πr3=23πr3\pi r^3 - \dfrac{1}{3}\pi r^3 = \dfrac{2}{3}\pi r^3, so the full sphere has volume V=43πr3V = \dfrac{4}{3}\pi r^3.

Proof sketch

At height z∈[−r,r]z \in [-r, r], a plane perpendicular to the axis cuts the sphere of radius rr in a disk of radius r2−z2\sqrt{r^2 - z^2} and area S(z)=π(r2−z2)S(z) = \pi(r^2 - z^2). Comparing this slice by Cavalieri's principle with a cylinder of radius rr and height 2r2r from which two cones of base radius rr and height rr have been removed (or integrating ∫−rrπ(r2−z2) dz\int_{-r}^{r} \pi(r^2 - z^2)\,dz) gives V=2πr3−2⋅13πr3=43πr3V = 2\pi r^3 - 2\cdot\dfrac{1}{3}\pi r^3 = \dfrac{4}{3}\pi r^3. Partitioning the sphere into thin pyramids of height rr with bases tiling the surface of area AA gives V=13ArV = \dfrac{1}{3}Ar, whence A=3Vr=4πr2A = \dfrac{3V}{r} = 4\pi r^2.

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Proved by

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Step-by-step proofs

No step-by-step proof yet for this theorem.

References

  1. Archimedes (translated by Thomas L. Heath) (1897). The Works of Archimedes (On the Sphere and Cylinder, Book I, Propositions 33–34)
  2. Reviel Netz, William Noel (2007). The Archimedes Codex