Math Detours · Grade 10 · Sharpen · 18

Volumes: why the cone is a third

Cavalieri's principle, the cone, the sphere via Archimedes' argument, and scaling: double the size, eight times the volume.

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Tonight's idea

You know the formulas: cone and pyramid are a third of base times height, and the sphere is \(\tfrac43\pi r^3\). Tonight is where the \(\tfrac13\) and the \(\tfrac43\) come from, using one idea.

Cavalieri's principle. Slice two solids by the same family of parallel planes. If at every height the cross-sections have equal area, the solids have equal volume.

Picture a stack of coins pushed into a leaning tower: same coins, same volume, every slice still one coin.

Why a third. A cube can be cut into three identical pyramids: pick one corner as the apex and use the three faces not touching it as the bases. Each has base \(s^2\), height \(s\), volume \(\tfrac13 s^3\). By Cavalieri, any pyramid or cone with base area \(B\) and height \(h\) then has volume \(\tfrac13 Bh\).

The sphere. Archimedes compared a hemisphere of radius \(r\) with a cylinder of radius \(r\) and height \(r\) that has a cone scooped out (apex at the centre of the base, opening upward), as in the figure. At height \(y\) the hemisphere's slice is a disc of area \(\pi(r^2 - y^2)\), and the scooped cylinder's slice is a ring of area \(\pi r^2 - \pi y^2\). Equal at every height, so the hemisphere is cylinder minus cone, \(\pi r^3 - \tfrac13\pi r^3 = \tfrac23\pi r^3\), and the sphere is \(\tfrac43\pi r^3\).

y √(r² − y²)hemisphere, radius r y ringcylinder minus cone rr
At height \(y\), the disc on the left and the ring on the right both have area \(\pi(r^2 - y^2)\).

Scaling. Enlarge any shape by \(k\): lengths multiply by \(k\), areas by \(k^2\), volumes by \(k^3\).

Watch one: the sphere in its tin

Sphere, radius 3: \(V = \tfrac43 \pi \cdot 27 = 36\pi\). The smallest cylinder that contains it has radius 3 and height 6, volume \(\pi \cdot 9 \cdot 6 = 54\pi\). Ratio \(\tfrac{36}{54} = \tfrac23\). Archimedes thought this the finest thing he ever proved and asked for a sphere in a cylinder on his tomb.

Your turn

  1. Find the volume of a square pyramid with base side 6 and height 5, and of a cone with radius 2 and height 9 (leave \(\pi\) in).
  2. A cylinder of radius 2 and height 7 is pushed sideways so its top is no longer above its base, but every cross-section is still a circle of radius 2. Find its volume and name the principle.
  3. Earth's radius is about 6,371 km and the Moon's about 1,737 km. Find the ratio of the radii and, without computing either volume, the ratio of the volumes.
  4. A model car is built at scale 1:24. How many times more paint does the real car need? How many times heavier is it? Then explain why an animal twice as tall as another of the same shape has bones under twice the stress.

Stretch: the sphere's surface, from its volume

Cut a sphere into thousands of thin pyramids with apex at the centre and tiny bases tiling the surface. Each has volume \(\tfrac13 \times \text{base} \times r\). Explain why \(V = \tfrac13 S r\), where \(S\) is the surface area, and deduce \(S\).

Answers

  1. Pyramid \(\tfrac13 \cdot 36 \cdot 5 = 60\). Cone \(\tfrac13 \pi \cdot 4 \cdot 9 = 12\pi\).
  2. \(\pi \cdot 4 \cdot 7 = 28\pi \approx 88.0\). Cavalieri: same cross-section at every height as the upright cylinder.
  3. Radii \(6371/1737 \approx 3.67\). Volumes \(3.67^3 \approx 49\): about 49 Moons fit inside Earth.
  4. Paint is area: \(24^2 = 576\) times. Weight is volume: \(24^3 = 13{,}824\) times. Doubling height makes an animal \(2^3 = 8\) times heavier, but bone strength goes with cross-section, only \(2^2 = 4\) times, so each square centimetre carries \(8/4 = 2\) times the load.

Stretch. \(V = \sum \tfrac13 (\text{base}_i) r = \tfrac13 r \sum \text{base}_i = \tfrac13 S r\), since the tiny bases add up to the surface. So \(S = 3V/r = 3 \cdot \tfrac43\pi r^3 / r = 4\pi r^2\).

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