Primary & Middle School Mathematics · Grades 1–9

70Solids and Volumes

Space geometry begins with a small family of solids — prisms, cylinders, pyramids, cones, spheres — and two questions: how much do they hold (volume), and what do you see when you slice them (sections)? The chapter closes with a fact of great practical importance: scaling a solid by $k$ multiplies its volume by $k^3$ .

70.1 The classical solids and their volumes

Theorem 70.1 (Volume formulas)

Write $B$ for the area of the base and $h$ for the height (the distance between the base and the opposite face or apex).

solid	volume
prism, cylinder	$V = B \times h$
[5pt] pyramid, cone	$V = \dfrac13\, B \times h$
[5pt] sphere of radius $r$	$V = \dfrac43\,\pi r^3$

For a cylinder and a cone of radius $r$ , the base area is $B = \pi r^2$ ; the sphere’s surface area is $4\pi r^2$ .

Proof. Admitted at this level. ∎

The solids of this chapter. For each, the volume involves a base area and a height — except the sphere, which only needs its radius.

Example 70.2

A cylindrical can has radius $4$ cm and height $10$ cm:

V = \pi r^2 h = \pi \times 16 \times 10 = 160\pi \approx 503 \text{ cm}^3 \approx 0.5 \text{ L}.

A cone with the same base and height holds one third of that: $\frac{160\pi}{3} \approx 168$ cm $^3$ . A sphere of radius $4$ cm: $V = \frac43 \pi \times 64 = \frac{256\pi}{3} \approx 268$ cm $^3$ .

Example 70.3 (A pyramid step by step)

A pyramid has a square base of side $6$ m and height $5$ m.

Base area: $B = 6^2 = 36$ m $^2$ .
Volume: $V = \frac13 \times 36 \times 5 = 60$ m $^3$ .

Mind the units: if the side were given in cm and the height in m, one of them would have to be converted first.

70.2 Sections by planes

Proposition 70.4 (Sections of the classical solids)

Cutting a solid by a plane produces a flat figure, its cross-section:

a prism or cylinder cut parallel to its base gives a copy of the base, at every height;
a pyramid or cone cut parallel to its base gives a reduction of the base: at distance $d$ from the apex, the scale factor is $k = \frac{d}{h}$ ;
a sphere of radius $r$ cut by a plane at distance $d < r$ from the center gives a circle of radius $\sqrt{r^2 - d^2}$ (by the Pythagorean theorem).

Proof. Admitted at this level. ∎

Left: cutting a cone parallel to its base at distance d from the apex gives a disk scaled by k = dh. Right: a plane at distance d from the center of a sphere cuts it in a circle of radius √r2 - d2. — Left: cutting a cone parallel to its base at distance $d$ from the apex gives a disk scaled by $k = \frac dh$ . Right: a plane at distance $d$ from the center of a sphere cuts it in a circle of radius $\sqrt{r^2 - d^2}$ .

Example 70.5

A cone has base radius $6$ and height $9$ . The section by a plane parallel to the base, at distance $3$ from the apex, is a disk scaled by $k = \frac39 = \frac13$ : its radius is $6 \times \frac13 = 2$ .

A sphere of radius $5$ is cut by a plane at distance $3$ from its center: the section is a circle of radius $\sqrt{25 - 9} = 4$ .

70.3 Scaling solids

Theorem 70.6 (Effect of a scaling on lengths, areas, volumes)

When a solid is enlarged or reduced by the scale factor $k > 0$ :

all lengths are multiplied by $k$ ;
all areas are multiplied by $k^2$ ;
all volumes are multiplied by $k^3$ .

Proof. Admitted at this level. ∎

Example 70.7

A model car at scale $\frac{1}{18}$ : lengths are those of the real car multiplied by $k = \frac{1}{18}$ , painted surface multiplied by $k^2 = \frac{1}{324}$ , and volume by $k^3 = \frac{1}{5832}$ .

Doubling the radius of a sphere ( $k = 2$ ) multiplies its volume by $2^3 = 8$ : check on the formula, $\frac43\pi (2r)^3 = \frac43 \pi \times 8r^3$ .

Example 70.8 (Truncated cone)

A cone of height $9$ and base radius $6$ (volume $\frac13 \pi \times 36 \times 9 = 108\pi$ ) is cut at distance $3$ from the apex, and the small cone above the cut is removed. The small cone is the reduction by $k = \frac13$ , so its volume is $108\pi \times \left(\frac13\right)^3 = 4\pi$ . The remaining solid (a truncated cone) has volume $108\pi - 4\pi = 104\pi \approx 327$ .

70.4 Exercises

Exercise 70.1 ★

Compute the volumes: a box (rectangular prism) of dimensions $4 \times 5 \times 12$ ; a cylinder of radius $3$ and height $7$ (exact value with $\pi$ , then rounded to the unit).

Solution

Solution of Exercise 70.1.

Box: $V = 4 \times 5 \times 12 = 240$ .

Cylinder: $V = \pi r^2 h = \pi \times 9 \times 7 = 63\pi \approx 198$ .

Exercise 70.2 ★

Compute the volume of a cone of radius $5$ and height $12$ , and of a pyramid with rectangular base $8 \times 3$ and height $10$ .

Solution

Solution of Exercise 70.2.

Cone: $V = \frac13 \pi r^2 h = \frac13 \pi \times 25 \times 12 = 100\pi \approx 314$ .

Pyramid: base area $B = 8 \times 3 = 24$ , so $V = \frac13 \times 24 \times 10 = 80$ .

Exercise 70.3 ★

Compute the volume and the surface area of a sphere of radius $6$ (exact values with $\pi$ ).

Solution

Solution of Exercise 70.3.

Volume: $V = \frac43 \pi \times 6^3 = \frac43 \pi \times 216 = 288\pi$ .

Surface area: $4\pi \times 6^2 = 144\pi$ .

Exercise 70.4 ★

A sphere of radius $10$ is cut by a plane at distance $8$ from its center. What is the radius of the section circle?

Solution

Solution of Exercise 70.4.

Radius of the section: $\sqrt{r^2 - d^2} = \sqrt{100 - 64} = \sqrt{36} = 6$ .

Exercise 70.5 ★★

A cone has base radius $8$ and height $12$ . A plane parallel to the base cuts it at distance $9$ from the apex. Compute the radius of the section, then the volume of the small cone above the cut.

Solution

Solution of Exercise 70.5.

Scale factor from the whole cone to the small one: $k = \frac{9}{12} = \frac34$ . Section radius: $8 \times \frac34 = 6$ .

Volume of the whole cone: $\frac13\pi \times 64 \times 12 = 256\pi$ . Volume of the small cone: $256\pi \times \left(\frac34\right)^3 = 256\pi \times \frac{27}{64} = 108\pi \approx 339$ .

Exercise 70.6 ★★

A cylindrical glass of inner radius $3$ cm contains water to a height of $10$ cm. A ball of radius $2$ cm is fully submerged in it. By how much does the water level rise? (The added volume spreads over the glass’s base area; give the exact rise, then round to the millimeter.)

Solution

Solution of Exercise 70.6.

Volume of the ball: $\frac43\pi \times 2^3 = \frac{32\pi}{3}$ cm $^3$ . This volume spreads over the base area $\pi \times 3^2 = 9\pi$ cm $^2$ , so the level rises by

\frac{32\pi/3}{9\pi} = \frac{32}{27} \approx 1.2 \text{ cm}

(about $12$ mm).

Exercise 70.7 ★★

A recipe fills a spherical mold of radius $6$ cm. You only have a spherical mold of radius $3$ cm. How many small molds can you fill? (Answer without computing either volume explicitly.)

Solution

Solution of Exercise 70.7.

The small mold is the reduction of the large one by $k = \frac36 = \frac12$ , so its volume is $\left(\frac12\right)^3 = \frac18$ of the large one: the recipe fills $8$ small molds.

Exercise 70.8 ★★

The Great Pyramid of Giza has a square base of side about $230$ m and a height of about $147$ m. Estimate its volume, and express it in millions of cubic meters.

Solution

Solution of Exercise 70.8.

$V = \frac13 \times 230^2 \times 147 = \frac13 \times 52\,900 \times 147 = 52\,900 \times 49 = 2\,592\,100$ m $^3$ — about $2.6$ million cubic meters.

Exercise 70.9 ★★★

A cone-shaped funnel of radius $6$ cm and height $12$ cm, apex down, is filled with water up to half of its height (measured from the apex).

What fraction of the funnel’s volume is filled?
If instead it is filled with half of its volume of water, show that the water height $d$ satisfies $\left(\frac{d}{12}\right)^3 = \frac12$ , and give $d$ to the millimeter ( $\sqrt[3]{0.5} \approx 0.7937$ ).

Solution

Solution of Exercise 70.9.

1. The water forms a cone scaled by $k = \frac12$ (apex down, half the height), so its volume is $\left(\frac12\right)^3 = \frac18$ of the funnel’s: one eighth, much less than half!

2. Water up to height $d$ forms a cone scaled by $k = \frac{d}{12}$ , of volume $k^3$ times the funnel’s. Half the volume means $k^3 = \frac12$ , i.e. $\left(\frac{d}{12}\right)^3 = \frac12$ . Then $\frac{d}{12} = \sqrt[3]{0.5} \approx 0.7937$ , so $d \approx 9.5$ cm: the second half of the volume occupies only the top $2.5$ cm — the wide part of the cone holds most of the water.

70.5 Problem: Archimedes’ tombstone

Problem 70.1

Weekend problem — the sphere is two thirds of its cylinder (twice over), the 1:2:3 stack, and the square–cube law that forbids giants

Archimedes proved many theorems, but one made him so proud that he asked for its figure to be carved on his tomb: a sphere nested in its tightest cylinder. A century later the Roman writer Cicero, searching the brambles near Syracuse, recognized the grave “by the sphere and the cylinder”. This problem retrieves what the carving encodes — a double two-thirds miracle — then follows volumes and surfaces (Theorem 70.1, Theorem 70.6) to a law that governs giants, ants and cooling planets.

Part I — The carving decoded. A sphere of radius $r$ sits exactly inside a cylinder: same radius, height $2r$ .

Compute the cylinder’s volume in terms of $r$ .
Compute the ratio of the sphere’s volume to the cylinder’s. Why might Archimedes have liked that the answer contains no $\pi$ and no $r$ ?
Now the surfaces: compare the sphere’s surface area with the cylinder’s total surface (lateral part plus the two lids). What ratio appears — again?
The empty space between sphere and cylinder has volume $2\pi r^3 - \frac43\pi r^3$ . Show that this leftover exactly equals the volume of two cones of radius $r$ and height $r$ .
A basketball of radius $12$ cm is sold in the tightest cylindrical box. Compute both volumes (to the nearest $100$ cm $^3$ ) and the percentage of the box that is empty.

Part II — Bowls, molds and moons.

A hemispherical bowl has radius $10$ cm. Compute its capacity, in cm $^3$ and in liters (to the deciliter).
The 1:2:3 stack: a cone, a hemisphere and a cylinder, all of radius $10$ cm and height $10$ cm. Compute the three volumes and verify the legendary proportion $1 : 2 : 3$ . (Archimedes would have appreciated this one, too.)
A chocolate sphere of radius $3$ cm is melted into a cylindrical mold of radius $3$ cm. What height does the chocolate reach? (Exact fraction, then to the millimeter.)
The Earth’s radius is about $6\,371$ km, the Moon’s about $1\,737$ km. Compute the ratio of the radii, then — with Theorem 70.6 — the ratio of the volumes: how many Moons would fit into a hollow Earth, by volume?
From the same theorem: what is the Earth–Moon ratio of surface areas? And in general, when a balloon’s radius doubles, what happens to its volume and to its surface? State the square–cube law: as a shape scales up, volumes outrun surfaces.

Part III — The square–cube law rules the world.

Galileo’s argument against giants: imagine a human scaled up by a factor $10$ , same proportions. By what factor does the weight grow (weight follows volume)? By what factor does the cross-section of the bones grow (an area)? By what factor, then, does the pressure on each square centimeter of bone grow — and what happens to the giant?
The same law in reverse explains ant heroics: strength follows muscle cross-section (an area), weight follows volume. For an animal $100$ times smaller in every direction, by what factors do weight and strength shrink, and by what factor does the strength-to-weight ratio improve?
Heat is produced by the body’s volume and lost through its surface. Show that for a sphere the surface-to-volume ratio is $\frac{3}{r}$ , and compute it for $r = 1$ and $r = 2$ . Which cools faster, a mouse or a bear — and why do babies need hats in winter?
Two spherical oranges of radii $4$ cm and $5$ cm are sold at the same price. Compute the ratio of their volumes: how much more orange does the big one give for the same money?
Finale: describe precisely what the tombstone figure encodes — the two “two thirds” of questions 2 and 3 — and answer Cicero in one sentence: why would a mathematician choose, over every conquest of his engineering genius, a sphere in a cylinder?

Solution

Solution of Problem 70.1.

1. $V_{\text{cyl}} = \pi r^2 \times 2r = 2\pi r^3$ .

2. $\dfrac{V_{\text{sphere}}}{V_{\text{cyl}}} = \dfrac{\frac43 \pi r^3}{2 \pi r^3} = \dfrac23$ . The $\pi$ and the $r^3$ cancel: the proportion is universal — true for a marble and for a planet. A relation between shapes, not between numbers: exactly the kind of truth worth carving in stone.

3. Sphere: $4\pi r^2$ . Cylinder: lateral $2\pi r \times 2r = 4\pi r^2$ , plus two lids $2 \times \pi r^2$ : total $6\pi r^2$ . Ratio: $\frac{4\pi r^2}{6\pi r^2} = \frac23$ — the same two thirds, for surfaces. (And a bonus: the sphere’s area exactly equals the cylinder’s lateral area.)

4. Leftover: $2\pi r^3 - \frac43\pi r^3 = \frac23 \pi r^3$ . Two cones of radius $r$ and height $r$ : $2 \times \frac13 \pi r^2 \times r = \frac23 \pi r^3$ . Equal.

5. Ball: $\frac43 \pi \times 12^3 \approx 7\,200$ cm $^3$ ; box: $2\pi \times 12^3 \approx 10\,900$ cm $^3$ . Empty: one third of the box, about $33\,\%$ — guaranteed by question 2, whatever the ball’s size.

6. Half a sphere: $\frac12 \times \frac43 \pi \times 10^3 = \frac23 \pi \times 1\,000 \approx 2\,094$ cm $^3 \approx 2.1$ L.

7. Cone: $\frac13 \pi \times 10^2 \times 10 = \frac{1\,000\pi}{3} \approx 1\,047$ cm $^3$ . Hemisphere: $\frac{2\,000\pi}{3} \approx 2\,094$ cm $^3$ . Cylinder: $1\,000\pi \approx 3\,142$ cm $^3$ . Ratios: $\frac{1000\pi}{3} : \frac{2000\pi}{3} : \frac{3000\pi}{3} = 1 : 2 : 3$ exactly.

8. $\frac43 \pi \times 27 = \pi \times 9 \times h$ gives $h = \frac43 \times 3 = 4$ cm exactly.

9. Radii: $\frac{6\,371}{1\,737} \approx 3.67$ . Volumes scale as the cube (Theorem 70.6): $3.67^3 \approx 49$ . About fifty Moons fit in the Earth.

10. Surfaces scale as the square: $3.67^2 \approx 13.5$ . Doubling a balloon’s radius multiplies its volume by $2^3 = 8$ but its surface by only $2^2 = 4$ : as things grow, volume (weight, content, heat produced) outruns surface (skin, material, heat lost) — the square–cube law.

11. Weight: $\times 10^3 = 1\,000$ . Bone cross-section: $\times 10^2 = 100$ . Pressure — weight per area of bone: $\times \frac{1000}{100} = 10$ . Bones built for human pressure receive ten times more: the giant’s skeleton snaps under its own weight. Giants are geometrically impossible; large animals need disproportionately thick bones (compare an elephant’s legs with a gazelle’s).

12. Weight: $\div 100^3 = 10^6$ . Strength (muscle cross-section): $\div 100^2 = 10^4$ . Strength-to-weight: $\times \frac{10^6}{10^4} = 100$ . The ant lifting fifty times its weight is not a super-athlete — it is merely small; a human shrunk to ant size could do likewise.

13. $\dfrac{S}{V} = \dfrac{4\pi r^2}{\frac43\pi r^3} = \dfrac{3}{r}$ : for $r = 1$ the ratio is $3$ , for $r = 2$ it is $1.5$ — halving the size doubles the surface available per unit of heat-producing volume. Small bodies bleed heat: the mouse must eat constantly, the bear can hibernate, and the baby — small sphere, large $\frac Sr$ — needs the hat.

14. $\left(\frac54\right)^3 = \frac{125}{64} \approx 1.95$ : the big orange holds nearly twice the fruit for the same price. Buy radius.

15. The carving says: sphere $= \frac23$ of the cylinder in volume, and $\frac23$ of it in total surface — two exact, universal proportions linking the roundest solid to the simplest one. Cicero’s answer: because a theorem is the only monument that neither armies nor centuries erode — Archimedes’ machines burned with Syracuse, but the two thirds are still exactly two thirds.