Why the volume formulas hold: dissection, slicing and Cavalieri

Prisms and cylinders. In an upright prism every cross section parallel to the base is a copy of the base, so one layer one unit thick holds B cubic units and h layers hold Bh. A leaning prism has the same cross sections at every height, so Cavalieri’s principle gives it the same volume. A cylinder of radius r has cross sections of area \pi r^2 at every height, exactly like a prism whose base has area \pi r^2 and whose height is the same, so its volume is \pi r^2 h.

Step 1: a triangular prism is three equal pyramids. This cut is in Euclid’s Elements, Book XII, Proposition 7. Call the bottom of the prism ABC and the top DEF, with D, E and F above A, B and C. Cut with the plane through A, B and F, and again with the plane through B, D and F. Three pyramids with triangular bases come out.

• ABCF and BDEF have equal volumes. Their bases ABC and DEF are the two ends of the prism, congruent, with area B. The apex F of the first lies in the top plane and the apex B of the second in the bottom plane, so both heights are h. Equal base areas and equal heights give equal volumes, by Cavalieri’s principle.
• ABDF and BDEF have equal volumes. Seen from F, their bases are ABD and BDE. These are the two halves of the side face ABED, a parallelogram, cut by its diagonal BD, and a diagonal cuts a parallelogram into two congruent triangles (Properties of a parallelogram). Both pyramids have apex F over the same plane, so the same height. Equal bases and equal heights again.

The three pyramids fill the prism and all have the same volume, so each is a third of it: \displaystyle \frac{1}{3}Bh. And every pyramid with a triangular base is one of these. Given a pyramid with base ABC and apex F, slide the triangle ABC in the direction from C to F until C lands on F; the prism it sweeps out has the same base and height, and the pyramid is its piece ABCF.

Step 2: every pyramid and every cone. A pyramid or cone with base area B and height h has cross sections \displaystyle B\left(1 - \frac{y}{h}\right)^2. So does a triangular pyramid with the same base area and height. By Cavalieri’s principle the two have the same volume, which is \displaystyle \frac{1}{3}Bh by Step 1. For a cone of radius r, B = \pi r^2 and \displaystyle V = \frac{1}{3}\pi r^2 h.

A limit picture of the cone. The cone can also be reached without comparing slices.

Put a regular polygon inside the cone’s base circle, corners on the circle, and build the pyramid on it with the cone’s apex. The pyramid sits inside the cone, and its volume is \displaystyle \frac{1}{3} of its base area times h. Each time the number of sides doubles, the polygon fills more of the circle, and its area closes in on \pi r^2. Pyramids on polygons drawn around the circle, sides touching it, hold more than the cone and close in on the same number from above. The cone is trapped between the two, so its volume is the number both close in on, \displaystyle \frac{1}{3}\pi r^2 h.

Step 3: the hemisphere. Stand a hemisphere of radius r on its flat face. Beside it on the same plane, stand a cylinder of radius r and height r, and hollow out of it a cone whose apex is the center of the cylinder’s bottom and whose base is the cylinder’s top. Call what is left the hollowed cylinder. Cut both solids at the same height y, where 0 \le y \le r.

The hemisphere’s slice is a disc. A point on its edge is at distance r from the center O of the flat face, and it sits y above that face, so by the Pythagorean theorem the disc’s radius s satisfies s^2 + y^2 = r^2. Its area is

The hollowed cylinder’s slice is a ring. Its outer edge is the cylinder, radius r. Its inner edge is the cone, whose radius grows from 0 at the bottom to r at height r. The part of the cone below height y is a scaled copy of the whole cone with factor \displaystyle \frac{y}{r}, so its radius at height y is \displaystyle r \cdot \frac{y}{r} = y. The ring’s area is

The slices match at every height, so by Cavalieri’s principle the two solids have the same volume:

The ball is two hemispheres, so \displaystyle V = \frac{4}{3}\pi r^3.

Archimedes’ ratio. The cylinder that just holds the ball has volume \pi r^2 \cdot 2r = 2\pi r^3. The cone with the same base and height has \displaystyle \frac{1}{3}\pi r^2 \cdot 2r = \frac{2}{3}\pi r^3. Next to the ball’s \displaystyle \frac{4}{3}\pi r^3, the three are \displaystyle \frac{2}{3}, \displaystyle \frac{4}{3} and \displaystyle \frac{6}{3} times \pi r^3: the ratio 1 : 2 : 3. Archimedes proved that the ball is two thirds of this cylinder, in his book On the Sphere and Cylinder.

The surface. Cover the sphere with a great many tiny patches and join the edge of each patch to the center. The ball falls into thin pieces that are very nearly pyramids, each with a patch as its base and height r. Adding their volumes, \displaystyle \frac{1}{3}r times each patch,

where S is the total area of the patches, the surface area. So \displaystyle S = \frac{3V}{r} = 4\pi r^2.

The lesson on the surface area of a sphere reaches 4\pi r^2 by a different road, comparing thin bands of the sphere with bands of the cylinder around it. Starting from there, the thin pyramids run the other way, \displaystyle V = \frac{1}{3} \cdot 4\pi r^2 \cdot r = \frac{4}{3}\pi r^3, a second route to the volume of the ball.

1. Base 81\ \text{cm}^2, height 9 cm, volume 243\ \text{cm}^3. The cube is 9^3 = 729 and the three congruent pyramids share it, \displaystyle \frac{729}{3} = 243; also \displaystyle \frac{1}{3} \cdot 81 \cdot 9 = 243.
2. \displaystyle \frac{256}{3} \approx 85.3\ \text{cm}^3. The cube is 8^3 = 512, shared by six congruent pyramids: \displaystyle \frac{512}{6} = \frac{256}{3}. Each has base 64 and height 4, and \displaystyle \frac{1}{3} \cdot 64 \cdot 4 = \frac{256}{3}.
3. 60\ \text{cm}^3 each. The prism is 15 \cdot 12 = 180, and the three pyramids are equal, so each is \displaystyle \frac{180}{3}.
4. 36\ \text{cm}^3. Put the six apexes together at one point with the bases facing out, one on each side: the bases are the faces of a cube of edge 6 whose center is 3 from every face, which is Example 3 run backwards. So one pyramid is \displaystyle \frac{6^3}{6} = 36, and \displaystyle \frac{1}{3} \cdot 36 \cdot 3 = 36.
5. 60\ \text{cm}^3, a pyramid with apex F and the four-sided base ABED. The piece cut off is a third of the prism, 30, and what is left is the other two pyramids, ABDF and BDEF, together.
6. 140\ \text{m}^3. \displaystyle \frac{1}{3} \cdot 42 \cdot 10; the shape of the base does not matter.
7. 48\pi \approx 150.8\ \text{cm}^3. \displaystyle \frac{1}{3}\pi \cdot 4^2 \cdot 9 = \frac{1}{3} \cdot 144\pi.
8. Cylinder 150\pi, cone 50\pi\ \text{cm}^3; three cone-fulls. \pi \cdot 25 \cdot 6 = 150\pi and the cone is a third of that.
9. 10 cm. \displaystyle \frac{1}{3} \cdot 60 \cdot h = 200, so 20h = 200.
10. 8 cm. \displaystyle \frac{1}{3}\pi \cdot 36 \cdot h = 96\pi, so 12h = 96.
11. Pyramid 24\ \text{cm}^3, cone 12\pi \approx 37.7\ \text{cm}^3. The square’s diagonal is a diameter, 4, so its side is \displaystyle 2\sqrt{2} and its area 8; \displaystyle \frac{1}{3} \cdot 8 \cdot 9 = 24, and \displaystyle \frac{1}{3}\pi \cdot 4 \cdot 9 = 12\pi. The square lies inside the circle, so the pyramid lies inside the cone.
12. Pyramid \displaystyle 48\sqrt{3} \approx 83.14\ \text{cm}^3; the cone holds about 17.39\ \text{cm}^3 more. The hexagon is six equilateral triangles of side 4, area \displaystyle 6 \cdot 4\sqrt{3} = 24\sqrt{3}, and \displaystyle \frac{1}{3} \cdot 24\sqrt{3} \cdot 6 = 48\sqrt{3}. The cone is \displaystyle \frac{1}{3}\pi \cdot 16 \cdot 6 = 32\pi \approx 100.53, and \displaystyle 32\pi - 48\sqrt{3} \approx 17.39.
13. Take F as the apex of both. Their bases, ABD and BDE, are the two halves of the parallelogram ABED cut by the diagonal BD, so they are congruent and have equal areas. Both bases lie in the plane of ABED, so the height from F is the same for both. Pyramids with equal base areas and equal heights have equal volumes, by Cavalieri’s principle.
14. At height y, each solid’s cross section is a copy of its base scaled by \displaystyle 1 - \frac{y}{h}, with area \displaystyle B\left(1 - \frac{y}{h}\right)^2. Equal B and equal h make the cross sections equal in area at every height, so Cavalieri’s principle gives equal volumes. The shape of the base never enters.
15. The cube only produces pyramids whose base is a square and whose apex is right above a corner of it, with height equal to the side; most pyramids are not that shape and cannot be cut out of a cube three at a time. What finishes the job is Cavalieri’s principle: every pyramid or cone has the cross sections \displaystyle B\left(1 - \frac{y}{h}\right)^2 of a pyramid already known to be \displaystyle \frac{1}{3}Bh, so it has the same volume.
16. \displaystyle \frac{9\pi}{2} \approx 14.1\ \text{cm}^3. \displaystyle \frac{4}{3}\pi \cdot 1.5^3 = \frac{4}{3}\pi \cdot 3.375 = 4.5\pi.
17. 486\pi \approx 1526.8\ \text{cm}^3. \displaystyle \frac{2}{3}\pi \cdot 9^3 = \frac{2}{3}\pi \cdot 729.
18. Both 64\pi \approx 201.1\ \text{cm}^2. The hemisphere’s slice has radius \displaystyle \sqrt{10^2 - 6^2} = 8; the ring is 100\pi - 36\pi, since the cone’s radius at height 6 is 6.
19. 6 cm. \displaystyle \frac{4}{3}\pi r^3 = 288\pi gives r^3 = 216.
20. Radius 4 cm, volume \displaystyle \frac{256\pi}{3} \approx 268.1\ \text{cm}^3. The can has radius r and height 2r, so 2\pi r^3 = 128\pi and r^3 = 64. The ball is two thirds of the can: \displaystyle \frac{2}{3} \cdot 128\pi.
21. The dome holds twice as much: 144\pi against 72\pi\ \text{m}^3. A hemisphere of radius 6 is 6 tall, so the cone is \displaystyle \frac{1}{3}\pi \cdot 36 \cdot 6 = 72\pi, and the dome is \displaystyle \frac{2}{3}\pi \cdot 216 = 144\pi.
22. 45\pi \approx 141.4\ \text{cm}^3. Turned over, the water is the cap of the hemisphere above height 3. The slab below height 3 matches the hollowed cylinder’s slab, \displaystyle \pi \cdot 36 \cdot 3 - \frac{1}{3}\pi \cdot 9 \cdot 3 = 108\pi - 9\pi = 99\pi, and the hemisphere is 144\pi, so the cap is 144\pi - 99\pi.
23. 256\pi \approx 804.2\ \text{cm}^2. \displaystyle V = \frac{4}{3}\pi \cdot 512 = \frac{2048\pi}{3}, and \displaystyle S = \frac{3V}{r} = \frac{2048\pi}{8}, which is 4\pi \cdot 8^2.
24. The part of the cone below height y is itself a cone with the same apex, a scaled copy of the whole cone with factor \displaystyle \frac{y}{r}, the ratio of the heights. The whole cone’s base has radius r, so the small cone’s base, which is the slice at height y, has radius \displaystyle r \cdot \frac{y}{r} = y.
25. The bigger ball is the smaller one enlarged by 2. The slice of the big ball at height 2y is the small ball’s slice at height y with every length doubled, so its area is 4 times as large. And the big ball’s slices are spread over twice the height. Four times the area at every level, over twice the height, gives 4 \cdot 2 = 8 times the volume.