Geometry · G.GMD.A.1 · G.GMD.A.2

Cavalieri's principle

A stack of coins pushed into a lean is still the same coins. How do you find the volume of a solid that leans, when no box of unit cubes fits it?

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Volume of prisms and cylindersMath 7 Cross sections of solidsMath 7

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Why the volume formulas hold: dissection, slicing and CavalieriGeometry

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Cavalieri's principle

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Why this exists

The volume of an upright prism comes from stacking. The base is one layer, the prism is that layer repeated all the way up, and so V = Bh (Volume of prisms and cylinders). The argument leans on the solid standing straight: each layer sits exactly on the one below, and a box of unit cubes fits.

Now push the stack sideways. A pile of coins, a deck of cards, a ream of paper: lean it over and it is still the same coins, the same cards, the same paper. Nothing was added and nothing was taken away, so the volume has not changed. But the leaning shape is no longer a box, no grid of unit cubes fits it neatly, and V = Bh was worked out for the upright stack only.

Cavalieri’s principle turns the coin stack into a tool. It compares two solids slice by slice, at every height, and says when their volumes must agree. With it, leaning prisms and cylinders get their volumes, pyramids and cones can be compared with one another, and the volume of a sphere comes within reach without calculus. It is named after Bonaventura Cavalieri, an Italian mathematician who worked with it in the 1630s.

Definition

A cross section of a solid is the flat shape where a plane cuts through it (Cross sections of solids). Choose a direction to call “up” and cut with planes at right angles to it. The cross section at height y has an area, written A(y).

Cavalieri’s principle. Suppose two solids lie between the same two parallel planes. If every plane parallel to those two cuts the solids in cross sections of equal area, then the two solids have equal volume.

In symbols, with heights measured from the lower plane and h the distance between the planes:

\begin{aligned} &A_1(y) = A_2(y) \text{ for every } y \text{ from } 0 \text{ to } h \\ &\Longrightarrow\quad V_1 = V_2 \end{aligned}

Figure 1Two solids between the same two planes. Every plane parallel to them cuts both in the same area.

The cross sections need equal areas, not equal shapes. A circle and a square of the same area count as a match, and so do two congruent squares turned to face different ways.

The same principle holds one dimension down. In the plane: suppose two regions lie between the same two parallel lines. If every line parallel to those two cuts the regions in segments of equal length, then the two regions have equal area.

Intuition

Figure 2The same ten coins, stacked straight and pushed into a lean. At any height, both stacks have one coin.

Each coin is a thin slice of the stack: a cross section with a little thickness. In the straight stack and the leaning one, the slice at any given height is the same coin, so the slices match one for one and the two stacks hold the same amount of metal. Sliding the coins changes where each slice sits, never how much of it there is.

Cavalieri’s principle is that picture with the coins made thinner and thinner, until the stack is a smooth solid. Two solids whose slices match in area at every height are, in effect, the same pile of slices arranged differently, and arranging slices differently does not change the total.

The slices do not even need the same shape. A round coin and a square tile of the same area and the same thickness hold the same amount of material, so a stack of one and a stack of the other, coin for tile, have the same volume.

Results

Cavalieri’s principle. Two solids between the same two parallel planes with A_1(y) = A_2(y) at every height y have V_1 = V_2. Two regions between the same two parallel lines, cut in segments of equal length by every parallel line, have equal area.

Leaning prisms and cylinders. A prism or cylinder that leans is called oblique. It has the same volume as an upright one with the same base area and height:

V = Bh

where B is the area of the base and h is the height, the perpendicular distance between the two base planes. The length of a slanted edge or side does not appear.

Pyramids and cones. For a pyramid or cone with base area B and height h, the cross section at height y has area

A(y) = B\left(1 - \frac{y}{h}\right)^2

This depends only on B and h. So any two pyramids or cones with equal base areas and equal heights have equal volumes, whatever the shape of the base and wherever the apex sits above it.

Triangles and parallelograms. Two triangles with equal bases on one line and their third vertices on a parallel line have equal areas. Two parallelograms with equal bases between the same two parallel lines have equal areas.

Why it is true

There are two things to see: why the principle holds, and why the cross sections of prisms, pyramids and triangles are what the results say they are.

The principle. Cut both solids with parallel planes into n thin layers, each of thickness \displaystyle \frac{h}{n}. A thin layer is very nearly a slab whose top and bottom are both the cross section at that height, so its volume is very nearly \displaystyle A(y) \cdot \frac{h}{n}. Layers of the two solids at the same height have cross sections of equal area, so they have very nearly equal volumes, and adding the layers up, so do the two solids. The thinner the layers, the smaller the error, and since it can be made smaller than any amount you name, the two volumes are equal.

This is an informal argument: “very nearly” is doing real work. Making it exact is what the definite integral is for. In calculus the volume is V = \int_0^h A(y)\,dy, which depends only on the function A, so two solids with the same A have the same volume (Volumes with known cross sections).

Cross sections of a prism or cylinder. The side edges of a prism are parallel and of equal length, so sliding the base in the direction of an edge carries it onto the top face, and partway along, onto the cross section at any height. A slide is a translation, which changes no length and no angle, so every cross section parallel to the base is congruent to the base and has area B. An upright prism with the same base area and height has A(y) = B too, and its volume is Bh by stacking. By the principle, the leaning one’s volume is also Bh. A cylinder is the same argument with a circle as the base.

Cross sections of a pyramid or cone. Cut a pyramid of height h at height y. The part above the cut is a smaller pyramid with the same apex, and it is a scaled copy of the whole: every length is multiplied by the ratio of the heights, \displaystyle \frac{h - y}{h} = 1 - \frac{y}{h}. Areas are multiplied by the square of a length factor (Scale factor, area and volume), so the cut, which is the base of the small pyramid, has area \displaystyle B\left(1 - \frac{y}{h}\right)^2. Nothing used the shape of the base or where the apex is, and a cone is a pyramid with a round base as far as this argument is concerned.

Triangles. The same argument one dimension down. In a triangle with base b and height h, the line at height y cuts off a smaller triangle at the top, similar to the whole with ratio \displaystyle 1 - \frac{y}{h} (Proportions in similar triangles), so the cut has length \displaystyle b\left(1 - \frac{y}{h}\right). That depends only on b and h, so two triangles with equal bases and equal heights have equal cuts at every height, and equal areas. In a parallelogram every cut is the base slid up a side, so it has the length of the base.

Worked examples

Example 1

A leaning cylinder

Figure 3 shows an oblique cylinder. Its circular ends have radius 3 cm, the perpendicular distance between them is 10 cm, and its slanted side is 12 cm long. Find its volume.

Figure 3The height is the dashed perpendicular, not the slanted side.

Solution

The cylinder is a circle slid along a slanted path, and sliding does not change the circle. So every cross section parallel to the ends is a circle of radius 3 cm, with area

A(y) = \pi \cdot 3^2 = 9\pi\ \text{cm}^2

An upright cylinder of radius 3 cm and height 10 cm, standing between the same two planes, has exactly the same cross sections. By Cavalieri’s principle the two volumes are equal, and the upright one’s volume is base area times height:

V = Bh = 9\pi \cdot 10 = 90\pi \approx 282.7\ \text{cm}^3

The 12 cm side plays no part. The slices only run from one end plane to the other, a distance of 10 cm; using 12 would give 108\pi, the volume of a stack taller than this one.

Example 2

A twisted column

A sculpture is a column 15 cm tall. Every horizontal cross section is a square of side 8 cm, but each square is turned a little further than the one below it, so the column twists as it rises. Find its volume.

Solution

The twist changes which way each square faces, not its area. At every height,

A(y) = 8^2 = 64\ \text{cm}^2

A straight square prism with an 8 cm by 8 cm base and a height of 15 cm has the same cross-section area at every height, between the same two planes. Cavalieri’s principle gives the twisted column the same volume as that prism:

V = 64 \cdot 15 = 960\ \text{cm}^3

This is a deck of cards in which each card has been turned rather than slid. Neither move changes the cards.

Example 3

A square prism to match a cylinder

A cylinder has radius 2 cm and height 7 cm. A square prism, also 7 cm tall and standing on the same plane, is to have cross sections of the same area at every height. How long must the side of its square be, and what is the volume of each solid?

Solution

Each cross section of the cylinder is a circle of area \pi \cdot 2^2 = 4\pi\ \text{cm}^2. Each cross section of the prism is a square of side s, with area s^2. Matching the areas:

\begin{aligned} s^2 &= 4\pi \\ s &= 2\sqrt{\pi} \approx 3.54\ \text{cm} \end{aligned}

taking the positive square root, since s is a length. Now every horizontal plane cuts the two solids in equal areas, so by Cavalieri’s principle their volumes are equal:

V = 4\pi \cdot 7 = 28\pi \approx 88.0\ \text{cm}^3

The slices are circles in one solid and squares in the other. The principle never asks the shapes to match, only the areas.

Example 4

Triangles between parallel lines

In Figure 4, two parallel lines are 5 units apart. Triangles ABC and DEF each have a base of length 8 on the lower line and their third vertex on the upper line. Show that every line parallel to the two cuts the triangles in segments of equal length, find that length at height 2, and find the area of each triangle.

Figure 4Two triangles with equal bases between the same parallel lines. The thick cuts have equal length at every height.

Solution

Take the parallel line at height y above the base, where 0 \le y \le 5. The part of triangle ABC above it is a smaller triangle with apex C. Its two sides lie along CA and CB, and its base is parallel to AB, so its angles match those of ABC (corresponding angles at parallel lines) and the two triangles are similar. Corresponding heights are in the same ratio as corresponding sides, so the ratio is \displaystyle \frac{5 - y}{5}, and the cut has length

8 \cdot \frac{5 - y}{5}

The same argument in triangle DEF gives the same length, because it used only the base, 8, and the height, 5. That DEF leans so far that F lies beyond E changes nothing.

At height 2 both cuts have length \displaystyle 8 \cdot \frac{3}{5} = 4.8.

Every parallel line cuts the two triangles in equal lengths, so by Cavalieri’s principle in the plane they have equal areas. Triangle ABC has area \displaystyle \frac{1}{2} \cdot 8 \cdot 5 = 20 square units, so triangle DEF has area 20 square units as well.

This is why the area of a triangle depends only on its base and height, and not on where the third vertex sits along the parallel line.

Example 5

A pyramid and a cone with the same volume

A square pyramid has a base 6 cm by 6 cm and a height of 10 cm. A cone, also 10 cm tall, has a circular base of the same area. Find the cone’s radius, compare the two cross sections at height 4 cm, and explain why the two solids have the same volume.

Figure 5A square pyramid and a cone with equal base areas and equal heights, cut by one plane.

Solution

The pyramid’s base has area 36\ \text{cm}^2. For the cone, \pi r^2 = 36, so

r = \frac{6}{\sqrt{\pi}} \approx 3.39\ \text{cm}

Cut both at height 4. The part of the pyramid above the cut is a smaller pyramid, a scaled copy of the whole with factor \displaystyle \frac{10 - 4}{10} = 0.6, the ratio of the heights. The cut is its base: a square of side 6 \cdot 0.6 = 3.6 cm, with area 3.6^2 = 12.96\ \text{cm}^2.

The part of the cone above the cut is a smaller cone, scaled by the same factor 0.6. Its base is a circle of radius 0.6r, with area

\pi (0.6r)^2 = 0.36 \cdot \pi r^2 = 0.36 \cdot 36 = 12.96\ \text{cm}^2

The two areas match. Nothing about height 4 was special: at any height y the factor is \displaystyle 1 - \frac{y}{10}, and both areas are \displaystyle 36\left(1 - \frac{y}{10}\right)^2. The solids stand between the same two planes with cross sections of equal area at every height, so by Cavalieri’s principle they have equal volumes.

Notice what this does not say: it does not say what the volume is. That takes one more comparison, which Volume of pyramids and cones makes.

Example 6

A leaning rod

A solid metal rod is an oblique cylinder of radius 2 cm with horizontal ends. The segment joining the centers of its two ends is 13 cm long, and the center of the top end is 5 cm to the side of the center of the bottom end, measured horizontally. Find the rod’s volume.

Solution

The volume needs the height, the vertical distance between the ends, and that is not given. The 13 cm segment, the vertical line down from the top center, and the horizontal line out from the bottom center make a right triangle, with the 13 cm segment as hypotenuse and the 5 cm offset as one leg. By the Pythagorean theorem, the height h satisfies

\begin{aligned} h^2 + 5^2 &= 13^2 \\ h^2 &= 169 - 25 = 144 \\ h &= 12 \end{aligned}

taking the positive root, since h is a length. Every horizontal cross section is a circle of radius 2 cm, with area 4\pi\ \text{cm}^2, so by Cavalieri’s principle the rod has the volume of an upright cylinder with that radius and a height of 12 cm:

V = 4\pi \cdot 12 = 48\pi \approx 150.8\ \text{cm}^3

Check. The upright cylinder of height 12 cm has exactly the same slices as the rod. Using the 13 cm length instead would give 52\pi, the volume of a taller stack than the rod is.

Boundaries

Every height, measured the same way. Both solids must sit between the same two parallel planes, and the cross sections compared must be cut by one plane parallel to those two, at the same height in both. Matching areas at a handful of heights proves nothing: two solids can agree at heights 0, 1, 2, \ldots and differ everywhere in between.

It runs one way only. Matching cross sections force equal volumes, but equal volumes do not force matching cross sections. A pyramid and the same pyramid turned upside down have the same volume, yet near the bottom one has a large cross section and the other a tiny one.

Areas in space, lengths in the plane. For solids it is the areas of the cross sections that must match; for regions in the plane, the lengths of the cuts. For solids, matching perimeters or matching widths of the cross sections are not enough: a circle of diameter 2 and a square of side 2 have the same width but areas \pi and 4.

Height is perpendicular height. In V = Bh for a leaning prism or cylinder, h is the distance between the two base planes, never the length of a slanted edge.

V = Bh needs parallel ends. A cylinder whose top has been cut off on a slant has circular cross sections of area B only up to where the slant begins. Above that the cuts are only parts of circles, so its volume is less than B times its greatest height.

The argument here is informal. The slab argument shows why the principle is right; a complete proof needs limits, which is what the integral supplies.

Where this goes next

The principle’s first job is the one done here: it carries V = Bh from upright prisms and cylinders over to leaning ones, and it shows that the volume of a pyramid or cone depends only on its base area and height.

Its larger job is deriving the volume formulas. Pyramids with equal base areas and heights have equal volumes, and that is the step that turns a prism cut into three pyramids into the factor \displaystyle \frac{1}{3} in the volume of a pyramid or cone. Comparing a half-sphere slice by slice with a cylinder that has a cone hollowed out of it gives the volume of a sphere.

In calculus the slab argument becomes exact. Adding up cross-section areas times thin thicknesses is a The definite integral, and Volumes with known cross sections uses it to find volumes of solids that no formula covers, including the solids made by spinning a region around a line.

Exercises

Leaning prisms and cylinders

An oblique cylinder has radius 5 cm and height 8 cm. Find its volume.
An oblique prism has a rectangular base 4 cm by 7 cm. Its bases are 9 cm apart, and each slanted edge is 11 cm long. Find its volume.
An oblique prism has a base that is a right triangle with legs 5 m and 12 m, and a height of 10 m. Find its volume.
Forty coins, each a cylinder of radius 1.2 cm and thickness 0.2 cm, are stacked and then pushed into a lean. Find the volume of the stack.
An oblique cylinder of radius 5 cm has volume 150\pi\ \text{cm}^3. Find its height.
An oblique prism with a square base has volume 405\ \text{cm}^3 and height 5 cm. Find the side of the base.

Same areas, different shapes

A cylinder has radius 4 cm and height 6 cm. A square prism of the same height has cross sections of the same area. Find the side of its square base and the volume of each solid.
Two solids lie between two parallel planes 12 cm apart. Every cross section of the first is a rectangle 3 cm by 5 cm. Every cross section of the second is a circle, all of one size, with the same area as the rectangle. Find the radius of the circles and the volume of each solid.
A solid is 10 cm tall, and every horizontal cross section is a triangle with base 6 cm and height 4 cm, though the triangles slide and turn from one height to the next. Find its volume.

In the plane

A parallelogram has base 9 and height 4. Use Cavalieri’s principle in the plane to compare it with a rectangle 9 by 4, and find its area.
A triangle has its base, of length 10, on one of two parallel lines that are 6 apart, and its third vertex on the other. Find the length of the segment cut from the triangle by the parallel line 4 above the base. Does the answer depend on where the third vertex is?

Pyramids and cones

A pyramid has a base of area 50\ \text{cm}^2 and a height of 10 cm. Find the areas of its cross sections at heights 0, 5 and 8 cm.
A square pyramid has a base 5 cm by 5 cm and a height of 9 cm. A cone has the same height and a base of the same area. Find the cone’s radius, and the area of each solid’s cross section at height 3 cm.

Harder

A rod is an oblique cylinder of radius 4 cm with horizontal ends. The segment joining the centers of its ends is 25 cm long, and the top center is 7 cm to the side of the bottom center, measured horizontally. Find the rod’s height and volume.
A solid upright cylinder has radius 5 cm and height 10 cm. A hole is drilled on a slant from the top face to the bottom face. The hole is an oblique cylinder whose horizontal cross sections are circles of radius 2 cm, and it lies entirely inside the solid. Find the volume of metal that is left.
A square pyramid has a base 6 cm by 6 cm and a height of 10 cm. A copy of it stands upside down on its apex, between the same two planes. Find the area of each solid’s cross section at height 2 cm. Do the two solids have the same volume, and does this contradict Cavalieri’s principle?

Reasoning

Explain why the length of the slanted side of an oblique cylinder does not enter its volume.
Two solids between the same two planes have cross sections of equal area at heights 0, 1, 2, \ldots, 10 cm, and nothing is known about other heights. Must their volumes be equal? Explain.
Explain why every cross section of a prism cut parallel to its base has the same area as the base.
Explain why an oblique cone and an upright cone with the same base radius and the same height have the same volume.

Answers

200\pi \approx 628.3\ \text{cm}^3. Every cross section is a circle of area 25\pi, the same as in an upright cylinder of height 8, so V = 25\pi \cdot 8.
252\ \text{cm}^3. B = 4 \cdot 7 = 28 and V = 28 \cdot 9. The 11 cm edge is not the height.
300\ \text{m}^3. \displaystyle B = \frac{1}{2} \cdot 5 \cdot 12 = 30 and V = 30 \cdot 10.
11.52\pi \approx 36.2\ \text{cm}^3. The stack is 40 \cdot 0.2 = 8 cm tall, and every cross section is a circle of area 1.2^2 \pi = 1.44\pi, leaning or not, so V = 1.44\pi \cdot 8.
6 cm. B = 25\pi, so 25\pi h = 150\pi.
9 cm. 5s^2 = 405, so s^2 = 81 and s = 9.
Side \displaystyle 4\sqrt{\pi} \approx 7.09 cm; each volume is 96\pi \approx 301.6\ \text{cm}^3. Matching areas, s^2 = 16\pi. The cross sections match at every height, so the volumes are equal: 16\pi \cdot 6.
Radius \displaystyle \sqrt{15/\pi} \approx 2.19 cm; each volume is 180\ \text{cm}^3. \pi r^2 = 15. Both solids have cross sections of area 15 at every height, the same as an upright rectangular prism 3 by 5 by 12, so by Cavalieri’s principle each has volume 15 \cdot 12.
120\ \text{cm}^3. Every cross section has area \displaystyle \frac{1}{2} \cdot 6 \cdot 4 = 12. A straight triangular prism with that base and height 10 has the same cross sections, so V = 12 \cdot 10.
Area 36. Put both between two parallel lines 4 apart, bases on the lower line. Every parallel line between them cuts the parallelogram in a copy of its base slid up the sides, of length 9, and cuts the rectangle in a segment of length 9 too. So the areas are equal, and the rectangle’s is 9 \cdot 4 = 36.
\displaystyle \frac{10}{3} \approx 3.33, and no. The part above the cut is a triangle similar to the whole with ratio \displaystyle \frac{6 - 4}{6} = \frac{1}{3}, so the cut is \displaystyle 10 \cdot \frac{1}{3}. Only the base and the height entered, not where the third vertex is.
50, 12.5 and 2\ \text{cm}^2. \displaystyle A(y) = 50\left(1 - \frac{y}{10}\right)^2, and the squared factors are 1, \displaystyle \frac{1}{4} and \displaystyle \frac{1}{25}.
\displaystyle r = \frac{5}{\sqrt{\pi}} \approx 2.82 cm; both cross sections have area \displaystyle \frac{100}{9} \approx 11.1\ \text{cm}^2. From \pi r^2 = 25. At height 3 the scale factor is \displaystyle \frac{9 - 3}{9} = \frac{2}{3}, so each area is \displaystyle 25 \cdot \frac{4}{9}.
Height 24 cm, volume 384\pi \approx 1206.4\ \text{cm}^3. h^2 + 7^2 = 25^2 gives h^2 = 576. Every horizontal cross section has area 16\pi, so V = 16\pi \cdot 24.
210\pi \approx 659.7\ \text{cm}^3. At every height the cross section is a disc of radius 5 with a disc of radius 2 removed, area 25\pi - 4\pi = 21\pi. An upright cylinder with a straight hole of radius 2 has the same cross sections, and its volume is 21\pi \cdot 10.
Upright: 23.04\ \text{cm}^2; upside down: 1.44\ \text{cm}^2. In the upright pyramid the part above the cut is scaled by 0.8, so the cut is a square of side 4.8. In the upside-down one the part below the cut is a small pyramid scaled by \displaystyle \frac{2}{10} = 0.2, so the cut has side 1.2. The volumes are equal, since the solids are congruent. There is no contradiction: the principle says matching cross sections force equal volumes, never the reverse.
Every cross section parallel to the ends is a congruent copy of the end, with area B, and the cross sections run only from one end plane to the other, a distance h. An upright cylinder with the same base and height has the same cross sections, so the same volume, Bh. The slant length measures how far the base was slid, not how tall the stack of slices is.
No. The principle needs equal areas at every height. An upright cylinder of radius 1 and height 10 and a stack of ten bead shapes, each 1 cm tall, as wide as the cylinder at its top and bottom but wider in between, have equal cross sections at every whole number of centimeters. The beads are wider everywhere else, so the stack of beads has the larger volume.
The side edges of a prism are parallel and of equal length, so sliding the base in the direction of an edge carries it onto any cross section parallel to the base. A slide is a translation, which keeps every length and angle, so the cross section is congruent to the base and has the same area.
In either cone, the part above the cut at height y is a smaller cone scaled from the whole by \displaystyle 1 - \frac{y}{h}, so the cross section is a circle of radius \displaystyle r\left(1 - \frac{y}{h}\right). Its area, \displaystyle \pi r^2\left(1 - \frac{y}{h}\right)^2, is the same in both cones at every height, because where the apex sits never entered. By Cavalieri’s principle the volumes are equal.