Math 8 · G.GMD.A.3 · 8.G.C.9

Volume of pyramids and cones

A pyramid and a box can stand on the same floor and reach the same height, but the pyramid narrows to a point. How much of the box does it fill?

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Volume of prisms and cylindersMath 7

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Composite solids and densityMath 8 Why the volume formulas hold: dissection, slicing and CavalieriGeometry

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Geometry11

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Volume of solids7

▴ 2 above Cavalieri's principle Volume of a cylinder

Volume of pyramids and cones

Composite solids and density Why the volume formulas hold: dissection, slicing and Cavalieri

When you reach for it

A pyramid is the solid you get from a flat polygon, its base, and one point not in the plane of the base, its apex, by joining every point of the base to the apex with a straight segment. A cone is built the same way on a circular base (see prisms and pyramids and cylinders and cones). Spires, tent roofs, funnels, piles of sand or grain, the tip of a pencil and a paper drinking cup are all pyramids or cones, or pieces of them.

Reach for this skill whenever the question is how much such a solid holds or contains: the air under a pyramid roof, the sand in a heap, the water in a cone-shaped cup or tank. It also runs backwards, to find how tall or how wide a cone must be to hold a given amount, and it measures shapes such as a bucket or a lampshade, which are cones with the tip cut off. It rests on the volume of a prism, base area times height, and for cones on the area of a circle, \pi r^2.

This lesson takes two days. Part 1 gives the one-third rule, says why it holds, and uses it to find volumes, including when only the slant height is known. Part 2 runs the rule backwards, compares solids of different sizes, and measures solids with the tip cut off and solids built from a cone and a cylinder. Examples 1 to 5 and Exercises 1 to 13 belong to Part 1; Examples 6 to 9 and Exercises 14 to 25 belong to Part 2.

The procedure

Part 1: one third of the prism

The height is measured straight down. The height h of a pyramid or cone is the perpendicular distance from the apex to the plane of the base: drop a line from the apex that meets the base plane at a right angle, and measure it. When the apex sits directly above the center of the base the solid is called a right pyramid or right cone; when it sits off to one side the solid leans, and the height is still that straight-down distance.

One third of the prism with the same base and height. Put a pyramid inside the prism that has the same base and the same height, as in Figure 1. The prism holds Bh, where B is the area of the base. The pyramid holds exactly one third of that. You can see it by experiment: fill a hollow pyramid with sand, pour it into the hollow prism, and it takes three pyramid-fulls to fill the prism to the brim. Why the volume formulas hold proves it, by cutting a cube into three identical pyramids and a triangular prism into three pyramids of equal volume.

Figure 1A pyramid and the prism with the same base and the same height. The pyramid holds exactly one third of the prism.

Why only the base area and the height matter. Slice a pyramid parallel to its base at distance d below the apex. The slice is the base shrunk toward the apex by the scale factor \displaystyle \frac{d}{h}, so its area is \displaystyle B \left(\frac{d}{h}\right)^2, since areas scale by the square of the scale factor. That area depends only on B, h and d: not on the shape of the base, and not on whether the apex sits over the center. So any two pyramids or cones with equal base areas and equal heights have slices of equal area at every level, and by Cavalieri’s principle they have equal volumes. One pyramid that is a third of its prism therefore settles every pyramid and every cone:

V = \frac{1}{3}Bh

For a cone the base is a circle of radius r, so B = \pi r^2 and

V = \frac{1}{3}\pi r^2 h

Put beside the prism, V = Bh, and the cylinder, V = \pi r^2 h, the only change is the factor \displaystyle \frac{1}{3}: it is what narrowing all the way to a point costs.

Find the base area first. B is an area, worked out from the base’s own shape: s^2 for a square of side s, lw for a rectangle, \displaystyle \frac{1}{2} \times \text{base} \times \text{height} for a triangle, \pi r^2 for a circle. A pyramid on a triangular base has two different heights in play, the height of the triangle, which goes into B, and the height of the pyramid, which goes into \displaystyle \frac{1}{3}Bh. They are different lengths and must be kept apart.

Use the height, not the slant height. The slant height \ell is measured along the sloping surface: for a cone, from the apex down the side to the edge of the base; for a right pyramid with a square base, from the apex down the middle of a face to the midpoint of a base edge. It is longer than the height, and it is the length used for surface area (Surface area of pyramids and cones), not for volume. When only the slant height is known, the height comes from a right triangle. In a right cone the height, a radius and the slant height form a right triangle with the right angle at the center of the base (Figure 2), so by the Pythagorean theorem

h^2 + r^2 = \ell^2

In a right square pyramid of side s the center of the base is \displaystyle \frac{s}{2} from the midpoint of each edge, so the same triangle has legs h and \displaystyle \frac{s}{2}, and \displaystyle h^2 + \left(\frac{s}{2}\right)^2 = \ell^2.

Figure 2The height h, the radius r and the slant height \ell of a cone form a right triangle, so h^2 + r^2 = \ell^2.

One unit, exact first. Every length must be in the same unit before multiplying, and the volume is in that unit cubed. Keep a cone’s volume as a number times \pi until the end, and round only when a decimal is wanted. A volume in cubic centimeters is the same number of milliliters, and 1000\ \text{cm}^3 is a liter (Volume and capacity).

Part 2: running the rule backwards, comparing, and cutting

Finding a length. Multiply both sides of \displaystyle V = \frac{1}{3}Bh by 3 to clear the fraction: 3V = Bh. Then divide by whichever factor is known:

h = \frac{3V}{B} \qquad B = \frac{3V}{h}

When B is a square’s area s^2 or a circle’s \pi r^2, finding B is not the end: take the square root of \displaystyle \frac{3V}{h} for s, or of \displaystyle \frac{3V}{\pi h} for r. Only the positive root is a length.

Comparing sizes. Multiply a cone’s radius by k and its height by m, and the new volume is

\frac{1}{3}\pi (kr)^2 (mh) = k^2 m \cdot \frac{1}{3}\pi r^2 h

so the volume is multiplied by k^2 m. When every length is multiplied by the same k, as in a scaled copy, that is k^2 \cdot k = k^3. The same holds for any pyramid, because scaling multiplies the base area by k^2 and the height by k (scale factor and volume). A cone of water standing in a cone-shaped glass is such a scaled copy, which is why a glass filled to half its height holds far less than half (Example 8).

Cutting off the tip. A bucket, a lampshade or a flowerpot is a cone with its tip cut off by a plane parallel to the base. Such a solid is called a frustum. Its volume is the whole cone minus the small cone that was cut away. The sides of the frustum, continued, meet at the missing apex, and the radius grows in proportion to the distance from that apex (the triangles in a side view are similar), which is how the height of the missing tip is found (Example 9).

Worked examples

Example 1

A square pyramid

A stone monument is a pyramid on a square base 9 m on each side, and its apex is 14 m above the ground. Find its volume.

Solution

The base is a square, so its area is B = 9^2 = 81\ \text{m}^2. The height is 14 m, the straight-up distance from the ground to the apex. Then

V = \frac{1}{3}Bh = \frac{1}{3} \cdot 81 \cdot 14 = 27 \cdot 14 = 378\ \text{m}^3

Dividing 81 by 3 first kept the numbers whole. Check against the prism with the same base and height: it holds 81 \cdot 14 = 1134\ \text{m}^3, and 378 is one third of 1134.

Example 2

A cone and its cylinder

A cone has radius 3 cm and height 10 cm. Find its volume exactly and to one decimal place, and compare it with the cylinder of the same radius and height.

Solution

The base is a circle of radius 3 cm, so B = \pi \cdot 3^2 = 9\pi\ \text{cm}^2. Then

V = \frac{1}{3} \cdot 9\pi \cdot 10 = 30\pi\ \text{cm}^3

and 30\pi \approx 94.2\ \text{cm}^3. The cylinder with the same base and height holds 9\pi \cdot 10 = 90\pi\ \text{cm}^3, which is 3 \cdot 30\pi: three cone-fulls of water fill it.

Example 3

A pyramid on a triangle

A glass paperweight is a pyramid whose base is a right triangle with legs 6 cm and 8 cm. Its apex is 5 cm above the base. Find its volume.

Solution

Two different heights are involved, so take them one at a time. The base is a right triangle, and its two legs are perpendicular, so one leg serves as the triangle’s base and the other as its height:

B = \frac{1}{2} \cdot 6 \cdot 8 = 24\ \text{cm}^2

The pyramid’s height is the 5 cm from the base up to the apex. Then

V = \frac{1}{3} \cdot 24 \cdot 5 = 40\ \text{cm}^3

Example 4

From the slant height of a pyramid

A right pyramid has a square base 24 cm on each side, and each triangular face has slant height 20 cm. Find the pyramid’s volume.

Solution

The rule needs the height, which is not given, so find it first. The pyramid is right, so the foot of the height is the center of the square, which is \displaystyle \frac{24}{2} = 12 cm from the midpoint of each base edge. The height, that 12 cm, and the slant height form a right triangle with the slant height as its hypotenuse:

h^2 + 12^2 = 20^2 \quad\text{so}\quad h^2 = 400 - 144 = 256

and h = 16 cm, the positive root since h is a length. The base area is 24^2 = 576\ \text{cm}^2, so

V = \frac{1}{3} \cdot 576 \cdot 16 = 192 \cdot 16 = 3072\ \text{cm}^3

Example 5

A wafer cone

A wafer cone for ice cream is 8 cm across the top, and its slant height, from the rim down the side to the point, is 12 cm. How much does it hold when filled level with the rim, to the nearest milliliter?

Solution

“Across” is a diameter, so r = 4 cm. The height comes from the right triangle of Figure 2:

h^2 + 4^2 = 12^2 \quad\text{so}\quad h = \sqrt{128} = 8\sqrt{2} \approx 11.31\ \text{cm}

Keep the exact root. With B = \pi \cdot 4^2 = 16\pi,

\begin{aligned} V &= \frac{1}{3} \cdot 16\pi \cdot 8\sqrt{2} \\ &= \frac{128\sqrt{2}}{3}\pi \approx 189.6\ \text{cm}^3 \end{aligned}

so the cone holds about 190 mL. Had the slant height 12 gone into the rule as the height, the answer would have been \displaystyle \frac{1}{3} \cdot 16\pi \cdot 12 = 64\pi \approx 201.1\ \text{cm}^3, too large, because the slant height is the hypotenuse and so longer than the height.

Example 6

The height of a sand pile

A heap of sand settles into a cone of radius 2 m holding 12\ \text{m}^3. How tall is the heap, to the nearest centimeter?

Solution

The base area is \pi \cdot 2^2 = 4\pi\ \text{m}^2, and the rule with the known values is

12 = \frac{1}{3} \cdot 4\pi \cdot h

Multiply both sides by 3 to clear the fraction, 36 = 4\pi h, then divide both sides by 4\pi, the base area, which is not zero:

h = \frac{36}{4\pi} = \frac{9}{\pi} \approx 2.86\ \text{m}

Check: \displaystyle \frac{1}{3} \cdot 4\pi \cdot \frac{9}{\pi} = \frac{36}{3} = 12, as it should be.

Example 7

The side of a square base

A solid concrete cap on a garden post is a pyramid with a square base. It is 6 cm tall and uses 288\ \text{cm}^3 of concrete. How long is each edge of its base?

Solution

Call the edge s, so B = s^2. Multiplying \displaystyle V = \frac{1}{3}Bh by 3 gives 3V = Bh, here

864 = s^2 \cdot 6

Dividing both sides by 6 leaves s^2 = 144. Two numbers square to 144, namely 12 and -12, but an edge is a length, so s = 12 cm.

Check: \displaystyle \frac{1}{3} \cdot 144 \cdot 6 = 288.

Example 8

A half-full glass

A glass is a cone 10 cm deep, standing on its point with the open circle at the top.

(a) It is filled to a depth of 5 cm. What fraction of the full glass does it hold?

(b) To what depth must it be filled to hold exactly half, to two decimal places?

Solution

(a) The water is itself a cone, with its point at the bottom of the glass. Its sloping side lies along the glass, and its top is a circle parallel to the rim, so in a side view the water’s triangle and the glass’s triangle share the angle at the point and have parallel tops: they are similar. If the glass has radius r at the rim, the water at depth d has radius \displaystyle r \cdot \frac{d}{10}. Every length of the water cone is the glass’s length times \displaystyle k = \frac{d}{10}, so its volume is the glass’s volume times k^3. At d = 5, \displaystyle k = \frac{1}{2} and

k^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8}

Half the depth holds only one eighth of the glass. The radius and the height of the water are both halved, and the radius counts twice.

(b) For half the volume, \displaystyle k^3 = \frac{1}{2}, so \displaystyle k = \frac{1}{\sqrt[3]{2}} and

d = \frac{10}{\sqrt[3]{2}} \approx 7.94\ \text{cm}

The top 2.06 cm of the glass holds as much as the bottom 7.94 cm, because the top is where the glass is widest.

Example 9

A bucket

A bucket is 24 cm deep, with radius 10 cm at the bottom and 15 cm at the top. How many liters does it hold, to one decimal place?

Figure 3The bucket in Example 9 is a large cone with a small cone cut off its tip. Continuing the sides finds the missing tip.

Solution

The bucket is not a cone, but it is a cone with the tip cut off, so continue its sides downward until they meet, as in Figure 3. Let the missing tip be x cm tall. In a side view, the radius grows in proportion to the distance from the apex, since the triangles are similar: radius 10 at distance x, radius 15 at distance x + 24. So

\frac{10}{x} = \frac{15}{x + 24}

Multiplying both sides by x(x + 24) gives 10x + 240 = 15x, so x = 48 cm. The whole cone has radius 15 and height 48 + 24 = 72; the tip has radius 10 and height 48. The bucket is the difference:

\begin{aligned} V &= \frac{1}{3}\pi \cdot 15^2 \cdot 72 - \frac{1}{3}\pi \cdot 10^2 \cdot 48 \\ &= 5400\pi - 1600\pi \\ &= 3800\pi \approx 11938.1\ \text{cm}^3 \end{aligned}

That is about 11.9 liters. Check with the frustum formula of Exercise 24, \displaystyle \frac{1}{3}\pi h (R^2 + Rr + r^2) with R = 15, r = 10, h = 24: 8\pi \cdot (225 + 150 + 100) = 8\pi \cdot 475 = 3800\pi.

Boundaries

The solid must narrow to a single point. The rule covers a flat base and every segment from it to one apex off its plane. The base may be any polygon, a circle, or any flat region with an area. A solid that stops short of a point, such as a bucket, is not a cone, and its volume is a difference of two cones once the missing tip has been found.

h is the perpendicular height. For a leaning pyramid or cone the rule still holds, but h is the straight-up distance from the apex to the plane of the base, not the length of any edge or side. The slant height is never the height.

The slant-height triangles need a right solid. h^2 + r^2 = \ell^2 holds for a right cone, and \displaystyle h^2 + \left(\frac{s}{2}\right)^2 = \ell^2 for a right pyramid on a square base. A rectangular base has two different slant heights, one for each pair of faces, and each uses half of the base edge it stands on. A leaning cone has no single slant height.

k^3 needs every length scaled by k. If the radius and the height change by different factors, the volume changes by k^2 m, not by a cube.

Units and decimals. Every length in one unit, the answer in that unit cubed. A decimal value of a cone’s volume is an approximation; the exact value is a number times \pi.

Backwards gives one length. Solving for a radius or a square’s side gives a square, and only the positive root is a length.

Where this goes next

The one-third rule is half of what the sphere needs: a ball holds the cylinder around it minus a cone (Volume of a sphere), shown by Cavalieri’s principle in Why the volume formulas hold: dissection, slicing and Cavalieri. Grain bins, pencils, rockets and ice cream cones are split into cones, cylinders and other pieces and added up (Composite solids and density), and the faces of a pyramid and the curved side of a cone, measured with the slant height, give its surface area.

Spinning a right triangle about one of its legs sweeps out a cone, the simplest solid of revolution after the cylinder. In calculus the one third is explained exactly: the slice at distance y below the apex has area \displaystyle B\frac{y^2}{h^2}, and adding up these slices is an integral (Volumes with known cross sections) whose value is \displaystyle \frac{1}{3}Bh, because the y^2 becomes \displaystyle \frac{h^3}{3}. And water draining from a cone-shaped tank lowers its level faster and faster as the cone narrows, the classic problem of related rates.

Exercises

Part 1: finding the volume

Give each cone’s volume exactly, in terms of \pi, and to one decimal place.

A pyramid has a square base 6 cm on each side and a height of 11 cm.
A pyramid has a rectangular base 8 m by 5 m and a height of 9 m.
A pyramid’s base is a triangle with base 10 cm and height 7 cm. The pyramid is 12 cm tall.
A cone has radius 6 cm and height 7 cm.
A cone has diameter 10 inches and height 12 inches.
A cone has radius 2.4 m and height 5 m.
A pyramid has a base of area 54\ \text{cm}^2. Its apex is 10 cm above the plane of the base, but not above the center of the base, so the pyramid leans. Find its volume.

Part 1: height and slant height

A right pyramid has a square base 40 cm on each side and slant height 29 cm. Find its height and its volume.
A right cone has radius 9 cm and slant height 41 cm. Find its height and its volume.
A right cone has height 4 cm and slant height 6 cm. Find its volume.
A roof is a pyramid over a square floor 9 m by 9 m, and its peak is 320 cm above the floor of the roof space. Find the volume of the roof space in cubic meters.

Part 1: reasoning

A hollow pyramid and a hollow prism have the same base and the same height. Say what fraction of the prism the pyramid fills, and describe an experiment with sand that shows it.
Explain why the slant height of a right pyramid on a square base is always longer than its height.

Part 2: working backwards

A cone has volume 50\pi\ \text{cm}^3 and radius 5 cm. Find its height.
A pyramid with a square base has volume 147\ \text{m}^3 and height 9 m. Find the length of each edge of its base.
A cone-shaped measuring scoop is 9 cm deep and holds 200 mL. Find the radius of its rim, exactly and to two decimal places.

Part 2: comparing and scaling

A cone’s radius is tripled and its height is halved. What happens to its volume?
Every length of a pyramid is doubled. What happens to its volume?
A cone-shaped glass 12 cm deep stands on its point and is filled to a depth of 8 cm. What fraction of the full glass does it hold?
A cone and a cylinder have the same radius and the same volume. How does the cone’s height compare with the cylinder’s?

Part 2: cut-off tips and combined solids

A flowerpot is 12 cm deep, with radius 6 cm at the bottom and 9 cm at the top. Find its volume, to one decimal place.
A pyramid has a square base 12 m on each side and a height of 15 m. The top 5 m is cut off by a plane parallel to the base. Find the volume of what is left, to one decimal place.
A grain bin is a cylinder of radius 3 m and height 5 m with a cone-shaped roof of the same radius and height 2 m. Find the bin’s total volume, to one decimal place.

Part 2: reasoning

A frustum of a cone has bottom radius r, a larger top radius R, and height h. Show that its volume is \displaystyle \frac{1}{3}\pi h (R^2 + Rr + r^2).
A cone and a pyramid have the same height, and the cone’s base circle just fits inside the pyramid’s square base, touching all four sides. Show that the cone holds \displaystyle \frac{\pi}{4} of the pyramid’s volume, and explain why this is the same fraction as for a cylinder standing in a box with a square base.

Answers

132\ \text{cm}^3. B = 36, and \displaystyle \frac{1}{3} \cdot 36 \cdot 11 = 12 \cdot 11 = 132.
120\ \text{m}^3. B = 8 \cdot 5 = 40, and \displaystyle \frac{1}{3} \cdot 40 \cdot 9 = 40 \cdot 3 = 120.
140\ \text{cm}^3. The triangle has area \displaystyle \frac{1}{2} \cdot 10 \cdot 7 = 35, and \displaystyle \frac{1}{3} \cdot 35 \cdot 12 = 35 \cdot 4 = 140.
84\pi \approx 263.9\ \text{cm}^3. \displaystyle \frac{1}{3}\pi \cdot 36 \cdot 7 = 12\pi \cdot 7 = 84\pi.
100\pi \approx 314.2\ \text{in}^3. The radius is 5 inches, so \displaystyle \frac{1}{3}\pi \cdot 25 \cdot 12 = 100\pi.
9.6\pi \approx 30.2\ \text{m}^3. \displaystyle \frac{1}{3}\pi \cdot 5.76 \cdot 5 = \frac{28.8}{3}\pi = 9.6\pi.
180\ \text{cm}^3. Leaning does not change the rule when h is the perpendicular height: \displaystyle \frac{1}{3} \cdot 54 \cdot 10 = 180.
h = 21 cm and V = 11200\ \text{cm}^3. Half the side is 20, so h^2 = 29^2 - 20^2 = 441, and \displaystyle V = \frac{1}{3} \cdot 1600 \cdot 21 = 1600 \cdot 7 = 11200.
h = 40 cm and V = 1080\pi \approx 3392.9\ \text{cm}^3. h^2 = 41^2 - 9^2 = 1600, and \displaystyle \frac{1}{3}\pi \cdot 81 \cdot 40 = 27 \cdot 40\pi = 1080\pi.
\displaystyle \frac{80}{3}\pi \approx 83.8\ \text{cm}^3. r^2 = 6^2 - 4^2 = 20, which is all the rule needs, and \displaystyle \frac{1}{3}\pi \cdot 20 \cdot 4 = \frac{80}{3}\pi.
86.4\ \text{m}^3. In meters the height is 3.2, so \displaystyle \frac{1}{3} \cdot 81 \cdot 3.2 = 27 \cdot 3.2 = 86.4.
One third. Fill the hollow pyramid level with sand and pour it into the hollow prism, and repeat: the prism is full to the brim after exactly three pyramid-fulls, so the pyramid holds \displaystyle \frac{1}{3}Bh where the prism holds Bh.
The height, the segment from the foot of the height to the midpoint of a base edge (of length \displaystyle \frac{s}{2}), and the slant height form a right triangle with the right angle at the foot of the height. The slant height is its hypotenuse, so \displaystyle \ell^2 = h^2 + \left(\frac{s}{2}\right)^2, which is more than h^2 because s > 0. So \ell > h.
h = 6 cm. \displaystyle 50\pi = \frac{1}{3}\pi \cdot 25 \cdot h, so 150 = 25h.
7 m. 3 \cdot 147 = 441 = s^2 \cdot 9, so s^2 = 49, and an edge is positive.
\displaystyle r = \sqrt{\frac{200}{3\pi}} \approx 4.61 cm. 200\ \text{mL} = 200\ \text{cm}^3, so 600 = \pi r^2 \cdot 9 and \displaystyle r^2 = \frac{600}{9\pi} = \frac{200}{3\pi}; take the positive root.
It is multiplied by \displaystyle \frac{9}{2}, that is 4.5. Here k = 3 and \displaystyle m = \frac{1}{2}, so \displaystyle k^2 m = \frac{9}{2}.
It is multiplied by 8. Every length is multiplied by 2, so the volume by 2^3 = 8.
\displaystyle \frac{8}{27} of it. The water is a cone similar to the glass with scale factor \displaystyle \frac{8}{12} = \frac{2}{3}, so its volume is \displaystyle \left(\frac{2}{3}\right)^3 = \frac{8}{27} of the glass’s.
The cone is 3 times as tall. From \displaystyle \frac{1}{3}\pi r^2 h_1 = \pi r^2 h_2, dividing by \pi r^2 gives \displaystyle \frac{1}{3}h_1 = h_2, so h_1 = 3h_2.
684\pi \approx 2148.8\ \text{cm}^3. The missing tip x satisfies \displaystyle \frac{6}{x} = \frac{9}{x + 12}, so 9x = 6x + 72 and x = 24. The whole cone is \displaystyle \frac{1}{3}\pi \cdot 81 \cdot 36 = 972\pi and the tip \displaystyle \frac{1}{3}\pi \cdot 36 \cdot 24 = 288\pi; the difference is 684\pi.
\displaystyle \frac{2080}{3} \approx 693.3\ \text{m}^3. The whole pyramid is \displaystyle \frac{1}{3} \cdot 144 \cdot 15 = 720. The top piece is a scaled copy with factor \displaystyle \frac{5}{15} = \frac{1}{3}, so its volume is \displaystyle \frac{720}{27} = \frac{80}{3}, and \displaystyle 720 - \frac{80}{3} = \frac{2080}{3}.
51\pi \approx 160.2\ \text{m}^3. The cylinder holds \pi \cdot 9 \cdot 5 = 45\pi and the roof \displaystyle \frac{1}{3}\pi \cdot 9 \cdot 2 = 6\pi.
Let the missing tip have height x. The radius is proportional to the distance from the apex, so \displaystyle \frac{r}{x} = \frac{R}{x + h}, giving Rx = rx + rh and \displaystyle x = \frac{rh}{R - r}, which is positive because R > r. The frustum is the whole cone minus the tip, so its volume is \displaystyle \frac{1}{3}\pi times R^2(x + h) - r^2 x, which is R^2 h + (R^2 - r^2)x. Since R^2 - r^2 = (R + r)(R - r), the term (R^2 - r^2)x equals (R + r)rh, so the bracket is R^2 h + Rrh + r^2 h, that is h(R^2 + Rr + r^2). The volume is \displaystyle \frac{1}{3}\pi h (R^2 + Rr + r^2).
A circle of radius r that touches all four sides of a square has the square’s side as its diameter, so the side is 2r and the square’s area is 4r^2. With height h, the pyramid holds \displaystyle \frac{1}{3} \cdot 4r^2 h and the cone \displaystyle \frac{1}{3}\pi r^2 h, and dividing gives \displaystyle \frac{\pi r^2}{4r^2} = \frac{\pi}{4}. The factor \displaystyle \frac{1}{3}h cancels, so the fraction is just the ratio of the base areas. The cylinder and the box are \pi r^2 h and 4r^2 h, the same two base areas times the same height, so their ratio is \displaystyle \frac{\pi}{4} too.