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Secondary 2 · Mathematics
3D shapes · Solids

Round Solids

Before you start, make sure you know: Circle and circumference (Secondary Year 1)
Before you start, make sure you know: Square root (Secondary Year 1)

A cone-shaped hat for an inn, a water tank on a rooftop, a soccer ball. How much cardstock do you need for the hat? How many liters fit in the tank? These are round solids, and on this page you will learn to answer questions like these.

Round solids and how they are generated

A round solid has at least one curved surface. It is not a polyhedron: its faces are not all polygons. The main ones are the cylinder, the cone, and the sphere.

The three main round solids: cylinder, cone, and sphere. cylinder cone sphere

All of them can be generated by rotating a plane figure one full turn around an axis. That is why they are also called solids of revolution:

  • A rectangle rotated about one of its sides forms a cylinder.

  • A right triangle rotated about one of its legs forms a cone.

  • A semicircle rotated about its diameter forms a sphere.

  • A right trapezoid rotated about its side perpendicular to the bases forms a truncated cone (a cone with its tip cut off).

The side of the rectangle that lies on the axis becomes the height. The side perpendicular to the axis becomes the radius of the base.

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Imagine you are a graphic designer creating a 3D logo. To do so, you rotate a rectangle with a base of $5 \text{ cm}$ and a height of $12 \text{ cm}$ around one of its…

Imagine you are a graphic designer creating a 3D logo. To do so, you rotate a rectangle with a base of $5 \text{ cm}$ and a height of $12 \text{ cm}$ around one of its longer sides (the $12 \text{ cm}$ side), as shown in the figure.

Rectangle with a 5 cm base and a 12 cm height with its left side, the longest one, marked as the axis of rotationAxis of rotation5 cm12 cm
  1. a) Draw the solid of revolution that is obtained.
  1. b) Identify the names of its main elements (base and height) and state how long each one is.

You can do it on paper and send a photo with your answer.

Solve
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Look at the three plane figures A, B and C. Each one makes a full turn around the dashed axis drawn next to it. Identify which of them produces a truncated cone…

Look at the three plane figures A, B and C. Each one makes a full turn around the dashed axis drawn next to it. Identify which of them produces a truncated cone (frustum). Briefly explain why you chose that option.

Three plane figures A (rectangle), B (right triangle) and C (right trapezoid), each with one side on a dashed vertical axis and arrows indicating a rotation around that axis A B C
Solve

Elements: radius, height, and slant height

Elements of a cylinder and of a cone: radius, height, and slant height. In the cone they form a right triangle. g = h r g h r

In the cone, $r$, $h$, and $g$ form a right triangle in which $g$ is the hypotenuse. By the Pythagorean theorem:

$$g^2 = r^2 + h^2$$

This works because the height drops vertically to the center of the base and forms a right angle with the radius.

Another useful relationship: the length of a circle's circumference is $C = 2\pi r$. If you know $C$, you can solve for the radius: $r = \dfrac{C}{2\pi}$.

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For a fancy-dress party, you want to make a conical hat. You have measured the radius of the base, which is $9 \text{ cm}$, and the height of the hat, which will be…

For a fancy-dress party, you want to make a conical hat. You have measured the radius of the base, which is $9 \text{ cm}$, and the height of the hat, which will be $12 \text{ cm}$. Use Pythagoras' theorem to calculate the length of the slant height ($g$) that you will need to draw the net on card.

h=12 r=9 g=?
Solve
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In a geography class, we look at a globe. If we know that the length of the equator (great circle) is approximately $62.8 \text{ cm}$, what is the radius of that sphere?…

In a geography class, we look at a globe. If we know that the length of the equator (great circle) is approximately $62.8 \text{ cm}$, what is the radius of that sphere? (Use $\pi \approx 3.14$).

r = ? L = 62.8 cm
Solve

The cylinder: net and area

If you cut open a can of corn and flatten it, you get two circles and a rectangle. That is the net of the cylinder.

Net of a cylinder: a rectangle of length 2πr and height h, with two circles of radius r. lateral area base r 2πr (length of the rectangle) h

The rectangle is $h$ tall and as long as the edge of the base, $2\pi r$. Its area is the lateral area:

$$A_L = 2\pi r h$$

For the total area, add the two bases, each with area $\pi r^2$: $A_T = 2\pi r h + 2\pi r^2$.

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A famous crisp brand uses cylindrical tubs. If the base radius of the tub is $4 \text{ cm}$ and its height is $20 \text{ cm}$, calculate the area of the paper label…

A famous crisp brand uses cylindrical tubs. If the base radius of the tub is $4 \text{ cm}$ and its height is $20 \text{ cm}$, calculate the area of the paper label surrounding the side of the tub (curved surface area). Take $\pi \approx 3.14$.

r=4 h=20 r=4 h=20 2·π·r Curved Surface Area A = 2·π·r·h
Solve
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A soda can is cylindrical in shape. The radius of its base is $3 \text{ cm}$ and its height is $12 \text{ cm}$. Calculate the volume of drink it can contain if it is…

A soda can is cylindrical in shape. The radius of its base is $3 \text{ cm}$ and its height is $12 \text{ cm}$. Calculate the volume of drink it can contain if it is filled completely. Express the result in cubic centimeters ($cm^3$).

Solve
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A soft drinks company wants to design a new cylindrical can. The base radius must be $3.5\text{ cm}$ and the height of the can $12\text{ cm}$. Calculate the total…

A soft drinks company wants to design a new cylindrical can. The base radius must be $3.5\text{ cm}$ and the height of the can $12\text{ cm}$. Calculate the total surface area of sheet metal needed to manufacture one can (not including the seals) and the volume of soft drink it can hold in milliliters ($1\text{ cm}^3 = 1\text{ ml}$).

r=3.5 cm h=12 cm
Solve

Capacity, liters, and mass

Volume also measures how much fits inside a container. The equivalences you need:

A cylindrical water tank with a radius of $1\text{ m}$ and a height of $1.5\text{ m}$ has $V = 3.14 \times 1^2 \times 1.5 = 4.71\text{ m}^3$, that is, $4{,}710\text{ L}$.

Sometimes you have to solve for a variable. If the volume $V$ and the height $h$ are given, $r = \sqrt{\dfrac{V}{\pi h}}$.

A hollow cylinder has a volume of material equal to that of the outer cylinder minus that of the inner hollow: $V = \pi R^2 h - \pi r^2 h$.

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Look at the net of a cylinder. It is known that the area of one of its circular bases is $28.26\text{ cm}^2$ and that the height of the lateral rectangle is…

Look at the net of a cylinder. It is known that the area of one of its circular bases is $28.26\text{ cm}^2$ and that the height of the lateral rectangle is $10\text{ cm}$. Calculate:

  1. a) The radius of the base.
  2. b) The lateral surface area of the cylinder.
  3. c) The total surface area.

(Use $\pi \approx 3.14$).

Curved Surface Area r A = 28.26 cm² h = 10 cm
Solve
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A water tank is in the shape of a cylinder with a height of $5\text{ m}$. If we want the tank to have a capacity of $100,000\text{ liters}$, what radius must the base of…

A water tank is in the shape of a cylinder with a height of $5\text{ m}$. If we want the tank to have a capacity of $100,000\text{ liters}$, what radius must the base of the tank have? (Remember that $1\text{ m}^3 = 1,000\text{ liters}$). Round the radius to two decimal places.

r = ? h = 5 m V = 100 000 L
Solve
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A steel steamroller cylinder has a diameter of $1.2\text{ meters}$ and a length of $2.5\text{ meters}$. Steel has a density of $7850\text{ kg/m}^3$. If the roller is…

A steel steamroller cylinder has a diameter of $1.2\text{ meters}$ and a length of $2.5\text{ meters}$. Steel has a density of $7850\text{ kg/m}^3$. If the roller is hollow and the steel thickness is $10\text{ cm}$ all around its circumference (excluding the side faces, which are considered negligible), calculate the total mass of the roller in metric tons.

Ø = 1.2 m L = 2.5 m e = 10 cm ρ = 7850 kg/m³
Solve

The cone: net and area

The net of a cone is a circle (the base) and a circular sector of radius $g$ whose arc is as long as the edge of the base, $2\pi r$. The area of that sector is $\frac{1}{2} \cdot g \cdot 2\pi r = \pi r g$.

Net of a cone: a circular sector of radius g and a circle of radius r. g sector: area πrg r base: area πr²
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An ice cream cone is shaped like a cone. If the area of the base is $25 \text{ cm}^2$ and the height of the cone is $10 \text{ cm}$, calculate its volume. Remember that…

An ice cream cone is shaped like a cone. If the area of the base is $25 \text{ cm}^2$ and the height of the cone is $10 \text{ cm}$, calculate its volume. Remember that the volume of a cone is one-third of the volume of a cylinder with the same base and height: $V = \frac{1}{3} \times A_{base} \times h$.

Solve
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A temporary tent for an event is being made in the shape of a perfect cone. The height of the tent from the ground to the top vertex is $4\text{ m}$ and the diameter of…

A temporary tent for an event is being made in the shape of a perfect cone. The height of the tent from the ground to the top vertex is $4\text{ m}$ and the diameter of the circular base is $6\text{ m}$. How many square meters of canvas were needed to cover only the lateral part of the tent? (Note: You must first find the slant height using the Pythagorean theorem).

h = 4m r = 3m g = ? d = 6m
Solve
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A perfume bottle is in the shape of a cylinder $10\text{ cm}$ high. Inside, the glass forms an inverted cone that reaches exactly to the bottom and whose base coincides…

A perfume bottle is in the shape of a cylinder $10\text{ cm}$ high. Inside, the glass forms an inverted cone that reaches exactly to the bottom and whose base coincides with the top lid of the cylinder. If the radius of the bottle is $3\text{ cm}$, calculate what percentage of the total volume of the cylinder is occupied by the perfume when the bottle is full. Does this percentage depend on the chosen radius and height? Reason your answer.

r = 3 cm
Solve

Rotating a triangle about its hypotenuse

Earlier you rotated a right triangle about a leg and got a cone. What happens if the axis is the hypotenuse? You get two cones joined at their base.

Right triangle with legs 6 and 8 rotating about its hypotenuse: two cones joined at the base form, with radius equal to the height onto the hypotenuse. 6 8 r = 4.8 axis

The two cones share the same base. Their radius is the height of the triangle onto the hypotenuse. The slant heights are the legs of the triangle.

To find that height, use the area of the triangle in two ways: $\dfrac{\text{leg}_1 \times \text{leg}_2}{2} = \dfrac{\text{hypotenuse} \times r}{2}$, so $r = \dfrac{\text{leg}_1 \times \text{leg}_2}{\text{hypotenuse}}$.

The total area is the sum of the lateral areas of the two cones: $A = \pi r g_1 + \pi r g_2 = \pi r (g_1 + g_2)$. There are no bases, because they are glued together.

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A solid of revolution is generated by rotating a right-angled triangle with legs $6\text{ cm}$ and $8\text{ cm}$ around its hypotenuse. a) Describe the resulting solid…

A solid of revolution is generated by rotating a right-angled triangle with legs $6\text{ cm}$ and $8\text{ cm}$ around its hypotenuse.

  1. a) Describe the resulting solid (you may draw it).
  2. b) Calculate the total surface area of the generated solid.

(Hint: The solid consists of two cones joined at their base. You will need to calculate the height of the triangle relative to the hypotenuse).

Rotation h C A B 6 cm 8 cm
Solve

The truncated cone

A lampshade or a plastic cup has the shape of a truncated cone (frustum): it has two circular bases, the larger one with radius $R$ and the smaller one with radius $r$.

Truncated cone with radii R and r, height h, and slant height g, and the right triangle with legs h and R minus r. R r h g R − r h g

The formula makes sense: the lateral net is a "strip" of width $g$ whose length is the average of the two edges, $\frac{2\pi R + 2\pi r}{2} = \pi(R + r)$.

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An interior design company manufactures lampshades in the shape of a frustum of a cone. The radius of the top base is $12\text{ cm}$, that of the bottom base is…

An interior design company manufactures lampshades in the shape of a frustum of a cone. The radius of the top base is $12\text{ cm}$, that of the bottom base is $20\text{ cm}$, and the height of the lampshade is $15\text{ cm}$. If the special fabric material costs $18.50\text{ \$/m}^2$, calculate the cost of the material needed to make 25 identical lampshades (note that the lampshade has no top or bottom bases, only the lateral surface).

r = 12 R = 20 h = 15 g
Solve

The sphere

The sphere has only one element: the radius $r$, the distance from the center to any point on the surface. Its diameter is $d = 2r$.

A hemisphere is half a sphere: its curved surface is $2\pi r^2$ and its volume is $\frac{2}{3}\pi r^3$. If you also count the flat base (a circle), add $\pi r^2$.

How much the water rises when you submerge a sphere

When you put an object in water, the level rises because the object displaces its own volume. That extra volume forms a layer of water shaped like the container:

$$V_{\text{sphere}} = A_{\text{base}} \times \text{rise}, \qquad \text{rise} = \frac{V_{\text{sphere}}}{A_{\text{base}}}$$
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A basketball has a diameter of $24 \text{ cm}$. What is the total surface area of leather needed to manufacture it? Remember that the radius is half the diameter and use…

A basketball has a diameter of $24 \text{ cm}$. What is the total surface area of leather needed to manufacture it? Remember that the radius is half the diameter and use the formula for the surface area of a sphere $A = 4 \times \pi \times r^2$. (Use $\pi \approx 3.14$).

r d = 24 cm
Solve
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A famous chocolate brand sells its chocolates in a box shaped like a sphere with a diameter of $10\text{ cm}$. If the box is filled to $75\%$ of its capacity with liquid…

A famous chocolate brand sells its chocolates in a box shaped like a sphere with a diameter of $10\text{ cm}$. If the box is filled to $75\%$ of its capacity with liquid chocolate, how many cubic centimeters of chocolate are inside? Express the result rounded to two decimal places.

D = 10 cm 75% full
Solve
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A glass sphere with a radius of $15\text{ cm}$ is submerged in a cylindrical container with a diameter of $32\text{ cm}$ containing water. If the sphere is completely…

A glass sphere with a radius of $15\text{ cm}$ is submerged in a cylindrical container with a diameter of $32\text{ cm}$ containing water. If the sphere is completely submerged to the bottom, by how many centimeters will the water level in the container rise? Express the result rounded to two decimal places.

r = 15 d = 32 Δh = ? Initial level Final level
Solve

Composite solids

Many real objects combine several solids: an ice cream cone with its scoop, a water tank with a cone-shaped bottom. To solve them:

  1. Break the object into known solids.

  2. Find the volume or the area of each one.

  3. Volume: add the parts. Area: add only the surfaces that remain visible; the faces where the solids are joined do not count.

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Calculate the volume of a wooden piece formed by joining a cylinder and a hemisphere. The cylinder has a height of $10\text{ cm}$ and its base radius is $4\text{ cm}$.…

Calculate the volume of a wooden piece formed by joining a cylinder and a hemisphere. The cylinder has a height of $10\text{ cm}$ and its base radius is $4\text{ cm}$. The hemisphere rests exactly on the top base of the cylinder.

r = 4 cm 4 cm 10 cm
Solve
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An ice cream cone is shaped like a cone. The wafer cone has a depth (height) of $12\text{ cm}$ and the radius of the opening is $3\text{ cm}$. If the ice cream fills the…

An ice cream cone is shaped like a cone. The wafer cone has a depth (height) of $12\text{ cm}$ and the radius of the opening is $3\text{ cm}$. If the ice cream fills the inside of the cone and also sticks out as a perfect scoop on top with the same radius as the cone, calculate the total volume of ice cream.

r=3 r=3 h=12
Solve
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A grain silo is to be constructed, consisting of a cylindrical body $4\text{ m}$ high and with a radius of $3\text{ m}$, topped at the bottom by a cone of the same…

A grain silo is to be constructed, consisting of a cylindrical body $4\text{ m}$ high and with a radius of $3\text{ m}$, topped at the bottom by a cone of the same radius and a slant height of $5\text{ m}$ to facilitate discharge. Calculate the total storage volume of the silo and the total surface area of sheet metal required to manufacture it (including the flat top lid of the cylinder).

R = 3m 4m g = 5m
Solve
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Calculate the surface area of an "ice cream cone" consisting of a right cone $12\text{ cm}$ high with a base radius of $5\text{ cm}$, topped by a perfect hemisphere that…

Calculate the surface area of an "ice cream cone" consisting of a right cone $12\text{ cm}$ high with a base radius of $5\text{ cm}$, topped by a perfect hemisphere that fits exactly onto the base of the cone. If we want to wrap the ice cream (cone and hemisphere) in protective paper, what minimum area of paper will we need?

h = 12 r = 5 r = 5 g
Solve

Cutting a round solid

If you cut a round solid with a plane, the figure that appears depends on the cut.

Plane cross sections of a cylinder: a cut parallel to the bases gives a circle; a cut through the axis gives a rectangle. cut parallel to the base: circle cut through the axis: rectangle

Radius of the cross-section circle in a sphere. If the sphere has radius $R$ and the plane is at distance $d$ from the center, a right triangle forms with hypotenuse $R$ and legs $d$ and $r$ (the radius of the circle cut):

$$r = \sqrt{R^2 - d^2}$$
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A wooden sphere with a radius of $13\text{ cm}$ is cut by a horizontal plane. If the distance from the center of the sphere to the cut is $5\text{ cm}$, what is the…

A wooden sphere with a radius of $13\text{ cm}$ is cut by a horizontal plane. If the distance from the center of the sphere to the cut is $5\text{ cm}$, what is the radius of the circle formed in the cross-section of the cut? (Apply the Pythagorean theorem in the right triangle that is formed).

5 r 13
Solve
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A wooden sphere with a radius of $20\text{ cm}$ is cut by a plane located $12\text{ cm}$ from the center. When the two pieces are separated, two spherical caps are…

A wooden sphere with a radius of $20\text{ cm}$ is cut by a plane located $12\text{ cm}$ from the center. When the two pieces are separated, two spherical caps are obtained. Calculate the total surface area (curved surface plus the flat circular base of the cross-section) of the smaller spherical cap.

(Hint: use the Pythagorean theorem to find the radius of the circular cross-section).

d = 12 R = 20 r = ? O R = 20
Solve

Go back to the beginning: for the cone-shaped hat you need the lateral area $\pi r g$ (with $g$ from Pythagoras), and for the water tank, the volume $\pi r^2 h$ converted to liters.

Summary

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