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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 (first cycle))
Before you start, make sure you know: Square root (Secondary Year 1 (first cycle))

A soda can, an ice cream cone, a baseball. None of them has flat faces like a polyhedron: they are round solids. A beverage company needs to know how much sheet metal it uses for each can and how much soda fits inside. By the end of this page you will know how to do those calculations.

The Three Round Solids

Illustration of the three main round solids: cylinder, cone, and sphere. cylinder cone sphere

None of them is a polyhedron: all of them have at least one curved surface. The cylinder and the cone also have flat bases, which are circles. The sphere has no flat faces at all.

These solids are also called solids of revolution, because they are generated by rotating a plane figure one full turn around an axis:

  • A rectangle rotating around one of its sides generates a cylinder.

  • A right triangle rotating around one of its legs generates a cone.

  • A semicircle rotating around its diameter generates a sphere.

  • A right trapezoid rotating around its side perpendicular to the bases generates a truncated cone (frustum).

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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 = hr ghr

Since $r$, $h$, and $g$ form a right triangle in the cone, the Pythagorean theorem holds:

$$g^2 = r^2 + h^2 \qquad\Rightarrow\qquad g = \sqrt{r^2 + h^2}$$

In a circle, the circumference is $L = 2\pi r$ and the area is $A = \pi r^2$. If you know one of the two, you can solve for $r$. For example, if $L = 31.4$ cm, then $r = \dfrac{31.4}{2 \times 3.14} = 5$ cm.

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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

Area of the Cylinder

If you cut the cylinder and "unroll" it, you get two circles and one rectangle. The rectangle is $2\pi r$ long (the perimeter of the base) and $h$ tall.

Flat net of a cylinder: a central rectangle of length 2πr and height h, with a circle of radius r above and another below. lateral rectangle length = 2πr h circle (base) circle (base)

Why? The lateral rectangle has area (length)(height) $= 2\pi r \cdot h$, and each base contributes $\pi r^2$.

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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 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
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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

Area of the Cone and the Truncated Cone

When a cone is unrolled onto a plane, the lateral surface is a circular sector of radius $g$ (the slant height), whose arc is as long as the perimeter of the base, $2\pi r$. Add the circular base.

A truncated cone (frustum) is what remains of a cone when you cut off its tip with a plane parallel to the base (for example, a lampshade or a cup). It has two circular bases, with radii $r_1$ and $r_2$, and its slant height $g$ is the slanted side.

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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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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 $18.50\text{ \text{RD\$}\ /m}^2$, 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: Area and Volume

The sphere has only one element: the radius $r$, the distance from the center to any point on the surface.

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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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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

Volume of the Cylinder and the Cone

Volume measures how much fits inside. In the cylinder it is the area of the base times the height. The cone takes up exactly one third of the cylinder that has the same base and the same height.

The units are cubic: $\text{cm}^3$, $\text{m}^3$. Remember that $1\ \text{cm}^3 = 1\ \text{mL}$ and $1\ \text{m}^3 = 1{,}000$ liters.

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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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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 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

Composite Solids

Many objects are the union of several solids: an ice cream cone with its scoop, a silo, a capsule. The strategy is to break it down: calculate each part separately and add. If a solid is hollow or a part is lost, subtract.

A useful case: a submerged solid displaces a volume of water equal to its own. If a sphere sinks in a cylindrical container, the water rises until $V_{sphere} = \pi R^2 \cdot \Delta h$, where $R$ is the radius of the container and $\Delta h$ is how much the water level rises.

Another case: a cone inside a cylinder with the same base and height takes up one third of the cylinder, that is, $33.\overline{3}\,\%$ regardless of the value of $r$ or $h$.

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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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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
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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

Cutting a Round Solid

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

Plane cross sections of a cylinder: circular cut parallel to the bases and rectangular cut perpendicular to the bases. cut parallel to the base: circle perpendicular cut: rectangle

Cross section of a sphere. If you cut a sphere of radius $R$ with a plane at distance $d$ from the center, the radius $r$ of the circle of the cut, $d$, and $R$ form a right triangle, with $R$ as the hypotenuse:

$$r^2 + d^2 = R^2 \qquad\Rightarrow\qquad r = \sqrt{R^2 - d^2}$$ Cut of a sphere by a plane at distance d from the center: the radius R, the distance d, and the radius r of the cross section form a right triangle. d r R

In the figure, the point $(225, 60)$ lies on the edge of the sphere: it is $\sqrt{75^2 + 40^2} \approx 85$ from the center, which is exactly the radius $R$.

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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

Combined Problems

Hard problems mix several ideas. A useful guideline:

  1. Draw the solid and mark $r$, $h$, and $g$. Convert diameters to radii.

  2. Decide whether you are asked for area (surface, cardboard, sheet metal, paint) or volume (capacity, material, mass).

  3. Calculate any missing data with Pythagoras.

  4. Break down if there are joined solids, or subtract if there are hollow parts.

  5. Check the units and that the result makes sense.

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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
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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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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

Back to the beginning: for the soda can, the sheet metal needed is its total area $2\pi r(h + r)$ and the soda that fits inside is its volume $\pi r^2 h$.

Summary

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