Develop the sphere-volume formula using cylinder and cone relationships

Ontario Grade 10 Mathematics

Measurement and Trigonometry

Use a cylinder and two cones to understand the volume of a sphere

A sphere is a round solid, like a ball. Its volume is the amount of space inside it. Instead of treating the sphere formula as something to memorize, we can build it from familiar solids. We will compare a sphere with a cylinder and two cones that have the same radius and height. The comparison will show where the sphere-volume formula comes from.

What you will learn

Review how to find the volume of a cylinder and a cone.
Describe how a sphere can be compared with a cylinder and two cones.
Develop the formula for the volume of a sphere.
Use the formula with a radius and explain what the answer means.

1. Grade 9 bridge: volumes you already know

Volume measures the space inside a three-dimensional shape. It is measured in cubic units, such as cubic centimetres. The base area tells us how much space one flat layer covers. Multiplying that area by the height gives the volume of a cylinder.

A cone has a circular base and narrows to a point. A cone with the same base and height as a cylinder has one-third of the cylinder’s volume. In these formulas,

r

means the radius of a circular base,

h

means height, and

\pi

is the number used in circle measurements. The radius is the distance from the centre of a circle to its edge.

V_{\text{cylinder}}=\pi r^2h,\quad V_{\text{cone}}=\frac{1}{3}\pi r^2h

A cylinder’s volume is its circular base area multiplied by its height.
A cone’s volume is one-third of the volume of a matching cylinder.

2. Compare a sphere with a cylinder and two cones

Imagine a sphere with radius

r

. Place it inside a cylinder whose circular ends just touch the top and bottom of the sphere. The cylinder has radius

r

and height

2r

, because the sphere reaches one radius above its centre and one radius below it.

Now imagine two cones inside that cylinder. Their pointed ends meet at the cylinder’s middle. Each cone has a base radius of

r

and a height of

r

. Together, the cones fill the parts near the cylinder’s top and bottom, leaving a central shape that matches the sphere’s volume.

One way to understand the match is to compare thin, horizontal layers at the same height. At each height, the sphere’s circular layer has the same area as the layer left inside the cylinder after removing the two cones. If two solids have matching layers all the way through, they have the same volume. This layer comparison is the reason the cylinder-minus-cones calculation gives the sphere’s volume.

The cylinder’s radius matches the sphere’s radius.
The cylinder’s height is the sphere’s diameter, or $2r$ .
Each cone has radius $r$ and height $r$ .

3. Develop the sphere formula

Find the volume of the whole cylinder first. Its height is

2r

, so its volume is

2\pi r^3

.

Each cone has volume one-third of a cylinder with radius

r

and height

r

. There are two cones, so together they have volume

\frac{2}{3}\pi r^3

.

Subtract the volume of both cones from the cylinder’s volume. The remaining volume is the sphere’s volume. Simplifying the subtraction gives the sphere formula. Here,

V_{\text{sphere}}

means the volume of the sphere.

V_{\text{sphere}}=2\pi r^3-2\left(\frac{1}{3}\pi r^3\right)=\frac{4}{3}\pi r^3

Use the same radius for the sphere, cylinder, and cones.
Subtract the volumes of both cones, not just one.
The result is expressed in cubic units.

4. Use the formula and practise

To use the formula, identify the sphere’s radius and substitute it for

r

. Then evaluate the expression. If the question asks for an approximate volume, use the given value of

\pi

or a calculator value. Keep the cubic units in your answer.

Try this independently: A sphere has radius

3

cm. Write the substitution into the sphere formula and find its volume in terms of

\pi

. Then give an approximate answer to one decimal place. Check that you used the radius, not the diameter.

A second practice prompt: A sphere has diameter

10

m. First determine its radius. Then write the formula with that radius and leave your answer in terms of

\pi

. This checks whether you can identify the measurement needed before substituting.

d=2r

The formula uses radius, even when a question gives diameter.
A diameter is twice the radius.
Volume answers use cubic units.

The comparison solids

Solid	Radius	Height	Volume
Cylinder	$r$	$2r$	$2\pi r^3$
One cone	$r$	$r$	$\frac{1}{3}\pi r^3$
Two cones	$r$ each	$r$ each	$\frac{2}{3}\pi r^3$
Sphere	$r$	—	$\frac{4}{3}\pi r^3$

Worked example

Build and use the formula

A sphere has radius

3

cm. Develop its volume from the matching cylinder and two cones, then calculate the sphere’s volume in terms of

\pi

.

Set the dimensions
The matching cylinder has the sphere’s radius and a height equal to its diameter. Each cone has the same radius and a height equal to the sphere’s radius.
$r=3,\quad h_{\text{cylinder}}=6,\quad h_{\text{cone}}=3$
Find the cylinder volume
Multiply the circular base area by the cylinder’s height. The result is the volume of the full comparison cylinder.
$V_{\text{cylinder}}=\pi(3)^2(6)=54\pi$
Find both cone volumes
A cone is one-third of a matching cylinder. Calculate one cone, then multiply by two because the comparison uses two cones.
$2V_{\text{cone}}=2\left(\frac{1}{3}\pi(3)^2(3)\right)=18\pi$
Subtract to get the sphere
The sphere has the same volume as the cylinder with both cone volumes removed. Subtract and include cubic centimetres.
$V_{\text{sphere}}=54\pi-18\pi=36\pi\text{ cm}^3$

Answer: The sphere’s volume is

36\pi\text{ cm}^3

.

Check: The general formula gives

\frac{4}{3}\pi(3)^3=36\pi\text{ cm}^3

, which matches the cylinder-minus-cones calculation.

Common mistakes and how to avoid them

Using the sphere’s diameter as

r

.

Correction: The formula uses radius. If the diameter is given, divide it by two first.

Subtracting only one cone from the cylinder.

Correction: The comparison uses two cones, one at each end of the cylinder. Subtract both cone volumes.

Using

r^2

instead of

r^3

in the final sphere formula.

Correction: The formula includes the square of the radius from the circular area and one more factor of

r

from the height.

Writing the answer in square units.

Correction: Volume is measured in cubic units, such as

\text{cm}^3

.

Lesson summary

A sphere of radius $r$ fits in a cylinder of radius $r$ and height $2r$ .
Two cones, each with radius $r$ and height $r$ , account for the part of the cylinder outside the sphere.
Subtracting the two cone volumes from the cylinder volume develops the sphere-volume formula.
Use the radius in the formula and give the result in cubic units.

Check your understanding

Question 1

A sphere has radius

2

cm. Which expression gives its volume?

$\frac{4}{3}\pi(2)^3$
$\frac{4}{3}\pi(2)^2$
$2\pi(2)^3$
$\frac{1}{3}\pi(2)^3$

Show answer and explanation

\frac{4}{3}\pi(2)^3

The sphere formula uses the radius raised to the third power. Substituting

r=2

gives

\frac{4}{3}\pi(2)^3

.

Question 2

A sphere’s diameter is

8

m. What radius should be used in the volume formula?

$16$ m
$8$ m
$4$ m
$2$ m

Show answer and explanation

4

m

The radius is half the diameter, so the radius is

4

m.

Question 3

Why are the volumes of two cones subtracted from the matching cylinder?

The two cones together have the same volume as the sphere.
The sphere’s volume matches the cylinder’s volume after the two cone volumes are removed.
Each cone has the same volume as the cylinder.
The cylinder has twice the volume of the sphere.

Show answer and explanation

The sphere’s volume matches the cylinder’s volume after the two cone volumes are removed.

The comparison uses matching horizontal layers. Removing both cones from the cylinder leaves a solid with the sphere’s volume.

Key terms

Volume: The amount of space inside a three-dimensional shape.
Radius: The distance from the centre of a circle or sphere to its surface.
Diameter: The distance across a circle or sphere through its centre; it is twice the radius.
Sphere: A round three-dimensional shape whose surface points are all the same distance from its centre.
Matching horizontal layers: Flat cross-sections taken at the same height that have equal areas.

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Published by DoAssignment. This reviewed lesson follows Ontario Grade 10 Mathematics (MFM2P), study topic M12. It is a study resource, not an official curriculum publication.

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