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M12 · Develop the sphere-volume formula using cylinder and cone relationships

Learn to develop the sphere-volume formula using cylinder and cone relationships through clear examples and targeted practice.

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

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, rr means the radius of a circular base, hh means height, and π\pi is the number used in circle measurements. The radius is the distance from the centre of a circle to its edge.
Vcylinder=πr2h,Vcone=13πr2hV_{\text{cylinder}}=\pi r^2h,\quad V_{\text{cone}}=\frac{1}{3}\pi r^2h

2. Compare a sphere with a cylinder and two cones

Imagine a sphere with radius rr. Place it inside a cylinder whose circular ends just touch the top and bottom of the sphere. The cylinder has radius rr and height 2r2r, 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 rr and a height of rr. 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.

3. Develop the sphere formula

Find the volume of the whole cylinder first. Its height is 2r2r, so its volume is 2πr32\pi r^3.
Each cone has volume one-third of a cylinder with radius rr and height rr. There are two cones, so together they have volume 23πr3\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, VsphereV_{\text{sphere}} means the volume of the sphere.
Vsphere=2πr3−2(13πr3)=43πr3V_{\text{sphere}}=2\pi r^3-2\left(\frac{1}{3}\pi r^3\right)=\frac{4}{3}\pi r^3

4. Use the formula and practise

To use the formula, identify the sphere’s radius and substitute it for rr. Then evaluate the expression. If the question asks for an approximate volume, use the given value of π\pi or a calculator value. Keep the cubic units in your answer.
Try this independently: A sphere has radius 33 cm. Write the substitution into the sphere formula and find its volume in terms of π\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 1010 m. First determine its radius. Then write the formula with that radius and leave your answer in terms of π\pi. This checks whether you can identify the measurement needed before substituting.
d=2rd=2r

The comparison solids

SolidRadiusHeightVolume
Cylinderrr2r2r2πr32\pi r^3
One conerrrr13πr3\frac{1}{3}\pi r^3
Two conesrr eachrr each23πr3\frac{2}{3}\pi r^3
Sphererr—43πr3\frac{4}{3}\pi r^3

Worked example

Build and use the formula

A sphere has radius 33 cm. Develop its volume from the matching cylinder and two cones, then calculate the sphere’s volume in terms of π\pi.
  1. 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,hcylinder=6,hcone=3r=3,\quad h_{\text{cylinder}}=6,\quad h_{\text{cone}}=3
  2. Find the cylinder volume
    Multiply the circular base area by the cylinder’s height. The result is the volume of the full comparison cylinder.
    Vcylinder=π(3)2(6)=54πV_{\text{cylinder}}=\pi(3)^2(6)=54\pi
  3. 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.
    2Vcone=2(13π(3)2(3))=18π2V_{\text{cone}}=2\left(\frac{1}{3}\pi(3)^2(3)\right)=18\pi
  4. Subtract to get the sphere
    The sphere has the same volume as the cylinder with both cone volumes removed. Subtract and include cubic centimetres.
    Vsphere=54π−18π=36π cm3V_{\text{sphere}}=54\pi-18\pi=36\pi\text{ cm}^3
Answer: The sphere’s volume is 36π cm336\pi\text{ cm}^3.
Check: The general formula gives 43π(3)3=36π cm3\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 rr.
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 r2r^2 instead of r3r^3 in the final sphere formula.
Correction: The formula includes the square of the radius from the circular area and one more factor of rr from the height.
Writing the answer in square units.
Correction: Volume is measured in cubic units, such as cm3\text{cm}^3.

Lesson summary

Check your understanding

Question 1

A sphere has radius 22 cm. Which expression gives its volume?
  1. 43π(2)3\frac{4}{3}\pi(2)^3
  2. 43π(2)2\frac{4}{3}\pi(2)^2
  3. 2π(2)32\pi(2)^3
  4. 13π(2)3\frac{1}{3}\pi(2)^3
Show answer and explanation
43π(2)3\frac{4}{3}\pi(2)^3
The sphere formula uses the radius raised to the third power. Substituting r=2r=2 gives 43π(2)3\frac{4}{3}\pi(2)^3.

Question 2

A sphere’s diameter is 88 m. What radius should be used in the volume formula?
  1. 1616 m
  2. 88 m
  3. 44 m
  4. 22 m
Show answer and explanation
44 m
The radius is half the diameter, so the radius is 44 m.

Question 3

Why are the volumes of two cones subtracted from the matching cylinder?
  1. The two cones together have the same volume as the sphere.
  2. The sphere’s volume matches the cylinder’s volume after the two cone volumes are removed.
  3. Each cone has the same volume as the cylinder.
  4. 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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About this lesson

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

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