# Volume of Cones

## V = πr^2h/3

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Volume of Cones

Sam and Jean have opened a small ice cream stand. They have two different sized cones they sell. A small is 4\begin{align*}4^{\prime \prime}\end{align*} in diameter and 4\begin{align*}4^{\prime \prime}\end{align*} in length. A large is 5\begin{align*}5^{\prime \prime}\end{align*} in diameter and 6\begin{align*}6^{\prime \prime}\end{align*} in length. Sam wants to charge twice as much for the larger cone because he says that the large cone has twice the volume. Is he correct?

In this concept, you will learn to calculate the volume of cones and truncated cones.

### Volume

Volume is the measure of how much space a three-dimensional figure takes up or holds. Volume is often what you think of when you talk about measuring liquid or liquid capacity. You use unit cubes to represent volume.

Cones are unusual because they are so much smaller at the top than they are at their base. A cone has exactly one-third the volume of a cylinder.

Since the base of a cone is a circle, the formula for finding the volume of a cone is:

V=13πr2h\begin{align*}V=\frac{1}{3} \pi r^2 h\end{align*}

Now you can see that you would find the area of the base, πr2\begin{align*}\pi r^2\end{align*}, multiply it by the height, h\begin{align*}h\end{align*}, and then multiply it by one-third or take one-third of the product of the base area and the height.

Let’s look at an example.

What is the volume of the cone below?

First, substitute what you know into the formula for the volume of a cone.

VV==13πr2h13π(3.5)2(22)\begin{align*}\begin{array}{rcl} V &=& \frac{1}{3} \pi r^2 h \\ V &=& \frac{1}{3} \pi (3.5)^2 (22) \end{array}\end{align*}

Next, use algebra to solve for the volume.

VVV===13π(3.5)2(22)13π(12.25)(22)282.2\begin{align*}\begin{array}{rcl} V &=& \frac{1}{3} \pi (3.5)^2 (22) \\ V &=& \frac{1}{3} \pi (12.25) (22) \\ V &=& 282.2 \end{array}\end{align*}

The volume of the cone is 282.2 cm3\begin{align*}282.2 \ cm^3\end{align*}.

The image below shows a truncated cone.

Notice that you have two radii to work with and the height of the truncated cone as well. You can use the following formula to calculate volume of a truncated cone:

V=13π(r21+r1r2+r22)h\begin{align*}V=\frac{1}{3} \pi (r_1^2+r_1r_2+r_2^2)h\end{align*}

Let’s look at an example.

What is the volume of a truncated cone with a top radius of 2 cm, a bottom radius of 4 cm and a height of 4.5 cm?

First, substitute what you know into the formula for the volume of a truncated cone.

VV==13π(r21+r1r2+r22)h13π(22+2×4+42)4.5\begin{align*}\begin{array}{rcl} V &=& \frac{1}{3} \pi (r_{1}^2+r_1r_2+r_2^2)h \\ V &=& \frac{1}{3} \pi (2^2+2 \times 4 + 4^2)4.5 \end{array}\end{align*}

Next, use algebra to solve for the volume.

VVV===13π(22+2×4+42)4.513π(28)4.5131.95\begin{align*}\begin{array}{rcl} V &=& \frac{1}{3} \pi (2^2+2 \times 4 + 4^2)4.5 \\ V &=& \frac{1}{3} \pi (28)4.5 \\ V &=& 131.95 \end{array}\end{align*}

The volume of the cone is 131.95 cm3\begin{align*}131.95 \ cm^3\end{align*}.

### Examples

#### Example 1

Earlier, you were given a problem about Sam and Jean’s ice cream venture.

Sam wants to charge double for the large cone that has a diameter of 5\begin{align*}5^{\prime \prime}\end{align*} and a height of 6\begin{align*}6^{\prime \prime}\end{align*} because he says it has double the volume of the small cone with a diameter of 4\begin{align*}4^{\prime \prime}\end{align*} and a height of 4\begin{align*}4^{\prime \prime}\end{align*}. Is Sam correct?

First, calculate the volume of the large cone.

VVVV====13πr2h13π(5)2(6)13π(25)(6)157.1\begin{align*}\begin{array}{rcl} V &=& \frac{1}{3} \pi r^2 h \\ V &=& \frac{1}{3} \pi (5)^2 (6) \\ V &=& \frac{1}{3} \pi (25)(6) \\ V &=& 157.1 \end{array}\end{align*}

Next, calculate the volume of the small cone.

VVVV====13πr2h13π(4)2(4)13π(16)(4)67.0\begin{align*}\begin{array}{rcl} V &=& \frac{1}{3} \pi r^2 h \\ V &=& \frac{1}{3} \pi (4)^2 (4) \\ V &=& \frac{1}{3} \pi (16)(4) \\ V &=& 67.0 \end{array} \end{align*}

Then, divide the volume of the large cone by the volume of the small cone to see if it is twice as big.

Vlarge coneVsmall coneVlarge coneVsmall cone==157.1672.34\begin{align*}\begin{array}{rcl} \frac{V_{\text{large cone}}}{V_{\text{small cone}}} &=& \frac{157.1}{67} \\ \frac{V_{\text{large cone}}}{V_{\text{small cone}}} &=& 2.34 \end{array} \end{align*}

The volume of ice cream in the large cone is 2.34 times the volume of the small cone. Sam is correct about charging twice the price (and could even charge a little more!)

#### Example 2

What is the height of a cone if the radius is 1.6 meters and volume is 20.1 cubic meters?

First, substitute what you know into the formula for the volume of the cone.

V20.1==13πr2h13π(1.6)2h\begin{align*}\begin{array}{rcl} V &=& \frac{1}{3} \pi r^2 h \\ 20.1 &=& \frac{1}{3} \pi (1.6)^2 h \end{array}\end{align*}

Next, use algebra to solve for the height, h\begin{align*}h\end{align*}.

20.120.120.120.12.68h=====13π(1.6)2h13π(2.56)h2.68h2.68h2.687.5\begin{align*}\begin{array}{rcl} 20.1 &=& \frac{1}{3} \pi (1.6)^2 h \\ 20.1 &=& \frac{1}{3} \pi (2.56) h \\ 20.1 &=& 2.68 h \\ \frac{20.1}{2.68} &=& \frac{2.68 h}{2.68} \\ h &=& 7.5 \end{array}\end{align*}

The height of the cone is 7.5 m\begin{align*}7.5 \ m\end{align*}.

#### Example 3

Find the volume of a cone with a radius of 2 inches and a height of 4 inches.

First, substitute what you know into the formula for the volume of a cone.

VV==13πr2h13π(2)2(4)\begin{align*}\begin{array}{rcl} V &=& \frac{1}{3} \pi r^2 h \\ V &=& \frac{1}{3} \pi (2)^2 (4) \end{array}\end{align*}

Next, use algebra to solve for the volume.

VVV===13π(2)2(4)13π(4)(4)16.76\begin{align*}\begin{array}{rcl} V &=& \frac{1}{3} \pi (2)^2 (4) \\ V &=& \frac{1}{3} \pi (4)(4)\\ V &=& 16.76 \end{array}\end{align*}

The volume of the cone is 16.76 in3\begin{align*}16.76 \ in^3\end{align*}.

#### Example 4

Find the volume of a cone with a radius of 5 cm and a height of 7 cm.

First, substitute what you know into the formula for the volume of a cone.

VV==13πr2h13π(5)2(7)\begin{align*}\begin{array}{rcl} V &=& \frac{1}{3} \pi r^2 h \\ V &=& \frac{1}{3} \pi (5)^2 (7) \end{array}\end{align*}

Next, use algebra to solve for the volume.

VVV===13π(5)2(7)13π(25)(7)183.26\begin{align*}\begin{array}{rcl} V &=& \frac{1}{3} \pi (5)^2 (7) \\ V &=& \frac{1}{3} \pi (25)(7) \\ V &=& 183.26 \end{array} \end{align*}

The volume of the cone is 183.26 cm3\begin{align*}183.26 \ cm^3\end{align*}.

#### Example 5

Find the volume of a cone with a radius of 3 m and a height of 8 m.

First, substitute what you know into the formula for the volume of a cone.

VV==13πr2h13π(3)2(8)\begin{align*}\begin{array}{rcl} V &=& \frac{1}{3} \pi r^2 h \\ V &=& \frac{1}{3} \pi (3)^2 (8) \end{array} \end{align*}

Next, use algebra to solve for the volume.

VVV===13π(3)2(8)13π(9)(8)75.40\begin{align*} \begin{array}{rcl} V &=& \frac{1}{3} \pi (3)^2 (8) \\ V &=& \frac{1}{3} \pi (9)(8) \\ V &=& 75.40 \end{array} \end{align*}

The volume of the cone is 75.40 m3\begin{align*}75.40 \ m^3\end{align*}.

### Review

Answer each of the following questions.

1. What is the formula for finding the volume of a cone?

2. True or false. A truncated cone is a cone without a vertex.

3. True or false. You can use the same formula to find the volume of a truncated cone as a regular cone.

4. What is the diameter of this cone?

5. What is the height of the cone?

6. What is the volume of the cone?

7. What is the diameter of this cone?

8. What is the height of the cone?

9. What is the volume of the cone?

10. What is the diameter of this cone?

11. What is the radius of this cone?

12. What is the height of the cone?

13. What is the volume of the cone?

Use what you have learned to solve each of the following problems.

14. A cone has a radius of 6 meters and a volume of 168π\begin{align*}168 \pi\end{align*}. What is its height?

15. The containers of icing for Tina’s cake decorator are cones. Each container has a radius of 2.4 inches and a height of 7 inches. If Tina buys containers of red, yellow, and blue icing, how much icing will she buy?

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### Vocabulary Language: English

TermDefinition
Cone A cone is a solid three-dimensional figure with a circular base and one vertex.
Volume Volume is the amount of space inside the bounds of a three-dimensional object.