Maria has a paperweight that is a glass sphere. The sphere is full of liquid. If the diameter of the paperweight is 6 inches, how much liquid does it contain?

In this concept, you will learn to calculate the volume of spheres.

**Volume**

**Volume** is the measure of three-dimensional space a figure takes up. You can also think of it as how much space the figure “holds.”

To find the volume of spheres, you can use pyramids. Imagine a pyramid with its base on the surface of the sphere and its point as the center of the sphere. The radius of the sphere would be the height of the pyramid.

The pyramid makes up a portion of the sphere’s volume. If you can fill the whole sphere with pyramids like this, you would know the volume of the sphere. It would be equal to the volumes of all the pyramids put together. How many pyramids would it take to fill a sphere? That depends on the surface area of the sphere.

You can combine the surface area of a sphere with the volume formula for a pyramid to calculate the volume of all the pyramids contained within the sphere.

The formula for the volume of a pyramid is \begin{align*}V = \frac{1}{3} Bh\end{align*}

You know that the height of the pyramid is the radius (\begin{align*}r\end{align*}

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

The volume of a sphere can be found using the formula

\begin{align*}V = \frac{4}{3} \pi r^3\end{align*}

Again, all you need to know is the radius of the sphere. You put the value in for \begin{align*}V\end{align*}

Let’s look at an example.

Find the volume of the sphere below.

First, substitute what you know into the volume formula.

\begin{align*}\begin{array}{rcl}
V &=& \frac{4}{3} \pi r^3\\
V &=& \frac{4}{3} \pi (6)^3
\end{array}\end{align*}

Next, use algebra to calculate the volume.

\begin{align*}\begin{array}{rcl}
V &=& \frac{4}{3} \pi (6)^3\\
V &=& \frac{4}{3} \pi (216)\\
V &=& 288 \pi\\
V &=& 904.78
\end{array}\end{align*}

The answer is 904.78.

The volume of the sphere is 904.78 m^{3}.

If you wanted to make your measurement more accurate, you could say that the volume is

.### Examples

#### Example 1

Earlier, you were given a problem about Maria’s liquid paperweight.

Maria has a paperweight with a diameter of 6 inches that is full of water, and she wants to find out how much liquid is inside.

First, find the radius of the sphere. Remember the radius is half the diameter.

\begin{align*}\begin{array}{rcl} r &=& \frac{d}{2}\\ r &=& \frac{6}{2}\\ r &=& 3 \end{array}\end{align*}

Next, substitute what you know into the volume formula.

\begin{align*}\begin{array}{rcl} V &=& \frac{4}{3} \pi r^3\\ V &=& \frac{4}{3} \pi (3)^3 \end{array}\end{align*}

Then, use algebra to calculate the volume.

\begin{align*}\begin{array}{rcl} V &=& \frac{4}{3} \pi (3)^3\\ V &=& \frac{4}{3} \pi (27)\\ V &=& 36 \pi\\ V &=& 113.10 \end{array}\end{align*}

The answer is 113.10.

The volume contained in Maria’s paperweight is 113.10 in^{3} or .

#### Example 2

Find the volume of the following sphere.

First, substitute what you know into the volume formula.

\begin{align*}\begin{array}{rcl} V &=& \frac{4}{3} \pi r^3\\ V &=& \frac{4}{3} \pi (8)^3 \end{array}\end{align*}

Next, use algebra to calculate the volume.

\begin{align*}\begin{array}{rcl} V &=& \frac{4}{3} \pi (8)^3\\ V &=& \frac{4}{3} \pi (512)\\ V &=& 682.67 \pi\\ V &=& 2144.66 \end{array}\end{align*}

The answer is 2144.66. The volume of the sphere is 2144.66 cm^{3}.

If you wanted to make your measurement more accurate, you could say that the volume is \begin{align*}682.67 \pi \ cm^3\end{align*}.

#### Example 3

Find the volume of a sphere with a radius of 4 inches.

First, substitute what you know into the volume formula.

\begin{align*}\begin{array}{rcl} V &=& \frac{4}{3} \pi r^3\\ V &=& \frac{4}{3} \pi (4)^3 \end{array}\end{align*}

Next, use algebra to calculate the volume.

\begin{align*}\begin{array}{rcl} V &=& \frac{4}{3} \pi (4)^3\\ V &=& \frac{4}{3} \pi (64)\\ V &=& 85.33 \pi\\ V &=& 268.08 \end{array}\end{align*}

The answer is 268.08.

The volume of the sphere is 268.08 in^{3}.

#### Example 4

Find the volume of a sphere with a radius of 5 ft.

First, substitute what you know into the volume formula.

\begin{align*}\begin{array}{rcl} V &=& \frac{4}{3} \pi r^3\\ V &=& \frac{4}{3} \pi (5)^3 \end{array}\end{align*}

Next, use algebra to calculate the volume.

\begin{align*}\begin{array}{rcl} V &=& \frac{4}{3} \pi (5)^3\\ V &=& \frac{4}{3} \pi (125)\\ V &=& 166.67 \pi\\ V &=& 523.60 \end{array}\end{align*}

The answer is 523.60.

The volume of the sphere is 523.60 ft^{3}.

#### Example 5

Find the volume of a sphere with a radius of 3.5 inches.

First, substitute what you know into the volume formula.

\begin{align*}\begin{array}{rcl} V &=& \frac{4}{3} \pi r^3\\ V &=& \frac{4}{3} \pi (3.5)^3 \end{array}\end{align*}

Next, use algebra to calculate the volume.

\begin{align*}\begin{array}{rcl} V &=& \frac{4}{3} \pi (3.5)^3\\ V &=& \frac{4}{3} \pi (42.875)\\ V &=& 57.17 \pi\\ V &=& 179.59 \end{array}\end{align*}

The answer is 179.59.

The volume of the sphere is 179.59 in^{3}.

### Review

Find the volume of each sphere. You may round to the nearest hundredth when necessary.

1. A sphere with a radius of 3 m.

2. A sphere with a radius of 2.5 m.

3. A sphere with a radius of 5 in.

4. A sphere with a radius of 6 in.

5. A sphere with a radius of 7 ft.

6. A sphere with a radius of 4.5 cm.

7. A sphere with a radius of 5.5 m.

8. A sphere with a radius of 13 mm.

9. A sphere with a diameter of 8 in.

10. A sphere with a diameter of 10 ft.

11. A sphere with a diameter of 3 m.

12. A sphere with a diameter of 13 m.

13. A sphere with a diameter of 22 ft.

Use what you have learned to solve each problem.

14. A sphere has a diameter of 12 feet. What is its volume?

15. Kelly has a perfume bottle in the shape of a sphere. The diameter of the bottle is 6 inches. How much perfume does Kelly have left if the bottle is only half full?

### Review (Answers)

To see the Review answers, open this PDF file and look for section 8.15.