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# 11.6: Surface Area and Volume of Spheres

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Find the surface area of a sphere.
• Find the volume of a sphere.

## Review Queue

1. List three spheres you would see in real life.
2. Find the area of a circle with a 6 cm radius.
3. Find the volume of a cylinder with the circle from #2 as the base and a height of 5 cm.

Know What? A regulation bowling ball is a sphere with a circumference of 27 inches. Find the radius of a bowling ball, its surface area and volume. You may assume the bowling ball does not have any finger holes. Round your answers to the nearest hundredth.

## Defining a Sphere

A sphere is the last of the three-dimensional shapes that we will find the surface area and volume of. Think of a sphere as a three-dimensional circle.

Sphere: The set of all points, in three-dimensional space, which are equidistant from a point.

The radius has an endpoint on the sphere and the other endpoint is the center.

The diameter must contain the center.

Great Circle: A cross section of a sphere that contains the diameter.

A great circle is the largest circle cross section in a sphere. The circumference of a sphere is the circumference of a great circle.

Every great circle divides a sphere into two congruent hemispheres.

Example 1: The circumference of a sphere is 26π feet\begin{align*}26 \pi \ feet\end{align*}. What is the radius of the sphere?

Solution: The circumference is referring to the circumference of a great circle. Use C=2πr\begin{align*}C = 2 \pi r\end{align*}.

2πrr=26π=13 ft.\begin{align*}2 \pi r &= 26\pi\\ r &= 13 \ ft.\end{align*}

## Surface Area of a Sphere

The best way to understand the surface area of a sphere is to watch the link by Russell Knightley, http://www.rkm.com.au/ANIMATIONS/animation-Sphere-Surface-Area-Derivation.html.

Surface Area of a Sphere: SA=4πr2\begin{align*}SA=4 \pi r^2\end{align*}.

Example 2: Find the surface area of a sphere with a radius of 14 feet.

Solution:

SA=4π(14)2=784π ft2\begin{align*}SA &= 4\pi (14)^2\\ &= 784 \pi \ ft^2\end{align*}

Example 3: Find the surface area of the figure below.

Solution: Be careful when finding the surface area of a hemisphere because you need to include the area of the base.

SA=πr2+124πr2=π(62)+2π(62)=36π+72π=108π cm2\begin{align*}SA &= \pi r^2+\frac{1}{2} 4 \pi r^2\\ &= \pi (6^2 )+2 \pi (6^2)\\ &= 36 \pi +72 \pi =108 \pi \ cm^2\end{align*}

Example 4: The surface area of a sphere is 100π in2\begin{align*}100 \pi \ in^2\end{align*}. What is the radius?

Solution:

SA100π255=4πr2=4πr2=r2=r\begin{align*}SA &= 4 \pi r^2\\ 100 \pi &= 4 \pi r^2\\ 25 &= r^2\\ 5 &= r\end{align*}

Example 5: Find the surface area of the following solid.

Solution: This solid is a cylinder with a hemisphere on top. It is one solid, so do not include the bottom of the hemisphere or the top of the cylinder.

\begin{align*}SA &=LA_{cylinder}+LA_{hemisphere}+A_{base \ circle}\\ &= \pi rh+\frac{1}{2} 4 \pi r^2+\pi r^2\\ &= \pi (6)(13)+2 \pi 6^2+\pi 6^2\\ &= 78 \pi +72 \pi +36 \pi\\ &= 186 \pi \ in^2 \qquad \qquad LA'' \ \text{stands for} \ lateral \ area.\end{align*}

## Volume of a Sphere

To see an animation of the volume of a sphere, see http://www.rkm.com.au/ANIMATIONS/animation-Sphere-Volume-Derivation.html by Russell Knightley.

Volume of a Sphere: \begin{align*}V=\frac{4}{3} \pi r^3\end{align*}.

Example 6: Find the volume of a sphere with a radius of 9 m.

Solution:

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

Example 7: A sphere has a volume of \begin{align*}14137.167 \ ft^3\end{align*}, what is the radius?

Solution:

\begin{align*}V &= \frac{4}{3} \pi r^3\\ 14137.167 &= \frac{4}{3} \pi r^3\\ \frac{3}{4 \pi} \cdot 14137.167 &= r^3\\ 3375 &= r^3\end{align*}

At this point, you will need to take the cubed root of 3375. Ask your teacher how to do this on your calculator.

\begin{align*}\sqrt[3]{3375}=15=r\end{align*}

Example 8: Find the volume of the following solid.

Solution:

\begin{align*}V_{cylinder} &= \pi 6^2 (13)=78 \pi\\ V_{hemisphere} &= \frac{1}{2} \left(\frac{4}{3} \pi 6^3\right)=36 \pi\\ V_{total} &= 78 \pi+36 \pi =114 \pi \ in^3\end{align*}

Know What? Revisited The radius would be \begin{align*}27=2 \pi r\end{align*}, or \begin{align*}r=4.30 \ inches\end{align*}. The surface area would be \begin{align*}4 \pi 4.3^2 \approx 232.35 \ in^2\end{align*}, and the volume would be \begin{align*}\frac{4}{3} \pi 4.3^3 \approx 333.04 \ in^3\end{align*}.

## Review Questions

• Questions 1-3 look at the definition of a sphere.
• Questions 4-17 are similar to Examples 1, 2, 4, 6 and 7.
• Questions 18-21 are similar to Example 3 and 5.
• Questions 22-25 are similar to Example 8.
• Question 26 is a challenge.
• Questions 27-29 are similar to Example 8.
• Question 30 analyzes the formula for the surface area of a sphere.
1. Are there any cross-sections of a sphere that are not a circle? Explain your answer.
2. List all the parts of a sphere that are the same as a circle.
3. List any parts of a sphere that a circle does not have.

Find the surface area and volume of a sphere with: (Leave your answer in terms of \begin{align*}\pi\end{align*})

1. a radius of 8 in.
2. a diameter of 18 cm.
3. a radius of 20 ft.
4. a diameter of 4 m.
5. a radius of 15 ft.
6. a diameter of 32 in.
7. a circumference of \begin{align*}26 \pi \ cm\end{align*}.
8. a circumference of \begin{align*}50 \pi \ yds\end{align*}.
9. The surface area of a sphere is \begin{align*}121 \pi \ in^2\end{align*}. What is the radius?
10. The volume of a sphere is \begin{align*}47916 \pi \ m^3\end{align*}. What is the radius?
11. The surface area of a sphere is \begin{align*}4 \pi \ ft^2\end{align*}. What is the volume?
12. The volume of a sphere is \begin{align*}36 \pi \ mi^3\end{align*}. What is the surface area?
13. Find the radius of the sphere that has a volume of \begin{align*}335 \ cm^3\end{align*}. Round your answer to the nearest hundredth.
14. Find the radius of the sphere that has a surface area \begin{align*}225 \pi \ ft^2\end{align*}.

Find the surface area of the following shapes. Leave your answers in terms of \begin{align*}\pi\end{align*}.

1. You may assume the bottom is open.

Find the volume of the following shapes. Round your answers to the nearest hundredth.

1. A sphere has a radius of 5 cm. A right cylinder has the same radius and volume. Find the height of the cylinder.

Tennis balls with a 3 inch diameter are sold in cans of three. The can is a cylinder. Round your answers to the nearest hundredth.

1. What is the volume of one tennis ball?
2. What is the volume of the cylinder?
3. Assume the balls touch the can on the sides, top and bottom. What is the volume of the space not occupied by the tennis balls?
4. How does the formula of the surface area of a sphere relate to the area of a circle?

## Review Queue Answers

1. Answers will vary. Possibilities are any type of ball, certain lights, or the 76/Unical orb.
2. \begin{align*}36 \pi\end{align*}
3. \begin{align*}180 \pi\end{align*}

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Subjects:
8 , 9 , 10
Date Created:
Feb 22, 2012