What if you were asked to geometrically consider a bowling ball? A regulation bowling ball is a sphere that weighs between 12 and 16 pounds. The maximum circumference of a bowling ball is 27 inches. Using this number, find the radius, surface area, and volume of the bowling ball. You may assume the bowling ball does not have any finger holes. Round your answers to the nearest hundredth. After completing this Concept, you'll be able to answer questions like these.

### Watch This

CK-12 Foundation: Chapter11SpheresA

Watch this video to learn more about the surface area of spheres.

Follow this link to watch a video about the volume of a sphere.

### Guidance

A **sphere** is the set of all points, in three-dimensional space, which are equidistant from a point. You can think of a sphere as a three-dimensional circle. A sphere has a **center**, **radius** and **diameter**, just like a circle. The **radius** has an endpoint on the sphere and the other is on the center. The **diameter** must contain the center. If it does not, it is a **chord**. The **great circle** is a plane that contains the diameter. It is the largest circle cross section in a sphere. There are infinitely many great circles. The circumference of a sphere is the circumference of a great circle. Every great circle divides a sphere into two congruent hemispheres, or two half spheres. Also like a circle, spheres can have tangent lines and secants. These are defined just like they are in a circle.

##### Surface Area

Surface area is a two-dimensional measurement that is the total area of all surfaces that bound a solid. The basic unit of area is the square unit. One way to find the formula for the surface area of a sphere is to look at a baseball. We can best *approximate* the cover of the baseball by 4 circles. The area of a circle is \begin{align*}\pi r^2\end{align*}, so the surface area of a sphere is \begin{align*}4 \pi r^2\end{align*}. While the covers of a baseball are not four perfect circles, they are stretched and skewed.

Another way to show 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. It is a great visual interpretation of the formula.

**Surface Area of a Sphere:** If \begin{align*}r\end{align*} is the radius, then the surface area of a sphere is \begin{align*}SA=4 \pi r^2\end{align*}.

##### Volume

To find the volume of any solid you must figure out how much space it occupies. The basic unit of volume is the cubic unit. A sphere can be thought of as a regular polyhedron with an infinite number of congruent regular polygon faces. As \begin{align*}n\end{align*}, the number of faces increases to an infinite number, the figure approaches becoming a sphere. So a sphere can be thought of as a polyhedron with an infinite number faces. Each of those faces is the base of a pyramid whose vertex is located at the center of the sphere. Each of the pyramids that make up the sphere would be congruent to the pyramid shown. The volume of this pyramid is given by \begin{align*}V=\frac{1}{3} Bh\end{align*}.

To find the volume of the sphere, you need to add up the volumes of an infinite number of infinitely small pyramids.

\begin{align*}V(all \ pyramids)& = V_1+V_2+V_3+ \ldots +V_n\\ & = \frac{1}{3} \left (B_1 h+B_2 h+B_3 h+ \ldots +B_n h \right )\\ & = \frac{1}{3} h(B_1+B_2+B_3+ \ldots +B_n )\end{align*}

The sum of all of the bases of the pyramids is the surface area of the sphere. Since you know that the surface area of the sphere is \begin{align*}4 \pi r^2\end{align*}, you can substitute this quantity into the equation above.

\begin{align*}=\frac{1}{3} h \left (4 \pi r^2 \right )\end{align*}

In the picture above, we can see that the height of each pyramid is the radius, so \begin{align*}h = r\end{align*}.

\begin{align*}& = \frac{4}{3} r( \pi r^2)\\ & = \frac{4}{3} \pi r^3\end{align*}

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. It is a slightly different interpretation than our derivation.

**Volume of a Sphere:** If a sphere has a radius \begin{align*}r\end{align*}, then the volume of a sphere is \begin{align*}V=\frac{4}{3} \pi r^3\end{align*}.

#### Example A

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

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

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

#### Example B

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

Use the formula, \begin{align*}r = 14 \ ft\end{align*}.

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

#### Example C

Find the volume of a sphere with a radius of 6 m.

Use the formula for the volume of a sphere.

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

Watch this video for help with the Examples above.

CK-12 Foundation: Chapter11SpheresB

#### Concept Problem Revisited

If the maximum circumference of a bowling ball is 27 inches, then the maximum radius would be \begin{align*}27=2 \pi r\end{align*}, or \begin{align*}r=4.30\end{align*} inches. Therefore, 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*}. The weight of the bowling ball refers to its density, how heavy something is. The volume of the ball tells us how much it can hold.

### Guided Practice

1. Find the surface area of the figure below.

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

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

**Answers:**

1. This is a hemisphere. Be careful when finding the surface area of a hemisphere because you need to include the area of the base. If the question asked for the *lateral surface area*, then your answer would *not* include the bottom.

\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*}

2. Plug in what you know to the formula and solve for \begin{align*}r\end{align*}.

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

3. Because we have a decimal, our radius might be an approximation. Plug in what you know to the formula and solve for \begin{align*}r\end{align*}.

\begin{align*}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. Depending on your calculator, you can use the \begin{align*}\sqrt[3]{x}\end{align*} function or \begin{align*}\land \left ( \frac{1}{3} \right )\end{align*}. The cubed root is the inverse of cubing a number, just like the square root is the inverse, or how you undo, the square of a number.

\begin{align*}\sqrt[3]{3375} = 15=r \qquad \text{The radius is} \ 15 \ ft.\end{align*}

### Explore More

- Are there any cross-sections of a sphere that are not a circle? Explain your answer.

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

- a radius of 8 in.
- a diameter of 18 cm.
- a radius of 20 ft.
- a diameter of 4 m.
- a radius of 15 ft.
- a diameter of 32 in.
- a circumference of \begin{align*}26 \pi \ cm\end{align*}.
- a circumference of \begin{align*}50 \pi \ yds\end{align*}.
- The surface area of a sphere is \begin{align*}121 \pi \ in^2\end{align*}. What is the radius?
- The volume of a sphere is \begin{align*}47916 \pi \ m^3\end{align*}. What is the radius?
- The surface area of a sphere is \begin{align*}4 \pi \ ft^2\end{align*}. What is the volume?
- The volume of a sphere is \begin{align*}36 \pi \ mi^3\end{align*}. What is the surface area?
- 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.
- Find the radius of the sphere that has a surface area \begin{align*}225 \pi \ ft^2\end{align*}.
- At the age of 81, Mr. Luke Roberts began collecting string. He had a ball of string 3 feet in diameter.
- Find the volume of Mr. Roberts’ ball of string in cubic inches.
- Assuming that each cubic inch weighs 0.03 pounds, find the weight of his ball of string.
- To the nearest inch, how big (diameter) would a 1 ton ball of string be? \begin{align*}(1 \ ton = 2000 \ lbs)\end{align*}

For problems 17-19, use the fact that the earth’s radius is approximately 4,000 miles.

- Find the length of the equator.
- Find the surface area of earth, rounding your answer to the nearest million square miles.
- Find the volume of the earth, rounding your answer to the nearest billion cubic miles.

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 11.7.