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# Spheres

## Volume and surface area of 3D circles.

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Spheres

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.

### 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 πr2\begin{align*}\pi r^2\end{align*}, so the surface area of a sphere is 4πr2\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 r\begin{align*}r\end{align*} is the radius, then the surface area of a sphere is SA=4πr2\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 n\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 V=13Bh\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.

V(all pyramids)=V1+V2+V3++Vn=13(B1h+B2h+B3h++Bnh)=13h(B1+B2+B3++Bn)\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 4πr2\begin{align*}4 \pi r^2\end{align*}, you can substitute this quantity into the equation above.

=13h(4πr2)\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 h=r\begin{align*}h = r\end{align*}.

=43r(πr2)=43πr3\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 r\begin{align*}r\end{align*}, then the volume of a sphere is V=43πr3\begin{align*}V=\frac{4}{3} \pi r^3\end{align*}.

#### Example A

The circumference of a sphere is 26π feet\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 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*}

#### Example B

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

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

SA=4π(14)2=784π ft2\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.

V=43π63=43π(216)=288π\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.

#### Concept Problem Revisited

If the maximum circumference of a bowling ball is 27 inches, then the maximum radius would be 27=2πr\begin{align*}27=2 \pi r\end{align*}, or r=4.30\begin{align*}r=4.30\end{align*} inches. Therefore, the surface area would be 4π4.32232.35 in2\begin{align*}4 \pi 4.3^2 \approx 232.35 \ in^2\end{align*}, and the volume would be 43π 4.33333.04 in3\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.

### Vocabulary

A sphere is the set of all points, in three-dimensional space, which are equidistant from a point. The radius has one endpoint on the sphere and the other endpoint at the center of that sphere. The diameter of a sphere must contain the center. The great circle is a plane that contains the diameter. A hemisphere is half of a sphere.

Surface area is a two-dimensional measurement that is the total area of all surfaces that bound a solid. Volume is a three-dimensional measurement that is a measure of how much three-dimensional space a solid occupies.

### Guided Practice

1. Find the surface area of the figure below.

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

3. A sphere has a volume of 14137.167 ft3\begin{align*}14137.167 \ ft^3\end{align*}, what is the radius?

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.

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*}

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

100π255=4πr2=r2=r\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 r\begin{align*}r\end{align*}.

14137.16734π14137.1673375=43πr3=r3=r3\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 x3\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*}

### Practice

1. 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*})

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*}.
15. At the age of 81, Mr. Luke Roberts began collecting string. He had a ball of string 3 feet in diameter.
1. Find the volume of Mr. Roberts’ ball of string in cubic inches.
2. Assuming that each cubic inch weighs 0.03 pounds, find the weight of his ball of string.
3. 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.

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

### Vocabulary Language: English Spanish

diameter

A chord that passes through the center of the circle. The length of a diameter is two times the length of a radius.

The radius of a circle is the distance from the center of the circle to the edge of the circle.

Sphere

A sphere is a round, three-dimensional solid. All points on the surface of a sphere are equidistant from the center of the sphere.

Volume

Volume is the amount of space inside the bounds of a three-dimensional object.

Cavalieri's Principle

States that if two solids have the same height and the same cross-sectional area at every level, then they will have the same volume.