Geometry (B): READ: Surface Area and Volume of a Sphere Reading

READ: Surface Area and Volume of a Sphere Reading

Spheres

Learning Objectives

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

Introduction

A sphere is a three-dimensional figure that has the shape of a ball.

Spheres can be characterized in three ways.

A sphere is the set of all points that lie a fixed distance $r$ from a single center point $O$ .
A sphere is the surface that results when a circle is rotated about any of its diameters.

A sphere results when you construct a polyhedron with an infinite number of faces that are infinitely small. To see why this is true, recall regular polyhedra.

As the number of faces on the figure increases, each face gets smaller in area and the figure comes more to resemble a sphere. When you imagine figure with an infinite number of faces, it would look like (and be!) a sphere.

Parts of a Sphere

As described above, a sphere is the surface that is the set of all points a fixed distance from a center point $O$ . Terminology for spheres is similar for that of circles.

The distance from $O$ to the surface of the sphere is $r$ , the radius.

The diameter, $d$ , of a sphere is the length of the segment connecting any two points on the sphere’s surface and passing through $O$ . Note that you can find a diameter (actually an infinite number of diameters) on any plane within the sphere. Two diameters are shown in each sphere below.

A chord for a sphere is similar to the chord of a circle except that it exists in three dimensions. Keep in mind that a diameter is a kind of chord—a special chord that intersects the center of the circle or sphere.

A secant is a line, ray, or line segment that intersects a circle or sphere in two places and extends outside of the circle or sphere.

A tangent intersects the circle or sphere at only one point.

In a circle, a tangent is perpendicular to the radius that meets the point where the tangent intersects with the circle. The same thing is true for the sphere. All tangents are perpendicular to the radii that intersect with them.

Finally, a sphere can be sliced by an infinite number of different planes. Some planes include point $O$ , the center of the sphere. Other points do not include the center.

Surface Area of a Sphere

You can infer the formula for the surface area of a sphere by taking measurements of spheres and cylinders. Here we show a sphere with a radius of $3$ and a right cylinder with both a radius and a height of $3$ and express the area in terms of $\pi$ .

Now try a larger pair, expressing the surface area in decimal form.

Look at a third pair.

Is it a coincidence that a sphere and a cylinder whose radius and height are equal to the radius of the sphere have the exact same surface area? Not at all! In fact, the ancient Greeks used a method that showed that the following formula can be used to find the surface area of any sphere (or any cylinder in which $r = h$ ).

The Surface Area of a Sphere is given by:

$A=4 \pi r^2$

Example 1

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

Use the formula.

$A &= 4 \pi r^2\\ &= 4 \pi (14)^2\\ &= 4 \pi (196)\\ &= 784 \pi\\ &= 2461.76 \ \text{square feet}$

Example 2

Find the surface area of the following figure in terms of $\pi$ .

The figure is made of one half sphere or hemisphere, and one cylinder without its top.

$A \text{(half sphere)} &= \frac{1}{2} A\text{(sphere)}\\ &= \frac{1}{2} \cdot 4\pi r^2\\ &= 2\pi (576)\\ &= 1152\pi \ \text{square cm}$

Now find the area of the cylinder without its top.

$A \text{(topless cylinder)} &= A\text{(cylinder)} - A\text{(top)}\\ &= 2(\pi r^2) + 2\pi r h - \pi r^2\\ &= \pi r^2 + 2 \pi r h\\ &= \pi (576) + 2\pi (24)(50)\\ &= 2976\pi \ \text{square cm}$

Thus, the total surface area is $1152 \pi + 2976 \pi = 4128 \pi \ \text{square cm}$

Volume of a Sphere

A sphere can be thought of as a regular polyhedron with an infinite number of congruent polygon faces. A series polyhedra with an increasing number of faces is shown.

As $n$ , 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 $O$ , the center of the sphere. This is shown below.

Each of the pyramids that make up the sphere would be congruent to the pyramid shown. The volume of this pyramid is given by:

$V \text{(each pyramid)} = \frac{1}{3}Bh$

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

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

Now, it should be obvious that the sum of all of the bases of the pyramids is simply the surface area of the sphere. Since you know that the surface area of the sphere is $4 \pi r^2$ , you can substitute this quantity into the equation above.

$V \text{(all pyramids)} &= \frac{1}{3} h \left( B_1 + B_2 + B_3 + \ldots + B_n \right)\\ &= \frac{1}{3} h (4 \pi r^2)$

Finally, as $n$ increases and the surface of the figure becomes more “rounded,” $h$ , the height of each pyramid becomes equal to $r$ , the radius of the sphere. So we can substitute $r$ for $h$ . This gives:

$V \text{(all pyramids)} &= \frac{1}{3} r (4 \pi r^2)\\ &= \frac{4}{3} \pi r^3$

We can write this as a formula.

Volume of a Sphere

Given a sphere that has radius $r$

$V= \frac{4}{3} \pi r^3$

Example 3

Find the volume of a sphere with radius $6.25\;\mathrm{m}$ .

Use the postulate above.

$V \text{(sphere)} &= \frac{4}{3} \pi r^3\\ &= \frac{4}{3} (3.14) (6.25)(6.25)(6.25)\\ &= 1022.14 \ \text{cubic m}$

Example 4

A sphere has a volume of $(85\pi)1/3$ . Find its diameter.

Use the postulate above. Convert $(85) 1/3$ to an improper fraction, $256/3.$

$V \text{(sphere)} &= \frac{4}{3} \pi r^3\\ \frac{256}{3}\pi &= \frac{4}{3} \pi r^3\\ \left(\frac{3}{4}\right)\left(\frac{256}{3}\right)\pi &= \pi r^3\\ \frac{192}{3} &= \pi r^3\\ 64 &= r^3\\ 4 &= r$

Since $r = 4$ , the diameter is $8\;\mathrm{units}$ .

The following questions are for your own review. The answers are listed below for you to check your work and understanding.

Review Questions

Find the radius of the sphere that has a volume of $335 \;\mathrm{cm}^3$ .

Determine the surface area and volume of the following shapes:

The radius of a sphere is $4$ . Find its volume and total surface area.
A sphere has a radius of $5$ . A right cylinder, having the same radius has the same volume. Find the height and total surface area of the cylinder.

In problems 6 and 7 find the missing dimension.

$& \mathrm{Sphere:}\ \mathrm{volume} = 296 \;\mathrm{cm}^3 \\ & \mathrm{Diameter} = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}$
$& \mathrm{Sphere:}\ \text{surface area is}\ 179 \;\mathrm{in}^2.\\ & \mathrm{Radius} = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}$
Tennis balls with a diameter of $3.5\;\mathrm{inches}$ are sold in cans of three. The can is a cylinder. 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?
A sphere has surface area of $36 \pi \;\mathrm{in}^2$ . Find its volume.
A giant scoop, operated by a crane, is in the shape of a hemisphere of $\mathrm{radius} = 21\;\mathrm{inches}$ . The scoop is filled with melted hot steel. When the steel is poured into a cylindrical storage tank that has a radius of $28\;\mathrm{inches}$ , the melted steel will rise to a height of how many inches?

Review Answers

$\;\mathrm{Radius} = 4.39 \;\mathrm{cm}$
$& \mathrm{Surface \ area} = 706.86 \;\mathrm{cm}^2 \\ & \mathrm{Volume} = 1767.15 \;\mathrm{cm}^3$
$& \mathrm{Surface \ area} = \text{surface \ area \ of \ hemisphere} + \;\mathrm{surface \ area \ of \ cone} = 678.58 \;\mathrm{in}^2\\ & \mathrm{Volume} = 2544.69 \;\mathrm{in}^3$
$& \mathrm{Volume} = 268.08 \;\mathrm{units}^3 \\ & \mathrm{Surface \ area} = 201.06 \;\mathrm{units}^2$
$\;\mathrm{Height} = 20/3 \;\mathrm{units \ total \ surface \ area} = 366.52 \;\mathrm{units}^2$
$\;\mathrm{Diameter} = 8.27\;\mathrm{inches}$
$\;\mathrm{Radius} = 3.77\;\mathrm{inches}$
$& \mathrm{Volume \ of \ cylinder} = 32.16 \pi \;\mathrm{in}^3\;\mathrm{volume\ of\ tennis\ balls} = 21.44 \pi \;\mathrm{in}^3\\ & \mathrm{Volume\ of\ space\ not\ occupied\ by\ tennis\ balls} = 33.67 \;\mathrm{in}^3$
$\;\mathrm{Volume} = 113.10 \;\mathrm{in}^3$
Height of molten steel in cylinder will be $7.88\;\mathrm{inches}$

Last modified: Wednesday, 7 July 2010, 11:18 AM

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