Unlocking the Secrets of Spheres: Volume, Surface Area, and Cavalieri's Principle

Using the formulas for the volume of a cylinder or volume of a cone, the formula to find the volume of a sphere can be obtained. Even cooler, in this lesson, the formula for finding the surface area of a sphere will be developed.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Circle
Volume
Cylinder
Volume of a cylinder
Cone
Volume of a cone

Discussion

Spheres

Discussion

Formula for the Volume of a Sphere

The formula for volume of a sphere has been developed and proven. It can now be put to use to solve a real-world problem!

Example

Earth's Radius

The volume of the Earth is approximately $259875159532$ cubic miles.

Assuming the Earth is spherical, what is the Earth's radius? Round the answer to one decimal place.

Hint

Solve the volume formula of the sphere for the radius and substitute the given volume.

Solution

Begin by writing the formula to find the volume of a sphere.

V = \frac{4}{3} π r^{3}

Since the volume is given and the radius is needed, solve the formula for

r .

V = \frac{4}{3} π r^{3}

Solve for

r

MultEqn

$LHS \cdot 3 = RHS \cdot 3$

3 V = 4 π r^{3}

DivEqn

$LHS / 4 π = RHS / 4 π$

\frac{3 V}{4 π} = r^{3}

RearrangeEqn

Rearrange equation

r^{3} = \frac{3 V}{4 π}

RadicalEqn

$3 LHS = 3 RHS$

3 r^{3} = 3 \frac{3 V}{4 π}

RootPowToNumber

$3 a^{3} = a$

r = 3 \frac{3 V}{4 π}

Next, to find the Earth's radius, substitute the given volume into the derived formula and simplify.

r = 3 \frac{3 V}{4 π}

Substitute

$V = 259875159532$

r = 3 \frac{3 ( 2 5 9 8 7 5 1 5 9 5 3 2 )}{4 π}

UseCalc

Use a calculator

r = 3958.755865 \dots

RoundDec

Round to $1$ decimal place(s)

r \approx 3958.8

Consequently, the Earth has a radius of approximately

3958.8

miles.

Knowing how to find the volume of different geometric solids helps to model some scenarios in the real world. Although geometric solids could not perfectly model real-world objects, these geometric solids can provide an approximation that yields valuable information about the scenario.

Example

Volume of a Pencil

Ramsha bought a standard pencil whose radius is $4$ millimeters and the length, not including the eraser, is $180$ millimeters. After a good sharpening, the tip turned into a $12$ millimeter high cone.

Assuming the eraser is half of a sphere, what is the volume of the pencil? Round the answer to two decimal places.

Hint

The pencil is made of a cone, a cylinder, and a hemisphere, all with the same radius. Therefore, the volume of the pencil equals the sum of the volumes of each solid.

Solution

The given pencil can be seen to consist of a cone, a cylinder, and half of a sphere — all with the same radius.

Consequently, the volume of the pencil equals the sum of the volumes of each of these solids.

Volume of a Cone	Volume of a Cylinder	Volume of a Hemisphere
$V_{1} = \frac{1}{3} π r^{2} h$	$V_{2} = π r^{2} h$	$V_{3} = \frac{2}{3} π r^{3}$

Volume of the Pencil's Tip

The tip of the pencil has a radius of

4

millimeters and a height of

12

millimeters. Substituting these values into the first formula will give the volume of the tip.

V_{1} = \frac{1}{3} π r^{2} h

SubstituteII

$r = 4$ , $h = 12$

V_{1} = \frac{1}{3} π (4)^{2} (12)

Simplify right-hand side

CalcPow

Calculate power

V_{1} = \frac{1}{3} π (16) (12)

Multiply

V_{1} = \frac{1 9 2}{3} π

CalcQuot

Calculate quotient

V_{1} = 64 π

The pencil's tip has a volume of

64 π

cubed millimeters.

Volume of the Pencil's Body

The body of the pencil has a radius of

4

millimeters. To find the height of this cylinder, subtract the height of the pencil's tip from the original length of the pencil.

180 mm - 12 mm = 168 mm

Next, substitute

r = 4

and

h = 168

into the formula for a cylinder's volume.

V_{2} = π r^{2} h

SubstituteII

$r = 4$ , $h = 168$

V_{2} = π (4)^{2} (168)

Simplify right-hand side

CalcPow

Calculate power

V_{2} = π (16) (168)

Multiply

V_{2} = 2688 π

It has been found that the pencil's body has a volume of

2688 π

cubed millimeters.

Volume of the Pencil's Eraser

The eraser is a hemisphere with a radius of

4 mm .

Therefore, to find its volume, substitute

r = 4

into the hemisphere volume formula.

V_{3} = \frac{2}{3} π r^{3}

Substitute

$r = 4$

V_{3} = \frac{2}{3} π (4)^{3}

Simplify right-hand side

CalcPow

Calculate power

V_{3} = \frac{2}{3} π (64)

Multiply

V_{3} = \frac{1 2 8}{3} π

Consequently, the pencil's eraser has a volume of

\frac{1 2 8}{3} π

cubed millimeters.

Pencil's Volume

Finally, the volume of the pencil is equal to the sum of the volume of its parts.

V_{P} = V_{1} + V_{2} + V_{3}

Substitute values

SubstituteValues

Substitute values

V_{P} = 64 π + 2688 π + \frac{1 2 8}{3} π

FactorOut

Factor out $π$

V_{P} = π (64 + 2688 + \frac{1 2 8}{3})

Rewrite

Rewrite $64$ as $\frac{1 9 2}{3}$

V_{P} = π (\frac{1 9 2}{3} + 2688 + \frac{1 2 8}{3})

Rewrite

Rewrite $2688$ as $\frac{8 0 6 4}{3}$

V_{P} = π (\frac{1 9 2}{3} + \frac{8 0 6 4}{3} + \frac{1 2 8}{3})

Simplify

AddFrac

Add fractions

V_{P} = π (\frac{1 9 2 + 8 0 6 4 + 1 2 8}{3})

AddTerms

Add terms

V_{P} = π (\frac{8 3 8 4}{3})

UseCalc

Use a calculator

V_{P} = 8779.704269 \dots

RoundDec

Round to $1$ decimal place(s)

V_{P} \approx 8779.7

In conclusion, the volume of the Ramsha's pencil is approximately

8779.7

cubed millimeters.

Discussion

Formula for the Surface Area of a Sphere

Consider the fact that baseballs commonly have a radius of $1.43$ inches. With this information, the volume of a baseball can be calculated, right? What about the surface area? How much leather is needed to make the lining of a baseball?

To answer these question, the following formula can be used.

Example

Surface Area of a Baseball

Find the amount of leather needed to make the lining of a baseball that has a radius of $1.43$ inches. Round the answer to one decimal place.

Hint

Use the formula to find the surface area of a sphere.

Solution

The amount of leather needed to make the lining of a baseball is the same as the surface area of the baseball. Therefore, the following formula can be used.

S = 4 π r^{2}

Next, substitute

r = 1.43

into the previous formula.

S = 4 π r^{2}

Substitute

r

for

1.43

and evaluate

Substitute

$r = 1.43$

S = 4 π (1.43)^{2}

CalcPow

Calculate power

S = 4 π (2.0449)

UseCalc

Use a calculator

S = 25.696971 \dots

RoundDec

Round to $1$ decimal place(s)

S = 25.7

Consequently, to make the lining of a baseball, about

25.7

square inches of leather are needed.

Covering of the Baseball

Notice that the surface area of a sphere with radius $r$ is four times the area of a circle with the same radius as the sphere. This relationship can be roughly seen in the covering of a baseball.

Pop Quiz

Finding the Volume or Surface Area of Different Spheres

In the following applet, calculate either the volume or the surface area of the given sphere and round the answer to two decimal places.

Illustration

Napkin Ring Problem

In talking about spheres, there is a fascinating case called the Napkin Ring Problem. Consider two solids that have spherical shapes, for example, a soccer ball and the planet Earth. Each is represented in the diagram, along with its respective diameter.

Imagine each of these objects is cut with a horizontal plane

9

centimeters above the center of the sphere. Then, imagine making a second cut, this time

9

centimeters below the center of each sphere. Both the soccer ball and the Earth will be left with their own respective solid with a circular base. Additionally, the heights of each solid are the same.

Next, bore a cylindrical hole through the center of each solid so that the radius of each cylinder is the radius of the corresponding circular base. The resulting solids are called napkin rings. Note that both cylinders have the same height.

The interesting fact about these two napkin rings is that, since they have the same height, they have exactly the same volume.

Why

Showing the Volumes are Equal

To show that the two napkin rings have the same volume, the Cavalieri's Principle will come into action. Start by considering a cross-section of each napkin ring at the same height.

The area of each cross-section is equal to the area of the outer circle minus the area of the inner circle. For a moment, focus all the attention on only one of the cross-sections. Let

h

be the height of the cylinder,

r

be the radius of the sphere, and

y

be the height above the center at which the cross-section was made.

From the diagram, the radius of the inner circle is $A B,$ which is equal to $O D .$ Since $△ O D E$ is a right triangle, by the Pythagorean Theorem, $O D$ can be found.

$O D = r^{2} - (\frac{h}{2})^{2} = A B$

Also, $A C$ is the radius of the outer circle and applying the Pythagorean Theorem to the right triangle $A O C,$ an expression for it can be deducted.

$A C = r^{2} - y^{2}$

Having the two radii, the area of the cross-section can be found.

A_{Cross - Section} = π A C^{2} - π A B^{2}

Substitute values and simplify

SubstituteII

$A C = r^{2} - y^{2}$ , $A B = r^{2} - (\frac{h}{2})^{2}$

A_{Cross - Section} = π (r^{2} - y^{2})^{2} - π ⎝ ⎜ ⎛ r^{2} - (\frac{h}{2})^{2} ⎠ ⎟ ⎞^{2}

PowSqrt

$(a)^{2} = a$

A_{Cross - Section} = π (r^{2} - y^{2}) - π (r^{2} - (\frac{h}{2})^{2})

FactorOut

Factor out $π$

A_{Cross - Section} = π (r^{2} - y^{2} - (r^{2} - (\frac{h}{2})^{2}))

Distr

Distribute $- 1$

A_{Cross - Section} = π (r^{2} - y^{2} - r^{2} + (\frac{h}{2})^{2})

SubTerms

Subtract terms

A_{Cross - Section} = π (- y^{2} + (\frac{h}{2})^{2})

CommutativePropAdd

\CommutativePropAdd

A_{Cross - Section} = π ((\frac{h}{2})^{2} - y^{2})

As might be seen, the area of the cross-section does not depend on the radius of the sphere, it depends only on the height of the napkin ring and the height at which the cross-section was made. Therefore, finding the area of the second cross-section will produce the same expression.

Consequently, since both napkin rings have the same height and the same cross-sectional area at every altitude, by Cavalieri's Principle, the two napkin rings have the same volume.

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Catch-Up and Review

Hint

Solution

Hint

Solution

Volume of the Pencil's Tip

Volume of the Pencil's Body

Volume of the Pencil's Eraser

Pencil's Volume

Hint

Solution

Covering of the Baseball

Why