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{{ printedBook.courseTrack.name }} {{ printedBook.name }} 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.**

Consider a hemisphere and a cone that have the same radius $r,$ and the height of the cone is also $r.$ Use the cone to fill up the hemisphere with water.
**full sphere** and the volume of the given cone?

What is the relationship between the volume of the

A sphere is the set of all points in the space that are equidistant from a given point called the *center* of the sphere.
*radius* of the sphere.

The distance from the center to any point on the sphere is called the

In the exploration, it was seen that it took two cones with radius $r$ and height $r$ to fill up a hemisphere of radius $r.$ Therefore, the volume of the hemisphere is twice the volume of the cone.
$V_{Hemisphere}=2⋅V_{Cone} $
Since the hemisphere is half of a sphere, the volume of a sphere is twice the volume of the hemisphere. Therefore, the volume of a sphere with radius $r$ is four times the volume of a cone with radius and height $r.$
Now, the formula for the volume of a sphere will be rigorously proven.

Additionally, the volume of the sphere can be connected to the volume of a cylinder. To do so, remember that the volume of a cone is one-third the volume of a cylinder with same radius and height.
$V_{Cone}=31 ⋅V_{Cylinder} $
Consequently, the volume of a sphere is four-thirds the volume of a cylinder with radius and height $r.$

Finally, the volume of a cylinder with height $r$ is given by the formula $πr_{3}.$ Substituting this expression into the last equation, an explicit formula to calculate the volume of a sphere is obtained.

$V_{Sphere}=34 πr_{3}$

The volume of a sphere with radius $r$ is four-thirds the product of pi and the radius cubed.

Cavalieri's Principle will be used to show that the formula for the volume of a sphere holds. For this purpose, consider a hemisphere and a right cylinder with a cone removed from its interior, each with the same radius and height.

Now, consider a plane that cuts the solids at a height $x$ and is parallel to the bases of the solids.
The area of each cross-section will be calculated one at a time.

Draw a right triangle with height $x,$ base $y,$ and hypotenuse $r.$ Here, $x$ is the distance between the center of the base of the hemisphere and the center of the cross sectional circle, $y$ is the radius of the cross sectional circle, and $r$ is the radius of the hemisphere.

Using the Pythagorean Theorem, an expression for $y$ can be found.$r_{2}=x_{2}+y_{2}$

Solve for $y$

SubEqn

$LHS−x_{2}=RHS−x_{2}$

$r_{2}−x_{2}=y_{2}$

RearrangeEqn

Rearrange equation

$y_{2}=r_{2}−x_{2}$

SqrtEqn

$LHS =RHS $

$y=±r_{2}−x_{2} $

$A_{H}=πy_{2}$

Substitute

$y=r_{2}−x_{2} $

$A_{H}=π⋅(r_{2}−x_{2} )_{2}$

PowSqrt

$(a )_{2}=a$

$A_{H}=π(r_{2}−x_{2})$

Now that the radii of the circles are known, the area $A_{C}$ of the cross-section can be calculated. It is the difference between the area $A_{G}$ of the greater circle and the area $A_{S}$ of the smaller circle.

$A_{C}=A_{G}−A_{S}$

SubstituteII

$A_{G}=πr_{2}$, $A_{S}=πx_{2}$

$A_{C}=πr_{2}−πx_{2}$

FactorOut

Factor out $π$

$A_{C}=π(r_{2}−x_{2})$

It can be stated that both solids have the same cross-sectional area at every altitude. $A_{H}=π(r_{2}−x_{2})=A_{C} $ Moreover, they have the same height. By Cavalieri's Principle, the hemisphere and the cylinder with a cone removed from its interior have the same volume.

If the volume of the cone is subtracted from the volume of the cylinder, the volume of the hemisphere can be found.$V_{hemisphere}=V_{cylinder}−V_{cone}$

SubstituteValues

Substitute values

$V_{hemisphere}=πr_{3}−31 πr_{3}$

Simplify right-hand side

NumberToFrac

$a=33⋅a $

$V_{hemisphere}=33πr_{3} −31 πr_{3}$

MoveRightFacToNum

$ca ⋅b=ca⋅b $

$V_{hemisphere}=33πr_{3} −3πr_{3} $

SubFrac

Subtract fractions

$V_{hemisphere}=32πr_{3} $

$V_{sphere} =2⋅V_{hemisphere}=2⋅32 πr_{3}=34 πr_{3} $

Last weekend, Tearrik installed a spherical water tank for his house.

If the tank has a radius of $2.15$ feet, what is the maximum amount of water it can hold? Round the answer to two decimal places.

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Find the volume of the tank.

Finding the largest amount of water the tank can hold is the same as the finding its volume. Since it has a spherical shape, its volume is four-thirds of pi multiplied by the radius cubed.
$V=34 πr_{3} $
It is given that the radius of the tank is $2.15$ feet. To find the volume, all that is needed is to substitute this value into the volume formula and solve.
Tearrik's tank can hold about $41.63$ cubic feet of water.

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.

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

Begin by writing the formula to find the volume of a sphere.
$V=34 πr_{3} $
Since the volume is given and the radius is needed, solve the formula for $r.$
Next, to find the Earth's radius, substitute the given volume into the derived formula and simplify.
Consequently, the Earth has a radius of approximately $3958.8$ miles.

$V=34 πr_{3}$

Solve for $r$

MultEqn

$LHS⋅3=RHS⋅3$

$3V=4πr_{3}$

DivEqn

$LHS/4π=RHS/4π$

$4π3V =r_{3}$

RearrangeEqn

Rearrange equation

$r_{3}=4π3V $

RadicalEqn

$3LHS =3RHS $

$3r_{3} =34π3V $

RootPowToNumber

$3a_{3} =a$

$r=34π3V $

$r=34π3V $

Substitute

$V=259875159532$

$r=34π3(259875159532) $

UseCalc

Use a calculator

$r=3958.755865…$

RoundDec

Round to ${\textstyle 1 \, \ifnumequal{1}{1}{\text{decimal}}{\text{decimals}}}$

$r≈3958.8$

Davontay needs to fill a cylindrical bucket in a school competition by throwing water balloons from a distance of $20$ feet. Each balloon has a spherical shape with a radius of $2.5$ inches. The radius of the bucket is $10$ inches, with a height of $15$ inches.
### Hint

### Solution

Using this information, to find the volume of the bucket, $r=10$ and $h=15$ can be substituted into the formula for the volume of a cylinder.
Next, substitute $r=2.5$ into the formula for the volume of a sphere to find the volume of each balloon.
Now, to determine how many water balloons will fill up the bucket, divide the volume of the bucket by the volume of a balloon.
$V_{S}V_{C} =362.5 πin_{3}1500πin_{3} $
Simplifying the right-hand side will give the number of balloons needed.
By multiplying the last equation by $V_{S}$ the equation $V_{C}=72V_{S}$ is obtained. This indicates that $72$ water balloons will fill up the bucket. Remember, what has been determined is the *minimum* number of balloons needed, but there could be a need for more if, for instance, Davontay misses some shots.

What is the minimum number of balloons needed to fill up the bucket?

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Find the volume of the bucket and the volume of each balloon. Then, divide the volume of the bucket by the volume of a balloon.

To determine the minimum number of balloons needed to fill up the bucket, first solve for the volume of the bucket and the volume of each balloon. To do so, recall the formulas to find the volume of a cylinder and a sphere.

Volume of a Cylinder | Volume of a Sphere |
---|---|

$V_{C}=πr_{2}h$ | $V_{S}=34 πr_{3}$ |

$V_{S}V_{C} =362.5 πin_{3}1500πin_{3} $

Simplify right-hand side

CrossCommonFac

Cross out common factors

$V_{S}V_{C} =362.5 πin_{3}1500πin_{3} $

CancelCommonFac

Cancel out common factors

$V_{S}V_{C} =362.5 1500 $

DivByFracD

$b/ca =ba⋅c $

$V_{S}V_{C} =62.53⋅1500 $

Multiply

Multiply

$V_{S}V_{C} =62.54500 $

CalcQuot

Calculate quotient

$V_{S}V_{C} =72$

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.

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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.

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

### 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.
The pencil's tip has a volume of $64π$ 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.
$180mm−12mm=168mm $
Next, substitute $r=4$ and $h=168$ into the formula for a cylinder's volume.
It has been found that the pencil's body has a volume of $2688π$ cubed millimeters. ### Volume of the Pencil's Eraser

The eraser is a hemisphere with a radius of $4mm.$ Therefore, to find its volume, substitute $r=4$ into the hemisphere volume formula.
Consequently, the pencil's eraser has a volume of $3128 π$ cubed millimeters. ### Pencil's Volume

Finally, the volume of the pencil is equal to the sum of the volume of its parts.
In conclusion, the volume of the Ramsha's pencil is approximately $8779.7$ cubed millimeters.

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 |
---|---|---|

$V_{1}=31 πr_{2}h$ | $V_{2}=πr_{2}h$ | $V_{3}=32 πr_{3}$ |

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

Substitute values

$V_{P}=π(64+2688+3128 )$

$V_{P}=π(3192 +38064 +3128 )$

$V_{P}≈8779.7$

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.

The surface area of a sphere with radius $r$ is four times pi multiplied by the radius squared.

The aim of this proof is to consider areas and volumes of known figures to approximate the ratio of the surface area of the sphere to the volume of the sphere. Suppose that a sphere is filled with $n$ congruent pyramids. Consider that the area of the base of one of those pyramids is $B.$

There is a formula for the volume of a pyramid using its height and the area of the base. Notice that since the base of the pyramid lies on the surface of the sphere, the height of the pyramid is the radius $r.$ $V_{pyramid}=31 Bh⇓V_{pyramid}=31 Br $ Now the ratio of the area of the base of the pyramid to its volume can be obtained by dividing $B$ over the formula for the volume.$V_{pyramid}A_{pyramid base} $

Evaluate

SubstituteExpressions

Substitute expressions

$31 BrB $

DivByFracD

$b/ca =ba⋅c $

$Br3B $

CancelCommonFac

Cancel out common factors

$B r3B $

SimpQuot

Simplify quotient

$r3 $

$n⋅V_{pyramid}n⋅A_{pyramid base} $

Evaluate

CancelCommonFac

Cancel out common factors

$n ⋅V_{pyramid}n ⋅A_{pyramid base} $

SimpQuot

Simplify quotient

$V_{pyramid}A_{pyramid base} $

SubstituteValues

Substitute values

$r3 $

$V_{sphere}SA_{sphere} =r3 $

MultEqn

$LHS⋅V_{sphere}=RHS⋅V_{sphere}$

$SA_{sphere}=r3 ⋅V_{sphere}$

Evaluate

SubstituteExpressions

Substitute expressions

$SA_{sphere}=r3 ⋅34 πr_{3}$

MultFrac

Multiply fractions

$SA_{sphere}=r(3)3(4πr_{3}) $

DivPow

$a_{n}a_{m} =a_{m−n}$

$SA_{sphere}=33(4πr_{2}) $

ReduceFrac

$ba =b/3a/3 $

$SA_{sphere}=4πr_{2}$

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.

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

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=4πr_{2} $
Next, substitute $r=1.43$ into the previous formula.
Consequently, to make the lining of a baseball, about $25.7$ square inches of leather are needed. ### Covering of the Baseball

$S=4πr_{2}$

Substitute $r$ for $1.43$ and evaluate

Substitute

$r=1.43$

$S=4π(1.43)_{2}$

CalcPow

Calculate power

$S=4π(2.0449)$

UseCalc

Use a calculator

$S=25.696971…$

RoundDec

Round to ${\textstyle 1 \, \ifnumequal{1}{1}{\text{decimal}}{\text{decimals}}}$

$S=25.7$

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.

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

In talking aboutspheres, 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.

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

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

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

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 $AB,$ which is equal to $OD.$ Since $△ODE$ is a right triangle, by the Pythagorean Theorem, $OD$ can be found.

$OD=r_{2}−(2h )_{2} =AB$

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

$AC=r_{2}−y_{2} $

$A_{Cross-Section}=πAC_{2}−πAB_{2}$

Substitute values and simplify

SubstituteII

$AC=r_{2}−y_{2} $, $AB=r_{2}−(2h )_{2} $

$A_{Cross-Section}=π(r_{2}−y_{2} )_{2}−π⎝⎜⎛ r_{2}−(2h )_{2} ⎠⎟⎞ _{2}$

PowSqrt

$(a )_{2}=a$

$A_{Cross-Section}=π(r_{2}−y_{2})−π(r_{2}−(2h )_{2})$

FactorOut

Factor out $π$

$A_{Cross-Section}=π(r_{2}−y_{2}−(r_{2}−(2h )_{2}))$

Distr

Distribute $-1$

$A_{Cross-Section}=π(r_{2}−y_{2}−r_{2}+(2h )_{2})$

SubTerms

Subtract terms

$A_{Cross-Section}=π(-y_{2}+(2h )_{2})$

CommutativePropAdd

Commutative Property of Addition

$A_{Cross-Section}=π((2h )_{2}−y_{2})$

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.

Briefly describe a circle and a sphere. Write the definitions side by side.

Circle | Sphere |
---|---|

A circle is the set of all the points in a plane that are equidistant from a given point. | A sphere is the set of all points in the space that are equidistant from a given point called the center of the sphere. |

As can be seen, the two definitions are identical except for the fact that a circle is a two-dimensional figure while a sphere is three-dimensional. But this is not all. There is one more gnarly relationship. To see it, consider a circle and draw a line containing one diameter.

Next, rotate the circle about this line to see what figure is formed. As can be seen, when a circle is rotated about a diameter, it generates a sphere.

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