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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.
Now, the formula for the volume of a sphere will be rigorously proven.

$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 $r.$
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 $π$

$A_{C}=π(r_{2}−x_{2})$

$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.
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} $

Since the volume is given and the radius is needed, solve the formula for $r.$
$V=34 πr_{3}$

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 $1$ decimal place(s)

$r≈3958.8$

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.
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} $

Simplifying the right-hand side will give the number of balloons needed.
$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.
### 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 |
---|---|---|

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

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

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.

To find the surface area of a sphere, this informal justification will use the areas and volumes of known figures. It includes approximating the ratio between a sphere's surface area and its volume. In considering a known figure, suppose that a sphere is filled with $n$ congruent pyramids. The area of the base of each pyramid is $B.$

The formula to find the volume of a pyramid uses the height and the area of its base. Here, since the base of the pyramid lies on the surface of the sphere, the height of the pyramid is equal to the radius $r$ of the sphere.$V_{pyramid}V_{pyramid} =31 Bh⇓=31 Br $

The ratio of the area of the base of the pyramid to its volume can be obtained by dividing $B$ by the formula for the volume.
$V_{pyramid}A_{pyramid base} $

Evaluate

SubstituteExpressions

Substitute expressions

$31 BrB $

DivByFracD

$b/ca =ba⋅c $

$Br3B $

CrossCommonFac

Cross out common factors

$B r3B $

CancelCommonFac

Cancel out common factors

$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 $

$SA_{sphere}V_{sphere} =n⋅A_{pyramid base}=n⋅V_{pyramid} $

Considering these relationships, the ratio of the surface area of the sphere to its volume is the same as the ratio of the areas of the bases of the pyramids to their volumes.
$V_{sphere}SA_{sphere} V_{sphere}SA_{sphere} =n⋅V_{pyramid}n⋅A_{pyramid base} ⇓=r3 $

Since the formula for the volume of a sphere is known, it can be substituted in this ratio to find the formula for the surface area of the sphere.
$V_{sphere}SA_{sphere} =r3 $