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Consider the three-dimensional figure given in the diagram. By dragging the point, the figure can be skewed. However, its height and the radius of the circles that make the base and the top of the cylinder remain the same. Think about the volume — space occupied — of the solid. Does the volume change when the solid is skewed?

Before moving on, the definition of a circle and some of its parts will be reviewed.

A circle is the set of all the points in a plane that are equidistant from a given point. There are a few particularly notable features of a circle.

- Center - The given point from which all points of the circle are equidistant. Circles are often named by their center point.
- Radius - A segment that connects the center and any point on the circle. Its length is usually represented algebraically by $r.$
- Diameter - A segment whose endpoints are on the circle and that passes through the center. Its length is usually represented algebraically by $d.$
- Circumference - The perimeter of a circle, usually represented algebraically by $C.$

circle $O,$since it is centered at $O.$

In any given circle, the lengths of any radius and any diameter are constant. They are called

Next, the formula for calculating the circumference of a circle will be discussed.

The circumference of a circle is calculated by multiplying its diameter by $π.$

$C=πd$

Since the diameter is twice the radius, the circumference of a circle can also be calculated by multiplying $2r$ by $π.$

$C=2πr$

Consider two circles and their respective diameters and circumferences.

By the Similar Circles Theorem, all circles are similar. Therefore, their corresponding parts are proportional.$C_{A}C_{B} =d_{A}d_{B} $

This proportion can be rearranged.
$C_{A}C_{B} =d_{A}d_{B} $

Rearrange equation

MultEqn

$LHS⋅C_{A}=RHS⋅C_{A}$

$C_{B}=d_{A}d_{B} (C_{A})$

DivEqn

$LHS/d_{B}=RHS/d_{B}$

$d_{B}C_{B} =d_{A}1 (C_{A})$

MoveRightFacToNumOne

$b1 ⋅a=ba $

$d_{B}C_{B} =d_{A}C_{A} $

$dC =π⇒C=πd$

Finally, the formula for calculating the area of a circle will be seen.

The area of a circle is the product of $π$ and the square of its radius.

A circle with radius $r$ will be divided into a number of equally sized sectors. Then, the top and bottom halves of the circle will be distinguished by filling them with different colors. Because the circumference of a circle is $2πr,$ the arc length of each semicircle is half this value, $πr.$

Now, the above sectors will be *unfolded*. By placing the sectors of the upper hemisphere as teeth pointing downwards and the sectors of the bottom hemisphere as teeth pointing upwards, a parallelogram-like figure can be formed. As such, the area of the figure below should be the same as the circle's area.

Here, the shorter sides become

$A=πr⋅r⇔A=πr_{2}$

It has been shown that the area of a circle is the product of $π$ and the square of its radius.

Izabella loves to use geometry in her art. Her school noticed her skills and hired her to paint the school's soccer field — for a substantial payment. The diagram shows how the field should look.

Izabella needs the field's measurements. She already knows the radius of the circle located at the center of the field to be $6.5$ meters. For logistical reasons, she wants to find the circumference and area of this circle. Help Izabella calculate these values to one decimal place.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Circumference <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.36687em;vertical-align:0em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":"meters","answer":{"text":["40.8"]}}

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Area <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.36687em;vertical-align:0em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":"square meters","answer":{"text":["132.7"]}}

The circumference is twice the product of $π$ and the radius. The area is the product of $π$ and the square of the radius.

The circumference of a circle with radius $r$ is twice the product of $π$ and $r.$ Furthermore, its area is the product of $π$ and $r_{2}.$ Since the radius is $6.5$ meters, the circumference and area can be calculated.

Formula | Substitute | Simplify | Approximate | |
---|---|---|---|---|

Circumference | $C=2πr$ | $C=2π(6.5)$ | $C=13π$ | $C≈40.8$ |

Area | $A=πr_{2}$ | $A=π(6.5_{2})$ | $A=42.25π$ | $A≈132.7$ |

The circumference and area of the circle located at the center of the soccer field are about $40.8$ meters and $132.7$ square meters, respectively.

Maya bought $20$ meters of fencing with the hopes of constructing a circular dog run for her dog Opie. Because Opie is a large Saint Bernard, she wants the area of the dog run to be at least $30$ square meters.

With a circumference of $20$ meters, is the area of the dog run at least $30$ square meters?{"type":"choice","form":{"alts":["Yes","No"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":0}

Start by finding the radius of the circle.

Because the dog run is in the shape of a circle, its area is the product of $π$ and the square of the radius. Therefore, to find the area of the circle, the first step is to find the radius. Since it is already known that the circumference is $20$ meters, the formula for the circumference will be used to find the radius.
The radius of Opie's circular dog run is about $3.18$ meters.
With this information, the area of the dog run can be calculated.
The area of Opie's dog run is about $31.8$ square meters. This is enough for him to have happy and healthy playtime!

$C=2πr $

Since Maya bought and used $20$ meters of fencing, the circumference measures $20$ meters. This value can be substituted in the above formula to find the radius $r.$
$C=2πr$

Substitute

$C=20$

$20=2πr$

Solve for $r$

DivEqn

$LHS/2=RHS/2$

$10=πr$

DivEqn

$LHS/π=RHS/π$

$π10 =r$

RearrangeEqn

Rearrange equation

$r=π10 $

UseCalc

Use a calculator

$r=3.183098…$

RoundSigDig

Round to $3$ significant digit(s)

$r≈3.18$

$A=πr_{2} $

To do so, $3.18$ will be substituted for $r$ in the formula for the area of a circle.
$A=πr_{2}$

Substitute

$r=3.18$

$A=π(3.18_{2})$

Evaluate right-hand side

CalcPow

Calculate power

$A=π(10.1124)$

Multiply

Multiply

$A=31.769041$

RoundSigDig

Round to $3$ significant digit(s)

$A≈31.8$

For the following questions, approximate the answers to one decimal place. Do *not* include units in the answer.

Now, two-dimensional figures will be left aside to move forward into solids, which are three-dimensional shapes.

Two solids with the same height and the same cross-sectional area at every altitude have the same volume. This means that, as long as their heights are equal, skewed versions of the same solid have the same volume.
### Proof

Informal Justification

If $V_{1}$ and $V_{2}$ are the volumes of the above solids, then they are equal.

$V_{1}=V_{2}$

This principle will be proven by using a set of *identical* coins. Consider a stack in which each of these coins is placed directly on top of each other. Consider also another stack where the coins lie on top of each other, but are not aligned.

The first stack can be considered as a right cylinder. Similarly, the second stack can be considered as an oblique cylinder, which is a *skewed* version of the first cylinder. Because the coins are identical, the cross-sectional areas of the cylinders at the same altitude are the same.

Since the coins are identical, they have the same volume. Furthermore, since the height is the same for both stacks, they both have the same number of coins. Therefore, both stacks — cylinders — have the same volume. This reasoning is strongly based on the assumption that the face of the coins have the same area.

A cylinder is a three-dimensional figure that has two circular bases that are parallel and equal in size, connected by a curved surface.

The axis of a cylinder is the segment that connects the center of the bases. The
If the axis of a cylinder is not perpendicular to the bases, it is called an oblique cylinder.

If the dimensions of a cylinder are known, then its volume and surface area can be calculated.

The volume of a cylinder is calculated by multiplying the base's area by its height.

If the radius of the circular base of a cylinder is $r,$ the volume can be calculated by the following formula.

$V=πr_{2}h$

Consider a rectangular prism and a right cylinder that have the same base area and height.

In this case, the cross-sections of the prism and the cylinder are congruent to their bases. Therefore, their cross-sectional areas at every altitude are equal.

By the Cavalieri's Principle, two solids with the same height and the same cross-sectional area at every altitude have the same volume. Therefore, the volume of the cylinder is the same as the volume of the prism.$V_{C}=V_{P} $

Furthermore, the volume of a prism can be calculated by multiplying the area of its base by its height.
$V_{P}=Bh $

By the Transitive Property of Equality, a formula for the volume of the cylinder can be written.
${V_{C}=V_{P}V_{P}=Bh ⇒V_{C}=Bh $

Finally, not only is $B$ the area of the base of the prism, but also the area of the base of the cylinder. Since the base of the cylinder is a circle, its area is the product of $π$ and the square of its radius $r.$ Therefore, $B=πr_{2}$ can be substituted in the above formula.
$V_{C}=Bhsubstitute V_{C}=πr_{2}h $

This formula applies to all cylinders because there is always a prism with the same base area and height. Also, by the Cavalieri's principle, this formula still holds true for oblique cylinders.
LaShay loves golfing. She is about to buy a new golf bag.

The the bag she is thinking about buying is a shaped like a cylinder with a height of $132$ centimeters and its radius is $15$ centimeters. For all of her golf clubs to fit in the bag, its volume must be at least $90000$ cubic centimeters. Therefore, she wants to calculate the volume of the bag. Help her do this, approximating the answer to three significant figures.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Volume <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.36687em;vertical-align:0em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.8238736249999999em;vertical-align:-0.009765625em;\"><\/span><span class=\"mord\"><span class=\"mord text\"><span class=\"mord Roboto-Regular\">cm<\/span><\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.8141079999999999em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">3<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","answer":{"text":["93300","93 300","93,300"]}}

The volume of a cylinder is the product of $π,$ the square of its radius, and its height.

The golf bag is in the shape of a cylinder. Therefore, to calculate its volume, the formula for the volume of a cylinder can be used.

It is known that $h=132$ and $r=15.$ These two values can be substituted into the formula.$V=πr_{2}h$

SubstituteII

$h=132$, $r=15$

$V=π(15_{2})(132)$

Evaluate right-hand side

CalcPow

Calculate power

$V=π(225)(132)$

Multiply

Multiply

$V=29700π$

UseCalc

Use a calculator

$V=93305.30181…$

RoundSigDig

Round to $3$ significant digit(s)

$V≈93300$

Kriz is making an experiment to complete a Chemistry project. He is using a test tube in the shape of a cylinder with a height of $75$ millimeters and a volume of $20000$ cube millimeters.

For this tube to suit the experiment, it radius must not be greater than $9$ millimeters. Help Kriz find the radius! Approximate the answer to one decimal place.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Radius <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.36687em;vertical-align:0em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":"millimeters","answer":{"text":["9.2"]}}

The volume of a cylinder is the product of $π,$ the square of the base's radius, and the height of the cylinder.
When solving the equation, only the principal root was considered. This is because the radius of a circle is always positive. Therefore, the radius of the cylinder's base is about $9.2$ millimeters. Since the radius is greater than $9$ millimeters, the tube does not suit the experiment.

$V=πr_{2}h $

It is known that the height and the volume of the cylinder are $75$ millimeters and $20000$ cube millimeters, respectively. These two values can be substituted into the formula for volume of a cylinder, which can then be solved for $r.$
$V=πr_{2}h$

SubstituteII

$h=75$, $V=20000$

$20000=πr_{2}(75)$

Solve for $r$

CommutativePropMult

Commutative Property of Multiplication

$20000=75πr_{2}$

DivEqn

$LHS/75π=RHS/75π$

$75π20000 =r_{2}$

SqrtEqn

$LHS =RHS $

$75π20000 =r$

RearrangeEqn

Rearrange equation

$r=75π20000 $

UseCalc

Use a calculator

$r=9.213177…$

RoundDec

Round to $1$ decimal place(s)

$r≈9.2$

Besides the radius, height, and volume, another essential characteristic of a cylinder is its surface area. The surface area of a solid is the measure of the total area that the surface of the solid occupies.

Consider a cylinder of height $h$ and radius $r.$

The surface area of this cylinder is given by the following formula.

$S=2πrh+2πr_{2}$

The cylinder's surface area can be seen as three separate parts that are top, bottom, and side. The area of the side has the shape of a rectangle of width $h.$ The top and the bottom are circles of radius $r.$

Since the top and the bottom are congruent circles, they have the same area.

$Area ofOne CircleA=πr_{2} Area ofTwo CirclesA=2πr_{2} $

To find the area of the rectangle, its length must be found first. The length of the rectangle that forms the lateral part of the cylinder is the circumference of the base, which is $2πr.$
Therefore, the area of the rectangle is the product of $h$ and $2πr.$

$Area of the RectangleA=2πrh $

The surface area of the cylinder $S$ is the sum of the areas of the rectangle and the circles. $S=2πrh+2πr_{2}$

Jordan's father is giving her a new baseball bat for her birthday. To avoid damage to the bat, Papa Jordan will store the bat in a box that has the shape of a cylinder with a height of $30$ inches and a radius of $1.6$ inches. Since this is a present for Jordan, Papa will use wrapping paper to make it even more special.

Ignoring any paper overlapping, what is the minimum area of paper that Jordan's father needs? Approximate the answer to one decimal place.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"square inches","answer":{"text":["317.7"]}}

The area of wrapping paper Jordan's father needs is the same as the surface area of the cylinder.

To find the minimum area of paper that Jordan's father needs, the surface area $S$ of the cylinder needs to be calculated.

$S=2πrh+2πr_{2} $

Here, $r$ and $h$ are the radius and the height of the cylinder, respectively. The radius is $1.6$ inches and the height is $30$ inches. Therefore, these values can be substituted in the above formula, which can then be simplified.
Find the volume or surface area of the cylinder.

The challenge presented at the beginning of this lesson asked whether the volume of a cylinder changes if the solid is skewed.
**not** change even if the cylinder is skewed.

The Cavalieri's Principle was studied in this lesson. This principle states that two solids with the same height and the same cross-sectional area at every altitude have the same volume. Therefore, if their heights are equal, skewed versions of the same solid have the same volume. This means that the volume of a cylinder does