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== 9.6#h(0.6em)Solve Geometry Applications: Volume and Surface Area

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Before you get started, take this readiness quiz.

Evaluate #math.equation(block: false, alt: "x cubed")[$x^(3)$] when #math.equation(block: false, alt: "x equals 5 .")[$x = 5 .$] #linebreak() If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "125")[$125$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Evaluate #math.equation(block: false, alt: "2 to the power x")[$2^(x)$] when #math.equation(block: false, alt: "x equals 5 .")[$x = 5 .$] #linebreak() If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "32")[$32$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the area of a circle with radius #math.equation(block: false, alt: "the fraction 7 over 2 .")[$frac(7, 2) .$] #linebreak() If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "the fraction 77 over 2")[$frac(77, 2)$]
]
]

In this section, we will finish our study of geometry applications. We find the volume and surface area of some three-dimensional figures. Since we will be solving applications, we will once again show our Problem-Solving Strategy for Geometry Applications.

+ #strong[Read] the problem and make sure you understand all the words and ideas. Draw the figure and label it with the given information.
+ #strong[Identify] what you are looking for.
+ #strong[Name] what you are looking for. Choose a variable to represent that quantity.
+ #strong[Translate] into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
+ #strong[Solve] the equation using good algebra techniques.
+ #strong[Check] the answer in the problem and make sure it makes sense.
+ #strong[Answer] the question with a complete sentence.


=== Find Volume and Surface Area of Rectangular Solids

A cheerleading coach is having the squad paint wooden crates with the school colors to stand on at the games.. The amount of paint needed to cover the outside of each box is the #strong[surface area], a square measure of the total area of all the sides. The amount of space inside the crate is the volume, a cubic measure.

#figure(figph[This is an image of a wooden crate.], alt: "This is an image of a wooden crate.", caption: [This wooden crate is in the shape of a rectangular solid.])

Each crate is in the shape of a #strong[rectangular solid]. Its dimensions are the length, width, and height. The rectangular solid shown in  has length #math.equation(block: false, alt: "4")[$4$] units, width #math.equation(block: false, alt: "2")[$2$] units, and height #math.equation(block: false, alt: "3")[$3$] units. Can you tell how many cubic units there are altogether? Let’s look layer by layer.

#figure(figph[A rectangular solid is shown. Each layer is composed of 8 cubes, measuring 2 by 4. The top layer is pink. The middle layer is orange. The bottom layer is green. Beside this is an image of the top layer that says “The top layer has 8 cubic units.” The orange layer is shown and says “The middle layer has 8 cubic units.” The green layer is shown and says, “The bottom layer has 8 cubic units.”], alt: "A rectangular solid is shown. Each layer is composed of 8 cubes, measuring 2 by 4. The top layer is pink. The middle layer is orange. The bottom layer is green. Beside this is an image of the top layer that says “The top layer has 8 cubic units.” The orange layer is shown and says “The middle layer has 8 cubic units.” The green layer is shown and says, “The bottom layer has 8 cubic units.”", caption: [Breaking a rectangular solid into layers makes it easier to visualize the number of cubic units it contains. This #math.equation(block: false, alt: "4")[$4$] by #math.equation(block: false, alt: "2")[$2$] by #math.equation(block: false, alt: "3")[$3$] rectangular solid has #math.equation(block: false, alt: "24")[$24$] cubic units.])

Altogether there are #math.equation(block: false, alt: "24")[$24$] cubic units. Notice that #math.equation(block: false, alt: "24")[$24$] is the #math.equation(block: false, alt: "length times width times height .")[$"length" #h(1.0em) × #h(1.0em) "width" #h(1.0em) × #h(1.0em) "height" "."$]

#figure(figph[The top line says V equals L times W times H. Beneath the V is 24, beneath the equal sign is another equal sign, beneath the L is a 4, beneath the W is a 2, beneath the H is a 3.], alt: "The top line says V equals L times W times H. Beneath the V is 24, beneath the equal sign is another equal sign, beneath the L is a 4, beneath the W is a 2, beneath the H is a 3.", caption: none)

The volume, #math.equation(block: false, alt: "V ,")[$V ,$] of any rectangular solid is the product of the length, width, and height.

#math.equation(block: true, alt: "V equals L W H")[$V = L W H$]We could also write the formula for volume of a rectangular solid in terms of the area of the base. The area of the base, #math.equation(block: false, alt: "B ,")[$B ,$] is equal to #math.equation(block: false, alt: "length times width .")[$"length" × "width" "."$]

#math.equation(block: true, alt: "B equals L times W")[$B = L · W$]We can substitute #math.equation(block: false, alt: "B")[$B$] for #math.equation(block: false, alt: "L times W")[$L · W$] in the volume formula to get another form of the volume formula.

#figure(figph[The top line says V equals red L times red W times H. Below this is V equals red parentheses L times W times H. Below this is V equals red capital B times h.], alt: "The top line says V equals red L times red W times H. Below this is V equals red parentheses L times W times H. Below this is V equals red capital B times h.", caption: none)

We now have another version of the volume formula for rectangular solids. Let’s see how this works with the #math.equation(block: false, alt: "4 times 2 times 3")[$4 #h(0.2em) × #h(0.2em) 2 #h(0.2em) × #h(0.2em) 3$] rectangular solid we started with.

#figure(figph[An image of a rectangular solid is shown. It is made up of cubes. It is labeled as 2 by 4 by 3. Beside the solid is V equals Bh. Below this is V equals Base times height. Below Base is parentheses 4 times 2. The next line says V equals parentheses 4 times 2 times 3. Below that is V equals 8 times 3, then V equals 24 cubic units.], alt: "An image of a rectangular solid is shown. It is made up of cubes. It is labeled as 2 by 4 by 3. Beside the solid is V equals Bh. Below this is V equals Base times height. Below Base is parentheses 4 times 2. The next line says V equals parentheses 4 times 2 times 3. Below that is V equals 8 times 3, then V equals 24 cubic units.", caption: none)

To find the #emph[surface area] of a rectangular solid, think about finding the area of each of its faces. How many faces does the rectangular solid above have? You can see three of them.

#math.equation(block: true, alt: "A sub front equals L times W, A sub side equals L times W, A sub top equals L times W; A sub front equals 4 times 3, A sub side equals 2 times 3, A sub top equals 4 times 2; A sub front equals 12, A sub side equals 6, A sub top equals 8")[$A_("front") = L × W &  &  & A_("side") = L × W &  &  & A_("top") = L × W \ A_("front") = 4 · 3 &  &  & A_("side") = 2 · 3 &  &  & A_("top") = 4 · 2 \ A_("front") = 12 &  &  & A_("side") = 6 &  &  & A_("top") = 8$]Notice for each of the three faces you see, there is an identical opposite face that does not show.

#math.equation(block: true, alt: "S equals open parenthesis front plus back close parenthesis + open parenthesis left side plus right side close parenthesis plus open parenthesis top plus bottom close parenthesis; S equals open parenthesis 2 times front close parenthesis plus open parenthesis 2 times left side close parenthesis plus open parenthesis 2 times top close parenthesis; S equals 2 times 12 plus 2 times 6 plus 2 times 8; S equals 24 plus 12 plus 16; S equals 52 sq. units")[$S = ( "front" + "back" ) "+" ( "left side" + "right side" ) + ( "top" + "bottom" ) \ S = ( 2 · "front" ) + ( "2" · "left side" ) + ( "2" · "top" ) \ S = 2 · 12 + 2 · 6 + 2 · 8 \ S = 24 + 12 + 16 \ S = 52 #h(0.2em) "sq. units"$]The surface area #math.equation(block: false, alt: "S")[$S$] of the rectangular solid shown in  is #math.equation(block: false, alt: "52")[$52$] square units.

In general, to find the surface area of a rectangular solid, remember that each face is a rectangle, so its area is the product of its two dimensions, either length and width, length and height, or width and height. Find the area of each face that you see and then multiply each area by two to account for the face on the opposite side.

#math.equation(block: true, alt: "S equals 2 L H plus 2 L W plus 2 W H")[$S = 2 L H + 2 L W + 2 W H$]#figure(figph[A rectangular solid is shown. The sides are labeled L, W, and H. One face is labeled LW and another is labeled WH.], alt: "A rectangular solid is shown. The sides are labeled L, W, and H. One face is labeled LW and another is labeled WH.", caption: [For each face of the rectangular solid facing you, there is another face on the opposite side. There are #math.equation(block: false, alt: "6")[$6$] faces in all.])

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Volume and Surface Area of a Rectangular Solid]

For a rectangular solid with length #math.equation(block: false, alt: "L ,")[$L ,$] width #math.equation(block: false, alt: "W ,")[$W ,$] and height #math.equation(block: false, alt: "H :")[$H :$]

#figure(figph[A rectangular solid is shown. The sides are labeled L, W, and H. Beside it is Volume: V equals LWH equals BH. Below that is Surface Area: S equals 2LH plus 2LW plus 2WH.], alt: "A rectangular solid is shown. The sides are labeled L, W, and H. Beside it is Volume: V equals LWH equals BH. Below that is Surface Area: S equals 2LH plus 2LW plus 2WH.", caption: none)
]

#examplebox("Example 1")[][
For a rectangular solid with length #math.equation(block: false, alt: "14")[$14$] cm, height #math.equation(block: false, alt: "17")[$17$] cm, and width #math.equation(block: false, alt: "9")[$9$] cm, find the ⓐ volume and ⓑ surface area.

#solutionbox[
Step 1 is the same for both ⓐ and ⓑ , so we will show it just once.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  [Step 1. #strong[Read] the problem. Draw the figure and #linebreak()  label it with the given information.], [#figure(figph[A cuboid shown in 3D perspective with dimensions: length 14, width 9, and height 17. Hidden edges are indicated by dashed lines.], alt: "A cuboid shown in 3D perspective with dimensions: length 14, width 9, and height 17. Hidden edges are indicated by dashed lines.", caption: none)],
))

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([ⓐ], []),
  [Step 2. #strong[Identify] what you are looking for.], [the volume of the rectangular solid],
  [Step 3. #strong[Name.] Choose a variable to represent it.], [Let #math.equation(block: false, alt: "V")[$V$]= volume],
  [Step 4. #strong[Translate.] #linebreak() Write the appropriate formula. #linebreak() Substitute.], [#linebreak() #math.equation(block: false, alt: "V equals L W H")[$V = L W H$] #linebreak() #math.equation(block: false, alt: "V equals 14 times 9 times 17")[$V = "14" ⋅ 9 ⋅ 17$]],
  [Step 5. #strong[Solve] the equation.], [#math.equation(block: false, alt: "V equals 2,142")[$V = 2,142$]],
  [Step 6. #strong[Check] #linebreak() We leave it to you to check your calculations.], [],
  [Step 7. #strong[Answer] the question.], [The volume is #math.equation(block: false, alt: "2,142")[$"2,142"$] cubic centimeters.],
))

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([ⓑ], []),
  [Step 2. #strong[Identify] what you are looking for.], [the surface area of the solid],
  [Step 3. #strong[Name.] Choose a variable to represent it.], [Let #math.equation(block: false, alt: "S")[$S$]= surface area],
  [Step 4. #strong[Translate.] #linebreak() Write the appropriate formula. #linebreak() Substitute.], [#linebreak() #math.equation(block: false, alt: "S equals 2 L H plus 2 L W plus 2 W H")[$S = 2 L H + 2 L W + 2 W H$] #linebreak() #math.equation(block: false, alt: "S equals 2 open parenthesis 14 times 17 close parenthesis plus 2 open parenthesis 14 times 9 close parenthesis plus 2 open parenthesis 9 times 17 close parenthesis")[$S = 2 ( 14 ⋅ 17 ) + 2 ( 14 ⋅ 9 ) + 2 ( 9 ⋅ 17 )$]],
  [Step 5. #strong[Solve the equation.]], [#math.equation(block: false, alt: "S equals 1,034")[$S = 1,034$]],
  [Step 6. #strong[Check:] Double-check with a calculator.], [],
  [Step 7. #strong[Answer] the question.], [The surface area is 1,034 square centimeters.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the ⓐ volume and ⓑ surface area of rectangular solid with the: length #math.equation(block: false, alt: "8")[$8$] feet, width #math.equation(block: false, alt: "9")[$9$] feet, and height #math.equation(block: false, alt: "11")[$11$]
feet.

#solutionbox[
+ ⓐ 792 cu. ft
+ ⓑ 518 sq. ft
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the ⓐ volume and ⓑ surface area of rectangular solid with the:
length #math.equation(block: false, alt: "15")[$15$] feet, width #math.equation(block: false, alt: "12")[$12$] feet, and height #math.equation(block: false, alt: "8")[$8$] feet.

#solutionbox[
+ ⓐ 1,440 cu. ft
+ ⓑ 792 sq. ft
]
]

#examplebox("Example 2")[][
A rectangular crate has a length of #math.equation(block: false, alt: "30")[$30$] inches, width of #math.equation(block: false, alt: "25")[$25$] inches, and height of #math.equation(block: false, alt: "20")[$20$] inches. Find its ⓐ volume and ⓑ surface area.

#solutionbox[
Step 1 is the same for both ⓐ and ⓑ , so we will show it just once.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  [Step 1. #strong[Read] the problem. Draw the figure and  #linebreak() label it with the given information.], [#figure(figph[A cuboid with dimensions 30 (length), 25 (width), and 20 (height) units. The front and top faces are visible, with dashed lines indicating hidden edges. It illustrates a basic 3D geometric shape with measurements.], alt: "A cuboid with dimensions 30 (length), 25 (width), and 20 (height) units. The front and top faces are visible, with dashed lines indicating hidden edges. It illustrates a basic 3D geometric shape with measurements.", caption: none)],
))

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([ⓐ], []),
  [Step 2. #strong[Identify] what you are looking for.], [the volume of the crate],
  [Step 3. #strong[Name.] Choose a variable to represent it.], [let #math.equation(block: false, alt: "V")[$V$]= volume],
  [Step 4. #strong[Translate.] #linebreak() Write the appropriate formula. #linebreak() Substitute.], [#linebreak() #math.equation(block: false, alt: "V equals L W H")[$V = L W H$] #linebreak() #math.equation(block: false, alt: "V equals 30 times 25 times 20")[$V = 30 ⋅ 25 ⋅ 20$]],
  [Step 5. #strong[Solve] the equation.], [#math.equation(block: false, alt: "V equals 15,000")[$V = 15,000$]],
  [Step 6. #strong[Check:] Double check your math.], [],
  [Step 7. #strong[Answer] the question.], [The volume is 15,000 cubic inches.],
))

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([ⓑ], []),
  [Step 2. #strong[Identify] what you are looking for.], [the surface area of the crate],
  [Step 3. #strong[Name.] Choose a variable to represent it.], [let #math.equation(block: false, alt: "S")[$S$]= surface area],
  [Step 4. #strong[Translate.] #linebreak() Write the appropriate formula. #linebreak() Substitute.], [#linebreak() #math.equation(block: false, alt: "S equals 2 L H plus 2 L W plus 2 W H")[$S = 2 L H + 2 L W + 2 W H$] #linebreak() #math.equation(block: false, alt: "S equals 2 open parenthesis 30 times 20 close parenthesis plus 2 open parenthesis 30 times 25 close parenthesis plus 2 open parenthesis 25 times 20 close parenthesis")[$S = 2 ( 30 ⋅ 20 ) + 2 ( 30 ⋅ 25 ) + 2 ( 25 ⋅ 20 )$]],
  [Step 5. #strong[Solve] the equation.], [#math.equation(block: false, alt: "S equals 3,700")[$S = 3,700$]],
  [Step 6. #strong[Check:] Check it yourself!], [],
  [Step 7. #strong[Answer] the question.], [The surface area is 3,700 square inches.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
A rectangular box has length #math.equation(block: false, alt: "9")[$9$] feet, width #math.equation(block: false, alt: "4")[$4$] feet, and height #math.equation(block: false, alt: "6")[$6$] feet. Find its ⓐ volume and ⓑ surface area.

#solutionbox[
+ ⓐ 216 cu. ft
+ ⓑ 228 sq. ft
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
A rectangular suitcase has length #math.equation(block: false, alt: "22")[$22$] inches, width #math.equation(block: false, alt: "14")[$14$] inches, and height #math.equation(block: false, alt: "9")[$9$] inches. Find its ⓐ volume and ⓑ surface area.

#solutionbox[
+ ⓐ 2,772 cu. in.
+ ⓑ 1,264 sq. in.
]
]


==== Volume and Surface Area of a Cube

A #strong[cube] is a rectangular solid whose length, width, and height are equal. See Volume and Surface Area of a Cube, below. Substituting, #emph[s] for the length, width and height into the formulas for volume and surface area of a rectangular solid, we get:

#math.equation(block: true, alt: "V equals L W H, S equals 2 L H plus 2 L W plus 2 W H; V equals s times s times s, S equals 2 s times s plus 2 s times s plus 2 s times s; V equals s cubed, S equals 2 s squared plus 2 s squared plus 2 s squared; S equals 6 s squared")[$V = L W H &  &  &  & S = 2 L H + 2 L W + 2 W H \ V = s · s · s &  &  &  & S = 2 s · s + 2 s · s + 2 s · s \ V = s^(3) &  &  &  & S = 2 s^(2) + 2 s^(2) + 2 s^(2) \  &  &  &  & S = 6 s^(2)$]So for a cube, the formulas for volume and surface area are #math.equation(block: false, alt: "V equals s cubed")[$V = s^(3)$] and #math.equation(block: false, alt: "S equals 6 s squared .")[$S = 6 s^(2) .$]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Volume and Surface Area of a Cube]

For any cube with sides of length #math.equation(block: false, alt: "s ,")[$s ,$]

#figure(figph[An image of a cube is shown. Each side is labeled s. Beside this is Volume: V equals s cubed. Below that is Surface Area: S equals 6 times s squared.], alt: "An image of a cube is shown. Each side is labeled s. Beside this is Volume: V equals s cubed. Below that is Surface Area: S equals 6 times s squared.", caption: none)
]

#examplebox("Example 3")[][
A cube is #math.equation(block: false, alt: "2.5")[$2.5$] inches on each side. Find its ⓐ volume and ⓑ surface area.

#solutionbox[
Step 1 is the same for both ⓐ and ⓑ , so we will show it just once.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  [Step 1. #strong[Read] the problem. Draw the figure and  #linebreak() label it with the given information.], [#figure(figph[A simple line drawing of a cube, with each visible side dimension labeled as 2.5, indicating equal length, width, and height. The cube is rendered in an isometric projection.], alt: "A simple line drawing of a cube, with each visible side dimension labeled as 2.5, indicating equal length, width, and height. The cube is rendered in an isometric projection.", caption: none)],
))

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([ⓐ], []),
  [Step 2. #strong[Identify] what you are looking for.], [the volume of the cube],
  [Step 3. #strong[Name.] Choose a variable to represent it.], [let #emph[V] = volume],
  [Step 4. #strong[Translate.] #linebreak() Write the appropriate formula.], [#linebreak() #math.equation(block: false, alt: "V equals s cubed")[$V = s^(3)$]],
  [Step 5. #strong[Solve.] Substitute and solve.], [#math.equation(block: false, alt: "V equals open parenthesis 2.5 close parenthesis cubed")[$V = attach(( 2.5 ), t: 3)$] #linebreak() #math.equation(block: false, alt: "V equals 15.625")[$V = 15.625$]],
  [Step 6. #strong[Check:] Check your work.], [],
  [Step 7. #strong[Answer] the question.], [The volume is 15.625 cubic inches.],
))

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([ⓑ], []),
  [Step 2. #strong[Identify] what you are looking for.], [the surface area of the cube],
  [Step 3. #strong[Name.] Choose a variable to represent it.], [let #emph[S] = surface area],
  [Step 4. #strong[Translate.] #linebreak() Write the appropriate formula.], [#linebreak() #math.equation(block: false, alt: "S equals 6 s squared")[$S = 6 s^(2)$]],
  [Step 5. #strong[Solve.] Substitute and solve.], [#math.equation(block: false, alt: "S equals 6 times open parenthesis 2.5 close parenthesis squared")[$S = 6 ⋅ attach(( 2.5 ), t: 2)$] #linebreak() #math.equation(block: false, alt: "S equals 37.5")[$S = 37.5$]],
  [Step 6. #strong[Check:] The check is left to you.], [],
  [Step 7. #strong[Answer] the question.], [The surface area is 37.5 square inches.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
For a cube with side 4.5 meters, find the ⓐ volume and ⓑ surface area of the cube.

#solutionbox[
+ ⓐ 91.125 cu. m
+ ⓑ 121.5 sq. m
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
For a cube with side 7.3 yards, find the ⓐ volume and ⓑ surface area of the cube.

#solutionbox[
+ ⓐ 389.017 cu. yd.
+ ⓑ 319.74 sq. yd.
]
]

#examplebox("Example 4")[][
A notepad cube measures #math.equation(block: false, alt: "2")[$2$] inches on each side. Find its ⓐ volume and ⓑ surface area.

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  [Step 1. #strong[Read] the problem. Draw the figure and  #linebreak() label it with the given information.], [#figure(figph[A wireframe drawing of a cube, indicating its dimensions are 2 units on each side. The three visible edges are labeled with the number '2', denoting its length, width, and height.], alt: "A wireframe drawing of a cube, indicating its dimensions are 2 units on each side. The three visible edges are labeled with the number '2', denoting its length, width, and height.", caption: none)],
))

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([ⓐ], []),
  [Step 2. #strong[Identify] what you are looking for.], [the volume of the cube],
  [Step 3. #strong[Name.] Choose a variable to represent it.], [let #emph[V] = volume],
  [Step 4. #strong[Translate.] #linebreak() Write the appropriate formula.], [#linebreak() #math.equation(block: false, alt: "V equals s cubed")[$V = s^(3)$]],
  [Step 5. #strong[Solve] the equation.], [#math.equation(block: false, alt: "V equals 2 cubed")[$V = 2^(3)$] #linebreak() #math.equation(block: false, alt: "V equals 8")[$V = 8$]],
  [Step 6. #strong[Check:] Check that you did the calculations  #linebreak() correctly.], [],
  [Step 7. #strong[Answer] the question.], [The volume is 8 cubic inches.],
))

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([ⓑ], []),
  [Step 2. #strong[Identify] what you are looking for.], [the surface area of the cube],
  [Step 3. #strong[Name.] Choose a variable to represent it.], [let #emph[S] = surface area],
  [Step 4. #strong[Translate.] #linebreak() Write the appropriate formula.], [#linebreak() #math.equation(block: false, alt: "S equals 6 s squared")[$S = 6 s^(2)$]],
  [Step 5. #strong[Solve] the equation.], [#math.equation(block: false, alt: "S equals 6 times 2 squared")[$S = 6 ⋅ 2^(2)$] #linebreak() #math.equation(block: false, alt: "S equals 24")[$S = 24$]],
  [Step 6. #strong[Check:] The check is left to you.], [],
  [Step 7. #strong[Answer] the question.], [The surface area is 24 square inches.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
A packing box is a cube measuring #math.equation(block: false, alt: "4")[$4$] feet on each side. Find its ⓐ volume and ⓑ surface area.

#solutionbox[
+ ⓐ 64 cu. ft
+ ⓑ 96 sq. ft
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
A wall is made up of cube-shaped bricks. Each cube is #math.equation(block: false, alt: "16")[$16$] inches on each side. Find the ⓐ volume and ⓑ surface area of each cube.

#solutionbox[
+ ⓐ 4,096 cu. in.
+ ⓑ 1536 sq. in.
]
]


=== Find the Volume and Surface Area of Spheres

A #strong[sphere] is the shape of a basketball, like a three-dimensional circle. Just like a circle, the size of a sphere is determined by its radius, which is the distance from the center of the sphere to any point on its surface. The formulas for the volume and surface area of a sphere are given below.

Showing where these formulas come from, like we did for a rectangular solid, is beyond the scope of this course. We will approximate #math.equation(block: false, alt: "π")[$π$] with #math.equation(block: false, alt: "3.14 .")[$3.14 .$]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Volume and Surface Area of a Sphere]

For a sphere with radius #math.equation(block: false, alt: "r :")[$r ":"$]

#figure(figph[An image of a sphere is shown. The radius is labeled r. Beside this is Volume: V equals four-thirds times pi times r cubed. Below that is Surface Area: S equals 4 times pi times r squared.], alt: "An image of a sphere is shown. The radius is labeled r. Beside this is Volume: V equals four-thirds times pi times r cubed. Below that is Surface Area: S equals 4 times pi times r squared.", caption: none)
]

#examplebox("Example 5")[][
A sphere has a radius #math.equation(block: false, alt: "6")[$6$] inches. Find its ⓐ volume and ⓑ surface area.

#solutionbox[
Step 1 is the same for both ⓐ and ⓑ , so we will show it just once.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  [Step 1. #strong[Read] the problem. Draw the figure and label #linebreak()  it with the given information.], [#figure(figph[A 3D illustration of a sphere, indicating a radius of 6 from its center to the perimeter. A dashed line represents the hidden part of the equatorial plane.], alt: "A 3D illustration of a sphere, indicating a radius of 6 from its center to the perimeter. A dashed line represents the hidden part of the equatorial plane.", caption: none)],
))

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([ⓐ], []),
  [Step 2. #strong[Identify] what you are looking for.], [the volume of the sphere],
  [Step 3. #strong[Name.] Choose a variable to represent it.], [let #emph[V] = volume],
  [Step 4. #strong[Translate.] #linebreak() Write the appropriate formula.], [#linebreak() #math.equation(block: false, alt: "V equals the fraction 4 over 3 π r cubed")[$V = frac(4, 3) π r^(3)$]],
  [Step 5. #strong[Solve.]], [#math.equation(block: false, alt: "V approximately equals the fraction 4 over 3 open parenthesis 3.14 close parenthesis 6 cubed")[$V ≈ frac(4, 3) ( 3.14 ) 6^(3)$] #linebreak() #math.equation(block: false, alt: "V approximately equals 904.32 cubic inches")[$V ≈ 904.32 #h(0.2em) "cubic inches"$]],
  [Step 6. #strong[Check:] Double-check your math on a calculator.], [],
  [Step 7. #strong[Answer] the question.], [The volume is approximately 904.32 cubic inches.],
))

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([ⓑ], []),
  [Step 2. #strong[Identify] what you are looking for.], [the surface area of the sphere],
  [Step 3. #strong[Name.] Choose a variable to represent it.], [let #emph[S] = surface area],
  [Step 4. #strong[Translate.] #linebreak() Write the appropriate formula.], [#linebreak() #math.equation(block: false, alt: "S equals 4 π r squared")[$S = 4 π r^(2)$]],
  [Step 5. #strong[Solve.]], [#math.equation(block: false, alt: "S approximately equals 4 open parenthesis 3.14 close parenthesis 6 squared")[$S ≈ 4 ( 3.14 ) 6^(2)$] #linebreak() #math.equation(block: false, alt: "S approximately equals 452.16")[$S ≈ 452.16$]],
  [Step 6. #strong[Check:] Double-check your math on a calculator], [],
  [Step 7. #strong[Answer] the question.], [The surface area is approximately 452.16 square inches.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the ⓐ volume and ⓑ surface area of a sphere with radius 3 centimeters.

#solutionbox[
+ ⓐ 113.04 cu. cm
+ ⓑ 113.04 sq. cm
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the ⓐ volume and ⓑ surface area of each sphere with a radius of #math.equation(block: false, alt: "1")[$1$] foot

#solutionbox[
+ ⓐ 4.19 cu. ft
+ ⓑ 12.56 sq. ft
]
]

#examplebox("Example 6")[][
A globe of Earth is in the shape of a sphere with radius #math.equation(block: false, alt: "14")[$14$] inches. Find its ⓐ volume and ⓑ surface area. Round the answer to the nearest hundredth.

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  [Step 1. #strong[Read] the problem. Draw a figure with the #linebreak()  given information and label it.], [#figure(figph[A stylized globe showing North and South America, Europe, and Africa. A dashed line represents the equator, and a horizontal line segment with the number '14' indicates a measurement from the center.], alt: "A stylized globe showing North and South America, Europe, and Africa. A dashed line represents the equator, and a horizontal line segment with the number '14' indicates a measurement from the center.", caption: none)],
))

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([ⓐ], []),
  [Step 2. #strong[Identify] what you are looking for.], [the volume of the sphere],
  [Step 3. #strong[Name.] Choose a variable to represent it.], [let #emph[V] = volume],
  [Step 4. #strong[Translate.] #linebreak() Write the appropriate formula. #linebreak() Substitute. (Use 3.14 for #math.equation(block: false, alt: "π")[$π$])], [#linebreak() #math.equation(block: false, alt: "V equals the fraction 4 over 3 π r cubed")[$V = frac(4, 3) π r^(3)$] #linebreak() #math.equation(block: false, alt: "V approximately equals the fraction 4 over 3 open parenthesis 3.14 close parenthesis 14 cubed")[$V ≈ frac(4, 3) ( 3.14 ) 14^(3)$]],
  [Step 5. #strong[Solve.]], [#math.equation(block: false, alt: "V approximately equals 11,488.21")[$V ≈ 11,488.21$]],
  [Step 6. #strong[Check:] We leave it to you to check your calculations.], [],
  [Step 7. #strong[Answer] the question.], [The volume is approximately 11,488.21 cubic inches.],
))

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([ⓑ], []),
  [Step 2. #strong[Identify] what you are looking for.], [the surface area of the sphere],
  [Step 3. #strong[Name.] Choose a variable to represent it.], [let #emph[S] = surface area],
  [Step 4. #strong[Translate.] #linebreak() Write the appropriate formula. #linebreak() Substitute. (Use 3.14 for #math.equation(block: false, alt: "π")[$π$])], [#linebreak() #math.equation(block: false, alt: "S equals 4 π r squared")[$S = 4 π r^(2)$] #linebreak() #math.equation(block: false, alt: "S approximately equals 4 open parenthesis 3.14 close parenthesis 14 squared")[$S ≈ 4 ( 3.14 ) 14^(2)$]],
  [Step 5. #strong[Solve.]], [#math.equation(block: false, alt: "S approximately equals 2461.76")[$S ≈ 2461.76$]],
  [Step 6. #strong[Check:] We leave it to you to check your calculations.], [],
  [Step 7. #strong[Answer] the question.], [The surface area is approximately 2461.76 square inches.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
A beach ball is in the shape of a sphere with radius of #math.equation(block: false, alt: "9")[$9$] inches. Find its ⓐ volume and ⓑ surface area.

#solutionbox[
+ ⓐ 3052.08 cu. in.
+ ⓑ 1017.36 sq. in.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
A Roman statue depicts Atlas holding a globe with radius of #math.equation(block: false, alt: "1.5")[$1.5$] feet. Find the ⓐ volume and ⓑ surface area of the globe.

#solutionbox[
+ ⓐ 14.13 cu. ft
+ ⓑ 28.26 sq. ft
]
]


=== Find the Volume and Surface Area of a Cylinder

If you have ever seen a can of soda, you know what a cylinder looks like. A #strong[cylinder] is a solid figure with two parallel circles of the same size at the top and bottom. The top and bottom of a cylinder are called the bases. The height #math.equation(block: false, alt: "h")[$h$] of a cylinder is the distance between the two bases. For all the cylinders we will work with here, the sides and the height, #math.equation(block: false, alt: "h")[$h$] , will be perpendicular to the bases.

#figure(figph[An image of a cylinder is shown. There is a red arrow pointing to the radius of the top labeling it r, radius. There is a red arrow pointing to the height of the cylinder labeling it h, height.], alt: "An image of a cylinder is shown. There is a red arrow pointing to the radius of the top labeling it r, radius. There is a red arrow pointing to the height of the cylinder labeling it h, height.", caption: [A cylinder has two circular bases of equal size. The height is the distance between the bases.])

Rectangular solids and cylinders are somewhat similar because they both have two bases and a height. The formula for the volume of a rectangular solid, #math.equation(block: false, alt: "V equals B h")[$V = B h$] , can also be used to find the volume of a cylinder.

For the rectangular solid, the area of the base, #math.equation(block: false, alt: "B")[$B$] , is the area of the rectangular base, length × width. For a cylinder, the area of the base, #math.equation(block: false, alt: "B ,")[$B ,$] is the area of its circular base, #math.equation(block: false, alt: "π r squared .")[$π r^(2) .$]  compares how the formula #math.equation(block: false, alt: "V equals B h")[$V = B h$] is used for rectangular solids and cylinders.

#figure(figph[In (a), a rectangular solid is shown. The sides are labeled L, W, and H. Below this is V equals capital Bh, then V equals Base times h, then V equals parentheses lw times h, then V equals lwh. In (b), a cylinder is shown. The radius of the top is labeled r, the height is labeled h. Below this is V equals capital Bh, then V equals Base times h, then V equals parentheses pi r squared times h, then V equals pi times r squared times h.], alt: "In (a), a rectangular solid is shown. The sides are labeled L, W, and H. Below this is V equals capital Bh, then V equals Base times h, then V equals parentheses lw times h, then V equals lwh. In (b), a cylinder is shown. The radius of the top is labeled r, the height is labeled h. Below this is V equals capital Bh, then V equals Base times h, then V equals parentheses pi r squared times h, then V equals pi times r squared times h.", caption: [Seeing how a cylinder is similar to a rectangular solid may make it easier to understand the formula for the volume of a cylinder.])

To understand the formula for the #strong[surface area] of a cylinder, think of a can of vegetables. It has three surfaces: the top, the bottom, and the piece that forms the sides of the can. If you carefully cut the label off the side of the can and unroll it, you will see that it is a rectangle.

#figure(figph[A cylindrical can of green beans is shown. The height is labeled h. Beside this are pictures of circles for the top and bottom of the can and a rectangle for the other portion of the can. Above the circles is C equals 2 times pi times r. The top of the rectangle says l equals 2 times pi times r. The left side of the rectangle is labeled h, the right side is labeled w.], alt: "A cylindrical can of green beans is shown. The height is labeled h. Beside this are pictures of circles for the top and bottom of the can and a rectangle for the other portion of the can. Above the circles is C equals 2 times pi times r. The top of the rectangle says l equals 2 times pi times r. The left side of the rectangle is labeled h, the right side is labeled w.", caption: [By cutting and unrolling the label of a can of vegetables, we can see that the surface of a cylinder is a rectangle. The length of the rectangle is the circumference of the cylinder’s base, and the width is the height of the cylinder.])

The distance around the edge of the can is the #strong[circumference] of the cylinder’s base it is also the length #math.equation(block: false, alt: "L")[$L$] of the rectangular label. The height of the cylinder is the width #math.equation(block: false, alt: "W")[$W$] of the rectangular label. So the area of the label can be represented as

#figure(figph[The top line says A equals l times red w. Below the l is 2 times pi times r. Below the w is a red h.], alt: "The top line says A equals l times red w. Below the l is 2 times pi times r. Below the w is a red h.", caption: none)

To find the total surface area of the cylinder, we add the areas of the two circles to the area of the rectangle.

#figure(figph[A rectangle is shown with circles coming off the top and bottom.], alt: "A rectangle is shown with circles coming off the top and bottom.", caption: none)

The surface area of a cylinder with radius #math.equation(block: false, alt: "r")[$r$] and height #math.equation(block: false, alt: "h ,")[$h ,$] is

#math.equation(block: true, alt: "S equals 2 π r squared plus 2 π r h")[$S = 2 π r^(2) + 2 π r h$]#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Volume and Surface Area of a Cylinder]

For a cylinder with radius #math.equation(block: false, alt: "r")[$r$] and height #math.equation(block: false, alt: "h :")[$h :$]

#figure(figph[A cylinder is shown. The height is labeled h and the radius of the top is labeled r. Beside it is Volume: V equals pi times r squared times h or V equals capital B times h. Below this is Surface Area: S equals 2 times pi times r squared plus 2 times pi times r times h.], alt: "A cylinder is shown. The height is labeled h and the radius of the top is labeled r. Beside it is Volume: V equals pi times r squared times h or V equals capital B times h. Below this is Surface Area: S equals 2 times pi times r squared plus 2 times pi times r times h.", caption: none)
]

#examplebox("Example 7")[][
A cylinder has height #math.equation(block: false, alt: "5")[$5$] inches and radius #math.equation(block: false, alt: "3")[$3$] inches. Find the ⓐ volume and ⓑ surface area.

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  [Step 1. #strong[Read] the problem. Draw the figure and label #linebreak()  it with the given information.], [#figure(figph[A simple black and white line drawing of a cylinder with its dimensions labeled: a radius of 3 units and a height of 5 units.], alt: "A simple black and white line drawing of a cylinder with its dimensions labeled: a radius of 3 units and a height of 5 units.", caption: none)],
))

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([ⓐ], []),
  [Step 2. #strong[Identify] what you are looking for.], [the volume of the cylinder],
  [Step 3. #strong[Name.] Choose a variable to represent it.], [let #emph[V] = volume],
  [Step 4. #strong[Translate.] #linebreak() Write the appropriate formula. #linebreak() Substitute. (Use 3.14 for #math.equation(block: false, alt: "π")[$π$])], [#linebreak() #math.equation(block: false, alt: "V equals π r squared h")[$V = π r^(2) h$] #linebreak() #math.equation(block: false, alt: "V approximately equals open parenthesis 3.14 close parenthesis 3 squared times 5")[$V ≈ ( 3.14 ) 3^(2) ⋅ 5$]],
  [Step 5. #strong[Solve.]], [#math.equation(block: false, alt: "V approximately equals 141.3")[$V ≈ 141.3$]],
  [Step 6. #strong[Check:] We leave it to you to check your calculations.], [],
  [Step 7. #strong[Answer] the question.], [The volume is approximately 141.3 cubic inches.],
))

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([ⓑ], []),
  [Step 2. #strong[Identify] what you are looking for.], [the surface area of the cylinder],
  [Step 3. #strong[Name.] Choose a variable to represent it.], [let #emph[S] = surface area],
  [Step 4. #strong[Translate.] #linebreak() Write the appropriate formula. #linebreak() Substitute. (Use 3.14 for #math.equation(block: false, alt: "π")[$π$])], [#linebreak() #math.equation(block: false, alt: "S equals 2 π r squared plus 2 π r h")[$S = 2 π r^(2) + 2 π r h$] #linebreak() #math.equation(block: false, alt: "S approximately equals 2 open parenthesis 3.14 close parenthesis 3 squared plus 2 open parenthesis 3.14 close parenthesis open parenthesis 3 close parenthesis 5")[$S ≈ 2 ( 3.14 ) 3^(2) + 2 ( 3.14 ) ( 3 ) 5$]],
  [Step 5. #strong[Solve.]], [#math.equation(block: false, alt: "S approximately equals 150.72")[$S ≈ 150.72$]],
  [Step 6. #strong[Check:] We leave it to you to check your calculations.], [],
  [Step 7. #strong[Answer] the question.], [The surface area is approximately 150.72 square inches.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the ⓐ volume and ⓑ surface area of the cylinder with radius 4 cm and height 7cm.

#solutionbox[
+ ⓐ 351.68 cu. cm
+ ⓑ 276.32 sq. cm
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the ⓐ volume and ⓑ surface area of the cylinder with given radius 2 ft and height 8 ft.

#solutionbox[
+ ⓐ 100.48 cu. ft
+ ⓑ 125.6 sq. ft
]
]

#examplebox("Example 8")[][
Find the ⓐ volume and ⓑ surface area of a can of soda. The radius of the base is #math.equation(block: false, alt: "4")[$4$] centimeters and the height is #math.equation(block: false, alt: "13")[$13$] centimeters. Assume the can is shaped exactly like a cylinder.

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  [Step 1. #strong[Read] the problem. Draw the figure and  #linebreak() label it with the given information.], [#figure(figph[An illustration of a soda can with a blue and pink abstract design. The can's height is labeled as 13 and its diameter as 4, indicating its geometric dimensions.], alt: "An illustration of a soda can with a blue and pink abstract design. The can's height is labeled as 13 and its diameter as 4, indicating its geometric dimensions.", caption: none)],
))

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([ⓐ], []),
  [Step 2. #strong[Identify] what you are looking for.], [the volume of the cylinder],
  [Step 3. #strong[Name.] Choose a variable to represent it.], [let #emph[V] = volume],
  [Step 4. #strong[Translate.] #linebreak() Write the appropriate formula. #linebreak() Substitute. (Use 3.14 for #math.equation(block: false, alt: "π")[$π$])], [#linebreak() #math.equation(block: false, alt: "V equals π r squared h")[$V = π r^(2) h$] #linebreak() #math.equation(block: false, alt: "V approximately equals open parenthesis 3.14 close parenthesis 4 squared times 13")[$V ≈ ( 3.14 ) 4^(2) ⋅ 13$]],
  [Step 5. #strong[Solve.]], [#math.equation(block: false, alt: "V approximately equals 653.12")[$V ≈ 653.12$]],
  [Step 6. #strong[Check:] We leave it to you to check.], [],
  [Step 7. #strong[Answer] the question.], [The volume is approximately 653.12 cubic centimeters.],
))

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([ⓑ], []),
  [Step 2. #strong[Identify] what you are looking for.], [the surface area of the cylinder],
  [Step 3. #strong[Name.] Choose a variable to represent it.], [let #emph[S] = surface area],
  [Step 4. #strong[Translate.] #linebreak() Write the appropriate formula. #linebreak() Substitute. (Use 3.14 for #math.equation(block: false, alt: "π")[$π$])], [#linebreak() #math.equation(block: false, alt: "S equals 2 π r squared plus 2 π r h")[$S = 2 π r^(2) + 2 π r h$] #linebreak() #math.equation(block: false, alt: "S approximately equals 2 open parenthesis 3.14 close parenthesis 4 squared plus 2 open parenthesis 3.14 close parenthesis open parenthesis 4 close parenthesis 13")[$S ≈ 2 ( 3.14 ) 4^(2) + 2 ( 3.14 ) ( 4 ) 13$]],
  [Step 5. #strong[Solve.]], [#math.equation(block: false, alt: "S approximately equals 427.04")[$S ≈ 427.04$]],
  [Step 6. #strong[Check:] We leave it to you to check your calculations.], [],
  [Step 7. #strong[Answer] the question.], [The surface area is approximately 427.04 square centimeters.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the ⓐ volume and ⓑ surface area of a can of paint with radius 8 centimeters and height 19 centimeters. Assume the can is shaped exactly like a cylinder.

#solutionbox[
+ ⓐ 3,818.24 cu. cm
+ ⓑ 1,356.48 sq. cm
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the ⓐ volume and ⓑ surface area of a cylindrical drum with radius 2.7 feet and height 4 feet. Assume the drum is shaped exactly like a cylinder.

#solutionbox[
+ ⓐ 91.5624 cu. ft
+ ⓑ 113.6052 sq. ft
]
]


=== Find the Volume of Cones

The first image that many of us have when we hear the word ‘cone’ is an ice cream cone. There are many other applications of cones (but most are not as tasty as ice cream cones). In this section, we will see how to find the volume of a cone.

In geometry, a #strong[cone] is a solid figure with one circular base and a vertex. The height of a cone is the distance between its base and the vertex.The cones that we will look at in this section will always have the height perpendicular to the base.

#figure(figph[An image of a cone is shown. The top is labeled vertex. The height is labeled h. The radius of the base is labeled r.], alt: "An image of a cone is shown. The top is labeled vertex. The height is labeled h. The radius of the base is labeled r.", caption: [The height of a cone is the distance between its base and the vertex.])

Earlier in this section, we saw that the volume of a cylinder is #math.equation(block: false, alt: "V equals π r squared h .")[$V = "π" r^(2) h .$] We can think of a cone as part of a cylinder.  shows a cone placed inside a cylinder with the same height and same base. If we compare the volume of the cone and the cylinder, we can see that the volume of the cone is less than that of the cylinder.

#figure(figph[An image of a cone is shown. There is a cylinder drawn around it.], alt: "An image of a cone is shown. There is a cylinder drawn around it.", caption: [The volume of a cone is less than the volume of a cylinder with the same base and height.])

In fact, the volume of a cone is exactly one-third of the volume of a cylinder with the same base and height. The volume of a cone is

#figure(figph[The formula V equals one-third times capital B times h is shown.], alt: "The formula V equals one-third times capital B times h is shown.", caption: none)

Since the base of a cone is a circle, we can substitute the formula of area of a circle, #math.equation(block: false, alt: "π r squared")[$"π" r^(2)$] , for #math.equation(block: false, alt: "B")[$B$]  to get the formula for volume of a cone.

#figure(figph[The formula V equals one-third times pi times r squared times h is shown.], alt: "The formula V equals one-third times pi times r squared times h is shown.", caption: none)

In this book, we will only find the volume of a cone, and not its surface area.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Volume of a Cone]

For a cone with radius #math.equation(block: false, alt: "r")[$r$] and height #math.equation(block: false, alt: "h")[$h$].

#figure(figph[An image of a cone is shown. The height is labeled h, the radius of the base is labeled r. Beside this is Volume: V equals one-third times pi times r squared times h.], alt: "An image of a cone is shown. The height is labeled h, the radius of the base is labeled r. Beside this is Volume: V equals one-third times pi times r squared times h.", caption: none)
]

#examplebox("Example 9")[][
Find the volume of a cone with height #math.equation(block: false, alt: "6")[$6$] inches and radius of its base #math.equation(block: false, alt: "2")[$2$] inches.

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Step 1. #strong[Read] the problem. Draw the figure and label it #linebreak()  with the given information.], [#figure(figph[A 3D image of a cone with its height shown as 6 and its base radius as 2. The circular base is shaded in purple.], alt: "A 3D image of a cone with its height shown as 6 and its base radius as 2. The circular base is shaded in purple.", caption: none)]),
  [Step 2. #strong[Identify] what you are looking for.], [the volume of the cone],
  [Step 3. #strong[Name.] Choose a variable to represent it.], [let #emph[V] = volume],
  [Step 4. #strong[Translate.] #linebreak() Write the appropriate formula. #linebreak() Substitute. (Use 3.14 for #math.equation(block: false, alt: "π")[$π$])], [#linebreak() #math.equation(block: false, alt: "V equals the fraction 1 over 3 π r squared h")[$V = frac(1, 3) #h(1em) π #h(1.9em) r^(2) #h(1.7em) h$] #linebreak() #math.equation(block: false, alt: "V approximately equals the fraction 1 over 3 3.14 open parenthesis 2 close parenthesis squared open parenthesis 6 close parenthesis")[$V ≈ frac(1, 3) #h(0.7em) 3.14 #h(1em) attach(( 2 ), t: 2) #h(1em) ( 6 )$]],
  [Step 5. #strong[Solve.]], [#math.equation(block: false, alt: "V approximately equals 25.12")[$V ≈ 25.12$]],
  [Step 6. #strong[Check:] We leave it to you to check your #linebreak()  calculations.], [],
  [Step 7. #strong[Answer] the question.], [The volume is approximately 25.12 cubic inches.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the volume of a cone with height #math.equation(block: false, alt: "7")[$7$] inches and radius #math.equation(block: false, alt: "3")[$3$] inches

#solutionbox[
65.94 cu. in.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the volume of a cone with height #math.equation(block: false, alt: "9")[$9$] centimeters and radius #math.equation(block: false, alt: "5")[$5$] centimeters

#solutionbox[
235.5 cu. cm
]
]

#examplebox("Example 10")[][
Marty’s favorite gastro pub serves french fries in a paper wrap shaped like a cone. What is the volume of a conic wrap that is #math.equation(block: false, alt: "8")[$8$] inches tall and #math.equation(block: false, alt: "5")[$5$] inches in diameter? Round the answer to the nearest hundredth.

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Step 1. #strong[Read] the problem. Draw the figure and label it with the given information. Notice here that the base is the circle at the top of the cone.], [#figure(figph[An inverted cone is depicted with a radius of 5 units and a height of 8 units. The circular top is shaded to indicate its open face.], alt: "An inverted cone is depicted with a radius of 5 units and a height of 8 units. The circular top is shaded to indicate its open face.", caption: none)]),
  [Step 2. #strong[Identify] what you are looking for.], [the volume of the cone],
  [Step 3. #strong[Name.] Choose a variable to represent it.], [let #emph[V] = volume],
  [Step 4. #strong[Translate.] Write the appropriate formula. Substitute. (Use 3.14 for #math.equation(block: false, alt: "π")[$π$], and notice that we were given the  distance across the circle, which is its diameter. The radius is 2.5 inches.)], [#linebreak() #math.equation(block: false, alt: "V equals the fraction 1 over 3 π r squared h")[$V = frac(1, 3) #h(1em) π #h(2.3em) r^(2) #h(2em) h$] #linebreak() #math.equation(block: false, alt: "V approximately equals the fraction 1 over 3 3.14 open parenthesis 2.5 close parenthesis squared open parenthesis 8 close parenthesis")[$V ≈ frac(1, 3) #h(0.7em) 3.14 #h(1em) attach(( 2.5 ), t: 2) #h(1em) ( 8 )$]],
  [Step 5. #strong[Solve.]], [#math.equation(block: false, alt: "V approximately equals 52.33")[$V ≈ 52.33$]],
  [Step 6. #strong[Check:] We leave it to you to check your calculations.], [],
  [Step 7. #strong[Answer] the question.], [The volume of the wrap is approximately 52.33 cubic inches.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
How many cubic inches of candy will fit in a cone-shaped piñata that is #math.equation(block: false, alt: "18")[$18$] inches long and #math.equation(block: false, alt: "12")[$12$] inches across its base? Round the answer to the nearest hundredth.

#solutionbox[
678.24 cu. in.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
What is the volume of a cone-shaped party hat that is #math.equation(block: false, alt: "10")[$10$] inches tall and #math.equation(block: false, alt: "7")[$7$] inches across at the base? Round the answer to the nearest hundredth.

#solutionbox[
128.2 cu. in.
]
]


=== Summary of Geometry Formulas

The following charts summarize all of the formulas covered in this chapter.

#figure(figph[A table is shown that summarizes all of the formulas in the chapter. The first cell is for Supplementary and Complementary Angles, and says that the measure of angle A plus the measure of angle B equals 180 degrees for supplementary angles A and B and the measure of angle C plus the measure of angle D equals 90 degrees for complementary angles C and D. There is an image of two angles A and B that together form a straight line and two angles C and D that together form a right angle. The next cell says Rectangular Solid and shows the formulas Volume equals LWH and Surface Area equals 2LH plus 2LW plus 2WH. An image of a rectangular solid with sides L, W, and H is shown. The next cell says Triangle. An image of a triangle is shown with sides a, b, and c, vertices A, B, and C, and height h. It says, “For triangle ABC, angle measures measure of angle A plus measure of angle B plus measure of angle C equal 180 degrees. Below this is Perimeter, P equals a plus b plus c. Below this is Area, A equals one-half bh. The next cell says Cube and shows an image of a cube with sides s. It says Volume V equals s cubed and Surface Area S equals 6 times s squared. The next cell says Similar Triangles. It shows two similar triangles ABC and XYZ. It says if triangle ABC is similar to triangle XYZ, then measure of angle A equals measure of angle X, measure of angle B equals measure of angle Y, and measure of angle C equals measure of angle Z. It then says a over x equals b over y equal c over z. The next cell says Sphere and shows an image of a sphere with radius r. It says volume V equals four-thirds times pi times r and Surface Area S equals 4 times pi times r squared. The next cell says Circle. There is an image with two radii labeled r and the diameter labeled d. It says Circumference C equals 2 pi times r and C equals pi times d. It says Area equals pi times r squared. The next cell says Cylinder and shows an image of a cylinder with height h and radius of the base r. It says Volume V equals pi times r squared times h. Below this is V equals Bh. Below that is Surface Area S equals 2 times pi times r squared plus 2 times pi times rh. The next cell says Rectangle and shows an image of a rectangle with sides W and L. It says Perimeter P equals 2L plus 2W, then Area A equals LW. The next cell says Cone and shows an image of a cone with height h and radius of the base r. It says Volume V equals one-third times pi times r squared times h. The last cell says Trapezoid and shows an image of a trapezoid with bases little b and capital B, and height h. It says Area A equals one-half times h times parentheses little b plus capital B.], alt: "A table is shown that summarizes all of the formulas in the chapter. The first cell is for Supplementary and Complementary Angles, and says that the measure of angle A plus the measure of angle B equals 180 degrees for supplementary angles A and B and the measure of angle C plus the measure of angle D equals 90 degrees for complementary angles C and D. There is an image of two angles A and B that together form a straight line and two angles C and D that together form a right angle. The next cell says Rectangular Solid and shows the formulas Volume equals LWH and Surface Area equals 2LH plus 2LW plus 2WH. An image of a rectangular solid with sides L, W, and H is shown. The next cell says Triangle. An image of a triangle is shown with sides a, b, and c, vertices A, B, and C, and height h. It says, “For triangle ABC, angle measures measure of angle A plus measure of angle B plus measure of angle C equal 180 degrees. Below this is Perimeter, P equals a plus b plus c. Below this is Area, A equals one-half bh. The next cell says Cube and shows an image of a cube with sides s. It says Volume V equals s cubed and Surface Area S equals 6 times s squared. The next cell says Similar Triangles. It shows two similar triangles ABC and XYZ. It says if triangle ABC is similar to triangle XYZ, then measure of angle A equals measure of angle X, measure of angle B equals measure of angle Y, and measure of angle C equals measure of angle Z. It then says a over x equals b over y equal c over z. The next cell says Sphere and shows an image of a sphere with radius r. It says volume V equals four-thirds times pi times r and Surface Area S equals 4 times pi times r squared. The next cell says Circle. There is an image with two radii labeled r and the diameter labeled d. It says Circumference C equals 2 pi times r and C equals pi times d. It says Area equals pi times r squared. The next cell says Cylinder and shows an image of a cylinder with height h and radius of the base r. It says Volume V equals pi times r squared times h. Below this is V equals Bh. Below that is Surface Area S equals 2 times pi times r squared plus 2 times pi times rh. The next cell says Rectangle and shows an image of a rectangle with sides W and L. It says Perimeter P equals 2L plus 2W, then Area A equals LW. The next cell says Cone and shows an image of a cone with height h and radius of the base r. It says Volume V equals one-third times pi times r squared times h. The last cell says Trapezoid and shows an image of a trapezoid with bases little b and capital B, and height h. It says Area A equals one-half times h times parentheses little b plus capital B.", caption: none)

#figure(figph[This image shows a row with three columns. The first column says Rectangular solid with the formula below that says volume: V equals LWH. Under this, it says Surface Area: S equals 2LH plus 2LW plus 2WH. An image shows an image of a rectangular solid with the sides labeled L , W and H. The middle column says Rectangle. Under this it says Perimeter P equals 2L plus 2W, then Area A equals LW.  An image of a rectangle with sides W and L. The right column says Cube. Under this it says “Volume: V equals s to the third power.” Under this is says “Surface area: S equals 6 times s squared. Below it is an image of a cube with three sides labeled “s”.], alt: "This image shows a row with three columns. The first column says Rectangular solid with the formula below that says volume: V equals LWH. Under this, it says Surface Area: S equals 2LH plus 2LW plus 2WH. An image shows an image of a rectangular solid with the sides labeled L , W and H. The middle column says Rectangle. Under this it says Perimeter P equals 2L plus 2W, then Area A equals LW. An image of a rectangle with sides W and L. The right column says Cube. Under this it says “Volume: V equals s to the third power.” Under this is says “Surface area: S equals 6 times s squared. Below it is an image of a cube with three sides labeled “s”.", caption: none)

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[ACCESS ADDITIONAL ONLINE RESOURCES]

- #link("https://openstaxcollege.org/l/24volcone")[Volume of a Cone]
]


=== Key Concepts

- #emph[Volume and Surface Area of a Rectangular Solid]
     - #math.equation(block: false, alt: "V equals L W H")[$V = L W H$]
- #math.equation(block: false, alt: "S equals 2 L H plus 2 L W plus 2 W H")[$S = 2 L H + 2 L W + 2 W H$]
- #emph[Volume and Surface Area of a Cube]
     - #math.equation(block: false, alt: "V equals s cubed")[$V = s^(3)$]
- #math.equation(block: false, alt: "S equals 6 s squared")[$S = 6 s^(2)$]
- #emph[Volume and Surface Area of a Sphere]
     - #math.equation(block: false, alt: "V equals the fraction 4 over 3 π r cubed")[$V = frac(4, 3) π r^(3)$]
- #math.equation(block: false, alt: "S equals 4 π r squared")[$S = 4 π r^(2)$]
- #emph[Volume and Surface Area of a Cylinder]
     - #math.equation(block: false, alt: "V equals π r squared h")[$V = π r^(2) h$]
- #math.equation(block: false, alt: "S equals 2 π r squared plus 2 π r h")[$S = 2 π r^(2) + 2 π r h$]
- #emph[Volume of a Cone]
     - For a cone with radius #math.equation(block: false, alt: "r")[$r$] and height #math.equation(block: false, alt: "h")[$h$]: #linebreak() 
Volume: #math.equation(block: false, alt: "V equals the fraction 1 over 3 π r squared h")[$V = frac(1, 3) π r^(2) h$]


==== Practice Makes Perfect

#strong[Find Volume and Surface Area of Rectangular Solids]

In the following exercises, find ⓐ the volume and ⓑ the surface area of the rectangular solid with the given dimensions.

length #math.equation(block: false, alt: "2")[$2$] meters, width #math.equation(block: false, alt: "1.5")[$1.5$] meters, height #math.equation(block: false, alt: "3")[$3$] meters

#solutionbox[
+ ⓐ 9 cu. m
+ ⓑ 27 sq. m
]

length #math.equation(block: false, alt: "5")[$5$] feet, width #math.equation(block: false, alt: "8")[$8$] feet, height #math.equation(block: false, alt: "2.5")[$2.5$] feet

length #math.equation(block: false, alt: "3.5")[$3.5$] yards, width #math.equation(block: false, alt: "2.1")[$2.1$] yards, height #math.equation(block: false, alt: "2.4")[$2.4$] yards

#solutionbox[
+ ⓐ 17.64 cu. yd.
+ ⓑ 41.58 sq. yd.
]

length #math.equation(block: false, alt: "8.8")[$8.8$] centimeters, width #math.equation(block: false, alt: "6.5")[$6.5$] centimeters, height #math.equation(block: false, alt: "4.2")[$4.2$] centimeters

In the following exercises, solve.

#strong[Moving van] A rectangular moving van has length #math.equation(block: false, alt: "16")[$16$] feet, width #math.equation(block: false, alt: "8")[$8$] feet, and height #math.equation(block: false, alt: "8")[$8$] feet. Find its ⓐ volume and ⓑ surface area.

#solutionbox[
+ ⓐ 1,024 cu. ft
+ ⓑ 640 sq. ft
]

#strong[Gift box] A rectangular gift box has length #math.equation(block: false, alt: "26")[$26$] inches, width #math.equation(block: false, alt: "16")[$16$] inches, and height
#math.equation(block: false, alt: "4")[$4$] inches. Find its ⓐ volume and ⓑ surface area.

#strong[Carton] A rectangular carton has length
#math.equation(block: false, alt: "21.3")[$21.3$] cm, width #math.equation(block: false, alt: "24.2")[$24.2$] cm, and height #math.equation(block: false, alt: "6.5")[$6.5$] cm. Find its ⓐ volume and ⓑ surface area.

#solutionbox[
+ ⓐ 3,350.49 cu. cm
+ ⓑ 1,622.42 sq. cm
]

#strong[Shipping container] A rectangular shipping container has length #math.equation(block: false, alt: "22.8")[$22.8$] feet, width #math.equation(block: false, alt: "8.5")[$8.5$] feet, and height #math.equation(block: false, alt: "8.2")[$8.2$] feet. Find its ⓐ volume and ⓑ surface area.

In the following exercises, find ⓐ the volume and ⓑ the surface area of the cube with the given side length.

#math.equation(block: false, alt: "5")[$5$] centimeters

#solutionbox[
+ ⓐ 125 cu. cm
+ ⓑ 150 sq. cm
]

#math.equation(block: false, alt: "6")[$6$] inches

#math.equation(block: false, alt: "10.4")[$10.4$] feet

#solutionbox[
+ ⓐ 1124.864 cu. ft.
+ ⓑ 648.96 sq. ft
]

#math.equation(block: false, alt: "12.5")[$12.5$] meters

In the following exercises, solve.

#strong[Science center] Each side of the cube at the Discovery Science Center in Santa Ana is #math.equation(block: false, alt: "64")[$64$] feet long. Find its ⓐ volume and ⓑ surface area.

#solutionbox[
+ ⓐ 262,144 cu. ft
+ ⓑ 24,576 sq. ft
]

#strong[Museum] A cube-shaped museum has sides #math.equation(block: false, alt: "45")[$45$] meters long. Find its ⓐ volume and ⓑ surface area.

#strong[Base of statue] The base of a statue is a cube with sides #math.equation(block: false, alt: "2.8")[$2.8$] meters long. Find its ⓐ volume and ⓑ surface area.

#solutionbox[
+ ⓐ 21.952 cu. m
+ ⓑ 47.04 sq. m
]

#strong[Tissue box] A box of tissues is a cube with sides 4.5 inches long. Find its ⓐ volume and ⓑ surface area.

#strong[Find the Volume and Surface Area of Spheres]

In the following exercises, find ⓐ the volume and ⓑ the surface area of the sphere with the given radius. Round answers to the nearest hundredth.

#math.equation(block: false, alt: "3")[$3$] centimeters

#solutionbox[
+ ⓐ 113.04 cu. cm
+ ⓑ 113.04 sq. cm
]

#math.equation(block: false, alt: "9")[$9$] inches

#math.equation(block: false, alt: "7.5")[$7.5$] feet

#solutionbox[
+ ⓐ 1,766.25 cu. ft
+ ⓑ 706.5 sq. ft
]

#math.equation(block: false, alt: "2.1")[$2.1$] yards

In the following exercises, solve. Round answers to the nearest hundredth.

#strong[Exercise ball] An exercise ball has a radius of #math.equation(block: false, alt: "15")[$15$] inches. Find its ⓐ volume and ⓑ surface area.

#solutionbox[
+ ⓐ 14,130 cu. in.
+ ⓑ 2,826 sq. in.
]

#strong[Balloon ride] The Great Park Balloon is a big orange sphere with a radius of #math.equation(block: false, alt: "36")[$36$] feet . Find its ⓐ volume and ⓑ surface area.

#strong[Golf ball] A golf ball has a radius of #math.equation(block: false, alt: "4.5")[$4.5$] centimeters. Find its ⓐ volume and ⓑ surface area.

#solutionbox[
+ ⓐ 381.51 cu. cm
+ ⓑ 254.34 sq. cm
]

#strong[Baseball] A baseball has a radius of #math.equation(block: false, alt: "2.9")[$2.9$] inches. Find its ⓐ volume and ⓑ surface area.

#strong[Find the Volume and Surface Area of a Cylinder]

In the following exercises, find ⓐ the volume and ⓑ the surface area of the cylinder with the given radius and height. Round answers to the nearest hundredth.

radius #math.equation(block: false, alt: "3")[$3$] feet, height #math.equation(block: false, alt: "9")[$9$] feet

#solutionbox[
+ ⓐ 254.34 cu. ft
+ ⓑ 226.08 sq. ft
]

radius #math.equation(block: false, alt: "5")[$5$] centimeters, height #math.equation(block: false, alt: "15")[$15$] centimeters

radius #math.equation(block: false, alt: "1.5")[$1.5$] meters, height #math.equation(block: false, alt: "4.2")[$4.2$] meters

#solutionbox[
+ ⓐ 29.673 cu. m
+ ⓑ 53.694 sq. m
]

radius #math.equation(block: false, alt: "1.3")[$1.3$] yards, height #math.equation(block: false, alt: "2.8")[$2.8$] yards

In the following exercises, solve. Round answers to the nearest hundredth.

#strong[Coffee can] A can of coffee has a radius of #math.equation(block: false, alt: "5")[$5$] cm and a height of #math.equation(block: false, alt: "13")[$13$] cm. Find its ⓐ volume and ⓑ surface area.

#solutionbox[
+ ⓐ 1,020.5 cu. cm
+ ⓑ 565.2 sq. cm
]

#strong[Snack pack] A snack pack of cookies is shaped like a cylinder with radius #math.equation(block: false, alt: "4")[$4$] cm and height #math.equation(block: false, alt: "3")[$3$] cm. Find its ⓐ volume and ⓑ surface area.

#strong[Barber shop pole] A cylindrical barber shop pole has a diameter of #math.equation(block: false, alt: "6")[$6$] inches and height of #math.equation(block: false, alt: "24")[$24$] inches. Find its ⓐ volume and ⓑ surface area.

#solutionbox[
+ ⓐ 678.24 cu. in.
+ ⓑ 508.68 sq. in.
]

#strong[Architecture] A cylindrical column has a diameter of #math.equation(block: false, alt: "8")[$8$] feet and a height of #math.equation(block: false, alt: "28")[$28$] feet. Find its ⓐ volume and ⓑ surface area.

#strong[Find the Volume of Cones]

In the following exercises, find the volume of the cone with the given dimensions. Round answers to the nearest hundredth.

height #math.equation(block: false, alt: "9")[$9$] feet and radius #math.equation(block: false, alt: "2")[$2$] feet

#solutionbox[
37.68 cu. ft
]

height #math.equation(block: false, alt: "8")[$8$] inches and radius #math.equation(block: false, alt: "6")[$6$] inches

height #math.equation(block: false, alt: "12.4")[$12.4$] centimeters and radius #math.equation(block: false, alt: "5")[$5$] cm

#solutionbox[
324.47 cu. cm
]

height #math.equation(block: false, alt: "15.2")[$15.2$] meters and radius #math.equation(block: false, alt: "4")[$4$] meters

In the following exercises, solve. Round answers to the nearest hundredth.

#strong[Teepee] What is the volume of a cone-shaped teepee tent that is #math.equation(block: false, alt: "10")[$10$] feet tall and #math.equation(block: false, alt: "10")[$10$] feet across at the base?

#solutionbox[
261.67 cu. ft
]

#strong[Popcorn cup] What is the volume of a cone-shaped popcorn cup that is #math.equation(block: false, alt: "8")[$8$] inches tall and #math.equation(block: false, alt: "6")[$6$] inches across at the base?

#strong[Silo] What is the volume of a cone-shaped silo that is #math.equation(block: false, alt: "50")[$50$] feet tall and #math.equation(block: false, alt: "70")[$70$] feet across at the base?

#solutionbox[
64,108.33 cu. ft
]

#strong[Sand pile] What is the volume of a cone-shaped pile of sand that is #math.equation(block: false, alt: "12")[$12$] meters tall and #math.equation(block: false, alt: "30")[$30$] meters across at the base?


==== Everyday Math

#strong[Street light post] The post of a street light is shaped like a truncated cone, as shown in the picture below. It is a large cone minus a smaller top cone. The large cone is #math.equation(block: false, alt: "30")[$30$] feet tall with base radius #math.equation(block: false, alt: "1")[$1$] foot. The smaller cone is #math.equation(block: false, alt: "10")[$10$] feet tall with base radius of #math.equation(block: false, alt: "0.5")[$0.5$] feet. To the nearest tenth,

+ ⓐ find the volume of the large cone.
+ ⓑ find the volume of the small cone.
+ ⓒ find the volume of the post by subtracting the volume of the small cone from the volume of the large cone.


#figure(figph[An image of a cone is shown. There is a dark dotted line at the top indicating a smaller cone.], alt: "An image of a cone is shown. There is a dark dotted line at the top indicating a smaller cone.", caption: none)

#solutionbox[
+ ⓐ 31.4 cu. ft
+ ⓑ 2.6 cu. ft
+ ⓒ 28.8 cu. ft
]

#strong[Ice cream cones] A regular ice cream cone is 4 inches tall and has a diameter of #math.equation(block: false, alt: "2.5")[$2.5$] inches. A waffle cone is #math.equation(block: false, alt: "7")[$7$] inches tall and has a diameter of #math.equation(block: false, alt: "3.25")[$3.25$] inches. To the nearest hundredth,

+ ⓐ find the volume of the regular ice cream cone.
+ ⓑ find the volume of the waffle cone.
+ ⓒ how much more ice cream fits in the waffle cone compared to the regular cone?


==== Writing Exercises

The formulas for the volume of a cylinder and a cone are similar. Explain how you can remember which formula goes with which shape.

#solutionbox[
Answers will vary.
]

Which has a larger volume, a cube of sides of #math.equation(block: false, alt: "8")[$8$] feet or a sphere with a diameter of #math.equation(block: false, alt: "8")[$8$] feet? Explain your reasoning.


==== Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

#figure(figph[A self-assessment table for students to rate their understanding of finding volume and surface area for rectangular solids, spheres, cylinders, and volume of cones. The options are Confidently, With some help, and No-I don't get it!], alt: "A self-assessment table for students to rate their understanding of finding volume and surface area for rectangular solids, spheres, cylinders, and volume of cones. The options are Confidently, With some help, and No-I don't get it!", caption: none)

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?