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== 10.7#h(0.6em)Volume and Surface Area


  #figure(figph[A 3D floor plan diagram.], alt: "A 3D floor plan diagram.", caption: [Volume is illustrated in this 3-dimensional view of an interior space. This gives a buyer a more realistic interpretation of space.])


  
    
=== Learning Objectives


    After completing this section, you should be able to:


    + Calculate the surface area of right prisms and cylinders.
+ Calculate the volume of right prisms and cylinders.
+ Solve application problems involving surface area and volume.


  
  Volume and surface area are two measurements that are part of our daily lives. We use volume every day, even though we do not focus on it. When you purchase groceries, volume is the key to pricing. Judging how much paint to buy or how many square feet of siding to purchase is based on surface area. The list goes on. An example is a three-dimensional rendering of a floor plan. These types of drawings make building layouts far easier to understand for the client. It allows the viewer a realistic idea of the product at completion; you can see the natural space, the volume of the rooms. This section gives you practical information you will use consistently. You may not remember every formula, but you will remember the concepts, and you will know where to go should you want to calculate volume or surface area in the future.


  We will concentrate on a few particular types of three-dimensional objects: right prisms and right cylinders. The adjective “right” refers to objects such that the sides form a right angle with the base. We will look at right rectangular prisms, right triangular prisms, right hexagonal prisms, right octagonal prisms, and right cylinders. Although, the principles learned here apply to all right prisms.


  
    
=== Three-Dimensional Objects


    In geometry, three-dimensional objects are called #strong[geometric solids]. Surface area refers to the flat surfaces that surround the solid and is measured in square units. Volume refers to the space inside the solid and is measured in cubic units. Imagine that you have a square flat surface with width and length. Adding the third dimension adds depth or height, depending on your viewpoint, and now you have a box. One way to view this concept is in the Cartesian coordinate three-dimensional space. The #math.equation(block: false, alt: "x")[$x$]-axis and the #math.equation(block: false, alt: "y")[$y$]-axis are, as you would expect, two dimensions and suitable for plotting two-dimensional graphs and shapes. Adding the #math.equation(block: false, alt: "z")[$z$]-axis, which shoots through the origin perpendicular to the #math.equation(block: false, alt: "x y")[$x y$]-plane, and we have a third dimension.


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    Here is another view taking the two-dimensional square to a third dimension.


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    To study objects in three dimensions, we need to consider the formulas for surface area and volume. For example, suppose you have a box  with a hinged lid that you want to use for keeping photos, and you want to cover the box with a decorative paper. You would need to find the surface area to calculate how much paper is needed. Suppose you need to know how much water will fill your swimming pool. In that case, you would need to calculate the volume of the pool. These are just a couple of examples of why these concepts should be understood, and familiarity with the formulas will allow you to make use of these ideas as related to right prisms and right cylinders.


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=== Right Prisms


    A #strong[right prism] is a particular type of three-dimensional object. It has a polygon-shaped base and a congruent, regular polygon-shaped top, which are connected by the height of its lateral sides, as shown. The lateral sides form a right angle with the base and the top. There are rectangular prisms, hexagonal prisms, octagonal prisms, triangular prisms, and so on.


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    Generally, to calculate surface area, we find the area of each side of the object and add the areas together. To calculate volume, we calculate the space inside the solid bounded by its sides.


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    #examplebox("Example 1")[Calculating Surface Area and Volume of a Rectangular Prism][
Find the surface area and volume of the rectangular prism that has a width of 10 cm, a length of 5 cm, and a height of 3 cm .


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      #solutionbox[
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        The volume is #math.equation(block: false, alt: "V equals 10 open parenthesis 5 close parenthesis open parenthesis 3 close parenthesis equals 150 cm cubed .")[$V = 10 ( 5 ) ( 3 ) = 150 #h(0.28em) "cm"^(3) .$]
]
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    In , we have three views of a right hexagonal prism. The regular hexagon is the base and top, and the lateral faces are the rectangular regions perpendicular to the base. We call it a right prism because the angle formed by the lateral sides to the base is #math.equation(block: false, alt: "90 degrees .")[$90^(∘) .$]


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)


    The first image is a view of the figure straight on with no rotation in any direction. The middle figure is the base or the top. The last figure shows you the solid in three dimensions. To calculate the surface area of the right prism shown in , we first determine the area of the hexagonal base and multiply that by 2, and then add the perimeter of the base times the height. Recall the area of a regular polygon is given as #math.equation(block: false, alt: "A equals the fraction 1 over 2 a p ,")[$A = frac(1, 2) a p ,$] where #math.equation(block: false, alt: "a")[$a$] is the apothem and #math.equation(block: false, alt: "p")[$p$] is the perimeter. We have that


    #math.equation(block: true, alt: "A sub b a s e equals the fraction 1 over 2 open parenthesis 5.2 close parenthesis open parenthesis 36 close parenthesis equals 93.6 cm squared")[$A_(b a s e) = frac(1, 2) ( 5.2 ) ( 36 ) = 93.6 #h(0.28em) "cm"^(2)$]
    Then, the surface area of the hexagonal prism is


    #math.equation(block: true, alt: "S A equals 2 open parenthesis 93.6 close parenthesis plus 36 open parenthesis 20 close parenthesis equals 907.2 in squared")[$S A = 2 ( 93.6 ) + 36 ( 20 ) = 907.2 #h(0.28em) "in"^(2)$]
    To find the volume of the right hexagonal prism, we multiply the area of the base by the height using the formula #math.equation(block: false, alt: "V equals B h .")[$V = B h .$] The base is #math.equation(block: false, alt: "93.6 cm squared ,")[$93.6 #h(0.28em) "cm"^(2) ,$] and the height is 20 #math.equation(block: false, alt: "cm")[$"cm"$]. Thus,


    #math.equation(block: true, alt: "V equals 93.6 open parenthesis 20 close parenthesis equals 1872 cm cubed .")[$V = 93.6 ( 20 ) = 1872 #h(0.28em) "cm"^(3) .$]
    
    #examplebox("Example 2")[Calculating the Surface Area of a Right Triangular Prism][
Find the surface area of the triangular prism .


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)


      
      #solutionbox[
The area of the triangular base is #math.equation(block: false, alt: "A sub b a s e equals the fraction 1 over 2 open parenthesis 12 close parenthesis open parenthesis 6 close parenthesis equals 36 in squared")[$A_(b a s e) = frac(1, 2) ( 12 ) ( 6 ) = 36 #h(0.28em) "in"^(2)$]. The perimeter of the base is #math.equation(block: false, alt: "p equals 12 plus 8.49 plus 8.49 equals 28.98 in .")[$p = 12 + 8.49 + 8.49 = 28.98 #h(0.28em) "in" .$] Then, the surface area of the triangular prism is #math.equation(block: false, alt: "S A equals 2 open parenthesis 36 close parenthesis plus 28.98 open parenthesis 10 close parenthesis equals 361.8 in squared .")[$S A = 2 ( 36 ) + 28.98 ( 10 ) = 361.8 #h(0.28em) "in"^(2) .$]
]
]


    
    #examplebox("Example 3")[Finding the Surface Area and Volume][
Find the surface area and the volume of the right triangular prism with an equilateral triangle as the base and height .


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  alt: "A right triangular prism. One of the sides of the triangle is labeled l equals 12 centimeters. The height of the triangle is labeled h equals 10.39 centimeters. The height of the prism is labeled h equals 10 centimeters.", caption: none,
)


      
      #solutionbox[
The area of the triangular base is #math.equation(block: false, alt: "A sub b a s e equals the fraction 1 over 2 open parenthesis 6 close parenthesis open parenthesis 10.39 close parenthesis equals 31.17 cm squared")[$A_(b a s e) = frac(1, 2) ( 6 ) ( 10.39 ) = 31.17 #h(0.28em) "cm"^(2)$]. Then, the surface area is #math.equation(block: false, alt: "S A equals 2 open parenthesis 31.17 close parenthesis plus 36 open parenthesis 10 close parenthesis equals 422.34 cm squared .")[$S A = 2 ( 31.17 ) + 36 ( 10 ) = 422.34 #h(0.28em) "cm"^(2) .$]


        The volume formula is found by multiplying the area of the base by the height. We have that #math.equation(block: false, alt: "V equals B times h equals open parenthesis 31.17 close parenthesis open parenthesis 10 close parenthesis equals 311.7 cm cubed .")[$V = B ⋅ h = ( 31.17 ) ( 10 ) = 311.7 #h(0.28em) "cm"^(3) .$]
]
]


    
    #examplebox("Example 4")[Determining Surface Area Application][
Katherine and Romano built a greenhouse made of glass with a metal roof . In order to determine the heating and cooling requirements, the surface area must be known. Calculate the total surface area of the greenhouse.


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)


      
      #solutionbox[
The area of the long side measures #math.equation(block: false, alt: "95 ft times 6.5 ft equals 1,235 ft squared .")[$95 #h(0.28em) "ft" × 6.5 #h(0.28em) "ft" = 1,235 #h(0.28em) "ft"^(2) .$] Multiplying by 2 gives #math.equation(block: false, alt: "2,470 ft squared .")[$2,470 #h(0.28em) "ft"^(2) .$] The front (minus the triangular area) measures #math.equation(block: false, alt: "22 ft times 6.5 ft equals 286 ft squared .")[$22 #h(0.28em) "ft" × 6.5 #h(0.28em) "ft" = 286 #h(0.28em) "ft"^(2) .$] Multiplying by 2 gives #math.equation(block: false, alt: "572 ft squared .")[$572 #h(0.28em) "ft"^(2) .$] The floor measures #math.equation(block: false, alt: "95 ft times 22 ft equals 2,090 ft squared .")[$95 #h(0.28em) "ft" × 22 #h(0.28em) "ft" = 2,090 #h(0.28em) "ft"^(2) .$] Each triangular region measures #math.equation(block: false, alt: "A equals the fraction 1 over 2 open parenthesis 22 close parenthesis open parenthesis 5 close parenthesis equals 55 ft squared .")[$A = frac(1, 2) ( 22 ) ( 5 ) = 55 #h(0.28em) "ft"^(2) .$] Multiplying by 2 gives #math.equation(block: false, alt: "110 ft squared .")[$110 #h(0.28em) "ft"^(2) .$] Finally, one side of the roof measures #math.equation(block: false, alt: "12.1 ft times 95 ft equals 1,149.5 ft squared .")[$12.1 #h(0.28em) "ft" × 95 #h(0.28em) "ft" = 1,149.5 #h(0.28em) "ft"^(2) .$] Multiplying by 2 gives #math.equation(block: false, alt: "2299 ft squared .")[$2299 #h(0.28em) "ft"^(2) .$] Add them up and we have #math.equation(block: false, alt: "S A equals 2,470 plus 572 plus 2,090 plus 110 plus 2,299 equals 7,541 ft squared .")[$S A = 2,470 + 572 + 2,090 + 110 + 2,299 = 7,541 #h(0.28em) "ft"^(2) .$]
]
]


    
  
  
    
=== Right Cylinders


    There are similarities between a prism and a cylinder. While a prism has parallel congruent polygons as the top and the base, a #strong[right cylinder] is a three-dimensional object with congruent circles as the top and the base. The lateral sides of a right prism make a #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] angle with the polygonal base, and the side of a cylinder, which unwraps as a rectangle, makes a #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] angle with the circular base.


    Right cylinders are very common in everyday life. Think about soup cans, juice cans, soft drink cans, pipes, air hoses, and the list goes on.


    In , imagine that the cylinder is cut down the 12-inch side and rolled out. We can see that the cylinder side when flat forms a rectangle. The #math.equation(block: false, alt: "S A")[$S A$] formula includes the area of the circular base, the circular top, and the area of the rectangular side. The length of the rectangular side is the circumference of the circular base. Thus, we have the formula for total surface area of a right cylinder.


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  content((74.43pt, 164.00pt), anchor: "south", text(size: 9pt)[r = 5 inches])
  content((16.00pt, 90.00pt), anchor: "east", text(size: 9pt)[12 inches])
})]),
  alt: "Two views of a right cylinder. The first view shows the top view of the right cylinder. The radius is marked r equals 5 inches. The second view shows the front view of the right cylinder. The height of the cylinder is labeled 12 inches.", caption: [Right Cylinder],
)


    #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[
The surface area of a right cylinder is given as #math.equation(block: false, alt: "S A equals 2 π r squared plus 2 π r h .")[$S A = 2 π r^(2) + 2 π r h .$]
]


    To find the volume of the cylinder, we multiply the area of the base with the height.


    #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[
The volume of a right cylinder is given as #math.equation(block: false, alt: "V equals π r squared h .")[$V = π r^(2) h .$]
]


    #examplebox("Example 5")[Finding the Surface Area and Volume of a Cylinder][
Given the cylinder in , which has a radius of 5 inches and a height of 12 inches, find the surface area and the volume.


      
      #solutionbox[
#strong[Step 1:] We begin with the areas of the base and the top. The area of the circular base is


        #math.equation(block: true, alt: "A sub b a s e equals π open parenthesis 5 close parenthesis squared equals 25 π equals 78.5 in squared .")[$A_(b a s e) = π attach(( 5 ), t: 2) = 25 π = 78.5 #h(0.28em) "in"^(2) .$]
        
        #strong[Step 2:] The base and the top are congruent parallel planes. Therefore, the area for the base and the top is


        #math.equation(block: true, alt: "A equals 2 open parenthesis 78.5 close parenthesis equals 157 in squared .")[$A = 2 ( 78.5 ) = 157 #h(0.28em) "in"^(2) .$]
        
        #strong[Step 3:] The area of the rectangular side is equal to the circumference of the base times the height:


        #math.equation(block: true, alt: "A equals 2 π open parenthesis 5 close parenthesis open parenthesis 12 close parenthesis equals 377 in squared")[$A = 2 π ( 5 ) ( 12 ) = 377 #h(0.28em) "in"^(2)$]
        
        #strong[Step 4:] We add the area of the side to the areas of the base and the top to get the total surface area. We have


        #math.equation(block: true, alt: "S A equals 157 plus 377 equals 534 in squared")[$S A = 157 + 377 = 534 #h(0.28em) "in"^(2)$]
        
        #strong[Step 5:] The volume is equal to the area of the base times the height. Then,


        #math.equation(block: true, alt: "V equals π open parenthesis 5 close parenthesis squared open parenthesis 12 close parenthesis equals 942.48 in cubed .")[$V = π attach(( 5 ), t: 2) ( 12 ) = 942.48 #h(0.28em) "in"^(3) .$]
]
]


    
  
  
    
=== Applications of Surface Area and Volume


    The following are just a small handful of the types of applications in which surface area and volume are critical factors. Give this a little thought and you will realize many more practical uses for these procedures.


    #examplebox("Example 6")[Applying a Calculation of Volume][
A can of apple pie filling has a radius of 4 cm and a height of 10 cm. How many cans are needed to fill a pie pan  measuring 22 cm in diameter and 3 cm deep?


        #figure(figph[A pie pan with its diameter and height marked 22 centimeters and 3 centimeters.], alt: "A pie pan with its diameter and height marked 22 centimeters and 3 centimeters.", caption: none)


      
      #solutionbox[
The volume of the can of apple pie filling is #math.equation(block: false, alt: "V equals π open parenthesis 4 close parenthesis squared open parenthesis 10 close parenthesis equals 502.7 cm cubed .")[$V = π attach(( 4 ), t: 2) ( 10 ) = 502.7 #h(0.28em) "cm"^(3) .$] The volume of the pan is #math.equation(block: false, alt: "V equals π open parenthesis 11 close parenthesis squared open parenthesis 3 close parenthesis equals 1140.4 cm cubed .")[$V = π attach(( 11 ), t: 2) ( 3 ) = 1140.4 #h(0.28em) "cm"^(3) .$] To find the number of cans of apple pie filling, we divide the volume of the pan by the volume of a can of apple pie filling. Thus, #math.equation(block: false, alt: "1140 divided by 502.7 equals 2.3.")[$1140 ÷ 502.7 = 2.3.$] We will need 2.3 cans of apple pie filling to fill the pan.
]
]


    
  
  
    
=== Optimization


    Problems that involve optimization are ones that look for the best solution to a situation under some given conditions. Generally, one looks to calculus to solve these problems. However, many geometric applications can be solved with the tools learned in this section. Suppose you want to make some throw pillows for your sofa, but you have a limited amount of fabric. You want to make the largest pillows you can from the fabric you have, so you would need to figure out the dimensions of the pillows that will fit these criteria. Another situation might be that you want to fence off an area in your backyard for a garden. You have a limited amount of fencing available, but you would like the garden to be as large as possible. How would you determine the shape and size of the garden? Perhaps you are looking for maximum volume or minimum surface area. Minimum cost is also a popular application of optimization. Let’s explore a few examples.


    #examplebox("Example 7")[Maximizing Area][
Suppose you have 150 meters of fencing that you plan to use for the enclosure of a corral on a ranch. What shape would give the greatest possible area?


      
      #solutionbox[
So, how would we start? Let’s look at this on a smaller scale. Say that you have 30 inches of string and you experiment with different shapes. The rectangle in  measures 12 in long by 3 in wide. We have a perimeter of #math.equation(block: false, alt: "P equals 2 open parenthesis 12 close parenthesis plus 2 open parenthesis 3 close parenthesis equals 30 in")[$P = 2 ( 12 ) + 2 ( 3 ) = 30 #h(0.28em) "in"$] and the area calculates as #math.equation(block: false, alt: "A equals 3 open parenthesis 12 close parenthesis equals 36 in squared .")[$A = 3 ( 12 ) = 36 #h(0.28em) "in"^(2) .$] The rectangle in , measures 8 in long and 7 in wide. The perimeter is #math.equation(block: false, alt: "P equals 2 open parenthesis 8 close parenthesis plus 2 open parenthesis 7 close parenthesis equals 30 in")[$P = 2 ( 8 ) + 2 ( 7 ) = 30 #h(0.28em) "in"$] and the area is #math.equation(block: false, alt: "A equals 8 open parenthesis 7 close parenthesis equals 56 in squared .")[$A = 8 ( 7 ) = 56 #h(0.28em) "in"^(2) .$] In , the square measures 7.5 in on each side. The perimeter is then #math.equation(block: false, alt: "P equals 4 open parenthesis 7.5 close parenthesis equals 30 in")[$P = 4 ( 7.5 ) = 30 #h(0.28em) "in"$] and the area is #math.equation(block: false, alt: "A equals 7.5 open parenthesis 7.5 close parenthesis equals 56.25 in squared .")[$A = 7.5 ( 7.5 ) = 56.25 #h(0.28em) "in"^(2) .$] If you want a circular corral as in , we would consider a circumference of #math.equation(block: false, alt: "30 equals 2 π r ,")[$30 = 2 π r ,$] which gives a radius of #math.equation(block: false, alt: "30 divided by 2 π equals 4.77 in")[$30 ÷ 2 π = 4.77 #h(0.28em) "in"$] and an area of #math.equation(block: false, alt: "A equals π open parenthesis 4.77 close parenthesis squared equals 71.5 in squared .")[$A = π attach(( 4.77 ), t: 2) = 71.5 #h(0.28em) "in"^(2) .$]


        #figure(
  stdfig([#grid(columns: 4, gutter: 1.5em,
  [#cetz.canvas({
  import cetz.draw: *
  content((100.00pt, 80.67pt), text(font: sys.inputs.at("title-font", default: "STIX Two Text"), size: 11pt)[Rectangle 12 in x 3 in])
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  [#cetz.canvas({
  import cetz.draw: *
  content((100.00pt, 174.00pt), text(font: sys.inputs.at("title-font", default: "STIX Two Text"), size: 11pt)[Rectangle 8 in x 7 in])
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  content((171.00pt, 80.00pt), anchor: "south", text(size: 9pt)[7 in])
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  line((20.00pt, 140.00pt), (20.00pt, 20.00pt), stroke: 0.7pt + black)
})],
  [#cetz.canvas({
  import cetz.draw: *
  content((100.00pt, 214.00pt), text(font: sys.inputs.at("title-font", default: "STIX Two Text"), size: 11pt)[Square 7.5 in])
  line((20.00pt, 20.00pt), (170.00pt, 20.00pt), stroke: 0.7pt + black)
  content((95.00pt, 29.00pt), anchor: "south", text(size: 9pt)[7.5 in])
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  line((20.00pt, 170.00pt), (20.00pt, 20.00pt), stroke: 0.7pt + black)
})],
  [#cetz.canvas({
  import cetz.draw: *
  content((100.00pt, 214.00pt), text(font: sys.inputs.at("title-font", default: "STIX Two Text"), size: 11pt)[Circle with radius 4.77 in])
  circle((100.00pt, 100.00pt), radius: 79.50pt, fill: none, stroke: 0.7pt + black)
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  content((139.75pt, 109.00pt), anchor: "south", text(size: 9pt)[r = 4.77 in])
  circle((100.00pt, 100.00pt), radius: 1.8pt, fill: black, stroke: none)
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  circle((179.50pt, 100.00pt), radius: 1.8pt, fill: black, stroke: none)
  content((183.50pt, 100.00pt), anchor: "west", text(size: 9pt)[A])
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)]),
  alt: "Four shapes. a. A rectangle with its length and width marked 12 inches and 3 inches. b. A rectangle with its length and width marked 8 inches and 7 inches. c. A square with its sides marked 7.5 inches. d. A circle with its radius marked r equals 4.77 inches.", caption: none,
)


        We see that the circular shape gives the maximum area relative to a circumference of 30 in. So, a circular corral with a circumference of 150 meters and a radius of 23.87 meters gives a maximum area of #math.equation(block: false, alt: "1,790.5 m squared .")[$1,790.5 #h(0.28em) "m"^(2) .$]
]
]


    
    #examplebox("Example 8")[Designing for Cost][
Suppose you want to design a crate built out of wood in the shape of a rectangular prism . It must have a volume of 3 cubic meters. The cost of wood is \$15 per square meter. What dimensions give the most economical design while providing the necessary volume?


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((24.00pt, 57.00pt), (81.16pt, 90.00pt), stroke: (paint: black, thickness: 0.6pt, dash: "dashed"))
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  content((52.58pt, 32.50pt), anchor: "north", text(size: 9pt)[1.44 meters])
  content((117.74pt, 40.50pt), anchor: "west", text(size: 9pt)[1.44 meters])
  content((16.00pt, 90.00pt), anchor: "east", text(size: 9pt)[1.44 meters])
})]),
  alt: "A crate in the shape of a rectangular prism with its length, width, and height marked 1.44 meters.", caption: none,
)


      
      #solutionbox[
To choose the optimal shape for this container, you can start by experimenting with different sizes of boxes that will hold 3 cubic meters. It turns out that, similar to the maximum rectangular area example where a square gives the maximum area, a cube gives the maximum volume and the minimum surface area.


        As all six sides are the same, we can use a simplified volume formula:


        #math.equation(block: true, alt: "V equals s cubed ,")[$V = s^(3) ,$]


        where #math.equation(block: false, alt: "s")[$s$] is the length of a side. Then, to find the length of a side, we take the cube root of the volume.


        We have


        #math.equation(block: true, alt: "3, equals, s cubed; the cube root of 3, equals, s; equals, 1.4422 m")[$3 & = & s^(3) \ root(3, 3) & = & s \  & = & 1.4422 #h(0.28em) "m"$]


        The surface area is equal to the sum of the areas of the six sides. The area of one side is #math.equation(block: false, alt: "A equals open parenthesis 1.4422 close parenthesis squared equals 2.08 m squared .")[$A = attach(( 1.4422 ), t: 2) = 2.08 #h(0.28em) "m"^(2) .$] So, the surface area of the crate is #math.equation(block: false, alt: "S A equals 6 open parenthesis 2.08 close parenthesis equals 12.5 m squared")[$S A = 6 ( 2.08 ) = 12.5 #h(0.28em) "m"^(2)$]. At \$15 a square meter, the cost comes to #math.equation(block: false, alt: "12.5 open parenthesis $ 15 close parenthesis equals $ 187.50.")[$12.5 ( "$" 15 ) = "$" 187.50.$] Checking the volume, we have #math.equation(block: false, alt: "V equals open parenthesis 1.4422 close parenthesis cubed equals 2.99 m cubed")[$V = attach(( 1.4422 ), t: 3) = 2.99 #h(0.28em) "m"^(3)$].
]
]


    
  
  
  
  
    
=== Key Terms


    - surface area
- volume
- right prism
- right cylinder


  
  
    
=== Key Concepts


    - A right prism is a three-dimensional object that has a regular polygonal face and congruent base such that that lateral sides form a #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] angle with the base and top. The surface area #math.equation(block: false, alt: "S A")[$S A$] of a right prism is found using the formula #math.equation(block: false, alt: "S A equals 2 B plus p h ,")[$S A = 2 B + p h ,$] where #math.equation(block: false, alt: "B")[$B$] is the area of the base, #math.equation(block: false, alt: "p")[$p$] is the perimeter of the base, and #math.equation(block: false, alt: "h")[$h$] is the height. The volume #math.equation(block: false, alt: "V")[$V$] of a right prism is found using the formula #math.equation(block: false, alt: "V equals B h ,")[$V = B h ,$] where #math.equation(block: false, alt: "B")[$B$] is the area of the base and #math.equation(block: false, alt: "h")[$h$] is the height.
- A right cylinder is a three-dimensional object with a circle as the top and a congruent circle is the base, and the side forms a #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] angle to the base and top. The surface area of a right cylinder is found using the formula #math.equation(block: false, alt: "S A equals 2 π r squared plus 2 π r h ,")[$S A = 2 π r^(2) + 2 π r h ,$] where #math.equation(block: false, alt: "r")[$r$] is the radius and #math.equation(block: false, alt: "h")[$h$] is the height. The volume is found using the formula #math.equation(block: false, alt: "V equals π r squared h ,")[$V = π r^(2) h ,$] where #math.equation(block: false, alt: "r")[$r$] is the radius and #math.equation(block: false, alt: "h")[$h$] is the height.


  
  
    
=== Formulas


    The formula for the surface area of a right prism is equal to twice the area of the base plus the perimeter of the base times the height, #math.equation(block: false, alt: "S A equals 2 B plus p h ,")[$S A = 2 B + p h ,$] where #math.equation(block: false, alt: "B")[$B$] is equal to the area of the base and top, #math.equation(block: false, alt: "p")[$p$] is the perimeter of the base, and #math.equation(block: false, alt: "h")[$h$] is the height.


    The formula for the volume of a rectangular prism, given in cubic units, is equal to the area of the base times the height, #math.equation(block: false, alt: "V equals B times h ,")[$V = B ⋅ h ,$] where #math.equation(block: false, alt: "B")[$B$] is the area of the base and #math.equation(block: false, alt: "h")[$h$] is the height.


    The surface area of a right cylinder is given as #math.equation(block: false, alt: "S A equals 2 π r squared plus 2 π r h .")[$S A = 2 π r^(2) + 2 π r h .$]


    The volume of a right cylinder is given as #math.equation(block: false, alt: "V equals π r squared h .")[$V = π r^(2) h .$]