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9.3 Surface Area, Volume, and Capacity

3 Surface Area

4 Suppose you wish to paint a box whose edges are each 3 ft, and you need to know how much paint to buy. To determine this you need to find the sum of the areas of all the faces—this is called the surface area.

5 Example 1 – Find the surface area of a box Find the amount of paint needed for a box with edges 3 ft. Solution: A box (cube) has 6 faces of equal area. 6  9 ft 2 = 54 ft 2 You need enough paint to cover 54 ft 2. Number of faces of cube Area of each face

6 Surface Area In the next example, we consider a box without a top and a can with a bottom but not a top.

7 Example 2 – Find the outside surface area Find the outside surface area.

8 Example 2 – Solution a. We find the sum of the areas of all the faces: Front: 30  50 = 1,500 Back: 1,500 Side: 80  30 = 2,400 Side: 2,400 Bottom: 80  50 = 4,000 Total: 11,800 cm 2 Same as front Sides are the same size.

9 Example 2 – Solution b. To find the surface area, find the area of a circle (the bottom) and think of the sides of the can as being “rolled out.” The length of the resulting rectangle is the circumference of the can and the width of the rectangle is the height of the can. cont’d

10 Example 2 – Solution Side: A = /w = (  d)w =  (6)(6) = 36  Bottom: A =  r 2 =  (3) 2 = 9  Surface area: 36  + 9  = 45   141.37167 The surface area is about 141 cm 2. cont’d

11 Volume

12 Volume To measure area, we covered a region with square units and then found the area by using a mathematical formula. A similar procedure is used to find the amount of space inside a solid object, which is called its volume. We can imagine filling the space with cubes.

13 Volume A cubic inch and a cubic centimeter are shown in Figure 9.12. If the solid is not a cube but is a box (called a rectangular parallelepiped) with edges of different lengths, the volume can be found similarly. Figure 9.12 Common units of measuring volume a. 1 cubic inch (1 cu in. or 1 in. 3 ) b. 1 cubic centimeter (1 cu cm, cc, or 1 cm 3 )

14 Volume

15 Example 4 – Find the volume of a box Find the volume of a box that measures 4 ft by 6 ft by 4 ft.

16 Example 4 – Solution There are 24 cubic feet on the bottom layer of cubes. Do you see how many layers of cubes will fill the solid? Since there are four layers with 24 cubes in each, the total is 4  24 = 96 The volume is 96 ft 3.

17 Volume This example leads us to the following formula for the volume of a box.

18 Example 5 – Find the volume of a solid Find the volume of each solid.

19 Example 5 – Solution a. V = s 3 = (10 cm) 3 = (10  10  10) cm 3 = 1,000 cm 3 b. V = /wh = (25 cm)(10 cm)(4 cm) = (25  10  4) cm 3 = 1,000 cm 3

20 Example 5 – Solution c. V = /wh = (11 in.)(7 in.)(3 in.) = (11  7  3) in. 3 Sometimes the dimensions for the volume we are finding are not all given in the same units. cont’d

21 Example 5 – Solution In such cases, you must convert all units to a common unit. The common conversions are as follows: 1 ft = 12 in. To convert feet to inches, multiply by 12. To convert inches to feet, divide by 12. 1 yd = 3 ft To convert yards to feet, multiply by 3. To convert feet to yards, divide by 3. 1 yd = 36 in. To convert yards to inches, multiply by 36. To convert inches to yards, divide by 36. cont’d

22 Capacity

23 Capacity One of the most common applications of volume involves measuring the amount of liquid a container holds, which we refer to as its capacity. For example, if a container is 2 ft by 2 ft by 12 ft, it is fairly easy to calculate the volume: 2  2  12 = 48 ft 3

24 Capacity

25 Capacity Some capacity statements from purchased products are listed in Table 9.1. Table 9.1 Capacities of Common Grocery Items, as Shown on Labels

26 Example 7 – Measure the quantity of a liquid Measure the amount of liquid in the measuring cup in Figure 9.14, both in the U.S. system and in the metric system. Figure 9.14 Standard measuring cup with both metric and U.S. measurements

27 Example 7 – Solution Metric: 240 mL U.S.: About 1 c or 8 oz

28 Capacity

29 Example 8 – Capacity of a container How much water would each of the following containers hold?

30 Example 8 – Solution a. V = 90 cm  80 cm  40 cm = 288,000 cm 3 Since each 1,000 cm 3 is 1 liter, This container would hold 288 liters.

31 Example 8 – Solution b. V = 7 in.  22 in.  6 in. = 924 in. 3 Since each 231 in. 3 is 1 gallon, This container would hold 4 gallons. cont’d

32 Example 8 – Solution Note that there is no need to estimate part a since the arithmetic is fairly easy. However, for part b, the estimate might be cont’d