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Formulas and examples for calculating the surface areas of different space figures, including right prisms, right cylinders, pyramids, right cones, and spheres.
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Typology: Summaries
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The surface area of a space figure is the total area of all the faces of the figure. In this section, we discuss the surface areas of some of the space fig- ures introduced in Section 41.
Right Prisms Let’s find the surface area of the right prism given in Figure 44.1.
Figure 44.
As you can see, we can dissamble the prism into six rectangles. The total area of these rectangles which is the surface area of the box is
SA = 2lw + 2h(w + l).
If we let A denote the area of the base and P the perimeter of the base then A = lw and P = 2(l + w). Thus,
SA = 2A + P h.
This formula is valid for any right prism with height h, area of base A, and perimeter of base P.
Right Cylinders To find the surface area of a right cylinder with height h and radius of base r we dissamble the cylinder into a rectangle and two circles as shown in Figure
Figure 44.
Thus, the surface area is the sum of the area of the rectangle together with the areas of the two circles. That is,
SA = 2πr^2 + 2πr · h = 2πr(r + h).
Example 44. A thin can has a height of 15 cm and a base radius of 7 cm. It is filled with water. Find the total surface area in contact with the water. (Take π = 22/7)
Solution.
SA = 2π(7)(7 + 15) = 2 ·
· 7 · 22 = 968 cm^2
Example 44. The lateral surface area of a solid cylinder is 880 cm^2 and its height is 10 cm. Find the circumference and area of the base of the cylinder. Take π = 227.
Solution. Since the lateral surface of a cylinder is height × perimeter of the base then P = 88010 = 88 cm. Now, the radius of the circle is r = (^882) π = 44 · 227 = 14 cm. Finally, the area of the base is A = π 142 = 227 · 196 = 616 cm^2
Pyramids The surface area of a pyramid is obtained by summing the areas of the faces.
Figure 44.
To find the surface area of the cone, we cut it along a slant height and open it out flat to obtain a circular sector shown in Figure 44.
Figure 44.
Thus, the lateral surface area is the area of the circular sector. But
Area of sector ABC Area of circle with center C =^
Arc length of AB Circumference of circle with center at C Area of sector ABC πl^2 =^
2 πr 2 πl Area of sector ABC = πrl
But
Total surface area = Area of base + lateral surface area.
Hence SA = πr^2 + πrl = πr(r + l) = πr(r +
r^2 + h^2 )
Spheres Archimedes discovered that a cylinder that circumscribes a sphere, as shown in Figure 44.6, has a lateral surface area equal to the surface area, SA, of the sphere.
Figure 44.
Thus, SA = 2 πrh = 2 πr × 2 r = 4 πr^2
Example 44. A solid sphere has a radius of 3 m. Calculate its surface area. Round your answer to a whole number. (Take π = 22/7)
Solution. SA = 4 πr^2 = 4 × 227 × 32 ≈ 113 m^2
Example 44. Find the radius of a sphere with a surface area of 64π m^2.
Problem 44. The Great Pyramid of Cheops is a right square pyramid with a height of 148 m and a square base with perimeter of 930 m. The slant height is 188 m. The basic shape of the Transamerica Building in San Fransisco is a right square pyramid that has a height of 260 m and a square base with a perimeter of 140 m. The altitude of slant height is 261 m. How do the lateral surface areas of the two structures compare?
Problem 44. Find the surface area of the following cone.
Problem 44. The Earth has a spherical shape of radius 6370 km. What is its surface area?
Problem 44. Suppose one right circular cylinder has radius 2 m and height 6 m and an- other has radius 6 m and height 2 m.
(a) Which cylinder has the greater lateral surface area? (b) Which cylinder has the greater total surface area?
Problem 44. The base of a right pyramid is a regular hexagon with sides of length 12 m. The height of the pyramid is 9 m. Find the total surface area of the pyramid.
Problem 44. A square piece of paper 10 cm on a side is rolled to form the lateral surface area of a right circular cylinder and then a top and bottom are added. What is the surface area of the cylinder?
Problem 44. The top of a rectangular box has an area of 88 cm^2. The sides have areas 32 cm^2 and 44 cm^2. What are the dimensions of the box?
Problem 44. What happens to the surface area of a sphere if the radius is tripled?
Problem 44. Find the surface area of a square pyramid if the area of the base is 100 cm^2 and the height is 20 cm.
Problem 44. Each region in the following figure revolves about the horizontal axis. For each case, sketch the three dimensional figure obtained and find its surface area.
Problem 44. The total surface area of a cube is 10,648 cm^2. Find the length of a diagonal that is not a diagonal of a face.
Problem 44. If the length, width, and height of a rectangular prism is tripled, how does the surface area change?
Problem 44. Find the surface area of the following figure.
Problem 44. A room measures 4 meters by 7 meters and the ceiling is 3 meters high. A liter of paint covers 40 square meters. How many liters of paint will it take to paint all but the floor of the room?