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Section 11.8 Surface Areas and Volumes of Spheres 647
Essential QuestionEssential Question How can you find the surface area and the
volume of a sphere?
Finding the Surface Area of a Sphere
Work with a partner. Remove the covering from a baseball or softball.
r
You will end up with two “figure 8” pieces of material, as shown above. From the amount of material it takes to cover the ball, what would you estimate the surface area S of the ball to be? Express your answer in terms of the radius r of the ball. S = Surface area of a sphere Use the Internet or some other resource to confirm that the formula you wrote for the surface area of a sphere is correct.
Finding the Volume of a Sphere
Work with a partner. A cylinder is circumscribed about a sphere, as shown. Write a formula for the volume V of the cylinder in terms of the radius r.
V = Volume of cylinder
When half of the sphere (a hemisphere ) is fi lled with sand and poured into the cylinder, it takes three hemispheres to fill the cylinder. Use this information to write a formula for the volume V of a sphere in terms of the radius r.
V = Volume of a sphere
Communicate Your AnswerCommunicate Your Answer
3. How can you find the surface area and the volume of a sphere? 4. Use the results of Explorations 1 and 2 to find the surface area and the volume of a sphere with a radius of (a) 3 inches and (b) 2 centimeters.
USING TOOLS
STRATEGICALLY
To be proficient in math, you need to identify relevant external mathematical resources, such as content located on a website.
Surface Areas and Volumes
of Spheres
r
r (^) 2 r
648 Chapter 11 Circumference, Area, and Volume
11.8 Lesson^ What You Will LearnWhat You Will Learn
Find surface areas of spheres. Find volumes of spheres.
Finding Surface Areas of Spheres A sphere is the set of all points in space equidistant from a given point. This point is called the center of the sphere. A radius of a sphere is a segment from the center to a point on the sphere. A chord of a sphere is a segment whose endpoints are on the sphere. A diameter of a sphere is a chord that contains the center.
C center
radius C
chord
diameter
As with circles, the terms radius and diameter also represent distances, and the diameter is twice the radius. If a plane intersects a sphere, then the intersection is either a single point or a circle. If the plane contains the center of the sphere, then the intersection is a great circle of the sphere. The circumference of a great circle is the circumference of the sphere. Every great circle of a sphere separates the sphere into two congruent halves called hemispheres.
To understand the formula for the surface area of a sphere, think of a baseball. The surface area of a baseball is sewn from two congruent shapes, each of which resembles two joined circles. So, the entire covering of the baseball consists of four circles, each with radius r. The area A of a circle with radius r is A = π r^2. So, the area of the covering can be approximated by 4 π r^2. This is the formula for the surface area of a sphere.
chord of a sphere, p. 648 great circle, p. 648 Previous sphere center of a sphere radius of a sphere diameter of a sphere hemisphere
Core VocabularyCore Vocabullarry
CoreCore ConceptConcept
Surface Area of a Sphere
The surface area S of a sphere is S = 4 π r^2 where r is the radius of the sphere.
great circle
hemispheres
r
S = 4 π r^2
r
leather covering
650 Chapter 11 Circumference, Area, and Volume
Finding the Volume of a Sphere
Find the volume of the soccer ball.
SOLUTION
V = 4 — 3 π r^3 Formula for volume of a sphere
= 4 — 3 π(4.5)^3 Substitute 4.5 for r.
= 121.5π Simplify. ≈ 381.70 Use a calculator.
The volume of the soccer ball is 121.5 π, or about 381.70 cubic inches.
Finding Volumes of Spheres The fi gure shows a hemisphere and a cylinder with a cone removed. A plane parallel to their bases intersects the solids z units above their bases.
r^2 − z^2
r
r
r z
Using the AA Similarity Theorem (Theorem 8.3), you can show that the radius of the cross section of the cone at height z is z. The area of the cross section formed by the plane is π( r^2 − z^2 ) for both solids. Because the solids have the same height and the same cross-sectional area at every level, they have the same volume by Cavalieri’s Principle.
V hemisphere = V cylinder − V cone
= π r^2 ( r ) − —^13 π r^2 ( r )
= 2 — 3 π r^3
So, the volume of a sphere of radius r is
(^2) ⋅ V hemisphere = (^2) ⋅^2 — 3 π r^3 = 4 — 3 π r^3.
CoreCore ConceptConcept
Volume of a Sphere
The volume V of a sphere is
V =
— 3
π r^3
where r is the radius of the sphere.
r
V = 43 π r^3
4.5 in.
Section 11.8 Surface Areas and Volumes of Spheres 651
Finding the Volume of a Sphere
The surface area of a sphere is 324 π square centimeters. Find the volume of the sphere.
SOLUTION
Step 1 Use the surface area to find the radius. S = 4 π r^2 Formula for surface area of a sphere 324 π = 4 π r^2 Substitute 324 π for S. 81 = r^2 Divide each side by 4 π. 9 = r Find the positive square root. The radius is 9 centimeters. Step 2 Use the radius to find the volume.
V = 4 — 3 π r^3 Formula for volume of a sphere
= 4 — 3 π(9)^3 Substitute 9 for r.
= 972 π Simplify. ≈ 3053.63 Use a calculator.
The volume is 972 π, or about 3053.63 cubic centimeters.
Finding the Volume of a Composite Solid
Find the volume of the composite solid.
SOLUTION
Volume of solid
Volume of cylinder
Volume of hemisphere
= π r^2 h − —^12 ( 4 — 3 π r^3 ) Write formulas.
= π(2)^2 (2) − 2 — 3 π(2)^3 Substitute.
= 8 π − 16 — 3 π Multiply.
= 24 — 3 π − —^163 π Rewrite fractions using least common denominator. = 8 — 3 π Subtract. ≈ 8.38 Use a calculator.
The volume is —^83 π, or about 8.38 cubic inches.
Monitoring ProgressMonitoring Progress (^) Help in English and Spanish at BigIdeasMath.com
4. The radius of a sphere is 5 yards. Find the volume of the sphere. 5. The diameter of a sphere is 36 inches. Find the volume of the sphere. 6. The surface area of a sphere is 576 π square centimeters. Find the volume of the sphere. 7. Find the volume of the composite solid at the left.
2 in.
2 in.
1 m
5 m
Section 11.8 Surface Areas and Volumes of Spheres 653
22. ERROR ANALYSIS Describe and correct the error in fi nding the volume of the sphere.
3 in.
V =^4 — 3 π (3)^3 = 36 π ≈ 113.10 in.^3
In Exercises 23–26, find the volume of the composite solid. (See Example 5.)
23.
9 in. 5 in.
12 ft
6 ft
10 cm
18 cm 26. 14 m 6 m
In Exercises 27–32, find the surface area and volume of the ball.
27. bowling ball 28. basketball
d = 8.5 in. C = 29.5 in.
29. softball 30. golf ball
C = 12 in. d = 1.7 in.
31. volleyball 32. baseball
C = 26 in. C = 9 in.
33. MAKING AN ARGUMENT You friend claims that if the radius of a sphere is doubled, then the surface area of the sphere will also be doubled. Is your friend correct? Explain your reasoning. 34. REASONING A semicircle with a diameter of 18 inches is rotated about its diameter. Find the surface area and the volume of the solid formed. 35. MODELING WITH MATHEMATICS A silo has the dimensions shown. The top of the silo is a hemispherical shape. Find the volume of the silo.
20 ft20 ft
60 ft
36. MODELING WITH MATHEMATICS Three tennis balls are stored in a cylindrical container with a height of 8 inches and a radius of 1.43 inches. The circumference of a tennis ball is 8 inches. a. Find the volume of a tennis ball. b. Find the amount of space within the cylinder not taken up by the tennis balls. 37. ANALYZING RELATIONSHIPS Use the table shown for a sphere.
Radius Surface area Volume 3 in. 36 π in. 2 36 π in. 3 6 in. 9 in. 12 in.
a. Copy and complete the table. Leave your answers in terms of π. b. What happens to the surface area of the sphere when the radius is doubled? tripled? quadrupled? c. What happens to the volume of the sphere when the radius is doubled? tripled? quadrupled?
38. MATHEMATICAL CONNECTIONS A sphere has a diameter of 4( x + 3) centimeters and a surface area of 784 π square centimeters. Find the value of x.
654 Chapter 11 Circumference, Area, and Volume
39. MODELING WITH MATHEMATICS The radius of Earth is about 3960 miles. The radius of the moon is about 1080 miles. a. Find the surface area of Earth and the moon. b. Compare the surface areas of Earth and the moon. c. About 70% of the surface of Earth is water. How many square miles of water are on Earth’s surface? 40. MODELING WITH MATHEMATICS The Torrid Zone on Earth is the area between the Tropic of Cancer and the Tropic of Capricorn. The distance between these two tropics is about 3250 miles. You can estimate the distance as the height of a cylindrical belt around the Earth at the equator.
Tropic of Cancer
Torrid Zone
Tropic of Capricorn
3250 mi (^) equator
a. Estimate the surface area of the Torrid Zone. (The radius of Earth is about 3960 miles.) b. A meteorite is equally likely to hit anywhere on Earth. Estimate the probability that a meteorite will land in the Torrid Zone.
41. ABSTRACT REASONING A sphere is inscribed in a cube with a volume of 64 cubic inches. What is the surface area of the sphere? Explain your reasoning. 42. HOW DO YOU SEE IT? The formula for the volume of a hemisphere and a cone are shown. If each solid has the same radius and r = h , which solid will have a greater volume? Explain your reasoning.
r r
h
V = 23 π r^3 V = 13 π r^2 h
43. CRITICAL THINKING Let V be the volume of a sphere, S be the surface area of the sphere, and r be the radius of the sphere. Write an equation for V in terms of r and S. ( Hint : Start with the ratio — V S . (^) ) 44. THOUGHT PROVOKING A spherical lune is the region between two great circles of a sphere. Find the formula for the area of a lune.
r θ
45. CRITICAL THINKING The volume of a right cylinder is the same as the volume of a sphere. The radius of the sphere is 1 inch. Give three possibilities for the dimensions of the cylinder. 46. PROBLEM SOLVING A spherical cap is a portion of a sphere cut off by a plane. The formula for the volume of a spherical cap is V = π h — 6 (3 a^2 + h^2 ), where a is the radius of the base of the cap and h is the height of the cap. Use the diagram and given information to fi nd the volume of each spherical cap. a. r = 5 ft, a = 4 ft b. r = 34 cm, a = 30 cm c. r = 13 m, h = 8 m d. r = 75 in., h = 54 in. 47. CRITICAL THINKING A sphere with a radius of 2 inches is inscribed in a right cone with a height of 6 inches. Find the surface area and the volume of the cone.
Maintaining Mathematical ProficiencyMaintaining Mathematical Proficiency
Solve the triangle. Round decimal answers to the nearest tenth. (Section 9.7)
48. A = 26 °, C = 35 °, b = 13 49. B = 102 °, C = 43 °, b = 21 50. a = 23, b = 24, c = 20 51. A = 103 °, b = 15, c = 24
Reviewing what you learned in previous grades and lessons
r
a
h
