This document provides information about surface areas and volumes of spheres. It defines key terms like great circle, pole, and hemisphere. It presents the formulas for calculating the surface area and volume of spheres and hemispheres. It then works through 6 examples applying these formulas to find surface areas and volumes given radius or diameter values. The examples demonstrate how to set up and solve the related equations.

1. Section 12-6
Surface Areas and Volumes of Spheres

2. Essential Questions
How do you find surface areas of spheres?
How do you find volumes of spheres?

3. Vocabulary
1. Great Circle:
2. Pole:
3. Hemisphere:

4. Vocabulary
1. Great Circle:
2. Pole:
3. Hemisphere:
A circle formed when a plane
intersects a sphere and the circle has the
same center as the sphere

5. Vocabulary
1. Great Circle:
2. Pole:
3. Hemisphere:
A circle formed when a plane
intersects a sphere and the circle has the
same center as the sphere
The endpoints of the diameter of a great
circle

6. Vocabulary
1. Great Circle:
2. Pole:
3. Hemisphere:
A circle formed when a plane
intersects a sphere and the circle has the
same center as the sphere
The endpoints of the diameter of a great
circle
One of the two congruent halves
of a sphere created by a great circle

7. Formulas
Surface Area of a Sphere:
Volume of a Sphere:

8. Formulas
Surface Area of a Sphere: SA = 4πr2
Volume of a Sphere:

9. Formulas
Surface Area of a Sphere: SA = 4πr2
Volume of a Sphere: V =
4
3
πr3

10. Example 1
Find the surface area of the sphere.

11. Example 1
Find the surface area of the sphere.
SA = 4πr2

12. Example 1
Find the surface area of the sphere.
SA = 4πr2
SA = 4π(4.5)2

13. Example 1
Find the surface area of the sphere.
SA = 4πr2
SA = 4π(4.5)2
SA ≈ 254.47 in2

14. Example 2
Find the surface area of a hemisphere with a
diameter of 8 mm to the nearest hundredth.

15. Example 2
Find the surface area of a hemisphere with a
diameter of 8 mm to the nearest hundredth.
SA = 1
2
i 4πr2
+ πr2

16. Example 2
Find the surface area of a hemisphere with a
diameter of 8 mm to the nearest hundredth.
SA = 1
2
i 4πr2
+ πr2
r = 1
2
d

17. Example 2
Find the surface area of a hemisphere with a
diameter of 8 mm to the nearest hundredth.
SA = 1
2
i 4πr2
+ πr2
r = 1
2
d
r = 1
2
(8)

18. Example 2
Find the surface area of a hemisphere with a
diameter of 8 mm to the nearest hundredth.
SA = 1
2
i 4πr2
+ πr2
r = 1
2
d
r = 1
2
(8)
r = 4 mm

19. Example 2
Find the surface area of a hemisphere with a
diameter of 8 mm to the nearest hundredth.
SA = 1
2
i 4πr2
+ πr2
r = 1
2
d
r = 1
2
(8)
r = 4 mm
SA = 2π(4)2
+ π(4)2

20. Example 2
Find the surface area of a hemisphere with a
diameter of 8 mm to the nearest hundredth.
SA = 1
2
i 4πr2
+ πr2
r = 1
2
d
r = 1
2
(8)
r = 4 mm
SA = 2π(4)2
+ π(4)2
SA ≈ 150.8 mm2

21. Example 3
Find the surface area of a sphere if the
circumference of the great circle is 14π inches.

22. Example 3
Find the surface area of a sphere if the
circumference of the great circle is 14π inches.
SA = 4πr2

23. Example 3
Find the surface area of a sphere if the
circumference of the great circle is 14π inches.
SA = 4πr2
C = 2πr

24. Example 3
Find the surface area of a sphere if the
circumference of the great circle is 14π inches.
SA = 4πr2
C = 2πr
14π = 2πr

25. Example 3
Find the surface area of a sphere if the
circumference of the great circle is 14π inches.
SA = 4πr2
C = 2πr
14π = 2πr
2π 2π

26. Example 3
Find the surface area of a sphere if the
circumference of the great circle is 14π inches.
SA = 4πr2
C = 2πr
14π = 2πr
2π 2π
r = 7 in.

27. Example 3
Find the surface area of a sphere if the
circumference of the great circle is 14π inches.
SA = 4πr2
C = 2πr
14π = 2πr
2π 2π
r = 7 in.
SA = 4π(7)2

28. Example 3
Find the surface area of a sphere if the
circumference of the great circle is 14π inches.
SA = 4πr2
C = 2πr
14π = 2πr
2π 2π
r = 7 in.
SA = 4π(7)2
SA ≈ 615.75 in2

29. Example 4
Find the volume of the sphere rounded to the
nearest hundredth.

30. Example 4
Find the volume of the sphere rounded to the
nearest hundredth.
V =
4
3
πr3

31. Example 4
Find the volume of the sphere rounded to the
nearest hundredth.
V =
4
3
πr3
V =
4
3
π(4.5)3

32. Example 4
Find the volume of the sphere rounded to the
nearest hundredth.
V =
4
3
πr3
V =
4
3
π(4.5)3
V ≈ 381.70 in3

33. Example 5
Find the volume of a hemisphere with a diameter
of 6 feet to the nearest hundredth.

34. Example 5
Find the volume of a hemisphere with a diameter
of 6 feet to the nearest hundredth.
V =
1
2
i
4
3
πr3

35. Example 5
Find the volume of a hemisphere with a diameter
of 6 feet to the nearest hundredth.
V =
1
2
i
4
3
πr3 r = 1
2
d

36. Example 5
Find the volume of a hemisphere with a diameter
of 6 feet to the nearest hundredth.
V =
1
2
i
4
3
πr3 r = 1
2
d
r = 1
2
(6)

37. Example 5
Find the volume of a hemisphere with a diameter
of 6 feet to the nearest hundredth.
V =
1
2
i
4
3
πr3 r = 1
2
d
r = 1
2
(6)
r = 3 ft

38. Example 5
Find the volume of a hemisphere with a diameter
of 6 feet to the nearest hundredth.
V =
1
2
i
4
3
πr3 r = 1
2
d
r = 1
2
(6)
r = 3 ft
V =
2
3
π(3)3

39. Example 5
Find the volume of a hemisphere with a diameter
of 6 feet to the nearest hundredth.
V =
1
2
i
4
3
πr3 r = 1
2
d
r = 1
2
(6)
r = 3 ft
V =
2
3
π(3)3
V ≈ 56.55 ft3

40. Example 6
Find the volume of a sphere that has a great
circle with an area of 9π square feet.

41. Example 6
Find the volume of a sphere that has a great
circle with an area of 9π square feet.
V =
4
3
πr3

42. Example 6
Find the volume of a sphere that has a great
circle with an area of 9π square feet.
V =
4
3
πr3
A = πr2

43. Example 6
Find the volume of a sphere that has a great
circle with an area of 9π square feet.
V =
4
3
πr3
A = πr2
9π = πr2

44. Example 6
Find the volume of a sphere that has a great
circle with an area of 9π square feet.
V =
4
3
πr3
A = πr2
9π = πr2
π π

45. Example 6
Find the volume of a sphere that has a great
circle with an area of 9π square feet.
V =
4
3
πr3
A = πr2
9π = πr2
π π
9 = r2

46. Example 6
Find the volume of a sphere that has a great
circle with an area of 9π square feet.
V =
4
3
πr3
A = πr2
9π = πr2
π π
r = 3 ft.
9 = r2

47. Example 6
Find the volume of a sphere that has a great
circle with an area of 9π square feet.
V =
4
3
πr3
A = πr2
9π = πr2
π π
r = 3 ft.
9 = r2
V =
4
3
π(3)3

48. Example 6
Find the volume of a sphere that has a great
circle with an area of 9π square feet.
V =
4
3
πr3
A = πr2
9π = πr2
π π
r = 3 ft.
9 = r2
V =
4
3
π(3)3
V ≈ 113.10 ft3

49. Problem Set

50. Problem Set
p. 868 #1-25 odd
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