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Section 11-4
Spheres
Essential Questions
How do you find surface areas of spheres?
How do you find volumes of spheres?
Vocabulary
1. Great Circle:
2. Pole:
3. Hemisphere:
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
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
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
Formulas
Surface Area of a Sphere:
Volume of a Sphere:
Formulas
Surface Area of a Sphere: SA = 4πr2
Volume of a Sphere:
Formulas
Surface Area of a Sphere: SA = 4πr2
Volume of a Sphere: V =
4
3
πr3
Example 1
Find the surface area of the sphere.
Example 1
Find the surface area of the sphere.
SA = 4πr2
Example 1
Find the surface area of the sphere.
SA = 4πr2
SA = 4π(4.5)2
Example 1
Find the surface area of the sphere.
SA = 4πr2
SA = 4π(4.5)2
SA ≈ 254.47 in2
Example 2
Find the surface area of a hemisphere with a
diameter of 8 mm to the nearest hundredth.
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
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
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)
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
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
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
Example 3
Find the surface area of a sphere if the
circumference of the great circle is 14π inches.
Example 3
Find the surface area of a sphere if the
circumference of the great circle is 14π inches.
SA = 4πr2
Example 3
Find the surface area of a sphere if the
circumference of the great circle is 14π inches.
SA = 4πr2
C = 2πr
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
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
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.
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
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
Example 4
Find the volume of the sphere rounded to the
nearest hundredth.
Example 4
Find the volume of the sphere rounded to the
nearest hundredth.
V =
4
3
πr3
Example 4
Find the volume of the sphere rounded to the
nearest hundredth.
V =
4
3
πr3
V =
4
3
π(4.5)3
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
Example 5
Find the volume of a hemisphere with a diameter
of 6 feet to the nearest hundredth.
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
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
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)
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
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
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
Example 6
Find the volume of a sphere that has a great
circle with an area of 9π square feet.
Example 6
Find the volume of a sphere that has a great
circle with an area of 9π square feet.
V =
4
3
πr3
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
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
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
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
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
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
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
Problem Set
Problem Set
p. 822 #1-9 all
“With self-discipline most anything is possible.”
- Theodore Roosevelt

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Geometry Section 11-4

  • 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
  • 50. Problem Set p. 822 #1-9 all “With self-discipline most anything is possible.” - Theodore Roosevelt