1. Volumes of Solids
Volumes By Discs & Washers
( slice ⊥ rotation )

2. Volumes of Solids
Volumes By Discs & Washers
( slice ⊥ rotation )
About the x axis

3. Volumes of Solids
Volumes By Discs & Washers
( slice ⊥ rotation )
About the x axis
y y = f ( x)
x

4. Volumes of Solids
Volumes By Discs & Washers
( slice ⊥ rotation )
About the x axis
y y = f ( x)
a b x

5. Volumes of Solids
Volumes By Discs & Washers
( slice ⊥ rotation )
About the x axis
y y = f ( x)
a ∆x b x

6. Volumes of Solids
Volumes By Discs & Washers
( slice ⊥ rotation )
About the x axis
y y = f ( x)
y
a ∆x b x
∆x

7. Volumes of Solids
Volumes By Discs & Washers
( slice ⊥ rotation )
About the x axis
y y = f ( x)
y = f ( x)
a ∆x b x
∆x

8. Volumes of Solids
Volumes By Discs & Washers
( slice ⊥ rotation )
About the x axis
y y = f ( x)
y = f ( x)
A( x ) = π [ f ( x ) ]
2
a ∆x b x
∆x

9. Volumes of Solids
Volumes By Discs & Washers
( slice ⊥ rotation )
About the x axis
y y = f ( x)
y = f ( x)
A( x ) = π [ f ( x ) ]
2
a ∆x b x ∆V = π [ f ( x ) ] ⋅ ∆x
2
∆x

10. Volumes of Solids
Volumes By Discs & Washers
( slice ⊥ rotation )
About the x axis
y y = f ( x)
y = f ( x)
A( x ) = π [ f ( x ) ]
2
a ∆x b x ∆V = π [ f ( x ) ] ⋅ ∆x
2
∆x
b
V = lim ∑ π [ f ( x ) ] ⋅ ∆x
2
∆x →0
x=a

11. Volumes of Solids
Volumes By Discs & Washers
( slice ⊥ rotation )
About the x axis
y y = f ( x)
y = f ( x)
A( x ) = π [ f ( x ) ]
2
a ∆x b x ∆V = π [ f ( x ) ] ⋅ ∆x
2
∆x
b
V = lim ∑ π [ f ( x ) ] ⋅ ∆x
2
∆x →0
x=a
b
= π ∫ [ f ( x ) ] dx
2
a

12. About the y axis
y y = f ( x)
x

13. About the y axis
y y = f ( x)
d
c
x

14. About the y axis
y y = f ( x)
d
∆y
c
x

15. About the y axis
y y = f ( x)
d
x
∆y ∆y
c
x

16. About the y axis
y y = f ( x)
d
x = g( y)
∆y ∆y
c
x

17. About the y axis
y y = f ( x)
d
x = g( y)
∆y ∆y
c
A( y ) = π [ g ( y ) ]
x 2

18. About the y axis
y y = f ( x)
d
x = g( y)
∆y ∆y
c
A( y ) = π [ g ( y ) ]
x 2
[ g ( y ) ] 2 ⋅ ∆y
∆V = π

19. About the y axis
y y = f ( x)
d
x = g( y)
∆y ∆y
c
A( y ) = π [ g ( y ) ]
x 2
[ g ( y ) ] 2 ⋅ ∆y
∆V = π
d
V = lim ∑ π [ g ( y ) ] ⋅ ∆y
2
∆y →0
y =c

20. About the y axis
y y = f ( x)
d
x = g( y)
∆y ∆y
c
A( y ) = π [ g ( y ) ]
x 2
[ g ( y ) ] 2 ⋅ ∆y
∆V = π
d
V = lim ∑ π [ g ( y ) ] ⋅ ∆y
2
∆y →0
y =c
d
= π ∫ [ g ( y ) ] dy
2
c

21. e.g. ( i ) Find the volume generated when the area between y = 4 − x 2 and
the x axis is rotated about the x axis

22. e.g. ( i ) Find the volume generated when the area between y = 4 − x 2 and
the x axis is rotated about the x axis
y
4
y = 4 − x2
-2 2 x

23. e.g. ( i ) Find the volume generated when the area between y = 4 − x 2 and
the x axis is rotated about the x axis
y
4
y = 4 − x2
∆x 2
-2 x

24. e.g. ( i ) Find the volume generated when the area between y = 4 − x 2 and
the x axis is rotated about the x axis
y
4
y = 4 − x2
∆x 2
-2 x
y = 4 − x2
∆x

25. e.g. ( i ) Find the volume generated when the area between y = 4 − x 2 and
the x axis is rotated about the x axis
y
4
y = 4 − x2
∆x 2
-2 x
y = 4 − x2
A( x ) = π ( 4 − x )
22
∆x

26. e.g. ( i ) Find the volume generated when the area between y = 4 − x 2 and
the x axis is rotated about the x axis
y
4
y = 4 − x2
∆x 2
-2 x
y = 4 − x2
A( x ) = π ( 4 − x )
22
∆V = π (16 − 8 x 2 + x 4 ) ⋅ ∆x
∆x

27. e.g. ( i ) Find the volume generated when the area between y = 4 − x 2 and
the x axis is rotated about the x axis
y
V = lim ∑ π (16 − 8 x 2 + x 4 ) ⋅ ∆x
2
4 ∆x →0
x = −2
y = 4− x 2
∆x 2
-2 x
y = 4 − x2
A( x ) = π ( 4 − x )
22
∆V = π (16 − 8 x 2 + x 4 ) ⋅ ∆x
∆x

28. e.g. ( i ) Find the volume generated when the area between y = 4 − x 2 and
the x axis is rotated about the x axis
y
V = lim ∑ π (16 − 8 x 2 + x 4 ) ⋅ ∆x
2
4 ∆x →0
x = −2
y = 4− x 2
2
= 2π ∫ (16 − 8 x 2 + x 4 ) dx
0
∆x 2
-2 x
y = 4 − x2
A( x ) = π ( 4 − x )
22
∆V = π (16 − 8 x 2 + x 4 ) ⋅ ∆x
∆x

29. e.g. ( i ) Find the volume generated when the area between y = 4 − x 2 and
the x axis is rotated about the x axis
y
V = lim ∑ π (16 − 8 x 2 + x 4 ) ⋅ ∆x
2
4 ∆x →0
x = −2
y = 4− x 2
2
= 2π ∫ (16 − 8 x 2 + x 4 ) dx
0
∆x 2
-2 x 2
16 x − 8 x 3 + 1 x 5 
= 2π 
5 0
 
3
y = 4 − x2
A( x ) = π ( 4 − x )
22
∆V = π (16 − 8 x 2 + x 4 ) ⋅ ∆x
∆x

30. e.g. ( i ) Find the volume generated when the area between y = 4 − x 2 and
the x axis is rotated about the x axis
y
V = lim ∑ π (16 − 8 x 2 + x 4 ) ⋅ ∆x
2
4 ∆x →0
x = −2
y = 4− x 2
2
= 2π ∫ (16 − 8 x 2 + x 4 ) dx
0
∆x 2
-2 x 2
16 x − 8 x 3 + 1 x 5 
= 2π 
5 0
 
3
32 − 64 + 32 − 0
= 2π  
 
35
y = 4 − x2
512π
= units 3
A( x ) = π ( 4 − x )
22 15
∆V = π (16 − 8 x 2 + x 4 ) ⋅ ∆x
∆x

31. ( ii ) Find the volume generated when the area enclosed by y = 4 − x 2 and
y = 3 is rotated about the line y = 3

32. ( ii ) Find the volume generated when the area enclosed by y = 4 − x 2 and
y = 3 is rotated about the line y = 3
y
4
y=3
x
y = 4 − x2

33. ( ii ) Find the volume generated when the area enclosed by y = 4 − x 2 and
y = 3 is rotated about the line y = 3
y
4
y=3
(1,3)
(-1,3)
x
y = 4 − x2

34. ( ii ) Find the volume generated when the area enclosed by y = 4 − x 2 and
y = 3 is rotated about the line y = 3
y
4
y=3
(-1,3) (1,3)
∆x
x
y = 4 − x2

35. ( ii ) Find the volume generated when the area enclosed by y = 4 − x 2 and
y = 3 is rotated about the line y = 3
y
4
y=3
(-1,3) (1,3)
∆x
x
y = 4 − x2
∆x

36. ( ii ) Find the volume generated when the area enclosed by y = 4 − x 2 and
y = 3 is rotated about the line y = 3
y
4
y=3
(-1,3) (1,3)
∆x
x
y = 4 − x2
y − 3 = 4 − x2 − 3
= 1− x2
∆x

37. ( ii ) Find the volume generated when the area enclosed by y = 4 − x 2 and
y = 3 is rotated about the line y = 3
y
4
y=3
(-1,3) (1,3)
∆x
x
y = 4 − x2
y − 3 = 4 − x2 − 3
= 1− x2
A( x ) = π (1 − x )
22
∆x

38. ( ii ) Find the volume generated when the area enclosed by y = 4 − x 2 and
y = 3 is rotated about the line y = 3
y
4
y=3
(-1,3) (1,3)
∆x
x
y = 4 − x2
y − 3 = 4 − x2 − 3
= 1− x2
A( x ) = π (1 − x )
22
∆V = π (1 − 2 x 2 + x 4 ) ⋅ ∆x ∆x

39. ( ii ) Find the volume generated when the area enclosed by y = 4 − x 2 and
y = 3 is rotated about the line y = 3
y
π (1 − 2 x 2 + x 4 ) ⋅ ∆x
1
∑
V = lim
4 ∆x →0
x = −1
y=3
(-1,3) (1,3)
∆x
x
y = 4 − x2
y − 3 = 4 − x2 − 3
= 1− x2
A( x ) = π (1 − x )
22
∆V = π (1 − 2 x 2 + x 4 ) ⋅ ∆x ∆x

40. ( ii ) Find the volume generated when the area enclosed by y = 4 − x 2 and
y = 3 is rotated about the line y = 3
y
π (1 − 2 x 2 + x 4 ) ⋅ ∆x
1
∑
V = lim
4 ∆x →0
x = −1
y=3
(-1,3) (1,3) 1
= 2π ∫ (1 − 2 x 2 + x 4 ) dx
∆x
0
x
y = 4 − x2
y − 3 = 4 − x2 − 3
= 1− x2
A( x ) = π (1 − x )
22
∆V = π (1 − 2 x 2 + x 4 ) ⋅ ∆x ∆x

41. ( ii ) Find the volume generated when the area enclosed by y = 4 − x 2 and
y = 3 is rotated about the line y = 3
y
π (1 − 2 x 2 + x 4 ) ⋅ ∆x
1
∑
V = lim
4 ∆x →0
x = −1
y=3
(-1,3) (1,3) 1
= 2π ∫ (1 − 2 x 2 + x 4 ) dx
∆x
0
x 1
 x − 2 x3 + 1 x5 
= 2π 
y = 4 − x2
5 0
3 
y − 3 = 4 − x2 − 3
= 1− x2
A( x ) = π (1 − x )
22
∆V = π (1 − 2 x 2 + x 4 ) ⋅ ∆x ∆x

42. ( ii ) Find the volume generated when the area enclosed by y = 4 − x 2 and
y = 3 is rotated about the line y = 3
y
π (1 − 2 x 2 + x 4 ) ⋅ ∆x
1
∑
V = lim
4 ∆x →0
x = −1
y=3
(-1,3) (1,3) 1
= 2π ∫ (1 − 2 x 2 + x 4 ) dx
∆x
0
x 1
 x − 2 x3 + 1 x5 
= 2π 
y = 4 − x2
5 0
3 
1 − 2 + 1 − 0
y − 3 = 4 − x2 − 3 = 2π  
 35 
= 1− x2
16π
= units 3
15
A( x ) = π (1 − x )
22
∆V = π (1 − 2 x 2 + x 4 ) ⋅ ∆x ∆x

43. ( iii ) The area between the curves y = x 2 and x = y 2 is rotated
about the line y axis. Find the volume generated.

44. ( iii ) The area between the curves y = x 2 and x = y 2 is rotated
about the line y axis. Find the volume generated.
y = x2
y
x = y2
x

45. ( iii ) The area between the curves y = x 2 and x = y 2 is rotated
about the line y axis. Find the volume generated.
y = x2
y
x = y2
(1,1)
x

46. ( iii ) The area between the curves y = x 2 and x = y 2 is rotated
about the line y axis. Find the volume generated.
y = x2
y
x = y2
(1,1)
∆y
x

47. ( iii ) The area between the curves y = x 2 and x = y 2 is rotated
about the line y axis. Find the volume generated.
y = x2
y
x = y2
(1,1)
∆y
x

48. ( iii ) The area between the curves y = x 2 and x = y 2 is rotated
about the line y axis. Find the volume generated.
y = x2
y
x = y2
(1,1)
∆y
x
1
x= y 2

49. ( iii ) The area between the curves y = x 2 and x = y 2 is rotated
about the line y axis. Find the volume generated.
y = x2
y
x = y2
(1,1)
∆y
x
1
x= y 2
x= y 2

50. ( iii ) The area between the curves y = x 2 and x = y 2 is rotated
about the line y axis. Find the volume generated.
y = x2
y
x = y2
(1,1)
∆y
x
1
x= y 2
x= y 2
 1  2 
A( y ) = π  y  − ( y ) 
22
2

  
 

51. ( iii ) The area between the curves y = x 2 and x = y 2 is rotated
about the line y axis. Find the volume generated.
y = x2
y
x = y2
(1,1)
∆y
x
1
x= y 2
x= y 2
 1  2 
A( y ) = π  y  − ( y ) 
22
2

  
 
∆V = π ( y − y 4 ) ⋅ ∆y

52. ( iii ) The area between the curves y = x 2 and x = y 2 is rotated
about the line y axis. Find the volume generated.
y = x2
y
V = lim ∑ π ( y − y 4 ) ⋅ ∆y
1
x= y 2
∆y → 0
y =0
(1,1)
∆y
x
1
x= y 2
x= y 2
 1  2 
A( y ) = π  y  − ( y ) 
22
2

  
 
∆V = π ( y − y 4 ) ⋅ ∆y

53. ( iii ) The area between the curves y = x 2 and x = y 2 is rotated
about the line y axis. Find the volume generated.
y = x2
y
V = lim ∑ π ( y − y 4 ) ⋅ ∆y
1
x= y 2
∆y → 0
y =0
(1,1) 1
= π ∫ ( y − y 4 ) dy
∆y
x 0
1
x= y 2
x= y 2
 1  2 
A( y ) = π  y  − ( y ) 
22
2

  
 
∆V = π ( y − y 4 ) ⋅ ∆y

54. ( iii ) The area between the curves y = x 2 and x = y 2 is rotated
about the line y axis. Find the volume generated.
y = x2
y
V = lim ∑ π ( y − y 4 ) ⋅ ∆y
1
x= y 2
∆y → 0
y =0
(1,1) 1
= π ∫ ( y − y 4 ) dy
∆y
x 0
1
1 y 2 − 1 y5 
=π
5 0
2 
1
x= y 2
x= y 2
 1  2 
A( y ) = π  y  − ( y ) 
22
2

  
 
∆V = π ( y − y 4 ) ⋅ ∆y

55. ( iii ) The area between the curves y = x 2 and x = y 2 is rotated
about the line y axis. Find the volume generated.
y = x2
y
V = lim ∑ π ( y − y 4 ) ⋅ ∆y
1
x= y 2
∆y → 0
y =0
(1,1) 1
= π ∫ ( y − y 4 ) dy
∆y
x 0
1
1 y 2 − 1 y5 
=π
5 0
2 
1
x= y 2
x= y 2
 1 − 1 − 0
=π 
 25 
3π
= units 3
 1  2 
A( y ) = π  y  − ( y )  10
22
2

  
 
∆V = π ( y − y 4 ) ⋅ ∆y

56. y
(iv) (1996)
1
x + y2 = 0 S
1 4 x
-1
The shaded region is bounded by the lines x = 1, y = 1 and y = -1 and
the curve x + y = 0. The region is rotated through 360about the line x
2
= 4 to form a solid.
When the region is rotated, the line segment S at height y sweeps out an
annulus.

57. a) Show that the area of the annulus at height y is equal to π ( y 4 + 8 y 2 + 7 )

58. a) Show that the area of the annulus at height y is equal to π ( y 4 + 8 y 2 + 7 )
∆y

59. a) Show that the area of the annulus at height y is equal to π ( y 4 + 8 y 2 + 7 )
r
∆y

60. a) Show that the area of the annulus at height y is equal to π ( y 4 + 8 y 2 + 7 )
R
∆y

61. 1
x + y2 = 0 S
1 4 x
-1

62. 1
x + y2 = 0 Sr
1 4 x
-1

63. 1
x + y2 = 0 S r = 4 −1
=3
1 4 x
-1

64. 1
x + y2 = 0 S r = 4 −1
=3
1 4 x
R
-1

65. 1
x + y2 = 0 S r = 4 −1
=3
1 4 x
R = 4− x
= 4 + y2
-1

66. a) Show that the area of the annulus at height y is equal to π ( y 4 + 8 y 2 + 7 )
∆y

67. a) Show that the area of the annulus at height y is equal to π ( y 4 + 8 y 2 + 7 )
R = 4− x
= 4 + y2
∆y

68. a) Show that the area of the annulus at height y is equal to π ( y 4 + 8 y 2 + 7 )
R = 4− x
= 4 + y2
∆y
r = 4 −1
=3

69. a) Show that the area of the annulus at height y is equal to π ( y 4 + 8 y 2 + 7 )
R = 4− x
= 4 + y2
∆y
{ }
r = 4 −1
A( y ) = π ( 4 + y ) − ( 3)
22 2
=3

70. a) Show that the area of the annulus at height y is equal to π ( y 4 + 8 y 2 + 7 )
R = 4− x
= 4 + y2
∆y
{ }
r = 4 −1
A( y ) = π ( 4 + y ) − ( 3)
22 2
=3
= π (16 + 8 y 2 + y 4 − 9 )
= π ( y 4 + 8 y 2 + 7)

71. b) Hence find the volume of the solid.

72. b) Hence find the volume of the solid.
∆V = π ( y 4 + 8 y 2 + 7 ) ⋅ ∆y

73. b) Hence find the volume of the solid.
∆V = π ( y 4 + 8 y 2 + 7 ) ⋅ ∆y
π ( y 4 + 8 y 2 + 7 ) ⋅ ∆y
1
∑
V = lim
∆y →0
y = −1

74. b) Hence find the volume of the solid.
∆V = π ( y 4 + 8 y 2 + 7 ) ⋅ ∆y
π ( y 4 + 8 y 2 + 7 ) ⋅ ∆y
1
∑
V = lim
∆y →0
y = −1
1
= 2π ∫ ( y 4 + 8 y 2 + 7 ) dy
0

75. b) Hence find the volume of the solid.
∆V = π ( y 4 + 8 y 2 + 7 ) ⋅ ∆y
π ( y 4 + 8 y 2 + 7 ) ⋅ ∆y
1
∑
V = lim
∆y →0
y = −1
1
= 2π ∫ ( y 4 + 8 y 2 + 7 ) dy
0
1
= 2π  y 5 + y 3 + 7 y 
1 8
5 0
 
3

76. b) Hence find the volume of the solid.
∆V = π ( y 4 + 8 y 2 + 7 ) ⋅ ∆y
π ( y 4 + 8 y 2 + 7 ) ⋅ ∆y
1
∑
V = lim
∆y →0
y = −1
1
= 2π ∫ ( y 4 + 8 y 2 + 7 ) dy
0
1
= 2π  y 5 + y 3 + 7 y 
1 8
5 0
 
3
1 + 8 + 7 − 0 
= 2π  
 
53
296π
= units 3
15

77. b) Hence find the volume of the solid.
∆V = π ( y 4 + 8 y 2 + 7 ) ⋅ ∆y
π ( y 4 + 8 y 2 + 7 ) ⋅ ∆y
1
∑
V = lim
∆y →0
y = −1
1
= 2π ∫ ( y 4 + 8 y 2 + 7 ) dy Exercise 3A;
0
2, 5, 7, 9, 13, 14, 15
1
= 2π  y 5 + y 3 + 7 y 
1 8
5 0
 
3 Exercise 3B;
1 + 8 + 7 − 0  1, 2, 5, 7, 9, 10, 11, 12
= 2π  
 
53
296π
= units 3
15