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# X2 T06 01 Discs & Washers

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### X2 T06 01 Discs & Washers

1. 1. Volumes of Solids Volumes By Discs & Washers ( slice ⊥ rotation )
2. 2. Volumes of Solids Volumes By Discs & Washers ( slice ⊥ rotation ) About the x axis
3. 3. Volumes of Solids Volumes By Discs & Washers ( slice ⊥ rotation ) About the x axis y y = f ( x) x
4. 4. Volumes of Solids Volumes By Discs & Washers ( slice ⊥ rotation ) About the x axis y y = f ( x) a b x
5. 5. Volumes of Solids Volumes By Discs & Washers ( slice ⊥ rotation ) About the x axis y y = f ( x) a ∆x b x
6. 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. 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. 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. 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. 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. 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. 12. About the y axis y y = f ( x) x
13. 13. About the y axis y y = f ( x) d c x
14. 14. About the y axis y y = f ( x) d ∆y c x
15. 15. About the y axis y y = f ( x) d x ∆y ∆y c x
16. 16. About the y axis y y = f ( x) d x = g( y) ∆y ∆y c x
17. 17. About the y axis y y = f ( x) d x = g( y) ∆y ∆y c A( y ) = π [ g ( y ) ] x 2
18. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 57. a) Show that the area of the annulus at height y is equal to π ( y 4 + 8 y 2 + 7 )
58. 58. a) Show that the area of the annulus at height y is equal to π ( y 4 + 8 y 2 + 7 ) ∆y
59. 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. 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. 61. 1 x + y2 = 0 S 1 4 x -1
62. 62. 1 x + y2 = 0 Sr 1 4 x -1
63. 63. 1 x + y2 = 0 S r = 4 −1 =3 1 4 x -1
64. 64. 1 x + y2 = 0 S r = 4 −1 =3 1 4 x R -1
65. 65. 1 x + y2 = 0 S r = 4 −1 =3 1 4 x R = 4− x = 4 + y2 -1
66. 66. a) Show that the area of the annulus at height y is equal to π ( y 4 + 8 y 2 + 7 ) ∆y
67. 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. 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. 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. 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. 71. b) Hence find the volume of the solid.
72. 72. b) Hence find the volume of the solid. ∆V = π ( y 4 + 8 y 2 + 7 ) ⋅ ∆y
73. 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. 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. 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. 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. 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