ISM_T11_C12_C - 808 Chapter 12 Vectors and the Geometry of...

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808 Chapter 12 Vectors and the Geometry of Space 58. 4x 4y z 4 59. 16y 9z 4x 60. z x y 1 # # # # # # # # œ œ œ 61. 9x 4y z 36 62. 4x 9z y 63. x y 16z 16 # # # # # # # # # œ œ œ 64. z 4y 9 65. z x y 66. y x z 1 # # # # # # # œ œ œ a b 67. x 4y 1 68. z 4x y 4 69. 4y z 4x 4 # # # # # # # œ œ œ 70. z 1 x 71. x y z 72. y z 1 œ œ œ # # # # # x 4
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Section 12.5 Lines and Planes in Space 809 73. yz 1 74. 36x 9y 4z 36 75. 9x 16y 4z œ œ œ # # # # # # 76. 4z x y 4 # # # œ 77. (a) If x 1 and z c, then x 1 A ab # # œ œ œ Ê œ Ê œ y y y 4 9 4 9 z 9 c x Š 9 c 9 4 9 c 9 1 œ œ 1 Š ‹ Š È È a b 9 c 2 9 c 3 3 9 2 9 c 1 (b) From part (a), each slice has the area , where 3 z 3. Thus V 2 9 z dz 2 9 z 9 9 2 1 1 a b # Ÿ Ÿ œ ' 0 3 a b 9 z dz 9z (27 9) 8 œ œ œ œ 4 4 z 4 9 9 3 9 1 1 1 ' 0 3 a b # $ ! 1 (c) 1 1 A x z x a b c c c y y a c z b c z œ Ê œ Ê œ È È a c z b c z c c 1 Š ‹ Š V 2 c z dz c z c . Note that if r a b c, Ê œ œ œ œ œ œ œ ' 0 c c 1 1 1 1 ab 2 ab z 2 ab 2 4 abc c c 3 c 3 3 a b ˆ # # # $ ! then V , which is the volume of a sphere. œ 4 r 3 1 78. The ellipsoid has the form 1. To determine c we note that the point (0 r h) lies on the surface x z R R c y œ ß ß # of the barrel. Thus, 1 c . We calculate the volume by the disk method: r h h R R c R r œ Ê œ # V y dz. Now, 1 y R 1 R 1 R z œ œ Ê œ œ œ 1 ' h h # # # # # # y R c c h R h z z R r z R r Š Š a b V R z dz R z z 2 R h R r h 2 Ê œ œ œ œ 1 1 1 1 ' h h h h Š Š Š a b # # # $ # # # " " R r R r 2R h r h h 3 h 3 3 3 R h r h, the volume of the barrel. If r R, then V 2 R h which is the volume of a cylinder of œ œ œ 4 2 3 3 1 1 1 # # # radius R and height 2h. If r 0 and h R, then V R which is the volume of a sphere. œ œ œ 4 3 1 $ 79. We calculate the volume by the slicing method, taking slices parallel to the xy-plane. For fixed z, x z a b c y œ gives the ellipse 1. The area of this ellipse is a b (see Exercise 77a). Hence x z z abz y c c c Š Š za zb c c œ œ 1 ˆ ‰ ˆ È È 1 the volume is given by V dz . Now the area of the elliptic base when z h is œ œ œ œ ' 0 h h 1 1 1 abz abz abh c 2c c ! A , as determined previously. Thus, V h (base)(altitude), as claimed. œ œ œ œ 1 1 1 abh abh abh c c c " " # # ˆ
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810 Chapter 12 Vectors and the Geometry of Space 80. (a) For each fixed value of z, the hyperboloid 1 results in a cross-sectional ellipse x z a b c y œ 1. The area of the cross-sectional ellipse (see Exercise 77a) is x y a c z b c z c c œ A(z) c z c z c z . The volume of the solid by the method of slices is œ œ 1 Š ‹ Š È È a b a b ab c c c # # # # # # 1 V A(z) dz c z dz c z z c h h 3c h œ œ œ œ œ ' ' 0 0 h h h 1 1 1 1 ab ab ab abh c c 3 c 3 3c a b a b ˆ # # # $ # $ # # " " ! (b) A A(0) ab and A A(h) c h , from part (a) V 3c h ! # # # # œ œ œ œ Ê œ 1 h 1 1 ab abh c 3c a b a b 2 1 2 2 ab c h (2A A ) œ œ œ œ 1 1 1 abh h abh c h h ab h 3 c 3 c 3 c 3 Š Š a b # # !
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