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 ² ² 61. 9x 4y z 36 62. 4x 9z y 63. x y 16z 16 # # # # # œ ± ² œ 64. z 4y 9 65. z x 66. y x z 1 # # ± œ œ ² ± ²²œ ab 67. x 4y 1 68. z 4x y 4 69. 4y z 4x 4 # # # ²œ œ± ² ± 70. z 1 x 71. x y z 72. y z 1 œ² ±²œ # # # 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 ²²œ 77. (a) If x 1 and z c, then x 1 A ab ± œ ±œ Ê ± œÊœ yy y 49 4 9 z9 c x # # # # Š‹ ’“ 9c 9 49 c 9 ˆ‰ 1 œœ 1 Š ÈÈ ab 29c 33 9 ±± ± # 1 (b) From part (a), each slice has the area , where 3 z 3. Thus V 2 9 z dz 29z 99 2 1 1 ± # # ²Ÿ Ÿ œ ² ' 0 3 9 9z (27 9) 8 œ² œ ² œ ² œ 44 z 4 3 9 11 1 ' 0 3 ab ’“ # $ ! $ 1 (c) 1 1 A xz x abc cc ac z bc z # ±±œÊ ± œÊ œ –— ## # 1 V 2 c c z c . Note that if r a b c, Êœ ² œ ² œ œ œ œ œ ' 0 c c 111 1 a b 2a b z b 2 4a b c 3 c 3 3 $ # $ ! then V , which is the volume of a sphere. œ 4r 3 1 $ 78. The ellipsoid has the form 1. To determine c we note that the point (0 r h) lies on the surface RRc y # ß ß # of the barrel. Thus, 1 c . We calculate the volume by the disk method: rh h R Rc R r # # # # ±œÊ œ # ± V y dz. Now, 1 y R 1 R 1 R z œ± œ Ê œ ² œ ² œ ² 1 ' h h # # # # ± ± y c h R h zz R r zR r # # # # # # # Š‹’ Š V R R z z 2 R h R r h 2 ² œ ² œ ² ² œ ± 1 ' h h h h Š ± # $ # # # ±" ± " Rr 2 R h r h h3 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 42 1 # radius R and height 2h. If r 0 and h R, then V R which is the volume of a sphere. œ 4 3 1 $ 79. We calculate the volume by the slicing method, taking slices parallel to the xy-plane. For fixed z, y # # ±œ gives the ellipse 1. The area of this ellipse is a b (see Exercise 77a). Hence z a b z y c # # Š‹ Š‹ za zb œ 1 1 the volume is given by V dz . Now the area of the elliptic base when z h is œ œ ' 0 h h 1 abz abz abh c2 c c ! A , as determined previously. Thus, V h (base)(altitude), as claimed. œ œ 1 abh abh abh c # ""
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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 xz abc y ## ### # ±²œ 1. The area of the cross-sectional ellipse (see Exercise 77a) is x y # # –— ac z bc z cc ## # Š‹ ±œ A(z) c z c z c z . The volume of the solid by the method of slices is œ± ± œ ± 1 ÈÈ ab a b c 1 # V A(z) dz c z dz c z z c h h 3c h œœ± œ ± œ ± œ ± '' 00 hh h 11 1 1 ab ab ab abh 3 c 3 3 c # # a b ± ‘ˆ # $ # $ "" ! (b) A A(0) ab and A A(h) c h , from part (a) V 3c h ! œœ ± Ê œ ± 1 h ab abh c3 c a b 21 2 2 a b c h ( 2 A A ) ± œ±œ ± ± œ ± 1 abh h abh c h h ab h 3c 3 c 3 c 3 Š ± # # ± ! 1 h (c) A A c 4c h (A 4A A ) m mh ± œ ± ʱ ± ˆ‰ ha b h a b h c4 4 c 6 # # ! # ab 4c h c h c 4c h c h 6c 2h ± ±± œ ± ± ± ± œ ± b a b a b h a b h 6c c 6 c 6 c ± a ba b a b 1 111 1 # # # #### # # 3c h V from part (a) œ 1 abh 3c # 81. y y , a parabola in the plane y vertex when 0 or c 0 x 0 œÊ œ ² œ œ ² œ Ê œ z x dz dz 2x cb a d x d x a y # # # 1 Vertex 0 y ; writing the parabola as x z we see that 4p p Êß ß œ ² ± œ ² Ê œ ² " # cy a y bc b c 4 c aa a # # # # 1 1 Focus 0 y ß
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This homework help was uploaded on 09/23/2007 for the course MATH 1910 taught by Professor Berman during the Spring '07 term at Cornell University (Engineering School).

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

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