ISM_T11_C06_E - Chapter 6 Practice Exercises(d A =2/< 7/>29...

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Chapter 6 Practice Exercises 413 (d) : A+=2/< 7/>29. V R (x) r (x) dx œ± ' a b 1 cd ## 4 d x ± 1 ' 1 2 ’“ ˆ‰ ˆ 74 x # # # $ 16 1 2x x dx ± ² 49 4 1 1 ' 1 2 ab ±$ ±' 1 6xx ²± 49 x 45 1 1 ±# # " & 16 2 1 1 ² ± ± ² ± 49 44 5 35 1 1 ± ˆ "" " # 16 ±² 49 4 4 160 5 1 1 """ (40 1 32) ± ²œ±œ 49 16 49 71 103 4 160 4 10 20 11 1 9. (a) : .3=5 7/>29. V x 1 dx (x 1) dx x œ ± œ ± 1 '' 55 Š‹ È #& # " x # 51 4 8 ± ± œ ± œ 1 ± ‘ˆ ˆ 25 24 # " (b) : A+=2/< 7/>29. R(y) 5, r(y) y 1 V R (y) r (y) dy 25 y 1 dy œœ ² Ê œ ± œ ± ² # # # c2 d2 25 y 2y 1 dy 24 y 2y dy 24y y 2 24 2 8 ± ± œ ± ± œ ± ± œ ± ± 1 1 22 a b %# $ # ±# y 53 23 2 2 & †† 32 3 (45 6 5) ± ± œ 1 2 32 1088 5 3 15 15 " (c) : .3=5 7/>29. R(y) 5 y 1 4 y œ± ² œ± V R (y) dy 4 y dy Êœ œ ± # 16 8y y dy ² 1 ' 2 2 #% 16y 2 32 ² ² 8y y 64 32 $ & # ±# 64 1 (15 10 3) ² ² œ 1 2 64 512 3 5 15 15 " 10. (a) : =2/66 7/>29. V 2 dy 2 y y dy ± c0 d4 Š shell shell radius height 4 y # 2 y dy 2 2 œ ± œ ± 1 ' 0 4 # % ! yy y 43 1 6 3 4 64 64 $$ % 64 2 13 # (b) : =2/66 7/>29. V 2 dx 2 x 2 x x dx 2 2x x dx 2 x ± œ ± œ ± ''' a00 b44 1 1 ˆ ˆ È shell shell radius height 4x $Î# # &Î# % ! $ 2 œ 1 4 64 128 1 5 1 (c) : =2/66 7/>29. V 2 dx 2 (4 x) 2 x 2 8x 4x 2x x dx ± ± œ ± ± ² ' a0 0 b4 4 111 ˆ ˆ È shell shell radius height "Î# $Î# # 2 x 2x x 2 8 32 32 64 1 ± ² œ ± ± ² œ ± ± ² 1 ˆ 16 4 x 16 4 64 4 4 2 3 3 5 3 3 5 3 $Î# # &Î# % ! $ 64 1 œ 1 46 4 1 (d) : =2/66 7/>29. V 2 dy 2 (4 y) y dy 2 4y y y dy œ œ ± ± œ ±±² ' 0 4 1 Š Š shell shell radius height 4 4 # $
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414 Chapter 6 Applications of Definite Integrals 2 4 y2 y d y22 y y 23 2 6 41 6 3 22 1 œ± ² œ ± ² œ ± ² œ ± ² œ 11 1 1 ' 0 4 Š‹ ˆ‰ ˆ ## $ % ! yy 43 1 6 3 3 3 8 3 2 $ % 1 11. : .3=5 7/>29. R(x) tan x, a 0, b V tan x dx sec x 1 dx [tan x x] œœ œ Ê œ œ ± œ 1 1 3 3 33 1 '' 00 ÎÎ Î$ ! ± ab È 12. : .3=5 7/>29. V (2 sin x) dx 4 4 sin x sin x dx 4 4 sin x dx ² ² 1 ' 0 1 ± # 1 cos 2x 4 x4 c o s x 4 4 0 ( 0400 ) 8 ( 9 1 6 ) œ ² ²± œ ±²± ± ²²± œ ± œ ± 1 1 1 ± ± ‘ˆ xs i n 2 x 9 4 # # ! 1 1 13. (a) : .3=5 7/>29. V x 2x dx x 4x 4x dx x x 16 œ±œ± ²œ ± ² œ ± ² 1 1 a b ’“ # %$# % $ # # ! x 4 32 32 53 5 3 & (6 15 10) ² œ 16 16 15 15 (b) : A+=2/< 7/>29. V 1 x 2x dx dx x dx ± ² " œ ² ± " œ # ± œ # ± œ ' 0 2 1 1 1 1 1 a b # " && & # ! #) x & 1 (c) : =2/66 7/>29. V 2 dx 2 (2 x) x 2x dx 2 (2 x) 2x x dx ± ± ± œ ± ± ' a0 0 b2 2 1 cd shell shell radius height 2 4x 2x 2x x dx 2 x 4x 4x dx 2 x 2x 2 4 8 œ ±±² œ ±² œ ± ² œ ± ² 1 1 a b $ $# # ! 3 2 3 % (36 32) œ 28 (d) : A+=2/< 7/>29. V 2 x 2x dx 2 dx 4 4 x 2x x 2x dx 8 ± ± ± ² ± ± 1 1 ' 0 2 ### # 4 4x 8x x 4x 4x dx 8 x 4x 8x 4 dx 8 œ ² ± œ ² ± 1 1 0 0 2 2 a b # %$ x 4x 4x 8 16 16 8 8 (32 40) 8 œ ± ² ² ±œ ±²²±œ ² ± œ 1 1 1 x3 2 7 2 4 0 3 2 55 5 5 5 5 & %# # ! 1 1 14. : .3=5 7/>29. V 2 4 tan x dx 8 sec x 8 [tan x x] 2 (4 ) ± œ ± œ ± 1 1 1 44 Î% ! 1 15. The material removed from the sphere consists of a cylinder and two "caps." From the diagram, the height of the cylinder is 2h, where h , i.e. h . Thus # ² $ œ# œ" È V h ft . To get the volume of a cap, cyl œ# $ œ' È # $ use the disk method and x y : V x dy ## # # " # œ cap ' 2 1 yd y y œ% ± œ % ± ' " # # " 2 ab’“ y 3 3 8 ft . Therefore, ± % ± œ 1 ± 8 3 "& $ 1 VV V f t .
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This note was uploaded on 10/13/2011 for the course MATHEMATIC 103 taught by Professor Thommas during the Spring '11 term at LCC Intl University.

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ISM_T11_C06_E - Chapter 6 Practice Exercises(d A =2/< 7/>29...

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