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img012mu6 - 67 50 — 0.5 = 0.125 x = 30 11,030 = 313(80 =...

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Unformatted text preview: 67. 50 — 0.5.: = 0.125; x = 30 11,030) = 313(80) = 10 Point of equilibrium: (80. 10] ID cs = I [(50 — 0.5;) — 10] a; D = [—42— + 4Gx]m=1600 mar) [10 — 0.125;] d: 0 DJ 2 [W _ 0.125%] mm 2 o 69. True Section 6.2 Volume: The Disc Method 1. V=1rJ; (-x+l)3dx=1r-L (xz-2x+l)dx=1r[—-x3+x] 1 z 2. V=1rJ (4-xz)3dx=1rf (fl-8x1+16)dx=w[ o a 3. V: wJIRJiFdx= 11"]:de = «[1; 1'2Jcb:=-1r£: (4‘x')dx—‘n[4x--a— 2 s V=«I[(zz)=—1x3)=ldx—wfwmax—,[———]'=2_" 4.v=wJ:(./Z— — _ _fi 11. 2—4 4 8=16—x3 111:8 1=2t2f v-—- «J::[(4——) —]dx(2)= =2wL fi[E-2-x‘+12]dx = 2v[—2;3 w — + lkxfi =21{123Ji_3_2fi+24J—] 80 -mfi 15 11' P—' 132.69 Section 6.2 Volume: The Disc Method —___—._—-—-—-—-——-_ 68- 1000 — 0.411!1 = 42: x = 2|) P100) = P1903 = Point of equilibrium: (m. 840) no CS =J’ [(1000 - 0.4x”) - Whiz n 3 =[160x — 91]m - 2133.33 3 n Ps=r°[s40-42x]¢£x D :11 = [840! - 21x11) = 84-00 70. True 1611' 35 13 ...
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