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img024bp1 - Section 6.2 Volume The Disc Method 25 8 10 80...

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Unformatted text preview: Section 6.2 Volume: The Disc Method 25 8 10 80 H— l- r —-I 1 l 3 (a) My) = '2'”! = 5(1 * 3})(%){l — 35‘) / \ 1—3; 1-3? = {in - fy)‘ \ J5 I , 2 _ Ji(1)_ J3 V‘To(1"/”_)dy_ 4 10 ‘40 "—l"”_" l l - 3 (d) AC3) = 51m!» = ”—32%: “ 35%)" = gt: - w 1 =3 _ a 1 = 3(L) = .1 V 21,0‘5’)” 210 20 H— |.- y—H 58. 'I'hecrosssectionsaresquarcs.3y symmetrywecansd S9. Assumn that the oil just hitsthetop of the cylinder, which upanimegral foraneighth of mcvulumeandmultiply mamthmmcvolume ofoilis omhalfofflievolume of by 8. the cylinder. Then we have = a = _ 2 2 . AU) 5 (VF: y J sin 20., = £31270“ 7 = 2 - V 850. J12)“. h=dsm7lf l r 511120“ = 8P” "3’31 1 Volume = Ehrrzh} = £3 3 = 3(g)1(d sin 70°) 2 2 sin 20" _ «d3 sin 70’ = 3 — —--—---8 sin 20, and calm“. 1| 60. Let Aitx) andAlbr) equal the areas of the mass sections of the two solids for a S x S 1:. Since A 1(3) = AZLI}, we have b b V: =[ 51h1d1=JI Maid-I: V: mummmesame. ...
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