# hw8 - HW#8 Solution 15.5 Problem 7(a Center of Mass Find...

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Unformatted text preview: HW #8 Solution 15.5 Problem 7. (a) Center of Mass: Find the center of mass of a solid of constant density bounded below by the paraboloid z = x2 + y2 and above by the plane z = 4. 7 42x2+y33y2i 4—x’ and x=i2 wheny=0. So, 2 f4: 4 2 4 2 4 M =j j [dazde 4 j j jdzdydz: 4 j j [dzdydz ~2 ﬂ/Zf2 x3+y3 o 0 x1+y3 0 0 9+)? 22 J-(4_x2_.yz)dydx: 4JJ(4—r2)rdrd(9 O 0 0 If 2 ﬂ' 4 2 2 4J‘(r2——%) d6: 4j4d9=8z O 0 0 27:2 4 27:2 7 4 2/r2 s 27! ‘ “ 32 4 MW: szrdzdrw: j Z—r we: “(gr—Lmrde: I—dﬁzé—Jr a oor2 002 r3 00 2 03 3 — Mn' 8 — — ' 8 So, z = M‘ = By symmetry, x = y = 0. So, the center of mass = (0, 0, —3—). (b) Find the plane z = c that divides the solid into two parts of equal volume. This plane does not pass through the center of mass. zzn/F (' 2nd? in _2 )4 _2 M24430, 47:: j j jrdzdrdaz j [(rcr—r3)drd9= [(C—J—MQJ“ .SO, 0 0 r3 0 0 0 2 4 2 c = Zx/E since c>0. Problem 20. The container is in the shape of the region bounded by y = O, z = 0, 2: = 4 — x2, and x = y2. The density of the liquid filling the region is 60-, y, Z) : kxy _ k a constant (see Exercise 14). work = migzi. 27:448 kg 2y; kg 2 W I g H lkxyzdzdm = —“ I J30“ - x2)2dydx = — J-xz (4 — x2 )de 0 0 o 2 4 0 0 0 16x3 816“ x7 2_256kg =k—gj(16x2—8x4+x6)dx=k—g — +——) .. 4 0 4 3 5 7 0 105 254:9 2% 2 3 5 2 M: J‘kx3r’dzdyr‘dx: J‘kxy(4—x2)dydx=£Jx2(4—x2)dx:ka —kx— 2% 0 0 0 0 0 20 3 10 0 15 Mrszé, MVZZSWQJM and Mfﬂszi '” 3 4 " 231 77 " 105 7 32k 8 256k) The work done is: W = mgd = (F) (g) (7) = 1058 . 15.5 Website Problem 21) 5c<—2x+10)3x 5 a 7 54 20(53) 7 1: I Icy/r dydx= I—dx=c I(2x‘ —20x“ +50x) dx=c (—— +25(5‘)) 0 0 0 2 0 2 3 625 6 1:c— :> c:——— 6 625 5 1X3)?“ dx Ix dx:(ﬁ) 0 3 0 625 625 _i 7 s ' ‘ 6 b) probability 2 —— yx dydx = _ 6" 625 625 625 C) average cost = 6 5 —2.x‘+10 6 5 —2X+10 2 5 — x’“ rdvd 2— “daz— 4—2 +103dx 6250] OJ( mm 5x 6250] ijy 5 x 6250Jx( x > 2 3 2 —1 3 2 2— Ix (—8x +120x —600x+1000)dx=—— x (x —15x +75x—125)dx 625 0 25 0 __ 5 __ 8 7 5 6 5 5 2—16jx7—15x6+75x5—125x4dx=—16(L—ISX +7 x —125X ) 625 0 625 8 7 6 5 0 _—16 (x8 _15x7 +75x6 125\$; _—16 (—2,343,750)_ 250 625 8 7 6 5 0 625 1680 7 15.6 Problem4. zr%3x/—4:r7 Ire/r _ 2 2 n% H jzdzrdrdazj {Wrdrdélzj J'(16r—4r3)drdt9 0 0 0 o 2 o 0 ~ 4—r 7T 2 4 21(8—6—7———4-)dt9=§—7£ 0 II“ it 15 Problem 17. D is the solid right cylinder whose base is the region in the xy—plane that lies inside the cardioid r : 1+ cos (9 and outside the circle r = l and whose top lies in the plane z = 4. 72.2 1+cos¢9 4 i i If(r,6,z)r dzdrd6_ 4% 1 0 Problem 22. 2” ”4 2 277% 2 27! %p4 3 j jj(pcos¢)p2sin¢dpd¢da=j Hp3cos¢sin¢dpd¢d6=j [Teesasingp d¢d6 0 0 0 0 O 0 0 0 0 27r% = I J4cos¢>sin¢ d¢d6 =hJ25in2(%) d6 = 2(27r)sin2(%) = 27r 00 Problem 41. Let D be the smaller cap cut from a solid ball of radius 2 units by a plane 1 unit from the center of the sphere. Express the volume of D as an iterated triple integral in (a) spherical, (b) cylindrical, and (c) rectangular coordinates. Then, ((1) find the volume by evaluating one of the three triple integrals. 1 cos (D (a) z=l=pcos¢:>p= =sec¢.So, sec¢£p£2. 2=sec¢:>¢=§ and (,7) startsfrom0.So, 03¢: Since it is a sphere, O S 6? S 27z. 2z% 2 spherical: IpZSin¢dpd¢d6 0 0 seed 7r 1 3 (b) p=2= r2+z2:>z=i\/4—r2 andz=l.So, issz4—r2. z=l=i\/4—r2 23=r32>r=iﬁsQ 09:6. (9 in spherical coordinate is the same as 6’ in cylindrical coordinate. 27! x5 cylindrical: J. J. J1” dz (11” d6 0 0 1 (c) 1gzgx/4—r2 :>lSzS1/4—x2—yz z=1=i‘/4—x2—y2 23=x2+yz:>.v=i 3—X2 J? 3—x: 4—\ —\ rectangular: I I I dz dV dx ((1) Cylindrical coordinate: Znﬁx/ﬁ erﬁ 2/! _ 2 3/2 2 M H jrdzdrcm: j I(J4—r2—l)rdrd6=I—Er—)——r— (zezijdezii 0 0 l 0 0 0 3 2 6 O 3 0 Problem45. z=—y=—rsin6 and r=3cosl9 3cosﬁ V = 2? MT irsjrlfdzdrdﬁz 2? 3C0Jairsinﬁdrdﬁ: 2i:3—rsin6? d6: 2]r‘—9cos3 (95in6d6 0 :7: yr 37V 0 0 ;;y 2 *2 > 2 0 * 2 =2COS46 =2 4 3% 4 U] ...
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