Question

Find the center mass of the solid bounded by planes x+y+z=1, x = 0, y = 0, and z = 0, assuming a mass density of p(x, y, z) = 15/2. (CCM, YCM, 2CM) =

Answer 1

Solution: o Given that: The planes: x+y+2=1, -) 2=1-3-4 x=0, put 2-0 y=o and, 1-X-Yo y=1-x 20 put yzo density P(x, y, 2) = 152 1-X=0 21 = 1 The limits ale! OLXZI, OFyzix, O2 z L 1-x-y Then, mass (m) SSS pdu 1-4-4 mas 15V2 dzdy de D 6 0 - = 15 1-X-4 3/2 وال 2 dydu D 32 6 1-Y 312 SI (i-x-4) dyda 0 (L- 512 du 0 512 = Tox2 (1-x)- da )-(-x)) - (1-x)}2} 4 CL-X - 1 S12 72 (1-x) ។ (1-x) dx . 0 U

2-4x2 (C1-x) 7/2 ) t UX (0-1) Mass (m) (-1) 7 &cm = Magz m - My z = SIS 1-X 1-x-y a (1587 ) dzayda Apdx v 0 (ir 1-x-y dydu 15 x 2 0 0 1-K 312 1882 ja (1-x-y) dyda 3 0 - - 10 ( (1-x-4) 2 j da Siz 6 B - vox 2 -(1-x2}an s x 1 $(6-2)-(i-a)) + 48*(1-3din | Cott put EI- dta-du 512 - 4 (dt) 5) XL-E les = -4 722) at

+12 t ul! 케 2 912 0 E-4 (1233) cy (1-23" (1-4) --4--)-( -4-187 - 18+14 63 - 16 Myz 63 16 7 18 564 cm Myz 63 8 M ya - MZX M 1- y where yp dz dydu Mzz = 0 1- (-4-9 y1552) dz dydu o O 312 |-r.-Y -15 s' y ( 2 ) dyda 0 0 3/2 y (inning) dydn 0 put t = l-x-y y-1-x-- Cl-x-t) dt da to dy - adt (1322 312 512 dedu L.Li.y=0, tal-x uliy=(-4,t=0 0

ESIL 512 10 H2 xt SI h 512 Sl2 du 0 712 10 I'l (1-x)(1-x) 52 (1-x)?! Jane 52 0 712 - 20 f ( x _ (xhan 2 l (1-x, 7/2 (1-x) -589/ -ax 0 = 20x (Co-oj=6*** ) 40 ( 25 ) 7 Max. 80 315 Then, 10 Чем MZu SO 315 7 70 M 77 315 315 ZCM Mх ч 1-X-4 where. Mxy si zodz dydu 6 9 ja visuaalyan 6 D | --4 = 15 l's 2 2 512 dydu O

trex- чарі А и (--9) ( dn -ti, о І -c«( ; ; (-х ч. а. ху", - о (-x) 1а ( 1 -А, о 6 + (+) (o-1 ) М 14 2cm Mч 8422 1 Ir 8 M 84 Zcm = 27 ។ ІІa (Хен, Чем, 2cm) - C то Зls +1» 504