Question

Many of the definitions we used for two-dimensional mass and moments can be extended to three dimensions rather easily. For example, mass p(x, y, z) DV would represent the mass of the solid Q where p(x, y, z) is the density at any point (x, y, z). Find the mass of the solid bounded laterally by the cylinder x² + y2 = 2x and bounded above and below by the cone z2 = +y?. Here, the density of the solid is proportional to the distance away from the z-axis.

Answer 1

If you have any doubts in the solution please ask me in comments here i used basic concept of cylinerical coordinates

of To bind mass the solid Bounded by the cylinder x² + y² = 2x & Bounded Below ļ ausone By 2²= x² + y² & density of solid is propotional to the distance away from the Z-axis . e(x, y, z) D x² + y2 P(x, y, z) rox²+y² --0 where Kis constant NOW given the solid Bounded by the cylinder 2e2+y2 = 2x (which has Center (1o) { radius 1 & Below By come a²=x²+4²

1 2 y Solld equired) x ? 2 اعا Projection on 2-4 plane. SO we have solid E = {(2,9, 2): -5x+y2 <2< 5x2+42 : -J22-22 sy 2x-p² 2 } < cx SSS P12,4, 2) av $ ss k Jxz+y2 cody dzdy dx E E assume NOW 2=2, y=rcino, x = Sess 7 Sino. cuordinates in to cylindricai We now change the wordinates 2²+ y²= 2x 22= 28caso . 7= 2coso dz dy dxe z dzdodo E = { 19,0.2) : -9 < 257 ; 0 <Ó <211, osts 2005o} e 20 2coso SSS €12,4, 2100 = s s krir dz do do 0 -7 E 2 T 20507 11 K ļļ Sredz do de 0 27 20010 275 2cos k kl 5 208 do do O O 21T 21 ) 20080 do = $k S costo do. 2 1 Bkx2 cos "odo .: Cos 420-0) = cosyo T1/2 OSS 218,9,21 av 32K S costo do cos" (1-0) = Como E

> Mass = SSS P(x, y, z) dva 32 K comodo. E T1/2 2 32% Ś (1+ CD 20) do 12 Sak 32 1+200520 + cos220 do 4 1/2 > M= sss erx, y, 2) dx/ 8k ? (1+2005 20 + 1+cosuo ) do 2 T1/2 UK 5 (2 + 46520 + 1 +(540) do M2 UK (3 + 410520 + cos40) do T/2 -uk T/2 [ 3x I 3x I + q (sin2o). + (Sinug). لا 4KX 317 6KT so Mass of solid GKIT where K is consteint Let K=1 6 IT Auswer mass of the Solid 18.8495