. Differential and integral calculus, an introductory course for colleges and engineering schools. As element of volume we take a thin slice parallel to the base. AH theparticles in such a slice will have the same density. The volume of suchan element of volume is 7r(r2 — z2)Az and its mass is x(r2 — z2) kzAz. 340 INTEGRAL CALCULUS §221 Hence z — 15 15 r r /r2z3 z5r „kfy-*)*dz -(T-j=) Therefore the center of mass is at the point (0, 0, tzt). Example 2. To find the center of mass of an octant of the spherex2 — y2 — z2 = r2 when the density is proportional to the distance fromthe yz-plane.

Captions are provided by our contributors.

## Image details

Contributor:

Reading Room 2020 / Alamy Stock Photo

Image ID:

2CEJ6AJ

File size:

7.1 MB (215.6 KB Compressed download)

Releases:

Model – no | Property – noDo I need a release?

Dimensions:

2329 x 1073 px | 39.4 x 18.2 cm | 15.5 x 7.2 inches | 150dpi

More information:

This image is a public domain image, which means either that copyright has expired in the image or the copyright holder has waived their copyright. Alamy charges you a fee for access to the high resolution copy of the image.

This image could have imperfections as it’s either historical or reportage.

. Differential and integral calculus, an introductory course for colleges and engineering schools. As element of volume we take a thin slice parallel to the base. AH theparticles in such a slice will have the same density. The volume of suchan element of volume is 7r(r2 — z2)Az and its mass is x(r2 — z2) kzAz. 340 INTEGRAL CALCULUS §221 Hence z — 15 15 r r /r2z3 z5r „kfy-*)*dz -(T-j=) Therefore the center of mass is at the point (0, 0, tzt). Example 2. To find the center of mass of an octant of the spherex2 — y2 — z2 = r2 when the density is proportional to the distance fromthe yz-plane. In this case n = kx, and we have x = Cx2dV CxydV Cxz Jy _ Jy _ Jy dV jxdV jxdV jxdVi. I Fig.l .Fig.2 JDetermining limits according to Fig. 1, we have x=r y=Vr2-xi z=-/r2-x2-2/2 x=r y=Vri-x1 Cx2dV = j j fx2 dz dy dx = fx2fVr2 – x2 – x=0 y=0 2=0 X=0 2/=0 y2dydx. J/=VrJ-x* fVr2-x2-y2 dy= ~[yVr2-x2-y2+(r2-x2)sm-1—^== = ^(r2-x2)./-o L v=o ~ §222 CENTERS OF MASS 341 To find CxydV it will be convenient to change the order of integrationand to proceed according to Fig. 2. We have VV2-22 x=Vri-y2-z2 y= Vr*-i l)ydydz 15 .5 fxydV-f f fyxdxdydz = lf j(, z=0 y=0 x=0 2=0 y = From symmetry it is evident that CxzdV = CxydV = —. Jv Jv 15 • As for the denominators in the expressions for x, y, z, we found in Art. 218, example, that | xdV= —•Jy 16 pp, , – 7rr5 Trr4 8 – – a:5 ^r4 16r/ , 1 Therefore x = -^- = lT; 2/=, = -^- = —^about-rj. Hence the center of mass is at the point (— r, —-, -— ) or (-— r, -r. ~r) • 15 Iott lo7r/ 15 3 3/ 222. Center of Mass of a Plane Area. The body whosecenter of mass we wish to find may be like a thin sheet of metal, whose thickness