. 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

CƠ HỘI VÀ THÁCH THỨC DÀNH CHO ĐT VIỆT NAM TẠI VÒNG LOẠI WORLD CUP 2026
CƠ HỘI VÀ THÁCH THỨC DÀNH CHO ĐT VIỆT NAM TẠI VÒNG LOẠI WORLD CUP 2026

. 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.

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. 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

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