. Differential and integral calculus, an introductory course for colleges and engineering schools. nx = b fixy ds I fixy V1 +x=a Ux = a 2d dx nyds fX fiyVl+y*dx Ux=a Example. Let the parabola y2 = 4 ax be revolved about OX; tofind the center of mass of that portion of the surface generated, which iscut off by the plane i = 3a, the density being constant. We have ds =\/: x + a dx, yds = 2 \ra^/x + a dx. * This equation is only approximately true, and the derivation of the for-mula? in the text is not rigorous, although the formulae themselves are accurate. 344 Hence INTEGRAL CALCULUS y/x + axdx
. Differential and integral calculus, an introductory course for colleges and engineering schools. nx = b fixy ds I fixy V1 +x=a Ux = a 2d dx nyds fX fiyVl+y*dx Ux=a Example. Let the parabola y2 = 4 ax be revolved about OX; tofind the center of mass of that portion of the surface generated, which iscut off by the plane i = 3a, the density being constant. We have ds =\/: x + a dx, yds = 2 \ra^/x + a dx. * This equation is only approximately true, and the derivation of the for-mula? in the text is not rigorous, although the formulae themselves are accurate. 344 Hence INTEGRAL CALCULUS y/x + axdx _ , , , .3~|3° 225 Vx + adx n [(* + o)«]; 5835 a. Problem 1. Obtain a formula for the center of mass of a solid of revo-lution. Problem 2. Devise a rigorous demonstration of formulae (D). 225. Center of Mass of Any Curved Surface. When the bodywhose center of mass is sought is a thin curved shell, whose thick-ness is so small that the shell may be regarded as a curved sur-face, the element of volume becomes the element of such a surface,and is, by Art. 216, -Tjfdxdy, ~dz Therefore where. Vr. Vr —tt dy dz, or —rj dz dx i dz dxdy V = M M = a tx Vr dz dxdy. By using one of the other elements ofsurface, we get other similar expressionsfor x, y, z. Example. To find the center of mass of thatportion of the surface of the homogeneous sphere■X x2 + y2 + z2 = a2 which lies above the x?/-plane and is cut out bythe cylinder x2 + y2 — ax = 0. 2 denotes the entire circular base of the cylinder. §225 CENTERS OF MASS 345 It is readily found that VR _ a = g Jj_~ z Va?-x2-y2dZHence. U Vc dxdy S x—a y=^ax—x2 Numerator = 2 a \ x I dx J J V a2 - x2 - y1 i=0 y=0 £=a y= Vax—x2 = 2a |x sin-1 = \ dx = 2a I xsin-1 v —%— dx,J L Va2-x2] Jo a +x 3=0 2/=0 This last integral may be evaluated by setting x = a tan2 0. We getfinally, Numerator of x = f a3. In exercise 1 of Art. 217 it is found that S, the area of that portion ofthe spherical surface whose center of mass is so
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