. . Find the center of mass of a thin plate of constant density 6 covering the region bounded by the parabola y = 2x2 and the line y = 8. The center of mass is located at (x y) = (Simplify your answer. Type an ordered pair.} 2. Find the center of mass of a thin plate of constant density 6 covering the given region. The region bounded by the parabola y = 3x 3x2 and the line y = 3x The center of mass is :|. (Type an ordered pair.) 3 Find the center of the mass of a thin plate of constant density 6 covering the region bounded by The center ofme mass is located at (i, 9) = _ 3 11: TI: . the x-axis and the curve y= , cos X. _ 7 5x5 (Type an ordered pair. Round to the nearest hundredth.) 2 41' 1008 = 3 Center of mass (@, y) is calculated as: x = M JJ. x dA 5 3x-2 2 = M x dyda 5 3x-2 2 = (ay da -3x 5 x3 + 6x2 dx 1 5 = M 4 + 203 0 CO 375 S. = 2.8125 1008 4 y = M y dA D 1 3x-2-2 = y dyda -3x 5 3x-2-2 dx M -3xD is the region bounded by curves: y = 3x - x- and y = -3x, as shown in graph-1 below. +4 +2 -2 +4 +8 - 2 -4 -6 -8 -10 -12 -14 -16 -18 Graph 1: Region of integration (D) Density is constant and is denoted as S x varies from 0 to 5 and y varies from y = -3x to y = 3x - 22 Therefore, 5 ~3x-2-2 M = 6 dydx -3x 5 3x - a2 - (-3ac) dac 5 = 8(3x2 3 0Answer ) Given curve is y= COST , THE XL 5/ 4 We consider ventical strips with width dix , length ly ? 7 dA = 3/ cosx , an mass dm= SdA = 2 & cosx dix (assume S = 1 ) 15/ 4 M - am 3 2 cost dix - T/4. . - Tu 3 12 2 The coordinates of Centre of mass are IT / 4 * = 3 (312 ) x. 2 cost du 2 -T/y By using integration by part. ( udv= UV- Syd = , x = 0 y = ( f () ) du 2 2 2 1 X 9 4 cosix dx. 2 - 1 /u 3 3 cosadn : + ~ 1. 2854 ~ 1029 Hence the centre of mass is located at ( x, y ) = (0, 1.29 )1 5x46sc3 1 1 m5 3:134 5 M5((? 7) )0 3 625 5 = 4.6875 1005 4 Thus center of mass of the lamina lies at point (2.8125, 4.6875)