Then the quality of the surrounding area m = ρ∫∫ DS =ρ∫∫∫ RDRD θ = ρ∫ (0->; 2π)dθ∫(0-& gt； A (1-cos θ)) RDR = 3 π a 2 ρ/2

Since the centroid line is symmetrical about x, the ordinate y0=0 of the centroid is 0.

x0 =ρ∫∫xds/m=ρ∫∫r^2cosθdrdθ/m =ρ∫(0-& gt； 2π)cosθdθ∫(0-& gt； One (1-cos θ)) r 2dr/m

=(-5πa^3ρ/4) / (3πa^2ρ/2)

=-5a/6

So the center of mass is (-5a/6,0).