设单位面积质量1,
得到此均质圆弧质量为:(a/(2π))*πR^2=(1/2)aR^2
显然,质心应在扇形的对称轴上,设其与圆心的距离为X
则:((1/2)aR^2)X=∫∫(r*cosα)*dr*rdα=∫∫(cosα)r^2drdα
(r从0到R,α从-a/2到a/2)
((1/2)aR^2)X=∫∫(cosα)r^2drdα=∫(cosα)dα ∫r^2dr =2sin(a/2)*(1/3)R^3
=(2/3)sin(a/2)R^3
X=(4R/3)sin(a/2)