均匀薄壳密度为ρ，形状为z=3/4-(x^2+y^2) {(x,y)|x^2+y^2<=3/4}

均匀薄壳密度为ρ，形状为z=3/4-(x^2+y^2) {(x,y)|x^2+y^2<=3/4}

∑为曲面z=3/4-(x²+y²)
求出偏导数为
zx=-2x，zy=-2y

∑在xoy面上的投影区域为
D：{(x,y)|x²+y²≤3/4}

M=∫∫[∑]ρdS
=∫∫[D]ρ√(1+zx²+zy²)·dxdy
=∫∫[D]ρ√(1+4x²+4y²)·dxdy
=∫[0~2π]dθ∫[0~√3/2]ρ√(1+4r²)·rdr
=2π·ρ·∫[0~√3/2]√(1+4r²)·rdr
=2π·ρ·1/8·∫[0~√3/2]√(1+4r²)·d(1+4r²)
=2π·ρ·1/12·√(1+4r²)³ |[0~√3/2]
=ρπ/6·(8-1)
=7ρπ/6

根据对称性，
质心的横坐标和纵坐标为0，
竖坐标为
z=1/M·∫∫[∑]zρdS
=1/M·∫∫[D](3/4-x²-y²)ρ√(1+zx²+zy²)·dxdy
=6/(7π)·∫∫[D](3/4-x²-y²)√(1+4x²+4y²)·dxdy
=6/(7π)·∫[0~2π]dθ∫[0~√3/2](3/4-r²)√(1+4r²)·rdr
=12/7·∫[0~√3/2](3/4-r²)√(1+4r²)·rdr
令u=1+4r²，则du=8rdr
r=0时，u=1
r=√3/2时，u=4
所以，
z=12/7·1/8·∫[1~4](1-u/4)√u·du
=3/14·{2/3·u√u-1/10·u²√u} |[1~4]
=3/14·(16/3-16/5-2/3+1/10)
=47/140