高等数学 详细过程 谢谢。
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不妨设面密度为 1.
z = √(a^2-x^2-y^2), ∂z/∂x = -x/√(a^2-x^2-y^2),
∂z/∂y = -y/√(a^2-x^2-y^2),
dS = √[1+(∂z/∂x)^2+(∂z/∂y)^2] = adxdy/√(a^2-x^2-y^2),
质量 M = ∫∫<∑>dS = ∫∫<Dxy>adxdy/√(a^2-x^2-y^2)
= a∫<0, π/2>dt∫<0, a>rdr/√(a^2-r^2)
= (πa/2)∫<0, a>(-1/2)d(a^2-r^2)/√(a^2-r^2)
= (πa/2)[-√(a^2-r^2)]<0, a> = (π/2)a^2
∫∫<∑>xdS = ∫∫<Dxy>axdxdy/√(a^2-x^2-y^2)
= a∫<0, π/2>dt∫<0, a>rcost rdr/√(a^2-r^2)
= a∫<0, π/2>costdt∫<0, a>r^2dr/√(a^2-r^2) (令 r = asinu)
= a^3∫<0, π/2>costdt∫<0, π/2>(sinu)^2du
= (1/2)a^3[sint]<0, π/2> [u-(1/2)sin2u]<0, π/2> = (π/4)a^3
质心横坐标 x0 = (π/4)a^3/[(π/2)a^2] = a/2
同理, y0 = z0 = a/2
z = √(a^2-x^2-y^2), ∂z/∂x = -x/√(a^2-x^2-y^2),
∂z/∂y = -y/√(a^2-x^2-y^2),
dS = √[1+(∂z/∂x)^2+(∂z/∂y)^2] = adxdy/√(a^2-x^2-y^2),
质量 M = ∫∫<∑>dS = ∫∫<Dxy>adxdy/√(a^2-x^2-y^2)
= a∫<0, π/2>dt∫<0, a>rdr/√(a^2-r^2)
= (πa/2)∫<0, a>(-1/2)d(a^2-r^2)/√(a^2-r^2)
= (πa/2)[-√(a^2-r^2)]<0, a> = (π/2)a^2
∫∫<∑>xdS = ∫∫<Dxy>axdxdy/√(a^2-x^2-y^2)
= a∫<0, π/2>dt∫<0, a>rcost rdr/√(a^2-r^2)
= a∫<0, π/2>costdt∫<0, a>r^2dr/√(a^2-r^2) (令 r = asinu)
= a^3∫<0, π/2>costdt∫<0, π/2>(sinu)^2du
= (1/2)a^3[sint]<0, π/2> [u-(1/2)sin2u]<0, π/2> = (π/4)a^3
质心横坐标 x0 = (π/4)a^3/[(π/2)a^2] = a/2
同理, y0 = z0 = a/2
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