物理高手请进！！！英文的！！！急急！！

半径r=6.50±0.20 cm
质量m=1.85±0.02 kg

相对误差δr=0.20/6.50=0.031
δm=0.02/1.85=0.011

V=4πr^3/3=4*3.14*6.5^3/3=1150 cm^3
δV=3δr=0.093

ρ=m/V=1.85/1150*10^-6=1.61*10^3 kg/m^3
δρ=δm+δV=0.104
ρ*δρ=1.61*10^3*0.104=0.17*10^3 kg/m^3

所以：该物体密度为(1.61±0.17)*10^3 kg/m^3.

Okay, I am not really good at explain this, but, normally, you need to know the volume of a sphere:
Volume = (4/3) (pi)(radius)^3
so, substitude the numbers in:
V= 4/3 X Pie X (6.50/100)cube which equals 1.1503...X10^-3
You then can calculate the density of the sphere by using the formulae: Density = mass / vol
so, 1.85kg divided 1.1503...X10^-3 by which equals 1608.21107...
Now, we can calculate the uncertainty in the density.
Absolute uncertainty of the density is given by:
Δρ = √[ (∂ρ/∂m)²∙(Δm)² + (∂ρ/∂r)² ∙ (ΔR)² ]
= √[ (m/((4/3)∙π∙R³))²