计算∫∫∑(1/2)ds，其中∑为球面x2+y2+z2=R2被平面z=a和
z=b(0﹤a﹤b﹤R)所截的部分．
∑在Oxy平面上的投影为 Dxy={(x，y)∣R2-b2≤x2+y2≤R2-a2} ={(r，θ)∣ 0≤θ≤2π,√R2-b2≤r≤√R2-b2} 又z=√(R2-x2-y2),∂z/∂x=-x/√(R2-x2-y2) ∂z/∂y=-y/√(R2-x2-y2). 所以∫∫∑(1/z)ds=∫∫Dxy1/√(R2-x2-y2) •√[1+(z/x)2+(z/y)2]dσ =∫∫DxyR/√(R2-x2-y2)dσ =R∫02πdθ∫√R2-b2√R2-a2 1/(R2-r2)•rdr =2πR•[-(1/2)]∫√R2-b2√R2-a2 1/(R2-r2)d(R2-r2) =-πR•ln(R2-r2)∣√R2-b2√R2-a2=πRln(b/a)
