【正确答案】因Ω(t)为球体，且被积函数为x²+y²+z²的函数，故用球面坐标系计算三重积分，对分子使用球坐标变换x=ρsinφcosθ，y=ρsinφsinθ，z=ρcosφ：
f(x²+y²+z²)dV=∫₀^2πdθ∫₀^πdφ∫₀^tf(ρ²)ρ²sinφdρ
=∫₀^2πdθ∫₀^πsinφdφ∫₀^tf(ρ²)ρ²dρ
=4π∫₀^tf(ρ²)ρ²dρ．
分母作极坐标变换x=rcosθ，y=rsinθ，得

【正确答案】F(t)一=2{∫₀^tf(r²)r²dr∫₀^tf(r²)dr—[∫₀^tf(r²)rdr]²}／[∫₀¹f(r²)dr∫₀¹rf(r²)dr]．
令g(t)=∫₀^tf(r²)r²dr∫₀^tf(r²)dr一[∫₀^trf(r²)dr]²，则g(0)=0．又因f(x)恒大于零，有
g'(t)=f(t²)t²∫₀^tf(r²)dr+f(t²)∫₀^tf(r²)r²dr一2f(t²)t∫₀^tf(r²)rdr
=f(t²)[∫₀^tf(t²)t²dr+∫₀^tf(r²)r²dr—2∫₀^tf(r²)rtdr]
=f(t²)∫₀^tf(t²)(t²一2rt+r²)dr
=f(t²)∫₀^tf(t²)(t一r)²dr＞0．
故g(t)在(0，+∞)内单调增加，又g(0)=0，所以当t＞0时有g(t)＞0，又∫₀^tf(r²)dr∫₀¹rf(r²)dr＞0，故当t＞0时

即