求函数$f(x,y,z)=x^2+y^2+z^2$在区域$Omega: x^2+y^2+z^2leqslant x+y+z$内的平均值.
解: 所求平均值可以通过函数在区域内的积分除以区域体积来计算.
此区域形状不容易直接看出, 稍作变形即可得到区域$Omega$的刻画为
[
(x-frac{1}{2})^2+(y-frac{1}{2})^2+(z-frac{1}{2})^2leqslantfrac{3}{4}.
]
区域$Omega$为一个球体, 易知体积为$frac{sqrt{3}pi}{2}$.
为计算函数的积分, 作如下积分变量替换:
egin{align*}
x= & frac{1}{2}+rsinvarphicos heta, \
y= & frac{1}{2}+rsinvarphisin heta, \
z= & frac{1}{2}+rcosvarphi,
end{align*}
计算出
[
J=frac{partial(x,y,z)}{partial(r,varphi, heta)}=r^2sinvarphi.
]
计算函数积分如下:
egin{align*}
I= & iiint_Omega(x^2+y^2+z^2){
m d}V \
= & int_{0}^{2pi}{
m d} hetaint_{0}^{pi}{
m d}varphiint_{0}^{frac{sqrt{3}}{2}}
ig[(frac{1}{2}+rsinvarphicos heta)^2+(frac{1}{2}+rsinvarphisin heta)^2
+(frac{1}{2}+rcosvarphi)^2ig]|J|{
m d}r\
= & int_{0}^{2pi}{
m d} hetaint_{0}^{pi}{
m d}varphiint_{0}^{frac{sqrt{3}}{2}}
ig[frac{3}{4}+rsinvarphicos heta+rsinvarphisin heta+rcosvarphi+r^2ig]
r^2sinvarphi{
m d}r\
= & frac{3sqrt{3}pi}{5}.
end{align*}
因此所求平均值为
[
frac{frac{3sqrt{3}pi}{5}}{frac{sqrt{3}pi}{2}}=frac{6}{5}.
]