📝 题目
例 14 试计算
$$ I = {\iiint }_{D}\left( {x + y - z}\right) \left( {-x + y + z}\right) \left( {x - y + z}\right) \mathrm{d}\left( {x,y,z}\right) , $$
这里的 $D$ 是闭区域
$$ 0 \leq x + y - z \leq 1, $$
$$ 0 \leq - x + y + z \leq 1, $$
$$ 0 \leq x - y + z \leq 1\text{ . } $$
💡 答案与解析
解 我们做变换
$$ \left\{ \begin{array}{l} u = x + y - z, \\ v = - x + y + z, \\ w = x - y + z. \end{array}\right. $$
计算雅可比行列式得
$$ \frac{\partial \left( {u,v,w}\right) }{\partial \left( {x,y,z}\right) } = \left| \begin{array}{rrr} 1 & 1 & - 1 \\ - 1 & 1 & 1 \\ 1 & - 1 & 1 \end{array}\right| = 4, $$
$$ \frac{\partial \left( {x,y,z}\right) }{\partial \left( {u,v,w}\right) } = \frac{1}{4}. $$
通过变元替换计算积分得
$$ I = \frac{1}{4}{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{uvw}\mathrm{\;d}u\mathrm{\;d}v\mathrm{\;d}w = \frac{1}{32}. $$