📝 题目
例 6 计算重积分
$$ I = {\iint }_{D}x\mathrm{\;d}x\mathrm{\;d}y $$
其中 $D$ 是以 $\left( {{x}_{1},{y}_{1}}\right) ,\left( {{x}_{2},{y}_{2}}\right) ,\left( {{x}_{3},{y}_{3}}\right)$ 为顶点,面积为 $A$ 的三角形.
💡 答案与解析
解 三角形为凸集, 它的点总可表示为
$$ \left\{ {\begin{array}{l} x = \mathop{\sum }\limits_{{i = 1}}^{3}{t}_{i}{x}_{i}, \\ y = \mathop{\sum }\limits_{{i = 1}}^{3}{t}_{i}{y}_{i}, \end{array}\;\left( {{t}_{1} \geq 0,{t}_{2} \geq 0,{t}_{3} \geq 0,{t}_{1} + {t}_{2} + {t}_{3} = 1}\right) }\right. $$
故启发我们作变换:
$$ \left\{ \begin{array}{l} x = {x}_{3} + \xi \left( {{x}_{1} - {x}_{3}}\right) + \eta \left( {{x}_{2} - {x}_{3}}\right) , \\ y = {y}_{3} + \xi \left( {{y}_{1} - {y}_{3}}\right) + \eta \left( {{y}_{2} - {y}_{3}}\right) \end{array}\right. $$
$$ \left( {D \rightarrow \xi \geq 0,\eta \geq 0,\xi + \eta \leq 1}\right) , $$
所以
$$ I = {\iint }_{\begin{matrix} {\xi \geq 0,\eta \geq 0} \\ {\xi + \eta \leq 1} \end{matrix}}\left\lbrack {{x}_{3} + \xi \left( {{x}_{1} - {x}_{3}}\right) + \eta \left( {{x}_{2} - {x}_{3}}\right) }\right\rbrack \left| \frac{\partial \left( {x,y}\right) }{\partial \left( {\xi ,\eta }\right) }\right| \mathrm{d}\xi \mathrm{d}\eta $$
(利用