📝 题目
例 2 计算积分 $\displaystyle{I = {\iint }_{{x}^{2} + {y}^{2} \leq 1}\left| {{3x} + {4y}}\right| \mathrm{d}x\mathrm{\;d}y}$ .
💡 答案与解析
解法 1 令 $x = r\cos \theta ,y = r\sin \theta$ ,则
$$ I = {\int }_{0}^{2\pi }\mathrm{d}\theta {\int }_{0}^{1} \mid 3\cos \theta + 4\sin \theta \mid {r}^{2}\mathrm{\;d}r $$
$$ = \frac{5}{3}{\int }_{0}^{2\pi }\left| {\cos \left( {\theta - {\theta }_{0}}\right) }\right| \mathrm{d}\theta \left( {{\theta }_{0} = \arccos \frac{3}{5}}\right) $$
$$ = \frac{5}{3}{\int }_{0}^{2\pi }\left| {\cos \theta }\right| \mathrm{d}\theta = \frac{20}{3}{\int }_{0}^{\frac{\pi }{2}}\cos \theta \mathrm{d}\theta = \frac{20}{3}. $$
解法 2 令 $\xi = \frac{3}{5}x + \frac{4}{5}y,\eta = - \frac{4}{5}x + \frac{3}{5}y$ ,变换的雅可比行列式为
$$ \left| \frac{\partial \left( {\xi ,\eta }\right) }{\partial \left( {x,y}\right) }\right| = 1,\;\text{ 故 }\;\left| \frac{\partial \left( {x,y}\right) }{\partial \left( {\xi ,\eta }\right) }\right| = 1. $$
且变换把区域 ${x}^{2} + {y}^{2} \leq 1$ 变为区域 ${\xi }^{2} + {\eta }^{2} \leq 1$ . 所以
$$ I = {\iint }_{{\xi }^{2} + {\eta }^{2} \leq 1}\left| {5\xi }\right| \mathrm{d}\xi \mathrm{d}\eta = {\int }_{-1}^{1}\mathrm{\;d}\xi {\int }_{-\sqrt{1 - {\xi }^{2}}}^{\sqrt{1 - {\xi }^{2}}}\left| {5\xi }\right| \mathrm{d}\eta $$
$$ = {\int }_{-1}^{1}\left| {5\xi }\right| \cdot 2\sqrt{1 - {\xi }^{2}}\mathrm{\;d}\xi = {20}{\int }_{0}^{1}\xi \sqrt{1 - {\xi }^{2}}\mathrm{\;d}\xi $$
$$ = \frac{20}{3}\text{ . } $$
评注 当然也可作 $\xi = {3x} + {4y},\eta = - {4x} + {3y}$ 的变换. 用这种方法, 可证一般的形式积分 $\forall$
$$ {\iint }_{{x}^{2} + {y}^{2} \leq 1}f\left( {{ax} + {by}}\right) \mathrm{d}x\mathrm{\;d}y = 2{\int }_{-1}^{1}\sqrt{1 - {\xi }^{2}}f\left( {\xi \sqrt{{a}^{2} + {b}^{2}}}\right) \mathrm{d}\xi . $$