方企勤 第六章 多元函数积分学 第6.1题
📝 题目
6.1.21 (1) $\displaystyle{\int }_{0}^{1}\mathrm{\;d}z{\int }_{0}^{z}\mathrm{\;d}y{\int }_{z - y}^{1 - y}f\left( {x,y}\right) \mathrm{d}x + {\int }_{0}^{1}\mathrm{\;d}z{\int }_{z}^{1}\mathrm{\;d}y{\int }_{0}^{1 - y}f\left( {x,y}\right) \mathrm{d}x$ ;
(2) $\displaystyle{\int }_{0}^{1}\mathrm{\;d}z{\int }_{-z}^{z}\mathrm{\;d}y{\int }_{-\sqrt{{z}^{2} - {y}^{2}}}^{\sqrt{{z}^{2} - {y}^{2}}}f\left( {x,y}\right) \mathrm{d}x$ .
💡 答案解析
6.1.21 (1) $\displaystyle{\int }_{0}^{1}\mathrm{\;d}z{\int }_{0}^{z}\mathrm{\;d}y{\int }_{z - y}^{1 - y}f\left( {x,y}\right) \mathrm{d}x + {\int }_{0}^{1}\mathrm{\;d}z{\int }_{z}^{1}\mathrm{\;d}y{\int }_{0}^{1 - y}f\left( {x,y}\right) \mathrm{d}x$ ;
(2) $\displaystyle{\int }_{0}^{1}\mathrm{\;d}z{\int }_{-z}^{z}\mathrm{\;d}y{\int }_{-\sqrt{{z}^{2} - {y}^{2}}}^{\sqrt{{z}^{2} - {y}^{2}}}f\left( {x,y}\right) \mathrm{d}x$ .