第13章 重积分 · 第2题

解 由于对称性, 只需求出第一卦限内的部分体积再乘以 8 :

$$ V = 8{\iiint }_{E}\mathrm{\;d}\left( {x,y,z}\right) , $$

这里

$$ E = \left\{ {\left. \left( {x,y,z}\right) \right| \;\begin{matrix} x,y,z \geq 0, \\ {x}^{2} + {y}^{2} \leq {a}^{2},{x}^{2} + {z}^{2} \leq {a}^{2} \end{matrix}}\right\} $$

$$ = \left\{ {\left( {x,y,z}\right) \mid \left( {x,y}\right) \in D,0 \leq z \leq \sqrt{{a}^{2} - {x}^{2}}}\right\} , $$

$$ D = \left\{ {\left( {x,y}\right) \mid x,y \geq 0,{x}^{2} + {y}^{2} \leq {a}^{2}}\right\} . $$

于是

$$ V = 8{\iint }_{D}\mathrm{\;d}\left( {x,y}\right) {\int }_{0}^{\sqrt{{a}^{2} - {x}^{2}}}\mathrm{\;d}z $$

$$ = 8{\iint }_{D}\sqrt{{a}^{2} - {x}^{2}}\mathrm{\;d}\left( {x,y}\right) $$

$$ = 8{\int }_{0}^{a}\mathrm{\;d}x{\int }_{0}^{\sqrt{{a}^{2} - {x}^{2}}}\sqrt{{a}^{2} - {x}^{2}}\mathrm{\;d}y $$

$$ = 8{\int }_{0}^{a}\left( {{a}^{2} - {x}^{2}}\right) \mathrm{d}x = \frac{16}{3}{a}^{3}. $$

另一种计算方案为

$$ V = 8{\iiint }_{E}\mathrm{\;d}\left( {x,y,z}\right) = 8{\int }_{0}^{a}\mathrm{\;d}x{\iint }_{{E}_{x}}\mathrm{\;d}\left( {y,z}\right) , $$

这里 ${E}_{x}$ 是一个正方形:

$$ {E}_{x} = \left\{ {\left( {y,z}\right) \left| {\;\begin{array}{l} 0 \leq y \leq \sqrt{{a}^{2} - {x}^{2}} \\ 0 \leq z \leq \sqrt{{a}^{2} - {x}^{2}} \end{array}}\right. }\right\} . $$

用这方案计算同样得到

$$ V = 8{\int }_{0}^{a}\left( {{a}^{2} - {x}^{2}}\right) \mathrm{d}x = \frac{16}{3}{a}^{3}. $$