方企勤 第六章 多元函数积分学 第6.5题
📝 题目
6.5.3 (1) 求
$$ f\left( {x,y,z}\right) = \frac{1}{\sqrt{{\left( x - a\right) }^{2} + {\left( y - b\right) }^{2} + {\left( z - c\right) }^{2}}}\;\left( {{a}^{2} + {b}^{2} + {c}^{2} \neq {R}^{2}}\right) $$
沿球面 ${x}^{2} + {y}^{2} + {z}^{2} = {R}^{2}$ 上各点外法线方向的方向导数;
(2)求球面的双层位势
$$ u\left( {a,b,c}\right) = {\iint }_{{x}^{2} + {y}^{2} + {z}^{2} = {R}^{2}}\frac{\left( {x - a}\right) \mathrm{d}y\mathrm{\;d}z + \left( {y - b}\right) \mathrm{d}z\mathrm{\;d}x + \left( {z - c}\right) \mathrm{d}x\mathrm{\;d}y}{{\left\lbrack {\left( x - a\right) }^{2} + {\left( y - b\right) }^{2} + {\left( z - c\right) }^{2}\right\rbrack }^{3/2}} $$
$$ \left( {{a}^{2} + {b}^{2} + {c}^{2} \neq {R}^{2}}\right) \text{ . } $$
💡 答案解析
6. 5.1 (1) ${4\pi }{R}^{3};\;\left( 2\right) \frac{2}{3}\pi {h}^{3};\;\left( 3\right) 1$ (利用变换求重积分).