方企勤 第六章 多元函数积分学 第5题
📝 题目
例 5 求下列线积分:
(1) $\displaystyle{\int }_{AB}\left( {{x}^{2} + {2xy} - {y}^{2}}\right) \mathrm{d}x + \left( {{x}^{2} - {2xy} - {y}^{2}}\right) \mathrm{d}y,A\left( {0,0}\right) ,B\left( {2,1}\right)$ ；
(2) $\displaystyle{\int }_{AB}\left( {{x}^{2} + y + z}\right) \mathrm{d}x + \left( {{y}^{2} + x + z}\right) \mathrm{d}y + \left( {{z}^{2} + x + y}\right) \mathrm{d}z,A(0,0$ , $0),B\left( {1,1,1}\right)$ .
💡 答案解析
解 (1) 令 $P = {x}^{2} + {2xy} - {y}^{2},Q = {x}^{2} - {2xy} - {y}^{2},\frac{\partial Q}{\partial x} = {2x} - {2y} =$ $\frac{\partial P}{\partial y}$ 在全平面成立,所以线积分在全平面上与路径无关,这时必有原函数存在. 为求被积表达式的原函数, 先求积分
$$ \int P\mathrm{\;d}x = \int \left( {{x}^{2} + {2xy} - {y}^{2}}\right) \mathrm{d}x = \frac{1}{3}{x}^{3} + {x}^{2}y - x{y}^{2}, $$
$$ \int \left\lbrack {Q - \frac{\partial }{\partial y}\int P\mathrm{\;d}x}\right\rbrack \mathrm{d}y = \int \left( {-{y}^{2}}\right) \mathrm{d}y = - \frac{1}{3}{y}^{3} + C, $$
所以原函数 $u = \frac{1}{3}{x}^{3} + {x}^{2}y - x{y}^{2} - \frac{1}{3}{y}^{3} + C$ ,因而
$$ {\int }_{AB}\left( {{x}^{2} + {2xy} - {y}^{2}}\right) \mathrm{d}x + \left( {{x}^{2} - {2xy} - {y}^{2}}\right) \mathrm{d}y $$
$$ = {\left. \left( \frac{1}{3}{x}^{3} + {x}^{2}y - x{y}^{2} - \frac{1}{3}{y}^{3} + C\right) \right| }_{\left( 0,0\right) }^{\left( 2,1\right) } $$
$$ = \frac{13}{3}\text{ . } $$
(2)记被积表达式为 $\omega$ ,则 $\omega$ 的外微分
$$ \mathrm{d}\omega = \left( {{2x}\mathrm{\;d}x + \mathrm{d}y + \mathrm{d}z}\right) \mathrm{d}x + \left( {\mathrm{d}x + {2y}\mathrm{\;d}y + \mathrm{d}z}\right) \mathrm{d}y $$
$$ + \left( {\mathrm{d}x + \mathrm{d}y + {2z}\mathrm{\;d}z}\right) \mathrm{d}z $$
$$ = \mathrm{d}y\mathrm{\;d}x + \mathrm{d}z\mathrm{\;d}x + \mathrm{d}x\mathrm{\;d}y + \mathrm{d}z\mathrm{\;d}y + \mathrm{d}x\mathrm{\;d}z + \mathrm{d}y\mathrm{\;d}z $$
$$ = 0\text{ , } $$
所以线积分在全空间上与路径无关. 为求 $\omega$ 的原函数,先求三个不定积分:
$$ \int P\mathrm{\;d}x = \int \left( {{x}^{2} + y + z}\right) \mathrm{d}x = \frac{1}{3}{x}^{3} + {xy} + {xz}; $$
$$ \int \left\lbrack {Q - \frac{\partial }{\partial y}\int P\mathrm{\;d}x}\right\rbrack \mathrm{d}y = \int \left( {{y}^{2} + z}\right) \mathrm{d}y = \frac{1}{3}{y}^{3} + {yz}; $$
$$ \int \left\lbrack {R - \frac{\partial }{\partial z}\int P\mathrm{\;d}x - \frac{\partial }{\partial z}\int \left\lbrack {Q - \frac{\partial }{\partial y}\int P\mathrm{\;d}x}\right\rbrack \mathrm{d}y}\right\rbrack \mathrm{d}z $$
$$ = \int {z}^{2}\mathrm{\;d}z = \frac{1}{3}{z}^{3} + C. $$
所以原函数为
$$ u = \frac{1}{3}\left( {{x}^{3} + {y}^{3} + {z}^{3}}\right) + {xy} + {yz} + {zx} + C. $$
因而
$$ {\int }_{AB}\left( {{x}^{2} + y + z}\right) \mathrm{d}x + \left( {{y}^{2} + z + x}\right) \mathrm{d}y + \left( {{z}^{2} + x + y}\right) \mathrm{d}z $$
$$ = {\left. \left\lbrack \frac{1}{3}\left( {x}^{3} + {y}^{3} + {z}^{3}\right) + xy + yz + zx\right\rbrack \right| }_{\left( 0,0,0\right) }^{\left( 1,1,1\right) } = 4. $$
评注 具体求 $u$ 时,先求不定积分 $\displaystyle{\int P\mathrm{\;d}x}$ . 再把 $Q$ 中含 $x$ 的项删去后,对变量 $y$ 求不定积分. 再把 $R$ 中含 $x$ 及含 $y$ 的项删去后,对变量 $z$ 求不定积分. 三部分相加即为所求的原函数.