【ID】1284 【题型】解答题 【类型】考研真题 【来源】2002年全国硕士研究生入学统一考试数学

$$I=\int_{L} \frac{1}{y}\left[1+y^{2} f(x y)\right] \mathrm{d} x+\frac{x}{y^{2}}\left[y^{2} f(x y)-1\right] \mathrm{d} y .$$
(1) 证明曲线积分 $I$ 与路径 $L$ 无关;
(2) 当 $a b=c d$ 时, 求 $I$ 的值.

(1) 记 $P(x, y)=\frac{1}{y}\left[1+y^{2} f(x y)\right], Q(x, y)=\frac{x}{y^{2}}\left[y^{2} f(x y)-1\right]$
\begin{aligned} \frac{\partial Q}{\partial x} &=\frac{\partial\left(\frac{x}{y^{2}}\left[y^{2} f(x y)-1\right]\right)}{\partial x}=\frac{\partial\left(\frac{x}{y^{2}}\right)}{\partial x} \times\left(\left[y^{2} f(x y)-1\right]\right)+\frac{x}{y^{2}} \times \frac{\partial\left(\left[y^{2} f(x y)-1\right]\right)}{\partial x} \\ &=\frac{1}{y^{2}} \times\left(\left[y^{2} f(x y)-1\right]\right)+\frac{x}{y^{2}} \times \frac{y^{2} \partial(f(x y)}{\partial x}=f(x y)-\frac{1}{y^{2}}+x \times f^{\prime}(x y) \frac{\partial(x y)}{\partial x} \\ &=f(x y)+x y f^{\prime}(x y)-\frac{1}{y^{2}} \frac{\partial P}{\partial y}=\frac{\partial\left(\frac{1}{y}\left[1+y^{2} f(x y)\right]\right)}{\partial y} \end{aligned}
\begin{aligned} &=\frac{\partial\left(\frac{1}{y}\right)}{\partial y}\left(\left[1+y^{2} f(x y)\right]\right)+\frac{1}{y} \frac{\partial\left(\left[1+y^{2} f(x y)\right]\right)}{\partial y} \\ &=-\frac{1}{y^{2}}\left(\left[1+y^{2} f(x y)\right]\right)+\frac{1}{y} \frac{\partial\left(y^{2}\right)}{\partial y} f(x y)+\frac{1}{y} \times \frac{\partial(f(x y))}{\partial y} \times y^{2} \\ &=-f(x y)-\frac{1}{y^{2}}+f(x y)+x y f^{\prime}(x y) \end{aligned}

(2)方法 1: 由该曲线积分与路径无关而只与端点有关所以用折线把两个端点连接起来. 先从 点 $(a, b)$ 到点 $(c, b)$, 再到点 $(c, d)$. 有
\begin{aligned} I &=\int_{a}^{c} \frac{1}{b}\left[1+b^{2} f(b x)\right] d x+\int_{b}^{d} \frac{c}{y^{2}}\left[y^{2} f(c y)-1\right] d y \\ &\left.=\frac{c-a}{b}+\int_{a}^{c} b f(b x)\right] d x+\int_{b}^{d} c f(c y) d y+\frac{c}{d}-\frac{c}{b} \end{aligned}

\begin{aligned} I &=\int_{L} \frac{1}{y^{2}}\left[1+y^{2} f(x y)\right] d x+\frac{x}{y^{2}}\left[y^{2} f(x y)-1\right] d y \\ &=\int_{L} \frac{y d x-x d y}{y^{2}}+\int_{L} f(x y)(y d x+x d y)=\int_{L} d\left(\frac{x}{y}\right)+\int_{L} f(x y) d(x y) \end{aligned}

\begin{aligned} &\int_{L} d\left(\frac{x}{y}\right)=\frac{x}{y} \mid \begin{array}{l} (c, d) \\ (a, b) \end{array}=\frac{c}{d}-\frac{a}{b} ; \\ &\int_{L} f(x y) d(x y)=F(x y) \mid \begin{array}{l} (c, d) \\ (a, b) \end{array}=F(c d)-F(a b)=0, \end{aligned}

\begin{aligned}
I &=\int_{a}^{d}\left[\frac{1}{y}\left(1+y^{2} f(k)\right)\left(-\frac{k}{y^{2}}\right)+\frac{k}{y^{2}}\left(y^{2} f(k)-1\right)\right] d y \\
&=\int_{b}^{d}\left(-\frac{2 k}{y^{3}}\right) d y=\frac{k}{y^{2}} \mid \frac{d}{b}=\frac{k}{d^{2}}-\frac{k}{b^{2}}=\frac{c d}{d^{2}}-\frac{a b}{b^{2}}=\frac{c}{d}-\frac{a}{b} .
\end{aligned}