下册 9.2 第二型曲线积分 第21题

\section*{解题过程：} （1）令 $u=x^{2}+y^{2}$ ，则 $\displaystyle \int_{L} f\left(x^{2}+y^{2}\right)(x \mathrm{~d} x+y \mathrm{~d} y)=\frac{1}{2} \int_{L} f(u) \mathrm{d} u$ ．又 $f(x)$ 在 $\mathbf{R}$ 上连续，故存在 $F(u)$使 $\mathrm{d} F(u)=f(u) \mathrm{d} u=f\left(x^{2}+y^{2}\right)(x \mathrm{~d} x+y \mathrm{~d} y)$ ，所以积分与路径无关． （2）等式 $f(t x, t y)=t^{-2} f(x, y)$ 两端对 $t$ 求导得 令 $t=1$ ，则 $$ \begin{aligned} x f_{x}^{\prime}(t x, t y)+y f_{y}^{\prime}(t x, t y) & =-2 t^{-3} f(x, y) \\ x f_{x}^{\prime}(x, y)+y f_{y}^{\prime}(x, y) & =-2 f(x, y) \end{aligned} $$ 记 $P(x, y)=y f(x, y), Q(x, y)=-x f(x, y)$ ，则 $$ \frac{\partial Q}{\partial x}=-f(x, y)-x f_{x}^{\prime}(x, y), \frac{\partial P}{\partial y}=f(x, y)+y f_{y}^{\prime}(x, y), $$ 于是 $\displaystyle \frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}$ ．故曲线积分与路径无关．$\oint_{L} y f(x, y) \mathrm{d} x-x f(x, y) \mathrm{d} y=0$ ． （3）记 $P=x^{2} f\left(x^{2}+y^{2}\right), Q=y^{2} f\left(x^{2}+y^{2}\right)$ ，则 $P_{y}=2 y x^{2} f^{\prime}\left(x^{2}+y^{2}\right), Q_{x}=2 x y^{2} f^{\prime}\left(x^{2}+y^{2}\right)$ ．由格林公式得 $$ \begin{aligned} \oint_{C} f\left(x^{2}+y^{2}\right)\left(x^{2} \mathrm{~d} x+y^{2} \mathrm{~d} y\right) & =\iint_{D_{x y}}\left(2 x y^{2} f^{\prime}\left(x^{2}+y^{2}\right)-2 x^{2} y f^{\prime}\left(x^{2}+y^{2}\right)\right) \mathrm{d} x \mathrm{~d} y \\ & =\iint_{D_{y y}} 2 x y(y-x) f^{\prime}\left(x^{2}+y^{2}\right) \mathrm{d} x \mathrm{~d} y \end{aligned} $$ 由对称性得 $$ I=\oint_{C} f\left(x^{2}+y^{2}\right)\left(x^{2} \mathrm{~d} x+y^{2} \mathrm{~d} y\right)=0 . $$