下册 9.2 第二型曲线积分 第46题

\section*{解题过程：} （1）记 $L$ 为 $x^{2}+y^{2}=r^{2}$ ，则 $\displaystyle \oint_{x^{2}+y^{2}=r^{2}} \frac{\partial f}{\partial n} \mathrm{~d} s=\oint_{L} \frac{\partial f}{\partial n} \mathrm{~d} s$ ． 因 $\displaystyle \frac{\partial f}{\partial n}=\frac{\partial f}{\partial x} \cos (n, x)+\frac{\partial f}{\partial y} \cos (n, y), \cos (n, x) \mathrm{d} s=\mathrm{d} y, \cos (n, y) \mathrm{d} s=-\mathrm{d} x$ ，所以 $$ \oint_{L} \frac{\partial f}{\partial n} \mathrm{~d} s=\oint_{L}\left(\frac{\partial f}{\partial x} \cos (n, x)+\frac{\partial f}{\partial y} \cos (n, y)\right) \mathrm{d} s=\oint_{L}-\frac{\partial f}{\partial y} \mathrm{~d} x+\frac{\partial f}{\partial x} \mathrm{~d} y . $$ 令 $x=r \cos \theta, y=r \sin \theta$ ，则 $$ \begin{aligned} & \frac{\partial f}{\partial r}=\frac{\partial f}{\partial x} \frac{\partial x}{\partial r}+\frac{\partial f}{\partial y} \frac{\partial y}{\partial r}=\frac{\partial f}{\partial x} \cos \theta+\frac{\partial f}{\partial y} \sin \theta \\ & \int_{0}^{2 \pi} \frac{\partial f}{\partial r} r \mathrm{~d} \theta=\int_{0}^{2 \pi} r\left(\frac{\partial f}{\partial x} \cos \theta+\frac{\partial f}{\partial y} \sin \theta\right) \mathrm{d} \theta=\int_{0}^{2 \pi}\left(\frac{\partial f}{\partial x} r \cos \theta \mathrm{~d} \theta+\frac{\partial f}{\partial y} r \sin \theta \mathrm{~d} \theta\right) \end{aligned} $$ 所以 $\displaystyle \oint_{L} \frac{\partial f}{\partial n} \mathrm{~d} s=\oint_{L}-\frac{\partial f}{\partial y} \mathrm{~d} x+\frac{\partial f}{\partial x} \mathrm{~d} y=\int_{0}^{2 \pi}\left(\frac{\partial f}{\partial y} r \cos \theta+\frac{\partial f}{\partial x} r \sin \theta\right) \mathrm{d} \theta=\int_{0}^{2 \pi} r \frac{\partial f}{\partial r} \mathrm{~d} \theta$ ． 由格林公式 $$ \begin{aligned} \oint_{L} \frac{\partial f}{\partial n} \mathrm{~d} s & =\oint_{L}-\frac{\partial f}{\partial y} \mathrm{~d} x+\frac{\partial f}{\partial x} \mathrm{~d} y=\iint_{x^{2}+y^{2}