下册 9.2 第二型曲线积分 第16题

\section*{解题过程：} （1）记 $P(x, y)=2 x \cos y-y^{2} \sin x, Q(x, y)=2 y \cos x-x^{2} \sin y$ ，则 $P_{y}=-2 x \sin y-2 y \sin x=Q_{x}$ ．故该积分与路线无关．因此 $$ \begin{aligned} & \int_{(0,0)}^{(x, y)}\left(2 x \cos y-y^{2} \sin x\right) \mathrm{d} x+\left(2 y \cos x-x^{2} \sin y\right) \mathrm{d} y \\ & =\int_{0}^{x} 2 x \mathrm{~d} x+\int_{0}^{y}\left(2 y \cos x-x^{2} \sin y\right) \mathrm{d} y=y^{2} \cos x+x^{2} \cos y \end{aligned} $$ （2）记 $\displaystyle P(x, y)=\frac{2 x\left(1-\mathrm{e}^{y}\right)}{\left(1+x^{2}\right)^{2}}, Q(x, y)=\frac{\mathrm{e}^{y}}{1+x^{2}}$ ，则 $\displaystyle P_{y}=\frac{-2 x \mathrm{e}^{y}}{\left(1+x^{2}\right)^{2}}=Q_{x}$ ．故该积分与路线无关． 由于 $\displaystyle \frac{2 x\left(1-\mathrm{e}^{y}\right)}{\left(1+x^{2}\right)^{2}} \mathrm{~d} x+\frac{\mathrm{e}^{y}}{1+x^{2}} \mathrm{~d} y=\mathrm{d}\left(\frac{\mathrm{e}^{y}}{1+x^{2}}\right),(x, y) \in \mathbf{R}^{2}$ ，所以 $\displaystyle \frac{\mathrm{e}^{y}}{1+x^{2}}+C$ 为所求的原函数． （3）记 $P=\mathrm{e}^{x}\left[\mathrm{e}^{y}(x-y+2)+y\right], Q=\mathrm{e}^{x}\left[\mathrm{e}^{y}(x-y)+1\right]$ ，由于 $\displaystyle \frac{\partial Q}{\partial x}=\mathrm{e}^{x}\left[\mathrm{e}^{y}(x-y+1)+1\right]=\frac{\partial P}{\partial y}$ 。从而原函数存在，原函数为 $$ u(x, y)=\int_{(0,0)}^{(x, y)} P \mathrm{~d} x+Q \mathrm{~d} y+C=\int_{0}^{x} \mathrm{e}^{x}(x+2) \mathrm{d} x+\int_{0}^{y} \mathrm{e}^{x}\left[\mathrm{e}^{y}(x-y)+1\right] \mathrm{d} y+C=(x-y+1) \mathrm{e}^{x+y}+y \mathrm{e}^{x}+C $$ （4）由于 $\displaystyle \frac{\partial}{\partial x}\left(x^{2}-2 x y-y^{2}\right)=2(x-y)=\frac{\partial}{\partial y}\left(x^{2}+2 x y-y^{2}\right)$ ，从而积分与路线无关。故其原函数为 $$ \begin{aligned} u(x, y) & =\int_{\left(x_{0}, y_{0}\right)}^{(x, y)}\left(x^{2}+2 x y-y^{2}\right) \mathrm{d} x+\left(x^{2}-2 x y-y^{2}\right) \mathrm{d} y+C \\ & =\int_{x_{0}}^{x}\left(x^{2}+2 y_{0} x-y_{0}^{2}\right) \mathrm{d} x+\int_{y_{0}}^{y}\left(x^{2}-2 x y-y^{2}\right) \mathrm{d} y+C \\ & =\left.\left(\frac{1}{3} x^{3}+y_{0} x^{2}-y_{0}^{2} x\right)\right|_{x_{0}} ^{x}+\left.\left(x^{2} y-x y^{2}-\frac{1}{3} y^{3}\right)\right|_{y_{0}} ^{y}+C \\ & =\frac{1}{3} x^{3}+x^{2} y-x y^{2}-\frac{1}{3} y^{3}+C \end{aligned} $$ （5）记 $P(x, y)=3 x^{2} y+8 x y^{2}, Q(x, y)=x^{3}+8 x^{2} y+12 y \mathrm{e}^{y}$ ，则 $$ P_{y}=3 x^{2}+16 x y, Q_{x}=3 x^{2}+16 x y, P_{y}=Q_{x} $$ 所以 $P \mathrm{~d} x+Q \mathrm{~d} y$ 是某个函数的全微分． 由于 $\mathrm{d}\left(x^{3} y+4 x^{2} y^{2}+12 y \mathrm{e}^{y}-12 \mathrm{e}^{y}\right)=\left(3 x^{2} y+8 x y^{2}\right) \mathrm{d} x+\left(x^{3}+8 x^{2} y+12 y \mathrm{e}^{y}\right) \mathrm{d} y$ ，所以所求原函数为 $$ u(x, y)=x^{3} y+4 x^{2} y^{2}+12(y-1) \mathrm{e}^{y}+C . $$ （6）记 $P=\mathrm{e}^{x} \sin y-2 y, Q=\mathrm{e}^{x} \cos y-2 x$ ，则 $P_{y}=\mathrm{e}^{x} \cos y-2=Q_{x}$ ，故 $P \mathrm{~d} x+Q \mathrm{~d} y$ 是某个函数的全微分． 由于 $\mathrm{d}\left(\mathrm{e}^{x} \sin y-2 x y\right)=\left(\mathrm{e}^{x} \sin y-2 y\right) \mathrm{d} x+\left(\mathrm{e}^{x} \cos y-2 x\right) \mathrm{d} y$ ，所以所求原函数为 $$ u(x, y)=\mathrm{e}^{x} \sin y-2 x y+C $$ （7）记 $\displaystyle P(x, y)=\frac{-y}{x^{2}+y^{2}}, Q(x, y)=\frac{x}{x^{2}+y^{2}}$ ．因为 $P, Q$ 在右半平面内具有一阶连续偏导数，且 $\displaystyle Q_{x}=\frac{y^{2}-x^{2}}{\left(x^{2}+y^{2}\right)^{2}}=P_{y}$ ，所以在右半平面内，$\displaystyle \frac{x \mathrm{~d} y-y \mathrm{~d} x}{x^{2}+y^{2}}$ 是某个函数的全微分． 取积分路线为在右半平面 $x>0$ 内从点 $A\left(x_{0}, y_{0}\right)$ 到点 $B\left(x, y_{0}\right)$ 再到点 $C(x, y)$ 的折线，则所求函数为 $$ u(x, y)=\int_{\left(x_{0}, y_{0}\right)}^{(x, y)} \frac{x \mathrm{~d} y-y \mathrm{~d} x}{x^{2}+y^{2}}=\int_{x_{0}}^{x} \frac{y_{0} \mathrm{~d} x}{x^{2}+y_{0}^{2}}+\int_{y_{0}}^{y} \frac{x \mathrm{~d} y}{x^{2}+y^{2}}=\arctan \frac{y}{x}+C $$ （8）因为 $$ \begin{aligned} \mathrm{d} u(x, y) & =\frac{x \mathrm{~d} y-y \mathrm{~d} x}{3 x^{2}-2 x y+y^{2}}=\frac{1}{3-2 \frac{y}{x}+\frac{y^{2}}{x^{2}}} \mathrm{~d}\left(\frac{y}{x}\right)=\frac{1}{\left(\frac{y}{x}-1\right)^{2}+(\sqrt{2})^{2}} \mathrm{~d}\left(\frac{y}{x}-1\right) \\ & =\frac{1}{\sqrt{2}} \mathrm{~d}\left\{\arctan \left[\frac{1}{\sqrt{2}}\left(\frac{y}{x}-1\right)\right]\right\} \end{aligned} $$ 故 $$ u(x, y)=\frac{1}{\sqrt{2}} \arctan \left[\frac{1}{\sqrt{2}}\left(\frac{y}{x}-1\right)\right]+C $$