下册 9.2 第二型曲线积分 第23题

\section*{解题过程：} （1）由题设条件有 $\displaystyle \frac{\partial Q}{\partial x}=2 x$ 。于是 $Q(x, y)=x^{2}+\varphi(y)$ ，其中 $\varphi(y)$ 待定。由条件 $$ \begin{aligned} & \int_{(0,0)}^{(t, 1)} 2 x y \mathrm{~d} x+Q(x, y) \mathrm{d} y=\int_{(0,0)}^{(1, t)} 2 x y \mathrm{~d} x+Q(x, y) \mathrm{d} y \\ & \left.\left(x^{2} y+\int_{0}^{y} \varphi(s) \mathrm{d} s\right)\right|_{(0,0)} ^{(t, 1)}=\left.\left(x^{2} y+\int_{0}^{y} \varphi(s) \mathrm{d} s\right)\right|_{(0,0)} ^{(1, t)} . \end{aligned} $$ 得 $$ \int_{0}^{1}\left(t^{2}+\varphi(y)\right) \mathrm{d} y=\int_{0}^{1}(1+\varphi(y)) \mathrm{d} y . $$ 化简得 两边对 $t$ 求导得 $2 t=1+\varphi(t)$ ，即 $\varphi(y)=2 y-1$ 。所以 $Q(x, y)=x^{2}+2 y-1$ ． （2）由题设条件有 $Q_{x}=\left(3 x^{2} y\right)_{y}=3 x^{2}$ ．因此 $Q(x, y)=x^{3}+\varphi(y)$ ． 另一方面，对任何实数 $z$ 成立等式 $\int_{(0,0)}^{(z, 1)} 3 x^{2} y \mathrm{~d} x+Q \mathrm{~d} y=\int_{(0,0)}^{(1, z)} 3 x^{2} y \mathrm{~d} x+Q \mathrm{~d} y$ 。故 $$ \int_{(0,0)}^{(0,1)} 3 x^{2} y \mathrm{~d} x+Q \mathrm{~d} y+\int_{(0,1)}^{(2,1)} 3 x^{2} y \mathrm{~d} x+Q \mathrm{~d} y=\int_{(0,0)}^{(1,0)} 3 x^{2} y \mathrm{~d} x+Q \mathrm{~d} y+\int_{(1,0)}^{(1,2)} 3 x^{2} y \mathrm{~d} x+Q \mathrm{~d} y . $$ 即 $$ \int_{0}^{1} Q(0, y) \mathrm{d} y+\int_{0}^{2} 3 x^{2} \cdot \mathrm{ld} x=\int_{0}^{1} 3 x^{2} \cdot 0 \mathrm{~d} x+Q(x, 0) \mathrm{d} 0+\int_{0}^{2}\left(3 \cdot 1^{2} \cdot y \mathrm{~d} 1+Q(1, y) \mathrm{d} y\right) . $$ 所以 $$ \int_{0}^{1} Q(0, y) \mathrm{d} y+\left.x^{3}\right|_{0} ^{2}=\int_{0}^{z} Q(1, y) \mathrm{d} y . $$ 把 $Q(x, y)=x^{3}+\varphi(y)$ 代入前式得 $$ \int_{0}^{1} \varphi(y) \mathrm{d} y+z^{3}=\int_{0}^{2}(1+\varphi(y)) \mathrm{d} y . $$ 等式两边对 $z$ 求导得 $3 z^{2}=1+\varphi(z)$ 。所以 $\varphi(y)=3 y^{2}-1$ 。从而 $\varphi(x, y)=x^{3}+\varphi(y)=x^{3}+3 y^{2}-1$ ．