题目
设一为圆周一,方向为逆时针，则由格林公式得一对错
设
为圆周
,方向为逆时针，则由格林公式得
- 对
- 错
题目解答
答案
根据格林公式，令
,
由上知，
;
即 
由格林公式知，
故本命题正确，本题答案为A.
解析
步骤 1：定义P和Q
令$P=-\dfrac {y}{{x}^{2}+{y}^{2}}$,$Q=\dfrac {x}{{x}^{2}+{y}^{2}}$，这是为了应用格林公式。
步骤 2：计算$\dfrac {\partial Q}{\partial x}$和$\dfrac {\partial P}{\partial y}$
$\dfrac {\partial Q}{\partial x}=\dfrac {({x}^{2}+{y}^{2})-2{x}^{2}}{{({x}^{2}+{y}^{2})}^{2}}=\dfrac {-{x}^{2}+{y}^{2}}{{({x}^{2}+{y}^{2})}^{2}}$
$\dfrac {\partial P}{\partial y}=\dfrac {-({x}^{2}+{y}^{2})+2{y}^{2}}{{({x}^{2}+{y}^{2})}^{2}}=\dfrac {-{x}^{2}+{y}^{2}}{{({x}^{2}+{y}^{2})}^{2}}$
步骤 3：验证$\dfrac {\partial Q}{\partial x}-\dfrac {\partial P}{\partial y}=0$
由上知，$\dfrac {\partial Q}{\partial x}-\dfrac {\partial P}{\partial y}=0$，这表明在圆周${x}^{2}+{y}^{2}={R}^{2}$上，$P$和$Q$满足格林公式的条件。
步骤 4：应用格林公式
由格林公式知，$\int \dfrac {xdy}{{x}^{2}+{y}^{2}}=\iint (\dfrac {\partial Q}{\partial x}-\dfrac {\partial P}{\partial y})dx\partial \overrightarrow {\partial y}=\iint 0dxdy=0$，这表明积分结果为0。
令$P=-\dfrac {y}{{x}^{2}+{y}^{2}}$,$Q=\dfrac {x}{{x}^{2}+{y}^{2}}$，这是为了应用格林公式。
步骤 2：计算$\dfrac {\partial Q}{\partial x}$和$\dfrac {\partial P}{\partial y}$
$\dfrac {\partial Q}{\partial x}=\dfrac {({x}^{2}+{y}^{2})-2{x}^{2}}{{({x}^{2}+{y}^{2})}^{2}}=\dfrac {-{x}^{2}+{y}^{2}}{{({x}^{2}+{y}^{2})}^{2}}$
$\dfrac {\partial P}{\partial y}=\dfrac {-({x}^{2}+{y}^{2})+2{y}^{2}}{{({x}^{2}+{y}^{2})}^{2}}=\dfrac {-{x}^{2}+{y}^{2}}{{({x}^{2}+{y}^{2})}^{2}}$
步骤 3：验证$\dfrac {\partial Q}{\partial x}-\dfrac {\partial P}{\partial y}=0$
由上知，$\dfrac {\partial Q}{\partial x}-\dfrac {\partial P}{\partial y}=0$，这表明在圆周${x}^{2}+{y}^{2}={R}^{2}$上，$P$和$Q$满足格林公式的条件。
步骤 4：应用格林公式
由格林公式知，$\int \dfrac {xdy}{{x}^{2}+{y}^{2}}=\iint (\dfrac {\partial Q}{\partial x}-\dfrac {\partial P}{\partial y})dx\partial \overrightarrow {\partial y}=\iint 0dxdy=0$，这表明积分结果为0。