题目
14、填空 已知点O(0,0)及点A(1,1)，且曲线积分int_(overline{OA)}(axcos y-y^2sin x)dx+(bycos x-x^2sin y)dy与路径无关，则常数a=____.
14、填空 已知点O(0,0)及点A(1,1)，且曲线积分$\int_{\overline{OA}}(ax\cos y-y^{2}\sin x)dx+(by\cos x-x^{2}\sin y)dy$与路径无关，则常数a=____.
题目解答
答案
设向量场 $\mathbf{F} = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j}$，其中
\[ P(x, y) = ax\cos y - y^2\sin x, \quad Q(x, y) = by\cos x - x^2\sin y. \]
由曲线积分与路径无关的条件，需满足
\[ \frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}. \]
计算得
\[ \frac{\partial Q}{\partial x} = -by\sin x - 2x\sin y, \]
\[ \frac{\partial P}{\partial y} = -ax\sin y - 2y\sin x. \]
令两式相等，整理得
\[ y\sin x(2 - b) = x\sin y(2 - a). \]
为使等式恒成立，应有
\[ 2 - b = 0, \quad 2 - a = 0, \]
解得
\[ b = 2, \quad a = 2. \]
**答案：** $\boxed{2}$
解析
步骤 1：定义向量场
设向量场 $\mathbf{F} = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j}$，其中 \[ P(x, y) = ax\cos y - y^2\sin x, \quad Q(x, y) = by\cos x - x^2\sin y. \]
步骤 2：应用路径无关条件
由曲线积分与路径无关的条件，需满足 \[ \frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}. \]
步骤 3：计算偏导数
计算得 \[ \frac{\partial Q}{\partial x} = -by\sin x - 2x\sin y, \] \[ \frac{\partial P}{\partial y} = -ax\sin y - 2y\sin x. \]
步骤 4：令偏导数相等
令两式相等，整理得 \[ y\sin x(2 - b) = x\sin y(2 - a). \]
步骤 5：求解常数
为使等式恒成立，应有 \[ 2 - b = 0, \quad 2 - a = 0, \] 解得 \[ b = 2, \quad a = 2. \]
设向量场 $\mathbf{F} = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j}$，其中 \[ P(x, y) = ax\cos y - y^2\sin x, \quad Q(x, y) = by\cos x - x^2\sin y. \]
步骤 2：应用路径无关条件
由曲线积分与路径无关的条件，需满足 \[ \frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}. \]
步骤 3：计算偏导数
计算得 \[ \frac{\partial Q}{\partial x} = -by\sin x - 2x\sin y, \] \[ \frac{\partial P}{\partial y} = -ax\sin y - 2y\sin x. \]
步骤 4：令偏导数相等
令两式相等，整理得 \[ y\sin x(2 - b) = x\sin y(2 - a). \]
步骤 5：求解常数
为使等式恒成立，应有 \[ 2 - b = 0, \quad 2 - a = 0, \] 解得 \[ b = 2, \quad a = 2. \]