2016年考研数学（一）考试试题 - 第10题回答

评分及理由

（1）得分及理由（满分4分）

该题考查向量场旋度的计算。旋度的计算公式为：

\[ \text{rot}\boldsymbol{A} = \nabla \times \boldsymbol{A} = \begin{vmatrix} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{vmatrix} \]

其中 \(\boldsymbol{A} = P\boldsymbol{i} + Q\boldsymbol{j} + R\boldsymbol{k}\)，本题中 \(P = x+y+z\)，\(Q = xy\)，\(R = z\)。

计算得：

\[ \text{rot}\boldsymbol{A} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right)\boldsymbol{i} + \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right)\boldsymbol{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)\boldsymbol{k} \]

代入计算：

• \(\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} = 0 - 0 = 0\)
• \(\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} = 1 - 0 = 1\)
• \(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = y - 1\)

因此标准答案为 \((0, 1, y-1)\)。

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