📝 题目
＊3．求下列向量场 $\boldsymbol{A}$ 的旋度： （1）$A=(2 z-3 y) i+(3 x-z) j+(y-2 x) k$ ； （2）$A=(z+\sin y) i-(z-x \cos y) j$ ； （3）$A=x^{2} \sin y i+y^{2} \sin (x z) j+x y \sin (\cos z) k$ ．
💡 答案与解析
[AI解答]
[AI解答]
**（1）** 给定向量场 $$ \boldsymbol{A} = (2z - 3y)\,\mathbf{i} + (3x - z)\,\mathbf{j} + (y - 2x)\,\mathbf{k} $$ 记 $$ P = 2z - 3y,\quad Q = 3x - z,\quad R = y - 2x $$ 旋度公式为 $$ \operatorname{rot}\boldsymbol{A} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \displaystyle\frac{\partial}{\partial x} & \displaystyle\frac{\partial}{\partial y} & \displaystyle\frac{\partial}{\partial z} \\ P & Q & R \end{vmatrix} $$ 计算各分量：
- $x$ 分量： $$ \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} = \frac{\partial}{\partial y}(y - 2x) - \frac{\partial}{\partial z}(3x - z) = 1 - (-1) = 2 $$
- $y$ 分量： $$ \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} = \frac{\partial}{\partial z}(2z - 3y) - \frac{\partial}{\partial x}(y - 2x) = 2 - (-2) = 4 $$ 注意公式中此项前面有负号，即 $y$ 分量为 $-\left(\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}\right)$，所以实际为 $-4$。
- $z$ 分量： $$ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \frac{\partial}{\partial x}(3x - z) - \frac{\partial}{\partial y}(2z - 3y) = 3 - (-3) = 6 $$
因此 $$ \operatorname{rot}\boldsymbol{A} = 2\mathbf{i} - 4\mathbf{j} + 6\mathbf{k} $$
---
**（2）** 给定 $$ \boldsymbol{A} = (z + \sin y)\,\mathbf{i} - (z - x\cos y)\,\mathbf{j} $$ 这里 $P = z + \sin y,\; Q = - (z - x\cos y) = -z + x\cos y,\; R = 0$。
计算旋度：
- $x$ 分量： $$ \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} = 0 - \frac{\partial}{\partial z}(-z + x\cos y) = 0 - (-1) = 1 $$
- $y$ 分量： $$ \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} = \frac{\partial}{\partial z}(z + \sin y) - 0 = 1 $$ 注意前面有负号，所以此项为 $-1$。
- $z$ 分量： $$ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \frac{\partial}{\partial x}(-z + x\cos y) - \frac{\partial}{\partial y}(z + \sin y) = \cos y - \cos y = 0 $$
因此 $$ \operatorname{rot}\boldsymbol{A} = \mathbf{i} - \mathbf{j} $$
---
**（3）** 给定 $$ \boldsymbol{A} = x^{2}\sin y\,\mathbf{i} + y^{2}\sin(xz)\,\mathbf{j} + xy\sin(\cos z)\,\mathbf{k} $$ 记 $$ P = x^{2}\sin y,\quad Q = y^{2}\sin(xz),\quad R = xy\sin(\cos z) $$
计算旋度：
- $x$ 分量： $$ \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} $$ 先求 $$ \frac{\partial R}{\partial y} = \frac{\partial}{\partial y}\big[xy\sin(\cos z)\big] = x\sin(\cos z) $$ 再求 $$ \frac{\partial Q}{\partial z} = \frac{\partial}{\partial z}\big[y^{2}\sin(xz)\big] = y^{2}\cos(xz)\cdot x = xy^{2}\cos(xz) $$ 所以 $$ x\text{分量} = x\sin(\cos z) - xy^{2}\cos(xz) $$
- $y$ 分量： $$ \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} $$ $$ \frac{\partial P}{\partial z} = \frac{\partial}{\partial z}(x^{2}\sin y) = 0 $$ $$ \frac{\partial R}{\partial x} = \frac{\partial}{\partial x}\big[xy\sin(\cos z)\big] = y\sin(\cos z) $$ 故 $$ \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} = 0 - y\sin(\cos z) = -y\sin(\cos z) $$ 前面加负号得 $y$ 分量为 $y\sin(\cos z)$。
- $z$ 分量： $$ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} $$ $$ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}\big[y^{2}\sin(xz)\big] = y^{2}\cos(xz)\cdot z = y^{2}z\cos(xz) $$ $$ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(x^{2}\sin y) = x^{2}\cos y $$ 所以 $$ z\text{分量} = y^{2}z\cos(xz) - x^{2}\cos y $$
因此 $$ \operatorname{rot}\boldsymbol{A} = \big[x\sin(\cos z) - xy^{2}\cos(xz)\big]\mathbf{i} + \big[y\sin(\cos z)\big]\mathbf{j} + \big[y^{2}z\cos(xz) - x^{2}\cos y\big]\mathbf{k} $$
---
**难度评级**：★★☆☆☆ （主要考查旋度定义与偏导计算，步骤固定，无复杂技巧，但第三小题计算稍繁）