同济高数 第11章 第11-7-*3题

[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
$$
\operatorname{rot}\boldsymbol{A} = 2\mathbf{i} - 4\mathbf{j} + 6\mathbf{k}
$$

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（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
$$
\operatorname{rot}\boldsymbol{A} = \big[x\sin(\cos z) - xy^{2}\cos(xz)\big]\mathbf{i} + \big[y\sin(\cos z)\big]\mathbf{j} + \big[y^{2}z\cos(xz) - x^{2}\cos y\big]\mathbf{k}
$$

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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）