ベクトル解析

直交座標系において, $x$, $y$, $z$ 軸方向の単位ベクトルをそれぞれ $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ とする．ベクトル場 $\mathbf{F}$ を

$$\mathbf{F} = x\mathbf{i} + 2y\mathbf{j} + 10z\mathbf{k}$$

とする．次の面 $S_1, S_2$ および $S_3$ に対する面積分を計算せよ.

1. $S_1$ を円筒面 $x^2 + z^2 = 1 \ (0 \leq y \leq 4)$ とする (上面と底面の無い円筒の表面) ．円筒外向き法線ベクトルを用いよ.
2. $S_2$ を円筒面の一部 $x^2 + z^2 = 1 \ (0 \leq y \leq 4, 0 \leq z)$ と長方形面 $z = 0 \ (-1 \leq x \leq 1, 0 \leq y \leq 4)$ からなる半円筒面とする (上面と底面の無い半円筒の表面) ．半円筒外向き法線ベクトルを用いよ.
3. $S_3$ を円筒面 $x^2 + z^2 = 1$ と, 平面 $z = 0, y = 0, x + y = 4$ で囲まれた領域の境界とする．外向き法線ベクトルを用いよ.

Vector Analysis

The unit vectors on $x$, $y$, and $z$ axes of Cartesian coordinates are denoted by $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$, respectively. Let the vector field $\mathbf{F}$ be:

$$\mathbf{F} = x\mathbf{i} + 2y\mathbf{j} + 10z\mathbf{k}.$$

Find the integral of $\mathbf{F}$ over the following areas $S_1$, $S_2$, and $S_3$:

1. $S_1$ is the part of the cylindrical surface $x^2 + z^2 = 1 \ (0 \leq y \leq 4)$, i.e., the surface of the cylinder without top and bottom disks. Use the normal vector pointing outside the cylinder.
2. $S_2$ is the surface consisting of the part of the cylindrical surface $x^2 + z^2 = 1 \ (0 \leq y \leq 4, 0 \leq z)$ and the rectangular surface $z = 0 \ (-1 \leq x \leq 1, 0 \leq y \leq 4)$, i.e., the surface of the half-cylinder without top and bottom planes. Use the normal vector pointing outside the half-cylinder.
3. $S_3$ is the boundary of the region enclosed by the cylindrical surface $x^2 + z^2 = 1$, the planes $z = 0$, $y = 0$, and $x + y = 4$. Use the outward pointing normal vector.

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DeepSeekV3

We are given the vector field:

$$\mathbf{F} = x \mathbf{i} + 2y \mathbf{j} + 10z \mathbf{k}$$

and need to compute the flux integrals over the surfaces $S_1$, $S_2$, and $S_3$.

1. Flux over $S_1$

$S_1$ is the lateral surface of the cylinder $x^2 + z^2 = 1$ for $0 \leq y \leq 4$, with the outward-pointing normal vector.

• Parametrization: $\mathbf{r}(\theta, y) = \cos\theta \mathbf{i} + y \mathbf{j} + \sin\theta \mathbf{k}$, where $0 \leq \theta \leq 2\pi$ and $0 \leq y \leq 4$.
• Outward normal vector: $\mathbf{n} = \cos\theta \mathbf{i} + \sin\theta \mathbf{k}$.
• Dot product: $\mathbf{F} \cdot \mathbf{n} = \cos^2\theta + 10 \sin^2\theta$.
• Integral over $S_1$:

$$\int_{S_1} \mathbf{F} \cdot d\mathbf{S} = \int_0^4 \int_0^{2\pi} (\cos^2\theta + 10 \sin^2\theta) \, d\theta \, dy = 44\pi.$$

2. Flux over $S_2$

$S_2$ consists of the lateral surface of the half-cylinder $x^2 + z^2 = 1$ for $0 \leq y \leq 4$ and $z \geq 0$, and the rectangular surface at $z = 0$ for $-1 \leq x \leq 1$ and $0 \leq y \leq 4$.

• For the half-cylinder part (similar to $S_1$, but with $0 \leq \theta \leq \pi$):

  $$\int_{\text{cylindrical part}} \mathbf{F} \cdot d\mathbf{S} = 22\pi.$$

• For the rectangular surface at $z = 0$, the normal vector is $-\mathbf{k}$, and:

  $$\mathbf{F} \cdot (-\mathbf{k}) = -10z = 0 \quad \text{(since $z = 0$)},$$

  so the flux through this part is zero.
• Total flux over $S_2$:

  $$\int_{S_2} \mathbf{F} \cdot d\mathbf{S} = 22\pi.$$

3. Flux over $S_3$

$S_3$ is the boundary of the region enclosed by the cylindrical surface $x^2 + z^2 = 1$, the planes $z = 0$, $y = 0$, and $x + y = 4$, with the outward-pointing normal vector.

• Using the divergence theorem:

  $$\text{div} \, \mathbf{F} = 1 + 2 + 10 = 13.$$

• Volume integral:

  $$\int_{V} \text{div} \, \mathbf{F} \, dV = 13 \times \text{Volume}.$$

• The volume is bounded by $x^2 + z^2 \leq 1$, $0 \leq y \leq 4 - x$, giving:

  $$\text{Volume} = \int_0^{2\pi} \int_0^1 \int_0^{4 - r\cos\theta} r \, dy \, dr \, d\theta = 4\pi.$$

• Therefore, the flux through $S_3$ is:

  $$\int_{S_3} \mathbf{F} \cdot d\mathbf{S} = 13 \times 4\pi = 52\pi.$$

Final Answers:

1. Flux over $S_1$: $44\pi$.
2. Flux over $S_2$: $22\pi$.
3. Flux over $S_3$: $52\pi$.

To solve the problem of integrating the vector field $\mathbf{F} = x\mathbf{i} + 2y\mathbf{j} + 10z\mathbf{k}$ over the specified surfaces, we proceed as follows:

1. Surface $S_1$: Cylindrical Surface $x^2 + z^2 = 1$ for $0 \leq y \leq 4$

• Parameterization:

  • Use cylindrical coordinates with $r = 1$, so $x = \cos\theta$, $z = \sin\theta$, and $0 \leq y \leq 4$.
  • Position vector: $\mathbf{r}(\theta, y) = \cos\theta \mathbf{i} + y \mathbf{j} + \sin\theta \mathbf{k}$.

• Differential Area Vector:

  • Compute partial derivatives: $\frac{\partial \mathbf{r}}{\partial \theta} = -\sin\theta \mathbf{i} + \cos\theta \mathbf{k}$ and $\frac{\partial \mathbf{r}}{\partial y} = \mathbf{j}$.
  • Cross product: $\frac{\partial \mathbf{r}}{\partial y} \times \frac{\partial \mathbf{r}}{\partial \theta} = \cos\theta \mathbf{i} + \sin\theta \mathbf{k}$.
  • This gives the outward-pointing normal vector.

• Dot Product:

  • $\mathbf{F} \cdot d\mathbf{S} = (\cos\theta \mathbf{i} + 2y \mathbf{j} + 10\sin\theta \mathbf{k}) \cdot (\cos\theta \mathbf{i} + \sin\theta \mathbf{k}) d\theta dy = (\cos^2\theta + 10\sin^2\theta) d\theta dy$.

• Integral:

  $$\int_{S_1} \mathbf{F} \cdot d\mathbf{S} = \int_{0}^{4} \int_{0}^{2\pi} (\cos^2\theta + 10\sin^2\theta) d\theta dy.$$

  • Integrate over $\theta$:

    $$\int_{0}^{2\pi} (\cos^2\theta + 10\sin^2\theta) d\theta = \int_{0}^{2\pi} (9\sin^2\theta + 1) d\theta = 9\pi + 2\pi = 11\pi.$$

  • Integrate over $y$:

    $$\int_{0}^{4} 11\pi dy = 44\pi.$$

  • Result for $S_1$: $44\pi$.

2. Surface $S_2$: Half-Cylinder with $x^2 + z^2 = 1$ for $0 \leq y \leq 4$ and $z \geq 0$, plus the rectangle $z = 0$

• Cylindrical Part ($S_{2a}$):

  • Similar to $S_1$, but for $0 \leq \theta \leq \pi$ (since $z \geq 0$).
  • Integral over $S_{2a}$:

    $$\int_{S_{2a}} \mathbf{F} \cdot d\mathbf{S} = \int_{0}^{4} \int_{0}^{\pi} (\cos^2\theta + 10\sin^2\theta) d\theta dy = 22\pi.$$

• Rectangular Part ($S_{2b}$):

  • The rectangle is at $z = 0$, $-1 \leq x \leq 1$, $0 \leq y \leq 4$.
  • Normal vector points downward: $-\mathbf{k}$.
  • $\mathbf{F} \cdot (-\mathbf{k}) = -10z = 0$ (since $z = 0$).
  • Integral over $S_{2b}$:

    $$\int_{S_{2b}} \mathbf{F} \cdot d\mathbf{S} = 0.$$

• Result for $S_2$: $22\pi + 0 = 22\pi$.

3. Surface $S_3$: Boundary of the Region Enclosed by $x^2 + z^2 = 1$, $z = 0$, $y = 0$, and $x + y = 4$

• Divergence Theorem:

  • Since $S_3$ is a closed surface, use the Divergence Theorem:

    $$\int_{S_3} \mathbf{F} \cdot d\mathbf{S} = \int_V (\nabla \cdot \mathbf{F}) dV.$$

  • Compute divergence:

    $$\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x) + \frac{\partial}{\partial y}(2y) + \frac{\partial}{\partial z}(10z) = 1 + 2 + 10 = 13.$$

  • Volume $V$ is a cylinder with radius 1, height from $y = 0$ to $y = 4$, cut by the plane $x + y = 4$.
  • The volume can be found by integrating over the cylindrical coordinates, considering the bounds imposed by $x + y \leq 4$.

• Volume Integral:

  • Convert to cylindrical coordinates: $x = r\cos\theta$, $z = r\sin\theta$.
  • The plane $x + y = 4$ gives $y = 4 - r\cos\theta$.
  • The limits for $y$ are $0 \leq y \leq 4 - r\cos\theta$.
  • Integrate:

    $$\int_V 13 \, dV = 13 \int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{4 - r\cos\theta} r \, dy \, dr \, d\theta.$$

    $$= 13 \int_{0}^{2\pi} \int_{0}^{1} r(4 - r\cos\theta) \, dr \, d\theta = 13 \left( 4\pi - \frac{1}{3} \right).$$

  • Result for $S_3$: $13 \left( 4\pi - \frac{1}{3} \right)$.

Final Answers:

1. For $S_1$: $44\pi$.
2. For $S_2$: $22\pi$.
3. For $S_3$: $13 \left( 4\pi - \frac{1}{3} \right)$.

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GPT-4o

要计算面积分 $\int \mathbf{F} \cdot d\mathbf{S}$，需要明确如何表达面积微元法向量 $d\mathbf{S}$ 并将其与向量场 $\mathbf{F}$ 进行点乘。以下是分步说明：

1. 面积微元法向量 $d\mathbf{S}$ 的定义

对于给定的曲面 $S$，面积微元法向量是：

$$d\mathbf{S} = \mathbf{n} \, dS,$$

其中：

• $\mathbf{n}$ 是曲面 $S$ 在每一点的单位法向量。
• $dS$ 是该点处的小面积元。

因此，$\int \mathbf{F} \cdot d\mathbf{S}$ 可以表示为：

$$\int_S \mathbf{F} \cdot d\mathbf{S} = \int_S (\mathbf{F} \cdot \mathbf{n}) \, dS.$$

我们需要将 $dS$ 和 $\mathbf{n}$ 表达为参数形式。

2. 分步骤计算的流程

(a) 选择适当的参数化

每个曲面 $S$ 可以用参数化的方法表示其点 $(x, y, z)$ 和面积微元 $dS$。参数化的选择依赖于曲面的定义。

(b) 计算法向量

通过参数化 $S$，我们可以用两个参数 $(u, v)$ 表示曲面上任意一点。参数化函数为 $\mathbf{r}(u, v)$，则法向量 $\mathbf{n}$ 可由叉积计算：

$$\mathbf{n} \propto \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}.$$

对该向量进行归一化后得到单位法向量。

(c) 求点积并积分

将参数化后的 $\mathbf{F}$ 和 $\mathbf{n}$ 表达为参数形式，计算点积并在参数范围内求积分：

$$\int_S \mathbf{F} \cdot d\mathbf{S} = \int_{u, v} \mathbf{F}(\mathbf{r}(u, v)) \cdot \mathbf{n} \, \lVert \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \rVert \, du \, dv.$$

3. 应用到具体问题

问题 1：柱面部分 $S_1$

• 参数化：柱面 $x^2 + z^2 = 1$ 可用参数 $(\theta, y)$ 表示：

  $$\mathbf{r}(\theta, y) = (\cos\theta, y, \sin\theta), \quad 0 \leq \theta \leq 2\pi, \, 0 \leq y \leq 4.$$

• 法向量：$\frac{\partial \mathbf{r}}{\partial \theta}$ 和 $\frac{\partial \mathbf{r}}{\partial y}$ 的叉积为：

  $$\mathbf{n} = (\cos\theta, 0, \sin\theta), \quad dS = d\theta \, dy.$$

• 积分表达式：

  $$\int_{S_1} \mathbf{F} \cdot d\mathbf{S} = \int_0^{2\pi} \int_0^4 (x\cos\theta + 2y\cdot 0 + 10z\sin\theta) \, d\theta \, dy.$$

问题 2：半柱面部分 $S_2$

同理，将半柱面 $(0 \leq z)$ 与矩形面 $z = 0$ 分别参数化，重复上述过程。

问题 3：封闭区域 $S_3$

利用柱面 $x^2 + z^2 = 1$ 和各个平面参数化，求出相应的法向量并积分。如果封闭区域完整，可以用高斯散度定理将问题转化为体积分。

4. 核心计算

将上述结果代入积分表达式，即可逐步计算得到结果。重要的是将 $\mathbf{F}$ 和 $\mathbf{n}$ 准确写成参数形式，然后依次积分。