R04-数学-向量分析

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 $x^2+z^2=1(0\le y\le 4)$ 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 $x^2+z^2=1(0\le y\le 4,0\le z)$ と長方形面 $z=0(-1\le x\le 1,0\le y\le 4)$ 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 $x^2+z^2=1(0\le y\le 4)$, 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 $x^2+z^2=1(0\le y\le 4,0\le z)$ and the rectangular surface $z=0(-1\le x\le 1,0\le y\le 4)$, 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.

---

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 $0\le y\le 4$, with the outward-pointing normal vector.

• Parametrization: $\mathbf{r}(\theta ,y)=\cos \theta \mathbf{i}+y\mathbf{j}+\sin \theta \mathbf{k}$, where $0\le \theta \le 2\pi$ and $0\le y\le 4$.
• 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 $0\le y\le 4$ and $z\ge 0$, and the rectangular surface at $z=0$ for $-1\le x\le 1$ and $0\le y\le 4$.

• For the half-cylinder part (similar to $S_1$, but with $0\le \theta \le \pi$): $${\int }_{\text{cylindrical part}}\mathbf{F}\cdot d\mathbf{S}=22\pi .$$
• For the rectangular surface at $z=0$, the normal vector is $-\mathbf{k}$, and: $$\mathbf{F}\cdot (-\mathbf{k})=-10z=0\text{(since z=0)},$$ 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 $x^2+z^2\le 1$, $0\le y\le 4-x$, giving: $$\text{Volume}={\int }_0^{2\pi }{\int }_0^1{\int }_0^{4-r\cos \theta }rdydrd\theta =4\pi .$$
• 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 $0\le y\le 4$

• Parameterization:

  • Use cylindrical coordinates with $r=1$, so $x=\cos \theta$, $z=\sin \theta$, and $0\le y\le 4$.
  • 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 $0\le y\le 4$ and $z\ge 0$, plus the rectangle $z=0$

• Cylindrical Part ($S_{2a}$):

  • Similar to $S_1$, but for $0\le \theta \le \pi$ (since $z\ge 0$).
  • 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 $z=0$, $-1\le x\le 1$, $0\le y\le 4$.
  • 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 $x+y\le 4$.

• 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 $0\le y\le 4-r\cos \theta$.
  • Integrate: $${\int }_{V}13dV=13{\int }_0^{2\pi }{\int }_0^1{\int }_0^{4-r\cos \theta }rdydrd\theta .$$ $$=13{\int }_0^{2\pi }{\int }_0^{1}r(4-r\cos \theta )drd\theta =13\left(4\pi -\frac{1}{3}\right).$$
  • 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 $S_3$: $13\left(4\pi -\frac{1}{3}\right)$.

---

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：

$${\int }_S\mathbf{F}\cdot d\mathbf{S}={\int }_{u,v}\mathbf{F}(\mathbf{r}(u,v))\cdot \mathbf{n}\Vert \frac{\partial \mathbf{r}}{\partial u}\times \frac{\partial \mathbf{r}}{\partial v}\Vert dudv.$$

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： $$\mathbf{r}(\theta ,y)=(\cos \theta ,y,\sin \theta ),0\le \theta \le 2\pi ,0\le y\le 4.$$
• 法向量：$\frac{\partial \mathbf{r}}{\partial \theta }$ 和 $\frac{\partial \mathbf{r}}{\partial y}$ 的叉积为： $$\mathbf{n}=(\cos \theta ,0,\sin \theta ),dS=d\theta dy.$$
• 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 $(0\le z)$ 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。