九州大学 システム情報科学府 情報理工学専攻・電気電子工学専攻 2024年9月実施 ベクトル解析

标签：

• Kyushu-University
• Surface-Normal
• Surface-Integral
• Divergence-Theorem

Author

祭音Myyura (co-authored with ChatGPT 5.5 Thinking)

Description

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

$$\mathbf{F}=2x\mathbf{i}+3y\mathbf{j}+6z\mathbf{k}.$$

Calculate the surface integral

$$\int_S \sqrt{4x^2+9y^2+36z^2}\,dS$$

over the closed surface

$$S:\ 8x^2+12y^2+24z^2=4$$

by answering the following questions.

(1) Find the outward-pointing normal vector $\mathbf{n}$ at each point $(x,y,z)$ on the closed surface $S$.

(2) Show

$$\mathbf{F}\cdot \mathbf{n} = \sqrt{4x^2+9y^2+36z^2}$$

on the closed surface $S$.

(3) Find

$$\int_S \sqrt{4x^2+9y^2+36z^2}\,dS.$$

题目描述

在直角坐标系中，$\mathbf i,\mathbf j,\mathbf k$ 分别为 $x,y,z$ 轴方向的单位向量。给定向量场

$$\mathbf F=2x\mathbf i+3y\mathbf j+6z\mathbf k$$

以及闭合椭球面

$$S:\ 8x^2+12y^2+24z^2=4.$$

通过下列步骤计算

$$\int_S\sqrt{4x^2+9y^2+36z^2}\,dS:$$

1. 求 $S$ 上每一点 $(x,y,z)$ 处的外向单位法向量 $\mathbf n$。
2. 证明在 $S$ 上
   $$\mathbf F\cdot\mathbf n =\sqrt{4x^2+9y^2+36z^2}.$$
3. 求上述曲面积分。

Kai

(1)

The closed surface $S$ is given by

$$8x^2+12y^2+24z^2=4.$$

Define

$$\phi(x,y,z)=8x^2+12y^2+24z^2.$$

Then the outward normal direction is given by the gradient

$$\nabla \phi = (16x,24y,48z).$$

Since

$$(16x,24y,48z) = 8(2x,3y,6z),$$

the outward-pointing unit normal vector is

$$\mathbf{n} = \frac{(2x,3y,6z)} {\sqrt{(2x)^2+(3y)^2+(6z)^2}}.$$

Therefore,

$$\boxed{ \mathbf{n} = \frac{(2x,3y,6z)} {\sqrt{4x^2+9y^2+36z^2}} }$$

on $S$.

(2)

We have

$$\mathbf{F}=(2x,3y,6z).$$

From part (1),

$$\mathbf{n} = \frac{(2x,3y,6z)} {\sqrt{4x^2+9y^2+36z^2}}.$$

Hence,

$$\begin{aligned} \mathbf{F}\cdot\mathbf{n} &= (2x,3y,6z)\cdot \frac{(2x,3y,6z)} {\sqrt{4x^2+9y^2+36z^2}} \\ &= \frac{4x^2+9y^2+36z^2} {\sqrt{4x^2+9y^2+36z^2}} \\ &= \sqrt{4x^2+9y^2+36z^2}. \end{aligned}$$

Therefore,

$$\boxed{ \mathbf{F}\cdot\mathbf{n} = \sqrt{4x^2+9y^2+36z^2} }.$$

(3)

By part (2),

$$\int_S \sqrt{4x^2+9y^2+36z^2}\,dS = \int_S \mathbf{F}\cdot\mathbf{n}\,dS.$$

By the divergence theorem,

$$\int_S \mathbf{F}\cdot\mathbf{n}\,dS = \iiint_V \operatorname{div}\mathbf{F}\,dV,$$

where $V$ is the solid ellipsoid enclosed by $S$.

Since

$$\mathbf{F}=(2x,3y,6z),$$

we have

$$\operatorname{div}\mathbf{F} = \frac{\partial}{\partial x}(2x) + \frac{\partial}{\partial y}(3y) + \frac{\partial}{\partial z}(6z) = 2+3+6=11.$$

Therefore,

$$\int_S \mathbf{F}\cdot\mathbf{n}\,dS = 11\operatorname{Vol}(V).$$

Now rewrite the ellipsoid:

$$8x^2+12y^2+24z^2=4.$$

Dividing by $4$, we get

$$2x^2+3y^2+6z^2=1.$$

Thus,

$$\frac{x^2}{1/2} + \frac{y^2}{1/3} + \frac{z^2}{1/6} = 1.$$

So the semi-axes are

$$a=\frac{1}{\sqrt{2}}, \qquad b=\frac{1}{\sqrt{3}}, \qquad c=\frac{1}{\sqrt{6}}.$$

Hence,

$$\operatorname{Vol}(V) = \frac{4}{3}\pi abc = \frac{4}{3}\pi \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{6}}.$$

Since

$$\sqrt{2}\sqrt{3}\sqrt{6} = \sqrt{36} = 6,$$

we get

$$\operatorname{Vol}(V) = \frac{4}{3}\pi\cdot \frac{1}{6} = \frac{2\pi}{9}.$$

Therefore,

$$\int_S \sqrt{4x^2+9y^2+36z^2}\,dS = 11\cdot \frac{2\pi}{9}.$$

Hence,

$$\boxed{ \int_S \sqrt{4x^2+9y^2+36z^2}\,dS = \frac{22\pi}{9} }.$$