Maths Analysis II - Homework #19 20220505

Author： Bi_Gu
发布时间：May 16, 2022
535 views
No comments
3398 words
Categories：课堂笔记数分作业数学分析

<button class="btn m-b-xs btn-light btn-roundedbtn-addon" onclick='window.open("https://cdn.dabigu.xyz/97423306_p0.jpg","_blank")'><i class="fa fa-camera"></i>$\quad$ **最忆是江南 by BODHI** </button>

# Maths Analysis II

## Homework #19 20220505

### 11.5.8

$$
\begin{align}
\iint_S\bigg(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}\bigg)\mathrm{d}x\mathrm{d}y&=\iiint_V\triangledown\cdot\vec{v}\mathrm{d}x\mathrm{d}y\mathrm{d}z=\iint_{\partial V}\vec{v}\cdot\mathrm{d}
S\\&=\oint_{\partial\vec{v}}\vec{v}\cdot\vec{n}\mathrm{d}l\\&=\oint_{\partial\vec{v}}\big(P(-\sin\theta)+Q\cos\theta\big)\mathrm{d}\theta\\&=\oint Q\mathrm{d}x-P\mathrm{d}y.
\end{align}
$$

**The proof is complete.**

### 11.5.9

$(1)$

$$
\oint_Ly\mathrm{d}x+z\mathrm{d}y+x\mathrm{d}z=3\oint_{\vec{AB}}y\mathrm{d}x+z\mathrm{d}y+x\mathrm{d}z=3\int_0^1(1-x)\mathrm{d}x=-\frac{3}{2}.
$$

$(2)$

$$
\begin{align}
&\quad\oint_L(y-z)\mathrm{d}x+(z-x)\mathrm{d}y+(x-y)\mathrm{d}z\\&=\iint_D\begin{vmatrix}\mathrm{d}y\mathrm{d}z&\mathrm{d}z\mathrm{d}x&\mathrm{d}x\mathrm{d}y\\\displaystyle\frac{\partial}{\partial x}&\displaystyle\frac{\partial}{\partial y}&\displaystyle\frac{\partial}{\partial z}\\y-z&z-x&x-y\end{vmatrix}\\&=-2\iint_D\mathrm{d}y\mathrm{d}z+\mathrm{d}z\mathrm{d}x+\mathrm{d}x\mathrm{d}y\\&=2\pi(a+h)a.
\end{align}
$$

$(3)$

$$
\begin{align}
&\quad\oint_L(y^2-z^2)\mathrm{d}x+(z^2-x^2)\mathrm{d}y+(x^2-y^2)\mathrm{d}z\\&=\iint_D\begin{vmatrix}\mathrm{d}y\mathrm{d}z&\mathrm{d}z\mathrm{d}x&\mathrm{d}x\mathrm{d}y\\\displaystyle\frac{\partial}{\partial x}&\displaystyle\frac{\partial}{\partial y}&\displaystyle\frac{\partial}{\partial z}\\y^2-z^2&z^2-x^2&x^2-y^2\end{vmatrix}\\&=-2\iint_D(y+z)\mathrm{d}y\mathrm{d}z+(z+x)\mathrm{d}z\mathrm{d}x+(x+y)\mathrm{d}x\mathrm{d}y\\&=\frac{7}{4}a^3.
\end{align}
$$

In which make $\begin{cases}x+y=u\\y-x=v\end{cases}$ and we have

$$
\iint(x+y)\mathrm{d}x\mathrm{d}y=\iint u\frac{\partial(x,y)}{\partial(u,v)}\mathrm{d}u\mathrm{d}v=\frac{1}{2}\int_{\frac{a}{2}}^{\frac{3a}{2}}\mathrm{d}u\int_{-u}^{u}u\mathrm{d}v=\frac{7}{24}a^3.
$$

$(4)$

$$
\begin{align}
&\quad\oint_Ly^2\mathrm{d}x+xy\mathrm{d}y+xz\mathrm{d}z\\&=\iint_D\begin{vmatrix}\mathrm{d}y\mathrm{d}z&\mathrm{d}z\mathrm{d}x&\mathrm{d}x\mathrm{d}y\\\displaystyle\frac{\partial}{\partial x}&\displaystyle\frac{\partial}{\partial y}&\displaystyle\frac{\partial}{\partial z}\\y^2&xy&xz\end{vmatrix}\\&=\iint_D-z\mathrm{d}z\mathrm{d}x-y\mathrm{d}x\mathrm{d}y\\&=2\int_0^{2\pi}\mathrm{d}\theta\int_0^1(1-\sin\theta)r\mathrm{d}r\\&=2\pi.
\end{align}
$$

$(5)$

$$
\begin{align}
&\quad\oint_L(y^2-y)\mathrm{d}x+(z^2-z)\mathrm{d}y+(x^2-x)\mathrm{d}z
\\&=\oint-y\mathrm{d}x-z\mathrm{d}y-x\mathrm{d}z\\&=\iint_D\mathrm{d}y\mathrm{d}z+\mathrm{d}z\mathrm{d}x+\mathrm{d}x\mathrm{d}y\\&=\sqrt{3}\pi a^2.
\end{align}
$$

$(6)$

$$
\begin{align}
&\quad\oint_L(y^2-z^2)\mathrm{d}x+(2z^2-x^2)\mathrm{d}y+(3x^2-y^2)\mathrm{d}z
\\&=\iint_D(-2y-4z)\mathrm{d}y\mathrm{d}z+(-6x-2z)\mathrm{d}z\mathrm{d}x+(-2x-2y)\mathrm{d}x\mathrm{d}y
\\&=-2\iint_D(y+2z)\mathrm{d}y\mathrm{d}z+(3x+z)\mathrm{d}z\mathrm{d}x
\\&=2\int_1^3\bigg(\int_{\frac{1-z}{2}}^{\frac{3-z}{2}}(3t+3z)\mathrm{d}t\bigg)\mathrm{d}z\\&=24.
\end{align}
$$

### 11.5.10

We have that

$$
\oint_Lx^2y^3\mathrm{d}x=\mathrm{d}y+z\mathrm{d}z=\iint_D-3x^2y^2\mathrm{d}x\mathrm{d}y,
$$

which is an invariant about $z$;

$\textbf{(1)}$ $x^2+y^2<R^2,z=0$;

$\textbf{(2)}$ $x^2+y^2<R^2,z=\sqrt{R^2-x^2-y^2}$ ; The result is the **same**.

### 11.5.11

$$
\oint\vec{c}\cdot\mathrm{d}\vec{l}=\iint\vec{v}\times\vec{c}\cdot\mathrm{d}\vec{S}=0.
$$

**The proof is complete.**

Last modification：May 16, 2022

Support if you like my work! qwq

版权声明：若文中无特殊说明，则本文为原创文章，版权归 Bi_Gu 所有。

本文链接：https://dabigu.xyz/index.php/archives/358/

所有原创文章采用知识共享署名-非商业性使用 4.0 国际许可协议进行许可。
您可以自由的转载和修改，但请务必注明文章来源并且不可用于商业目的。