Maths Analysis II - Homework #17 20220428

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发布时间：May 5, 2022
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# Maths Analysis II

## Homework #17 20220428

### 11.4.1

$(5)$

$$
\begin{align}
\iint_Sx^2\mathrm{d}y\mathrm{d}z+y^2\mathrm{d}z\mathrm{d}x+z^2\mathrm{d}x\mathrm{d}y&=3\iint_Sx^2\mathrm{d}y\mathrm{d}z\\&=3\int_0^1\mathrm{d}y\int_0^{1-y}(1-y-z)^2\mathrm{d}z\\&=\frac{1}{4}.

\end{align}
$$

$(7)$

$$
\begin{align}
&\quad\iint_Sxz^2\mathrm{d}y\mathrm{d}z+x^2y\mathrm{d}z\mathrm{d}x+y^2z\mathrm{d}x\mathrm{d}y\\&=3\iint_Sy^2z\mathrm{d}x\mathrm{d}y\\&=12\int_0^{\frac{\pi}{2}}\mathrm{d}\phi\int_0^{\frac{\pi}{2}}(a\sin\theta\sin\phi)^2\cdot a\cos\theta\cdot a^2\sin\theta\cos\theta\mathrm{d}\theta
\\&=\frac{2}{5}\pi a^5.
\end{align}
$$

$(8)$

$$
\begin{align}
&\quad\iint_Sf(x)\mathrm{d}y\mathrm{d}z+g(y)\mathrm{d}z\mathrm{d}x+h(z)\mathrm{d}x\mathrm{d}y\\&=bc\big(f(a)-f(0)\big)+ac\big(g(b)-g(0)\big)+ab\big(h(c)-h(0)\big).

\end{align}
$$

### 11.4.2

$$
\begin{align}
&\quad\iint_S(x^2-yz)\mathrm{d}y\land\mathrm{d}z-2x^2y\mathrm{d}z\land\mathrm{d}x+z\mathrm{d}x\land\mathrm{d}y
\\&=bc\cdot a^2-ca\cdot 2a^2b+ab\cdot c\\&=abc(2a+1)(1-a).
\end{align}
$$

### 11.3.3

Let $\vec{r}=(a\cos\theta,b\sin\theta)$, then $\mathrm{d}\vec{S}=(-a\sin\theta,b\cos\theta)\mathrm{d}\theta$;

Meanwhile $\vec{F}=(-kx,-ky)=(-ka\cos\theta,-kb\sin\theta)$. So that

$$
\begin{align}
W&=\oint_L\vec{F}\mathrm{d}\vec{S}=\int_0^{\frac{\pi}
{2}}ka^2\cos\theta\sin\theta-kb^2\sin\theta\cos\theta\mathrm{d}\theta
\\&=\frac{1}{2}k(a^2-b^2).
\end{align}
$$

### 11.3.4

$(2)$

$$
\begin{align}
&\quad\oint_L(xy+x+y)\mathrm{d}x+(xy+x-y)\mathrm{d}y\\&=\iint_D\bigg(\frac{\partial}{\partial x}(xy+x-y)-\frac{\partial}{\partial y}(xy+x+y)\bigg)\mathrm{d}x\mathrm{d}y
\\&=\iint_D(y-x)\mathrm{d}x\mathrm{d}y\\&=0.
\end{align}
$$

$(4)$

$$
\begin{align}
&\quad\oint_L\sqrt{x^2+y^2}\mathrm{d}x+y\bigg(xy+\ln(x+\sqrt{x^2+y^2})\bigg)\mathrm{d}y\\&=\iint_D\bigg(\frac{\partial}{\partial x}y\big(xy+\ln(x+\sqrt{x^2+y^2})\big)-\frac{\partial}{\partial y}\sqrt{x^2+y^2}\bigg)\mathrm{d}x\mathrm{d}y
\\&=\int_{-1}^1y^2\mathrm{d}y\int_{y^2+1}^2\mathrm{d}x\\&=\frac{4}{15}.
\end{align}
$$

$(6)$

$$
\begin{align}
&\quad\int_{AMO}(e^x\sin y-my)\mathrm{d}x+(e^x\cos y-m)\mathrm{d}y
\\&=\iint_D\bigg(\frac{\partial}{\partial x}(e^x\cos y-m)-\frac{\partial}{\partial y}\sqrt{x^2+y^2}\bigg)\mathrm{d}x\mathrm{d}y-\oint_{OA}\sim\mathrm{d}l\\&=\iint_Dm\mathrm{d}x\mathrm{d}y\\&=\pi mab.
\end{align}
$$

### 11.3.5

$(1)$

$$
\begin{align}
\iint_D\mathrm{d}x\mathrm{d}y&=\frac{1}{2}\oint_Lx\mathrm{d}y-\mathrm{y}\mathrm{d}x\\&=\frac{1}{2}\int_0^{2\pi}a\cos^3t(3a\sin^2t\cos t)-a\sin^3t(-3a\sin t\cos^2t)
\\&=\frac{1}{2}\int_0^{2\pi}3a^2\sin^2t\cos^2t\mathrm{d}t
\\&=\frac{3}{4}\pi a^2.
\end{align}
$$

$(2)$

$$
\int_0^{2\pi}y\mathrm{d}x=\int_0^{2\pi}(1-\cos t)^2\mathrm{d}t=\frac{1}{2}\sin 2t-2\sin t+\frac{3}{2}t\bigg|_0^{2\pi}=3\pi.
$$

### 11.3.6

Let $x=r\cos t,y=r\sin t$. Then we have

$$
\begin{align}
\int_L\frac{-y\mathrm{d}x+x\mathrm{d}y}{x^2+y^2}&=\int_L\frac{-r\sin t(r\cos t-r\sin t)+r\cos t(r\sin t+r\cos t)}{r^2}\mathrm{d}t
\\&=\int_L\mathrm{d}t\\&=\Delta t.
\end{align}
$$

$(1)$ Since that $L:(-a,0)\rightarrow(a,0)\Leftrightarrow t:\pi\rightarrow 0$, we have that

$$
\int_L\frac{-y\mathrm{d}x+x\mathrm{d}y}{x^2+y^2}=0-\pi=-\pi.
$$

$(2)$ Since that $L:(-1,0)\rightarrow (3,0)\Leftrightarrow t:\pi\rightarrow 0$, we have that

$$
\int_L\frac{-y\mathrm{d}x+x\mathrm{d}y}{x^2+y^2}=0-\pi=-\pi.
$$

### 11.3.7

$(1)$

$$
\begin{align}
\oint_L\frac{\partial f}{\partial \vec{r}_0}\mathrm{d}s&=\oint_L\triangledown f\cdot\frac{\vec{n}}{|\vec{n}|}\mathrm{d}s=\oint_L\triangledown f\cdot\mathrm{d}\vec{n}
\\&=\oint_L-\frac{\partial f}{\partial y}\mathrm{d}x+\frac{\partial f}{\partial x}\mathrm{d}y\\&=\iint_D\bigg(\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}\bigg)\mathrm{d}x\mathrm{d}y\\&=\iint_D\Delta f\mathrm{d}x\mathrm{d}y.
\end{align}
$$

**The proof is complete.**

$(2)$

$$
\begin{align}
\oint_L\cos(\vec{a},\vec{n})\mathrm{d}s&=\oint_L\frac{\vec{a}\cdot\vec{n}}{|\vec{a}||\vec{n}|}\mathrm{d}s=\frac{1}{|\vec{a}|}\int_L\vec{a}\cdot\mathrm{d}\vec{n}
\\&=\frac{1}{|\vec{a}|}\int_L-a_y\mathrm{d}x+a_x\mathrm{d}y
\\&=-\frac{a_x}{|\vec{a}|}\int\mathrm{d}x+\frac{a_y}{|\vec{a}|}\int\mathrm{d}y\\&=0.
\end{align}
$$

**The proof is complete.**

$(3)$

$$
\begin{align}
\oint_L\bigg(v\frac{\partial u}{\partial \vec{n}}-u\frac{\partial v}{\partial\vec{n}}\bigg)\mathrm{d}s&=v\oint_L\frac{\partial u}{\partial\vec{n}}\mathrm{d}s-u\oint_L\frac{\partial v}{\partial\vec{n}}\mathrm{d}s
\\&=v\iint_D\Delta u\mathrm{d}x\mathrm{d}y-u\iint_D\Delta v\mathrm{d}x\mathrm{d}y
\\&=\iint_D(v\Delta u-u\Delta v)\mathrm{d}x\mathrm{d}y.
\end{align}
$$

**The proof is complete.**

Last modification：May 6, 2022

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Maths Analysis II - Homework #17 20220428

Bi_Gu • 2022 年 05 月 05 日

# Maths Analysis II

## Homework #17 20220428

### 11.4.1

$(5)$

\end{align}
$$

$(7)$

$(8)$

$$
\begin{align}
&\quad\iint_Sf(x)\mathrm{d}y\mathrm{d}z+g(y)\mathrm{d}z\mathrm{d}x+h(z)\mathrm{d}x\mathrm{d}y\\&=bc\big(f(a)-f(0)\big)+ac\big(g(b)-g(0)\big)+ab\big(h(c)-h(0)\big).

\end{align}
$$

### 11.4.2

### 11.3.3

Let $\vec{r}=(a\cos\theta,b\sin\theta)$, then $\mathrm{d}\vec{S}=(-a\sin\theta,b\cos\theta)\mathrm{d}\theta$;

Meanwhile $\vec{F}=(-kx,-ky)=(-ka\cos\theta,-kb\sin\theta)$. So that

$$
\begin{align}
W&=\oint_L\vec{F}\mathrm{d}\vec{S}=\int_0^{\frac{\pi}
{2}}ka^2\cos\theta\sin\theta-kb^2\sin\theta\cos\theta\mathrm{d}\theta
\\&=\frac{1}{2}k(a^2-b^2).
\end{align}
$$

### 11.3.4

$(2)$

$(4)$

$(6)$

### 11.3.5

$(1)$

$(2)$

$$
\int_0^{2\pi}y\mathrm{d}x=\int_0^{2\pi}(1-\cos t)^2\mathrm{d}t=\frac{1}{2}\sin 2t-2\sin t+\frac{3}{2}t\bigg|_0^{2\pi}=3\pi.
$$

### 11.3.6

Let $x=r\cos t,y=r\sin t$. Then we have

$$
\begin{align}
\int_L\frac{-y\mathrm{d}x+x\mathrm{d}y}{x^2+y^2}&=\int_L\frac{-r\sin t(r\cos t-r\sin t)+r\cos t(r\sin t+r\cos t)}{r^2}\mathrm{d}t
\\&=\int_L\mathrm{d}t\\&=\Delta t.
\end{align}
$$

$(1)$ Since that $L:(-a,0)\rightarrow(a,0)\Leftrightarrow t:\pi\rightarrow 0$, we have that

$$
\int_L\frac{-y\mathrm{d}x+x\mathrm{d}y}{x^2+y^2}=0-\pi=-\pi.
$$

$(2)$ Since that $L:(-1,0)\rightarrow (3,0)\Leftrightarrow t:\pi\rightarrow 0$, we have that

$$
\int_L\frac{-y\mathrm{d}x+x\mathrm{d}y}{x^2+y^2}=0-\pi=-\pi.
$$

### 11.3.7

$(1)$

**The proof is complete.**

$(2)$

**The proof is complete.**

$(3)$

**The proof is complete.**

Maths Analysis II - Homework #17 20220428

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