Maths Analysis II - Homework #12 20220324

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# Maths Analysis II

## Homework #12 20220324

### 9.6.2

$$
\triangledown\times\boldsymbol{v}=\triangledown\times(\boldsymbol{w}\times\boldsymbol{r})=\begin{vmatrix}\boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\w_2z-w_3y&w_3x-w_1z&w_1y-w_2x\end{vmatrix}.
$$

**Explanation by Physics**: The changing speed of the direction of $\boldsymbol{v}$ is in direct proportion with $w$.

### 9.6.3

$(1)$

$$
\triangledown\cdot\boldsymbol{v}=6x+3y^2+z^2+xy-6xz=6+16-14=8.
$$

$(2)$

$$
\triangledown\cdot\boldsymbol{v}=2x\sin y+2y\sin xz-xy\sin z\cos(\cos z).
$$

### 9.6.5

$(1)$

$$
\triangledown\times\boldsymbol{v}=\begin{vmatrix}\boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\y^2&z^2&x^2\end{vmatrix}=(-2z,-2x,-2y).
$$

$(2)$

### 9.6.7

Since that $\displaystyle\triangledown f(r)=\frac{\partial f}{\partial r}\triangledown r=f'(r)\frac{\boldsymbol{r}}{r}$, we have that

$(1)$

$$
\triangledown(\boldsymbol{w}\cdot f(r)\boldsymbol{r})=(\boldsymbol{r}\cdot\boldsymbol{w})\triangledown f(r)+f(r)\triangledown(\boldsymbol{r}\cdot\boldsymbol{w})=(\boldsymbol{r}\cdot\boldsymbol{w})\frac{f'(r)}{r}\boldsymbol{r}+f(r)\boldsymbol{w}.
$$

$(2)$

$$
\triangledown\cdot(\boldsymbol{w}\times f(r)\boldsymbol{r})=\triangledown f(r)\cdot(\boldsymbol{w}\times\boldsymbol{r})+f(r)\triangledown\cdot(\boldsymbol{w}\times\boldsymbol{r})=\frac{f'(r)}{r}\boldsymbol{r}\cdot(\boldsymbol{w}\times\boldsymbol{r})=0.
$$

$(3)$

$$
\begin{align}
\triangledown\times(\boldsymbol{w}\times f(r)\boldsymbol{r})&=\triangledown f(r)\times(\boldsymbol{w}\times\boldsymbol{r})+f(r)\triangledown\times(\boldsymbol{w}\times\boldsymbol{r})=\frac{f'(r)}{r}\boldsymbol{r}\times(\boldsymbol{w}\times\boldsymbol{r})+2f(r)\boldsymbol{w}
\\&=\frac{f'(r)}{r}(r^2\boldsymbol{w}-(\boldsymbol{r}\cdot\boldsymbol{w})\boldsymbol{r})+2f(r)\boldsymbol{w}
\\&=(rf'(r)+2f(r))\boldsymbol{w}-\frac{f'(r)(\boldsymbol{r}\cdot\boldsymbol{w})}{r}\boldsymbol{r}.
\end{align}
$$

### 9.6.8

$(1)$

$$
\begin{align}
\triangledown(\phi+\psi)&=(\frac{\partial}{\partial x}(\phi+\psi),\frac{\partial}{\partial y}(\phi+\psi),\frac{\partial}{\partial z}(\phi+\psi))\\
&=(\frac{\partial}{\partial x}\phi,\frac{\partial}{\partial y}\phi,\frac{\partial}{\partial z}\phi)+(\frac{\partial}{\partial x}\psi,\frac{\partial}{\partial y}\psi,\frac{\partial}{\partial z}\psi)
\\&=\triangledown\phi+\triangledown\psi.
\end{align}
$$

$(2)$

$$
\begin{align}
\triangledown\cdot(\boldsymbol{a}+\boldsymbol{b})&=\begin{pmatrix}\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\end{pmatrix}\begin{pmatrix}\boldsymbol{a}_x+\boldsymbol{b}_x\\\boldsymbol{a}_y+\boldsymbol{b}_y\\\boldsymbol{a}_z+\boldsymbol{b}_z\end{pmatrix}\\
&=\begin{pmatrix}\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\end{pmatrix}\begin{pmatrix}\boldsymbol{a}_x\\\boldsymbol{a}_y\\\boldsymbol{a}_z\end{pmatrix}+\begin{pmatrix}\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\end{pmatrix}\begin{pmatrix}\boldsymbol{b}_x\\\boldsymbol{b}_y\\\boldsymbol{b}_z\end{pmatrix}
\\&=\triangledown\cdot\boldsymbol{a}+\triangledown\cdot\boldsymbol{b}.
\end{align}
$$

$(3)$

$$
\begin{align}
\triangledown\times(\boldsymbol{a}+\boldsymbol{b})&=\begin{vmatrix}\boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\\boldsymbol{a}_x+\boldsymbol{b}_x&\boldsymbol{a}_y+\boldsymbol{b}_y&\boldsymbol{a}_z+\boldsymbol{b}_z\end{vmatrix}\\&=\begin{vmatrix}\boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\\boldsymbol{a}_x&\boldsymbol{a}_y&\boldsymbol{a}_z\end{vmatrix}+\begin{vmatrix}\boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\\boldsymbol{b}_x&\boldsymbol{b}_y&\boldsymbol{b}_z\end{vmatrix}
\\&=\triangledown\times\boldsymbol{a}+\triangledown\times\boldsymbol{b}.
\end{align}
$$

$(4)$

$$
\begin{align}
\triangledown(\phi\psi)&=(\frac{\partial\phi\psi}{\partial x},\frac{\partial\phi\psi}{\partial y},\frac{\partial\phi\psi}{\partial z})=(\psi\frac{\partial\phi}{\partial x}+\phi\frac{\partial\psi}{\partial x},\psi\frac{\partial\phi}{\partial y}+\phi\frac{\partial\psi}{\partial y},\psi\frac{\partial\phi}{\partial z}+\phi\frac{\partial\psi}{\partial z})
\\&=\phi\triangledown\psi+\psi\triangledown\phi.
\end{align}
$$

$(5)$

$$
\begin{align}
\triangledown\cdot\phi\boldsymbol{a}&=(\frac{\partial\phi\boldsymbol{a}_x}{\partial x},\frac{\partial\phi\boldsymbol{a}_y}{\partial y},\frac{\partial\phi\boldsymbol{a}_z}{\partial z})=(\boldsymbol{a}_x\frac{\partial\phi}{\partial x}+\phi\frac{\partial\boldsymbol{a}_x}{\partial x},\boldsymbol{a}_y\frac{\partial\phi}{\partial x}+\phi\frac{\partial\boldsymbol{a}_y}{\partial y}，\boldsymbol{a}_z\frac{\partial\phi}{\partial x}+\phi\frac{\partial\boldsymbol{a}_z}{\partial z})
\\&=\boldsymbol{a}\cdot\triangledown\phi+\phi\triangledown\cdot\boldsymbol{a}.
\end{align}
$$

$(6)$

$$
\begin{align}
\triangledown\cdot(\boldsymbol{a}\times\boldsymbol{b})&=\begin{vmatrix}\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\\boldsymbol{a}_x&\boldsymbol{a}_y&\boldsymbol{a}_z\\\boldsymbol{b}_x&\boldsymbol{b}_y&\boldsymbol{b}_z\end{vmatrix}\\&=\frac{\partial}{\partial x}(\boldsymbol{a}_y\boldsymbol{b}_z-\boldsymbol{a}_z\boldsymbol{b}_y)+\frac{\partial}{\partial y}(\boldsymbol{a}_z\boldsymbol{b}_x-\boldsymbol{a}_x\boldsymbol{b}_z)+\frac{\partial}{\partial z}(\boldsymbol{a}_x\boldsymbol{b}_y-\boldsymbol{a}_y\boldsymbol{b}_x)
\\&=\boldsymbol{b}_z\frac{\partial\boldsymbol{a}_y}{\partial x}+\boldsymbol{a}_y\frac{\partial\boldsymbol{b}_z}{\partial x}-\boldsymbol{a}_z\frac{\partial\boldsymbol{b}_y}{\partial x}-\boldsymbol{b}_y
\frac{\partial\boldsymbol{a}_z}{\partial x}
\\&+\boldsymbol{b}_x\frac{\partial\boldsymbol{a}_z}{\partial y}+\boldsymbol{a}_z\frac{\partial\boldsymbol{b}_x}{\partial y}-\boldsymbol{a}_x\frac{\partial\boldsymbol{b}_z}{\partial y}-\boldsymbol{b}_z
\frac{\partial\boldsymbol{a}_x}{\partial y}
\\&+\boldsymbol{b}_y\frac{\partial\boldsymbol{a}_x}{\partial z}+\boldsymbol{a}_x\frac{\partial\boldsymbol{b}_y}{\partial z}-\boldsymbol{a}_y\frac{\partial\boldsymbol{b}_x}{\partial z}-\boldsymbol{b}_x
\frac{\partial\boldsymbol{a}_y}{\partial z}
\\&=\begin{vmatrix}\boldsymbol{b}_x&\boldsymbol{b}_y&\boldsymbol{b}_z\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\\boldsymbol{a}_x&\boldsymbol{a}_y&\boldsymbol{a}_z\end{vmatrix}+\begin{vmatrix}\boldsymbol{a}_x&\boldsymbol{a}_y&\boldsymbol{a}_z\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\\boldsymbol{b}_x&\boldsymbol{b}_y&\boldsymbol{b}_z\end{vmatrix}
\\&=\boldsymbol{b}\cdot\triangledown\times\boldsymbol{a}-\boldsymbol{a}\cdot\triangledown\times\boldsymbol{b}.
\end{align}
$$

$(7)$

$$
\begin{align}
\triangledown\times(\phi\boldsymbol{a})&=\triangledown\times\begin{vmatrix}\boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\\phi\boldsymbol{a}_x&\phi\boldsymbol{a}_y&\phi\boldsymbol{a}_z\end{vmatrix}
\\&=\begin{pmatrix}\boldsymbol{a}_z\frac{\partial\phi}{\partial y}+\phi\frac{\partial\boldsymbol{a}_z}{\partial y}-\boldsymbol{a}_y\frac{\partial\phi}{\partial z}-\phi\frac{\partial\boldsymbol{a}_y}{\partial z}\\\boldsymbol{a}_x\frac{\partial\phi}{\partial z}+\phi\frac{\partial\boldsymbol{a}_x}{\partial z}-\boldsymbol{a}_z\frac{\partial\phi}{\partial x}-\phi\frac{\partial\boldsymbol{a}_z}{\partial x}\\\boldsymbol{a}_y\frac{\partial\phi}{\partial x}+\phi\frac{\partial\boldsymbol{a}_y}{\partial x}-\boldsymbol{a}_x\frac{\partial\phi}{\partial y}-\phi\frac{\partial\boldsymbol{a}_x}{\partial y}\end{pmatrix}
\\&=\begin{vmatrix}\boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k}\\\frac{\partial\phi}{\partial x}&\frac{\partial\phi}{\partial y}&\frac{\partial\phi}{\partial z}\\\boldsymbol{a}_x&\boldsymbol{a}_y&\boldsymbol{a}_z\end{vmatrix}+\phi\begin{vmatrix}\boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\\boldsymbol{a}_x&\boldsymbol{a}_y&\boldsymbol{a}_z\end{vmatrix}
\\&=\triangledown\phi\times\boldsymbol{a}+\phi\triangledown\times\boldsymbol{a}.
\end{align}
$$

**The proof is complete.**

### 9.6.10

$$
\begin{align}
\triangledown\times(\triangledown\times\boldsymbol{v})&=\triangledown\times\begin{vmatrix}\boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\\boldsymbol{v}_x&\boldsymbol{v}_y&\boldsymbol{v}_z\end{vmatrix}=\begin{vmatrix}\boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\\frac{\partial\boldsymbol{v}_z}{\partial y}-\frac{\partial\boldsymbol{v}_y}{\partial z}&\frac{\partial\boldsymbol{v}_x}{\partial z}-\frac{\partial\boldsymbol{v}_z}{\partial x}&\frac{\partial\boldsymbol{v}_y}{\partial x}-\frac{\partial\boldsymbol{v}_x}{\partial y}\end{vmatrix}\\
&=\begin{pmatrix}
\frac{\partial^2\boldsymbol{v}_y}{\partial y\partial x}-\frac{\partial^2\boldsymbol{v}_z}{\partial y^2}-\frac{\partial^2\boldsymbol{v}_x}{\partial z^2}+
\frac{\partial^2\boldsymbol{v}_z}{\partial z\partial x}
\\
\frac{\partial^2\boldsymbol{v}_z}{\partial z\partial y}-\frac{\partial^2\boldsymbol{v}_y}{\partial z^2}-\frac{\partial^2\boldsymbol{v}_x}{\partial x\partial z}+
\frac{\partial^2\boldsymbol{v}_y}{\partial x^2}
\\
\frac{\partial^2\boldsymbol{v}_x}{\partial x\partial z}-\frac{\partial^2\boldsymbol{v}_y}{\partial x^2}-\frac{\partial^2\boldsymbol{v}_z}{\partial y^2}+
\frac{\partial^2\boldsymbol{v}_y}{\partial y\partial z}\end{pmatrix}
\\&=\begin{pmatrix}
\frac{\partial}{\partial x}(\frac{\partial\boldsymbol{v}_x}{\partial x}+\frac{\partial\boldsymbol{v}_y}{\partial y}+\frac{\partial\boldsymbol{v}_z}{\partial z})
\\
\frac{\partial}{\partial y}(\frac{\partial\boldsymbol{v}_x}{\partial x}+\frac{\partial\boldsymbol{v}_y}{\partial y}+\frac{\partial\boldsymbol{v}_z}{\partial z})
\\
\frac{\partial}{\partial z}(\frac{\partial\boldsymbol{v}_x}{\partial x}+\frac{\partial\boldsymbol{v}_y}{\partial y}+\frac{\partial\boldsymbol{v}_z}{\partial z})
\end{pmatrix}-\begin{pmatrix}
\frac{\partial^2\boldsymbol{v}_x}{\partial x^2}+\frac{\partial^2\boldsymbol{v}_x}{\partial y^2}+\frac{\partial^2\boldsymbol{v}_x}{\partial z^2}\\
\frac{\partial^2\boldsymbol{v}_y}{\partial x^2}+\frac{\partial^2\boldsymbol{v}_y}{\partial y^2}+\frac{\partial^2\boldsymbol{v}_y}{\partial z^2}\\
\frac{\partial^2\boldsymbol{v}_z}{\partial x^2}+\frac{\partial^2\boldsymbol{v}_z}{\partial y^2}+\frac{\partial^2\boldsymbol{v}_z}{\partial z^2}
\end{pmatrix}\\&=
\triangledown(\triangledown\cdot\boldsymbol{v})-\triangledown^2\boldsymbol{v}.
\end{align}
$$

**The proof is complete.**

### 9.c.13

According to **Lagrange Mean Value Theorem**, there exists $\xi$ $\mathrm{s.t.}$

$$
f(x,y,z)-f(x+y+z,0,0)=-\frac{\partial f(\xi)}{\partial x}(y+z)+\frac{\partial f(\xi)}{\partial y}y+\frac{\partial f(\xi)}{\partial z}z=0
$$

So that $f(x,y,z)>0$.

**The proof is complete.**

### 9.c.14

First we consider $f(x,y)$ inside $D$. Its **Hess Determinant**

$$
H=\frac{\partial^2 f}{\partial x^2}\frac{\partial ^2 f}{\partial y^2}-(\frac{\partial^2 f}{\partial x\partial y})^2=4x-4y^2
$$

From $\begin{cases}f'_x=0\\f'_y=0\end{cases}$ we have that $\begin{cases}2x+y^2-1=0\\2xy=0\end{cases}\Rightarrow\begin{cases}x=\frac{1}{2}\\y=0\end{cases}\text{ or }\begin{cases}x=0\\y=\pm1\end{cases}$

At $(\frac{1}{2},0)$, $H>0$, it is a minimum point where $f=-\frac{1}{4}$;

At $(0,\pm1)$, $H<0$, is not an extreme point.

Then we consider $f(x,y)$ on the edge, where $x^2+y^2=2$.

$f(x,y)=x^2+xy^2-x=-x^3+x^2+x$ which gets the maximum point at $x=1$ where $f(1,\pm1)=1$.

### 10.1.1

$(2)$

$$
\int_0^2\mathrm{d}x\int_{2x}^{6-x}f(x,y)\mathrm{d}y=\int_0^6\mathrm{d}y\int_0^{\text{min}\{\frac{y}{2},6-y\}}f(x,y)\mathrm{d}x.
$$

$(4)$

$$
\int_a^b\mathrm{d}y\int_y^bf(x,y)\mathrm{d}x=\int_a^b\mathrm{d}x\int_a^xf(x,y)\mathrm{d}y.
$$

$(5)$

$$
\int_0^1\mathrm{d}x\int_0^xf(x,y)\mathrm{d}y+\int_1^2\mathrm{d}x\int_0^{2-x}f(x,y)\mathrm{dy}=\int_0^1\mathrm{d}y\int_{y}^{2-y}f(x,y)\mathrm{d}y.
$$

$(6)$

$$
\int_0^1\mathrm{d}y\int_{\frac{1}{2}}^{1}f(x,y)\mathrm{d}x+\int_1^2\mathrm{d}y\int_{\frac{1}{2}}^{\frac{1}{y}}f(x,y)\mathrm{d}x=\int_{\frac{1}{2}}^{1}\mathrm{d}x+\int_0^{\frac{1}{x}}f(x,y)\mathrm{d}y.
$$

### 10.1.2

$(2)$

$$
\iint_D\sin(x+y)\mathrm{d}x\mathrm{d}y=\int_0^{\pi}\int_0^{\pi}\sin(x+y)\mathrm{d}x\mathrm{d}y=\int_0^{\pi}2\cos\theta\mathrm{d}y=0.
$$

$(4)$

$$
\iint_D(x+y)\mathrm{d}x\mathrm{d}y=\int_0^a\mathrm{d}x\int_{0}^{\sqrt{1-x^2}}(x+y)\mathrm{d}y=\frac{1}{2}a-\frac{1}{6}a^3+\frac{1}{3}(1-a^2)^{\frac{3}{2}}-\frac{1}{3}.
$$

$(6)$

$$
\iint_D\frac{\sin y}{y}\mathrm{d}x\mathrm{d}y=\int_0^1\frac{\sin y}{y}(y-y^2)\mathrm{d}y=\int_0^1(\sin y-y)\mathrm{d}y=1-\sin1.
$$

$(8)$

$$
\iint_D|\cos(x+y)|\mathrm{d}x\mathrm{d}y=2\int_{0}^{\frac{\pi}{2\sqrt{2}}}t\cos(\sqrt{2}t)\mathrm{d}t=\frac{\pi-2}{2}.
$$

### 10.1.3

$(1)$ Since that $f$ is **even**,

$$
\iint_D(x^2+y^2)\mathrm{d}x\mathrm{d}y=4\int_0^1\int_0^1(x^2+y^2)\mathrm{d}x\mathrm{d}y=\frac{8}{3}.
$$

$(2)$ Since that $f$ is **odd**,

$$
\iint_D\sin x\sin y\mathrm{d}x\mathrm{d}y=0.
$$

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Maths Analysis II - Homework #12 20220324

Bi_Gu • 2022 年 03 月 30 日

# Maths Analysis II

## Homework #12 20220324

### 9.6.2

**Explanation by Physics**: The changing speed of the direction of $\boldsymbol{v}$ is in direct proportion with $w$.

### 9.6.3

$(1)$

$$
\triangledown\cdot\boldsymbol{v}=6x+3y^2+z^2+xy-6xz=6+16-14=8.
$$

$(2)$

$$
\triangledown\cdot\boldsymbol{v}=2x\sin y+2y\sin xz-xy\sin z\cos(\cos z).
$$

### 9.6.5

$(1)$

$(2)$

### 9.6.7

Since that $\displaystyle\triangledown f(r)=\frac{\partial f}{\partial r}\triangledown r=f'(r)\frac{\boldsymbol{r}}{r}$, we have that

$(1)$

$(2)$

$(3)$

### 9.6.8

$(1)$

$(2)$

$(3)$

$(4)$

$(5)$

$(6)$

$(7)$

**The proof is complete.**

### 9.6.10

**The proof is complete.**

### 9.c.13

According to **Lagrange Mean Value Theorem**, there exists $\xi$ $\mathrm{s.t.}$

$$
f(x,y,z)-f(x+y+z,0,0)=-\frac{\partial f(\xi)}{\partial x}(y+z)+\frac{\partial f(\xi)}{\partial y}y+\frac{\partial f(\xi)}{\partial z}z=0
$$

So that $f(x,y,z)>0$.

**The proof is complete.**

### 9.c.14

First we consider $f(x,y)$ inside $D$. Its **Hess Determinant**

$$
H=\frac{\partial^2 f}{\partial x^2}\frac{\partial ^2 f}{\partial y^2}-(\frac{\partial^2 f}{\partial x\partial y})^2=4x-4y^2
$$

At $(\frac{1}{2},0)$, $H>0$, it is a minimum point where $f=-\frac{1}{4}$;

At $(0,\pm1)$, $H<0$, is not an extreme point.

Then we consider $f(x,y)$ on the edge, where $x^2+y^2=2$.

$f(x,y)=x^2+xy^2-x=-x^3+x^2+x$ which gets the maximum point at $x=1$ where $f(1,\pm1)=1$.

### 10.1.1

$(2)$

$$
\int_0^2\mathrm{d}x\int_{2x}^{6-x}f(x,y)\mathrm{d}y=\int_0^6\mathrm{d}y\int_0^{\text{min}\{\frac{y}{2},6-y\}}f(x,y)\mathrm{d}x.
$$

$(4)$

$$
\int_a^b\mathrm{d}y\int_y^bf(x,y)\mathrm{d}x=\int_a^b\mathrm{d}x\int_a^xf(x,y)\mathrm{d}y.
$$

$(5)$

$$
\int_0^1\mathrm{d}x\int_0^xf(x,y)\mathrm{d}y+\int_1^2\mathrm{d}x\int_0^{2-x}f(x,y)\mathrm{dy}=\int_0^1\mathrm{d}y\int_{y}^{2-y}f(x,y)\mathrm{d}y.
$$

$(6)$

### 10.1.2

$(2)$

$$
\iint_D\sin(x+y)\mathrm{d}x\mathrm{d}y=\int_0^{\pi}\int_0^{\pi}\sin(x+y)\mathrm{d}x\mathrm{d}y=\int_0^{\pi}2\cos\theta\mathrm{d}y=0.
$$

$(4)$

$$
\iint_D(x+y)\mathrm{d}x\mathrm{d}y=\int_0^a\mathrm{d}x\int_{0}^{\sqrt{1-x^2}}(x+y)\mathrm{d}y=\frac{1}{2}a-\frac{1}{6}a^3+\frac{1}{3}(1-a^2)^{\frac{3}{2}}-\frac{1}{3}.
$$

$(6)$

$$
\iint_D\frac{\sin y}{y}\mathrm{d}x\mathrm{d}y=\int_0^1\frac{\sin y}{y}(y-y^2)\mathrm{d}y=\int_0^1(\sin y-y)\mathrm{d}y=1-\sin1.
$$

$(8)$

$$
\iint_D|\cos(x+y)|\mathrm{d}x\mathrm{d}y=2\int_{0}^{\frac{\pi}{2\sqrt{2}}}t\cos(\sqrt{2}t)\mathrm{d}t=\frac{\pi-2}{2}.
$$

### 10.1.3

$(1)$ Since that $f$ is **even**,

$$
\iint_D(x^2+y^2)\mathrm{d}x\mathrm{d}y=4\int_0^1\int_0^1(x^2+y^2)\mathrm{d}x\mathrm{d}y=\frac{8}{3}.
$$

$(2)$ Since that $f$ is **odd**,

$$
\iint_D\sin x\sin y\mathrm{d}x\mathrm{d}y=0.
$$

Maths Analysis II - Homework #12 20220324

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