\documentclass[a4paper]{article}
\usepackage[utf8]{inputenc}
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\usepackage{setspace}
\usepackage{amsmath}
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\title{Chapter 3\\Vector Analysis}
\author{solutions by Hikari}
\date{July 2021}
\begin{document}

\maketitle

\newcommand{\V}{\mathbf}
\newcommand{\M}{\mathrm}
\newcommand{\VE}{\hat{\V{e}}}
\newcommand{\del}{\boldsymbol{\nabla}}
\newcommand{\pdv}[2]{\frac{\partial#1}{\partial#2}}
\newcommand{\ppdv}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}
\newcommand{\dv}[2]{\frac{d #1}{d #2}}

\section*{3.2 Vectors in 3-D Space}

\paragraph{3.2.1}
$\V{P}\times \V{Q}=\sum_i \hat{\V{e}}_i\sum_{jk}\varepsilon_{ijk}P_j Q_k$.  $P_z=Q_z=0$, so $\varepsilon_{ijk}P_j Q_k\neq0$ only when $i=z$. So $\sum_i \hat{\V{e}}_i\sum_{jk}\varepsilon_{ijk}P_j Q_k=\hat{\V{e}}_z(P_xQ_y-P_yQ_x)\neq0$ because $\V{P}$ and $\V{Q}$ are nonparallel.

\paragraph{3.2.2}
$(\V{A}\times \V{B})\cdot(\V{A}\times \V{B})=(A_xB_y-A_yB_x)^2+(A_xB_z-A_zB_x)^2+(A_yB_z-A_zB_y)^2=A_x^2B_y^2+A_x^2B_z^2+A_y^2B_x^2+A_y^2B_z^2+A_z^2B_x^2+A_z^2B_y^2-2A_xB_xA_yB_y-2A_xB_xA_zB_z-2A_yB_yA_zB_z$. $(AB)^2-(\V{A}\cdot \V{B})^2=(A_x^2+A_y^2+A_z^2)(B_x^2+B_y^2+B_z^2)-(A_xB_x+A_yB_y+A_zB_z)^2=A_x^2B_y^2+A_x^2B_z^2+A_y^2B_x^2+A_y^2B_z^2+A_z^2B_x^2+A_z^2B_y^2-2A_xB_xA_yB_y-2A_xB_xA_zB_z-2A_yB_yA_zB_z$, so $(\V{A}\times \V{B})\cdot(\V{A}\times \V{B})=(AB)^2-(\V{A}\cdot \V{B})^2$.

\paragraph{3.2.3}
$\sin(\theta+\psi)=\frac{|\V{P}\times\V{Q}|}{|\V{P}||\V{Q}|}=|(-\sin{\theta}\cos{\psi}-\cos{\theta}\sin{\psi})\hat{e_z}|=\sin{\theta}\cos{\psi}+\cos{\theta}\sin{\psi}$.

$\cos{(\theta+\psi)}=\frac{\V{P}\cdot\V{Q}}{|\V{P}||\V{Q}|}=\cos{\theta}\cos{\psi}-\sin{\theta}\sin{\psi}$.

\paragraph{3.2.4}
(a) $\V{U}\times\V{V}=-3\hat{\V{e}}_y-3\hat{\V{e}}_z$ is perpendicular with $\V{U}$ and $\V{V}$.

(b) $\frac{\V{U}\times\V{V}}{|\V{U}\times\V{V}|}=\frac{1}{\sqrt{2}}\VE_y+\frac{1}{\sqrt{2}}\VE_z$

\paragraph{3.2.5}
All the four vectors are in the same plane, so both $\V{a}\times\V{b}$ and $\V{c}\times\V{d}$ are perpendicular to the plane, so $\V{a}\times\V{b}$ and $\V{c}\times\V{d}$ are parallel, so $(\V{a}\times\V{b})\times(\V{c}\times\V{d})=0$.

\paragraph{3.2.6}
The area of the triangle $=\frac{1}{2}|\V{B}||\V{C}|\sin\alpha=\frac{1}{2}|\V{A}||\V{C}|\sin\beta=\frac{1}{2}|\V{A}||\V{B}|\sin\gamma$. Devided by $|\V{A}||\V{B}||\V{C}|/2$, we get $\frac{\sin\alpha}{|\V{A}|}=\frac{\sin\beta}{|\V{B}|}=\frac{\sin\gamma}{|\V{C}|}$.

\paragraph{3.2.7}
$\VE_x\times\V{B}=2\VE_z-4\VE_y$, $\VE_y\times\V{B}=4\VE_x-\VE_z$, $\VE_z\times\V{B}=\VE_y-2\VE_x$ by the experiments.  $\VE_x\cdot(\VE_y\times \V{B})=\V{B}\cdot(\VE_x\times\VE_y)=\V{B}\cdot(\VE_z)=\V{B}_z=4$, $\VE_y\cdot(\VE_z\times \V{B})=\V{B}\cdot(\VE_y\times\VE_z)=\V{B}\cdot(\VE_x)=\V{B}_x=1$, $\VE_z\cdot(\VE_x\times \V{B})=\V{B}\cdot(\VE_z\times\VE_x)=\V{B}\cdot(\VE_y)=\V{B}_y=2$. So $\V{B}=\VE_x+2\VE_y+4\VE_z$.

\paragraph{3.2.8}
(a) $\V{A}\cdot\V{B}\times\V{C}=(\VE_x+\VE_y)\cdot(-\VE_x+\VE_y-\VE_z)=0$. It is true because $\V{B}$, $\V{C}$ and $\V{B}\times\V{C}$ are in the same plane, so the volume of the parallelepiped is zero.

(b) $\V{A}\times(\V{B}\times\V{C})=(\VE_x+\VE_y)\times(-\VE_x+\VE_y-\VE_z)=-\VE_x+\VE_y+2\VE_z$.

\paragraph{3.2.9}
$\V{a}\times(\V{b}\times\V{c})+\V{b}\times(\V{c}\times\V{a})+\V{c}\times(\V{a}\times\V{b})=\V{b}(\V{a}\cdot\V{c})-\V{c}(\V{a}\cdot\V{b})+\V{c}(\V{b}\cdot\V{a})-\V{a}(\V{b}\cdot\V{c})+\V{a}(\V{c}\cdot\V{b})-\V{b}(\V{c}\cdot\V{a})=0$.

\paragraph{3.2.10}
(a) $\V{A}_r=\hat{\V{r}}(\V{A}\cdot\hat{\V{r}})$ is quite obvious by the definition.

(b) $\V{A}_t=\V{A}-\V{A}_r=\V{A}(\hat{\V{r}}\cdot\hat{\V{r}})-\hat{\V{r}}(\V{A}\cdot\hat{\V{r}})=-\hat{\V{r}}\times(\hat{\V{r}}\times\V{A})$.

\paragraph{3.2.11}
If $\V{A}$, $\V{B}$, and $\V{C}$ are coplanar, then $\V{B}\times\V{C}$ is perpendicular to  $\V{B}$, $\V{C}$ and therefore perpendicular to $\V{A}$, so $\V{A}\cdot\V{B}\times\V{C}=0$. If $\V{A}\cdot\V{B}\times\V{C}=0$, then $\V{A}$ is perpendicular to $\V{B}\times\V{C}$, but $\V{B}\times\V{C}$ is perpendicular to the plane of $\V{B}$ and $\V{C}$, so $\V{A}$, $\V{B}$, and $\V{C}$ are coplanar.

\paragraph{3.2.12}
$\V{A}\cdot\V{B}\times\V{C}=-120$, $\V{A}\times(\V{B}\times\V{C})=-60\VE_x-40\VE_y+50\VE_z$, $\V{C}\times(\V{A}\times\V{B})=24\VE_x+88\VE_y-62\VE_z$, $\V{B}\times(\V{C}\times\V{A})=36\VE_x-48\VE_y+12\VE_z$.

\paragraph{3.2.13}
$(\V{A}\times\V{B})\cdot(\V{C}\times\V{D})=(\V{B}\times(\V{C}\times\V{D}))\cdot\V{A}=(\V{C}(\V{B}\cdot\V{D})-\V{D}(\V{B}\cdot\V{C}))\cdot\V{A}=(\V{A}\cdot\V{C})(\V{B}\cdot\V{D})-(\V{A}\cdot\V{D})(\V{B}\cdot\V{C})$.

\paragraph{3.2.14}
$(\V{A}\times\V{B})\times(\V{C}\times\V{D})=\V{C}(\V{A}\times\V{B}\cdot\V{D})-\V{D}(\V{A}\times\V{B}\cdot\V{C})=(\V{A}\cdot\V{B}\times\V{D})\V{C}-(\V{A}\cdot\V{B}\times\V{C})\V{D}$.

\paragraph{3.2.15}
(a) $\V{F}_2=q_2\V{v}_2\times\V{B}=\frac{\mu_0}{4\pi}\frac{q_1q_2}{r^2}\V{v}_2\times(\V{v}_1\times\hat{\V{r}})$.

(b) With $\V{v}_1$ replaced by $\V{v}_2$, $\V{v}_2$ replaced by $\V{v}_1$, $\hat{\V{r}}$ replaced by $-\hat{\V{r}}$, $\V{F}_1=-\frac{\mu_0}{4\pi}\frac{q_1q_2}{r^2}\V{v}_1\times(\V{v}_2\times\hat{\V{r}})$.

(c) $\hat{\V{r}}$ are perpendicular to $\V{v}_2,\V{v}_1$, and $\V{v}_2,\V{v}_1$ are parallel, so $\V{F}_2=-\frac{\mu_0}{4\pi}\frac{q_1q_2}{r^2}v_2v_1\hat{\V{r}}=-\V{F}_1$.

\section*{3.3 Coordinate Transformations}

\paragraph{3.3.1}
\[
\begin{pmatrix}
\cos(\varphi_1+\varphi_2)&\sin(\varphi_1+\varphi_2)\\
-\sin(\varphi_1+\varphi_2)&\cos(\varphi_1+\varphi_2)
\end{pmatrix}=
\begin{pmatrix}
\cos\varphi_2&\sin\varphi_2\\
-\sin\varphi_2&\cos\varphi_2
\end{pmatrix}
\begin{pmatrix}
\cos\varphi_1&\sin\varphi_1\\
-\sin\varphi_1&\cos\varphi_1
\end{pmatrix}
\]

\paragraph{3.3.2}
Let the three reflecting surfaces be parallel to $xy$, $xz$, $yz$ surfaces. Let the direction vector of the incident light be $(k_1,k_2,k_3)$. Then after each reflection, one of the coordinate changes sign, so the direction vector of reflected light is $(-k_1,-k_2,-k_3)$, parallel to the incident light.

\paragraph{3.3.3}
$(\V{x}')^T\V{y}'=\V{x}^T\M{S}^T\M{S}\V{y}=\V{x}^T\V{y}$ because $\M{S}$ is orthogonal.

\paragraph{3.3.4}
(a) $\det\M{S}=1$

(b) $\V{a}\cdot\V{b}=-1$, $\M{S}\V{a}\cdot\M{S}\V{b}=(0.8, 0.12, 1.16)\cdot(1.2, 0.68, -1.76)=-1$

(c) $\V{a}\times\V{b}=(-2, 1, 2)$, $\M{S}(\V{a}\times\V{b})=(-1, 2.8, 0.4)$, $\M{S}\V{a}\times\M{S}\V{b}=(-1, 2.8, 0.4)$. It is as expected because pseudovectors transform as vectors when the orthogonal transformation is not a reflection ($\det (\M{S})=-1$).

\paragraph{3.3.5}
(a) $\det (\M{S})=-1$

(b) $(\M{S}\V{a})\times(\M{S}\V{b})=(-0.4, -1.64, -2.48)$; $\M{S}(\V{a}\times\V{b})=(0.4, 1.64, 2.48)$. The sign changes.

(c) $(\M{S}\V{a}\times\M{S}\V{b})\cdot\M{S}\V{c}=-3$; $\M{S}((\V{a}\times\V{b})\cdot\V{c})=(\V{a}\times\V{b})\cdot\V{c}=3$. The sign changes.

(d) $\M{S}\V{a}\times(\M{S}\V{b}\times\M{S}\V{c})=(-0.4, -8.84, 7.12)$; $\M{S}(\V{a}\times(\V{b}\times\V{c}))=(-0.4, -8.84, 7.12)$. The sign does not change.

(e) $\V{a}\times\V{b}$ is a pseudovector, $(\V{a}\times\V{b})\cdot\V{c}$ is a pseudoscalar, $\V{a}\times(\V{b}\times\V{c})$ is a vector.

\section*{3.4 Rotations in $\mathbb{R}^3$}

\paragraph{3.4.1}
The corresponding transformation matrix is
\[
\begin{pmatrix}
\cos\psi&\sin\psi&0\\
-\sin\psi&\cos\psi&0\\
0&0&1\\
\end{pmatrix}
\begin{pmatrix}
1&0&0\\
0&\cos\theta&\sin\theta\\
0&-\sin\theta&\cos\theta\\
\end{pmatrix}
\begin{pmatrix}
\cos\varphi&\sin\varphi&0\\
-\sin\varphi&\cos\varphi&0\\
0&0&1\\
\end{pmatrix}
\]
Make the substitution by $\cos\varphi=-\sin\alpha$, $\sin\varphi=\cos\alpha$, $\cos\theta=\cos\beta$, $\sin\theta=\sin\beta$, $\cos\psi=\sin\gamma$, $\sin\psi=-\cos\gamma$. The matrix becomes
\[
\begin{pmatrix}
\sin\gamma&-\cos\gamma&0\\
\cos\gamma&\sin\gamma&0\\
0&0&1\\
\end{pmatrix}
\begin{pmatrix}
1&0&0\\
0&\cos\beta&\sin\beta\\
0&-\sin\beta&\cos\beta\\
\end{pmatrix}
\begin{pmatrix}
-\sin\alpha&\cos\alpha&0\\
-\cos\alpha&-\sin\alpha&0\\
0&0&1\\
\end{pmatrix}
\]
\[=
\begin{pmatrix}
\sin\gamma&-\cos\gamma&0\\
\cos\gamma&\sin\gamma&0\\
0&0&1\\
\end{pmatrix}
\begin{pmatrix}
-\sin\alpha&\cos\alpha&0\\
-\cos\beta\cos\alpha&-\cos\beta\sin\alpha&\sin\beta\\
\sin\beta\cos\alpha&\sin\beta\sin\alpha&\cos\beta\\
\end{pmatrix}
\]
\[=
\begin{pmatrix}
-\sin\gamma\sin\alpha+\cos\gamma\cos\beta\cos\alpha&
\sin\gamma\cos\alpha+\cos\gamma\cos\beta\sin\alpha&
-\cos\gamma\sin\beta\\
-\cos\gamma\sin\alpha-\sin\gamma\cos\beta\cos\alpha&
\cos\gamma\cos\alpha-\sin\gamma\cos\beta\sin\alpha&
\sin\gamma\sin\beta\\
\sin\beta\cos\alpha&
\sin\beta\sin\alpha&
\cos\beta\\
\end{pmatrix}
\]
which is the same as the transformation matrix in Eq. (3.37).

\paragraph{3.4.2}
(In the original system, the North Pole is in the $x_3$-axis direction, and the middle point of Prime Meridian($0^\circ,0^\circ$) is in the $x_2$-axis direction) Rotate $\alpha=70^\circ$ around $x_3$-axis to align $x_1$-axis with $20^\circ$ west, and rotate $\beta=60^\circ$ around $x_2$-axis to align the North Pole with $30^\circ$ north, and Rotate $\gamma=-80^\circ$ around $x_3$-axis to align the $10^\circ$ west Meridian with the Meridian in new system). Calculating from Eq. (3.37), the transformation matrix is
\[
\begin{pmatrix}
0.9551&-0.2552&-0.1504\\
0.0052&0.5221&-0.8529\\
0.2962&0.8138&0.5000\\
\end{pmatrix}
\]

\paragraph{3.4.3}
All the trigonometric function except $\cos\beta$ change sign. Substituting, we found that the rotation matrix $\M{S}$ remains unchanged.

\paragraph{3.4.4}
$\M{S}(\alpha,\beta,\gamma)=\M{S}_3(\gamma)\M{S}_2(\beta)\M{S}_1(\alpha)$. Note that $\M{S}_i^{-1}(x)=\tilde{\M{S}}_i(x)$ (orthogonality) and $\M{S}_i(x)^{-1}=\M{S}_i(-x)$ (property of rotation matrix). So
\[\M{S}^{-1}(\alpha,\beta,\gamma)=\M{S}_1^{-1}(\alpha)\M{S}_2^{-1}(\beta)\M{S}_3^{-1}(\gamma)=\tilde{\M{S}}_1(\alpha)\tilde{\M{S}}_2(\beta)\tilde{\M{S}}_3(\gamma)=\tilde{(\M{S}_3(\gamma)\M{S}_2(\beta)\M{S}_1(\alpha))}=\tilde{\M{S}}(\alpha,\beta,\gamma)\]
\[\M{S}^{-1}(\alpha,\beta,\gamma)=\M{S}_1^{-1}(\alpha)\M{S}_2^{-1}(\beta)\M{S}_3^{-1}(\gamma)=\M{S}_1(-\alpha)\M{S}_2(-\beta)\M{S}_3(-\gamma)=\M{S}_3(-\alpha)\M{S}_2(-\beta)\M{S}_1(-\gamma)=\M{S}(-\gamma,-\beta,-\alpha)\]

\paragraph{3.4.5}
(a) Decompose $\V{r}$ into $\V{r}_{\parallel}$ parallel to $\hat{\V{n}}$ and $\V{r_{\perp}}$ perpendicular to $\hat{\V{n}}$.
$\V{r}_{\parallel}=\hat{\V{n}}(\hat{\V{n}}\cdot\V{r})$, $\V{r_{\perp}}=\V{r}-\hat{\V{n}}(\hat{\V{n}}\cdot\V{r})$.
After rotation, $\V{r}_{\parallel}$ remains unchanged, and $\V{r_{\perp}}$ becomes $\frac{\V{r_{\perp}}}{|\V{r_{\perp}}|}\times\hat{\V{n}}(|\V{r_{\perp}}|\sin\V{\Phi})$ in the $\VE_{\V{\Phi}}$ direction and $\frac{\V{r_{\perp}}}{|\V{r_{\perp}}|}|\V{r_{\perp}}|\cos\V{\Phi}$ in the $\VE_{\V{r}_{\perp}}$ direction. So
\[\V{r}'=\V{r}_{\parallel}+\V{r_{\perp}}\times\hat{\V{n}}\sin\V{\Phi}+\V{r_{\perp}}\cos\V{\Phi}=\hat{\V{n}}(\hat{\V{n}}\cdot\V{r})+(\V{r}-\hat{\V{n}}(\hat{\V{n}}\cdot\V{r}))\times\hat{\V{n}}\sin\V{\Phi}+(\V{r}-\hat{\V{n}}(\hat{\V{n}}\cdot\V{r}))\cos\V{\Phi}\]
\[
=\V{r}\cos\V{\Phi}+\V{r}\times\hat{\V{n}}\sin\V{\Phi}+\hat{\V{n}}(\hat{\V{n}}\cdot\V{r})(1-\cos\V{\Phi})
\]

(b) Let $\V{r}=(x,y,z)$, then
\[
\V{r}'=
\begin{pmatrix}x'\\y'\\z'\end{pmatrix}=
\begin{pmatrix}x\cos\V{\Phi}\\y\cos\V{\Phi}\\z\cos\V{\Phi}\end{pmatrix}+
\begin{pmatrix}y\sin\V{\Phi}\\-z\sin\V{\Phi}\\0\end{pmatrix}+
\begin{pmatrix}0\\0\\z(1-\cos\V{\Phi})\end{pmatrix}=
\begin{pmatrix}x\cos\V{\Phi}+y\sin\V{\Phi}\\-z\sin\V{\Phi}+y\cos\V{\Phi}\\z\end{pmatrix}
\]
\[
\begin{pmatrix}x'\\y'\\z'\end{pmatrix}=
\begin{pmatrix}
\cos\V{\Phi}&\sin\V{\Phi}&0\\
-\sin\V{\Phi}&\cos\V{\Phi}&0\\
0&0&1
\end{pmatrix}
\begin{pmatrix}x\\y\\z\end{pmatrix}
\]
which is identical with Eq. (3.55).

(c)
$\V{r}\cdot(\V{r}\times\hat{\V{n}})=0$, $(\V{r}\times\hat{\V{n}})\cdot\hat{\V{n}}=0$ because they are perpendicular. Let $\theta$ be the angle between $\V{r}$ and $\hat{\V{n}}$, then $|\V{r}\times\hat{\V{n}}|=r\sin\theta$, $|\hat{\V{n}}\cdot\V{r}|=r\cos\theta$. \[r'^2=\V{r}'\cdot\V{r}'=r^2\cos^2\Phi+(\V{r}\times\hat{\V{n}})^2\sin^2\Phi+(\hat{\V{n}}\cdot\V{r})^2(1-\cos\Phi)^2+2(\hat{\V{n}}\cdot\V{r})^2(\cos\Phi)(1-\cos\Phi)\]
\[
=r^2\cos^2\Phi+r^2\sin^2\theta\sin^2\Phi+r^2\cos^2\theta(1-\cos\Phi)^2+2r^2\cos^2\theta(\cos\Phi)(1-\cos\Phi)
\]
\[=r^2\cos^2\Phi+r^2\sin^2\theta\sin^2\Phi+r^2\cos^2\theta(1-\cos^2\Phi)
\]
\[
=r^2\cos^2\Phi+r^2\sin^2\Phi
\]
\[
=r^2
\]

\section*{3.5 Differential Vector Operators}

\paragraph{3.5.1}
(a) $\del\V{S}=-3(x^2+y^2+z^2)^{-5/2}(x\VE_x+y\VE_y+z\VE_z)=-3(14)^{-5/2}(1\VE_x+2\VE_y+3\VE_z)$ at point $(1,2,3)$.

(b) $\del\V{S}=3\cdot14^{-5/2}\cdot14^{1/2}=\frac{3}{196}$

(c) $(\frac{-1}{\sqrt{14}},\frac{-2}{\sqrt{14}},\frac{-3}{\sqrt{14}})$

\paragraph{3.5.2}
(a) $(\pdv{f}{x},\pdv{f}{y},\pdv{f}{z})\cdot(dx,dy,dz)=df=0$ when $f(x,y,z)=constant$, so $(\pdv{f}{x},\pdv{f}{y},\pdv{f}{z})$ is perpendicular to the equipotential surfaces. $\del(x^2+y^2+z^2)=2x\VE_x+2y\VE_y+2z\VE_z=2\VE_x+2\VE_y+2\VE_z$ at point $(1,1,1)$, and its unit vector is $\frac{\VE_x+\VE_y+\VE_z}{\sqrt{3}}$.

(b) $(x-1,y-1,z-1)\cdot(1,1,1)=0$, so the tangent surface is $x+y+z=3$.

\paragraph{3.5.3}
$r_{12}=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$, so $\del_1r_{12}=\pdv{r_{12}}{x}\VE_x+\pdv{r_{12}}{y}\VE_y+\pdv{r_{12}}{z}\VE_z=\frac{x_1-x_2}{r_{12}}\VE_x+\frac{y_1-y_2}{r_{12}}\VE_y+\frac{z_1-z_2}{r_{12}}\VE_z=\frac{\V{r_{12}}}{r_{12}}$ which is the unit vector in the direction of $\V{r_{12}}$.

\paragraph{3.5.4}
\[\V{F}=F_x\VE_x+F_y\VE_y+F_z\VE_z\]
\[
d\V{F}=(\pdv{F_x}{x}dx+\pdv{F_x}{y}dy+\pdv{F_x}{z}dz+\pdv{F_x}{t}dt)\VE_x+(\pdv{F_y}{x}dx+\pdv{F_y}{y}dy+\pdv{F_y}{z}dz+\pdv{F_y}{t}dt)\VE_y+(\pdv{F_z}{x}dx+\pdv{F_z}{y}dy+\pdv{F_z}{z}dz+\pdv{F_z}{t}dt)\VE_z
\]
\[=
(dx\pdv{}{x}+dy\pdv{}{y}+dz\pdv{}{z})F_x\VE_x+(dx\pdv{}{x}+dy\pdv{}{y}+dz\pdv{}{z})F_y\VE_y+(dx\pdv{}{x}+dy\pdv{}{y}+dz\pdv{}{z})F_z\VE_z+(\pdv{F_x}{t}\VE_x+\pdv{F_y}{t}\VE_y+\pdv{F_z}{t}\VE_z)dt
\]
\[
=(dx\pdv{}{x}+dy\pdv{}{y}+dz\pdv{}{z})(F_x\VE_x+F_y\VE_y+F_z\VE_z)+\pdv{\V{F}}{t}dt
\]
\[
=(d\V{r}\cdot\del)\V{F}+\pdv{\V{F}}{t}dt
\]

\paragraph{3.5.5}
$\del(uv)=\pdv{uv}{x}\VE_x+\pdv{uv}{y}\VE_y+\pdv{uv}{z}\VE_z=(\pdv{u}{x}v+u\pdv{v}{x})\VE_x+(\pdv{u}{y}v+u\pdv{v}{y})\VE_y+(\pdv{u}{z}v+u\pdv{v}{z})\VE_z$\\
$=(\pdv{u}{x}\VE_x+\pdv{u}{y}\VE_y+\pdv{u}{z}\VE_z)v+u(\pdv{v}{x}\VE_x+\pdv{v}{y}\VE_y+\pdv{v}{z}\VE_z)=v\del u+u\del v$

\paragraph{3.5.6}
(a) $\Dot{\V{r}}=\frac{d\V{r}}{dt}=-r\omega\sin{\omega t}\,\VE_x+r\omega\cos{\omega t}\,\VE_y$, so
\[\V{r}\times\Dot{\V{r}}=(r\cos{\omega t}\,\VE_x+r\sin{\omega t}\,\VE_t)\times(-r\omega\sin{\omega t}\,\VE_x+r\omega\cos{\omega t}\,\VE_y)=r\omega^2(\cos^2{\omega t}+\sin^2{\omega t})\VE_z=r\omega^2\VE_z\]

(b) $\Ddot{\V{r}}=-r\omega^2\cos{\omega t}\,\VE_x-r\omega^2\sin{\omega t}\,\VE_y$, so $\Ddot{\V{r}}+\omega^2\V{r}=0$.

\paragraph{3.5.7}
$\V{A'}=\M{S}\V{A}$, which means $A'_i=\sum_{j}\M{S}_{ij}A_j$. Because $\M{S}_{ij}$ is independent of $t$, so $\frac{dA'_i}{dt}=\sum_{j}\M{S}_{ij}\frac{dA_j}{dt}$, which means $\frac{d\V{A'}}{dt}=\M{S}\frac{d\V{A}}{dt}$.

\paragraph{3.5.8}
(a) $\frac{d}{dt}(\V{A}\cdot\V{B})=\frac{d}{dt}(A_xB_x+A_yB_y+A_zB_z)=\frac{dA_x}{dt}B_x+A_x\frac{dB_x}{dt}+\frac{dA_y}{dt}B_y+A_y\frac{dB_y}{dt}+\frac{dA_z}{dt}B_z+A_z\frac{dB_z}{dt}$\\
$=\frac{d\V{A}}{dt}\cdot\V{B}+\V{A}\cdot\frac{d\V{B}}{dt}$
\medskip

(b)
\[
\frac{d}{dt}(\V{A}\times\V{B})=\frac{d}{dt}\bigg( (A_yB_z-A_zB_y)\VE_x+(A_zB_x-A_xB_z)\VE_y+(A_xB_y-A_yB_x)\VE_z \bigg)
\]
\[
=
\scriptstyle\left(\dv{A_y}{t}B_z+A_y\dv{B_z}{t}-\dv{A_z}{t}B_y-A_z\dv{B_y}{t}\right)\VE_x+\left(\dv{A_z}{t}B_x+A_z\dv{B_x}{t}-\dv{A_x}{t}B_z-A_x\dv{B_z}{t}\right)\VE_y+\left(\dv{A_x}{t}B_y+A_x\dv{B_y}{t}-\dv{A_y}{t}B_x-A_y\dv{B_x}{t}\right)\VE_z
\]
\[
=
\scriptstyle\left(\dv{A_y}{t}B_z-\dv{A_z}{t}B_y\right)\VE_x+\left(\dv{A_z}{t}B_x-\dv{A_x}{t}B_z\right)\VE_y+\left(\dv{A_x}{t}B_y-\dv{A_y}{t}B_x\right)\VE_z+
\left(A_y\dv{B_z}{t}-A_z\dv{B_y}{t}\right)\VE_x+\left(A_z\dv{B_x}{t}-A_x\dv{B_z}{t}\right)\VE_y+\left(A_x\dv{B_y}{t}-A_y\dv{B_x}{t}\right)\VE_z
\]

\[
=\dv{\V{A}}{t}\times\V{B}+\V{A}\times\dv{\V{B}}{t}
\]

\paragraph{3.5.9}
$\frac{d}{dx}a_ib_j=\frac{da_i}{dx}b_j+a_i\frac{db_j}{dx}$, so we can decompose $\del\cdot(\V{a}\times\V{b})$ into $\del_a\cdot(\V{a}\times\V{b})+\del_b\cdot(\V{a}\times\V{b})$, with $\del_a$ operating on $\V{a}$ only, and $\del_b$ operating on $\V{b}$ only. Then by the symmetry of scalar triple product, $\del_a\cdot(\V{a}\times\V{b})=\V{b}\cdot(\del_a\times\V{a})=\V{b}\cdot(\del\times\V{a})$, and $\del_b\cdot(\V{a}\times\V{b})=\V{a}\cdot(\V{b}\times\del_b)=-\V{a}\cdot(\del_b\times\V{b})=-\V{a}\cdot(\del\times\V{b})$. So $\del\cdot(\V{a}\times\V{b})=\V{b}\cdot(\del\times\V{a})-\V{a}\cdot(\del\times\V{b})$.

\paragraph{3.5.10}
$\V{L}=\V{r}\times(-i\del)=(x\VE_x+y\VE_y+z\VE_z)\times(-i\pdv{}{x}\VE_x-i\pdv{}{y}\VE_y-i\pdv{}{z}\VE_z)=\\-i\left(y\pdv{}{z}-z\pdv{}{y} \right)\VE_x-i\left(z\pdv{}{x}-x\pdv{}{z} \right)\VE_y-i\left(x\pdv{}{y}-y\pdv{}{x} \right)\VE_z$

\paragraph{3.5.11}
\[L_xL_y-L_yL_x\]
\[
=-\left(y\pdv{}{z}-z\pdv{}{y} \right)\left(z\pdv{}{x}-x\pdv{}{z} \right)+\left(z\pdv{}{x}-x\pdv{}{z} \right)\left(y\pdv{}{z}-z\pdv{}{y} \right)
\]
\[
=-\left(y\pdv{}{x}+yz\ppdv{}{z}{x}-yx\pdv{^2}{z^2}-z^2\ppdv{}{y}{x}+xz\ppdv{}{y}{z} \right)+\left(zy\ppdv{}{x}{z}-z^2\ppdv{}{x}{y}-xy\pdv{^2}{z^2}+x\pdv{}{y}+xz\ppdv{}{z}{y} \right)
\]
\[
=x\pdv{}{y}-y\pdv{}{x}=iL_z
\]

\paragraph{3.5.12}
The problem is ill-defined because the way of vector multiplication has not been specified(scalar product, cross product or tensor product). If $[\V{a},\V{b}]=0$, $[\V{a},\V{L}]=0$, $[\V{b},\V{L}]=0$ means $[a_i,b_j]=0$, $[a_i,L_j]=0$, $[b_i,L_j]=0$ for all $i,j\in\{x,y,z\}$, then $a$ and $b$, $a$ and $L$, $b$ and $L$ commute, but $L$ and $L$ do not commute ($[L_i,L_j]=iL_k$). So
\[[\V{a}\cdot\V{L},\V{b}\cdot\V{L}]\]
\[=[a_xL_x+a_yL_y+a_zL_z,b_xL_x+b_yL_y+b_zL_z]\]
\[=\sum_{i=1}^3\sum_{j=1}^3[a_iL_i,b_jL_j]\]
\[
=\sum_{i=1}^3\sum_{j=1}^3 (a_iL_ib_jL_j-b_jL_ja_iL_i)\]
\[
=\sum_{i=1}^3\sum_{j=1}^3(a_ib_jL_iL_j-b_ja_iL_jL_i)
\]
\[=
\scriptstyle a_xb_y(L_xL_y-L_yL_x)+a_xb_z(L_xL_z-L_zL_x)+a_yb_x(L_yL_x-L_xL_y)+a_yb_z(L_yL_z-L_zL_y)+a_zb_x(L_zL_x-L_xL_z)+a_zb_y(L_zL_y-L_yL_z)
\]
\[
=a_xb_yiL_z-a_xb_ziL_y-a_yb_xiL_z+a_yb_ziL_x+a_zb_xiL_y-a_zb_yiL_x
\]
\[
=i(a_yb_z-a_zb_y)L_x+i(a_zb_x-a_xb_z)L_y+i(a_xb_y-a_yb_x)L_z
\]
\[
=i(\V{a}\times\V{b})\cdot\V{L}
\]

\paragraph{3.5.13}
A stream line of a vector field should be parallel to the vector at every point in the space. So $\frac{dy}{dx}=\frac{b_y}{b_x}=\frac{x}{-y}$, where $y=y(x)$ is a stream line. Solving  the differential equation, $xdx+ydy=0$, $\frac{x^2}{2}+\frac{y^2}{2}=k'$, $x^2+y^2=k$, which is a circle. The direction of the stream line at $(1,1)$ is $-\VE_x+\VE_y$, which is counterclockwise relative $(0,0)$, the center of the circle.

\section*{3.6 Differential Vector Operators: Further Properties}

\paragraph{3.6.1}
By the identity in Exercise 3.5.9, $\del\cdot(\V{u}\times\V{v})=\V{v}\cdot(\del\times\V{u})-\V{u}\cdot(\del\times\V{v})=0$ because $\del\times\V{u}=0$ and $\del\times\V{v}=0$ (irrotational).

\paragraph{3.6.2}
$\del\cdot(\V{A}\times\V{r})=\V{r}\cdot(\del\times\V{A})-\V{A}\cdot(\del\times\V{r})=0$ because both $\V{A}$ and the position vector $\V{r}=x\VE_x+y\VE_y+z\VE_z$ are irrotational, so their curl vanish.

\paragraph{3.6.3}
The linear velocity $\V{v}=\boldsymbol{\omega}\times\V{r}$, so $\del\cdot\V{v}=\del\cdot(\boldsymbol{\omega}\times\V{r})=\V{r}\cdot(\del\times\boldsymbol{\omega})-\boldsymbol{\omega}\cdot(\del\times\V{r})=0$ because the curl of constant vector $\boldsymbol{\omega}$ and position vector $\V{r}$ are zero.

\paragraph{3.6.4}
$\del\times\V{V}\neq0$ but $\del\times(g\V{V})=g(\del\times\V{V})+(\del g)\times\V{V}=0$. So $\V{V}\cdot\big(g(\del\times\V{V})+(\del g)\times\V{V} \big)=g\V{V}\cdot(\del\times\V{V})+\V{V}\cdot((\del g)\times\V{V})=0$. But $\V{V}\cdot((\del g)\times\V{V})=0$ (perpendicular) and $g\neq 0$, so $\V{V}\cdot(\del\times\V{V})=0$.

\paragraph{3.6.5}
All the terms of $\V{A}\times\V{B}$ have the form $A_iB_j\VE_k$, and $\pdv{}{x}(A_iB_j)=\pdv{A_i}{x}B_j+A_i\pdv{B_j}{x}$. So we can separate $\del\times(\V{A}\times\V{B})$ into $\del_A\times(\V{A}\times\V{B})$ and $\del_B\times(\V{A}\times\V{B})$, with $\del_A$ acting only on $\V{A}$, and $\del_B$ acting only on $\V{B}$. Using the BAC-CAB rule, but noting that $\del_A$ must go before $\V{A}$ and after $\V{B}$, $\del_B$ go before $\V{B}$ and after $\V{A}$, we get
\[
\del\times(\V{A}\times\V{B})=\del_A\times(\V{A}\times\V{B})+\del_B\times(\V{A}\times\V{B})
\]
\[
=(\V{B}\cdot\del_A)\V{A}-\V{B}(\del_A\cdot\V{A})+\V{A}(\del_B\cdot\V{B})-(\V{A}\cdot\del_B)\V{B}
\]
\[
=(\V{B}\cdot\del)\V{A}-\V{B}(\del\cdot\V{A})+\V{A}(\del\cdot\V{B})-(\V{A}\cdot\del)\V{B}
\]

\paragraph{3.6.6}
\[[(\V{A}\times\del)\times\V{B}]_x=A_z\pdv{B_z}{x}-A_x\pdv{B_z}{z}-A_x\pdv{B_y}{y}+A_y\pdv{B_y}{x}
\]
\[[(\V{B}\times\del)\times\V{A}]_x=B_z\pdv{A_z}{x}-B_x\pdv{A_z}{z}-B_x\pdv{A_y}{y}+B_y\pdv{A_y}{x}
\]
\[
[\V{A}(\del\cdot\V{B})]_x=A_x\pdv{B_x}{x}+A_x\pdv{B_y}{y}+A_x\pdv{B_z}{z}
\]
\[
[\V{B}(\del\cdot\V{A})]_x=B_x\pdv{A_x}{x}+B_x\pdv{A_y}{y}+B_x\pdv{A_z}{z}
\]
Summing together we get
\[
A_x\pdv{B_x}{x}+B_x\pdv{A_x}{x}+A_y\pdv{B_y}{x}+B_y\pdv{A_y}{x}+A_z\pdv{B_z}{x}+B_z\pdv{A_z}{x}
\]
which is $\pdv{(\V{A}\cdot\V{B})}{x}=[\del(\V{A}\cdot\V{B})]_x$. The same is for $y$ and $z$ components. Therefore,

\[\del(\V{A}\cdot\V{B})=
(\V{A}\times\del)\times\V{B}+(\V{B}\times\del)\times\V{A}+\V{A}(\del\cdot\V{B})+\V{B}(\del\cdot\V{A})
\]

\paragraph{3.6.7}
To distinguish between the two $\V{A}$, let the first $\V{A}$ be $\V{A}_1$ and the second be $\V{A}_2$. Applying BAC-CAB rule,  $\V{A}_1\times(\del\times\V{A}_2)=\del_2(\V{A}_1\cdot \V{A}_2)-(\V{A}_1\cdot\del)\V{A}_2$ with $\del_2$ acting only on $\V{A}_2$. Noting that $\del(\V{A}_1\cdot \V{A}_2)=\del_1(\V{A}_1\cdot \V{A}_2)+\del_2(\V{A}_1\cdot \V{A}_2)$ and $\del_1(\V{A}_1\cdot \V{A}_2)=\del_2(\V{A}_1\cdot \V{A}_2)$, we have $\del_2(\V{A}_1\cdot \V{A}_2)=\frac{\del(\V{A}_1\cdot \V{A}_2)}{2}$, so \[\V{A}\times(\del\times\V{A})=\frac{1}{2}\del(\V{A}\cdot \V{A})-(\V{A}\cdot\del)\V{A}\]

\paragraph{3.6.8}
\[\del(\V{A}\cdot\V{B}\times\V{r})=\del(\V{r}\cdot\V{A}\times\V{B})\]\[=\del(x[\V{A}\times\V{B}]_x+y[\V{A}\times\V{B}]_y+z[\V{A}\times\V{B}]_z)\]\[=[\V{A}\times\V{B}]_x\VE_x+[\V{A}\times\V{B}]_y\VE_y+[\V{A}\times\V{B}]_z\VE_z\]\[=\V{A}\times\V{B}\]
because $\V{A}\times\V{B}$ is constant, so $\pdv{[\V{A}\times\V{B}]_j}{x_i}=0$

\paragraph{3.6.9}
\[
[\del\times(\del\times\V{V})]_x=\ppdv{V_y}{y}{x}-\pdv{^2V_x}{z^2}-\pdv{^2V_x}{z^2}+\ppdv{V_z}{z}{x}
\]
\[
[\del(\del\cdot\V{V})]_x=\pdv{^2V_x}{x^2}+\ppdv{V_y}{x}{y}+\ppdv{V_z}{x}{z}
\]
\[
-[\del\cdot\del\V{V}]_x=-\pdv{^2V_x}{x^2}-\pdv{^2V_y}{y^2}-\pdv{^2V_z}{z^2}
\]
So $[\del\times(\del\times\V{V})]_x=[\del(\del\cdot\V{V})]_x-[\del\cdot\del\V{V}]_x$, as well as the $y$ and $z$ components. Therefore
\[\del\times(\del\times\V{V})=\del(\del\cdot\V{V})-\del\cdot\del\V{V}\]

\paragraph{3.6.10}
\[
[\del\times(\varphi\del\varphi)]_x=\pdv{}{y}\left(\varphi\pdv{\varphi}{z}\right)-\pdv{}{z}\left(\varphi\pdv{\varphi}{y}\right)
\]
\[
=\pdv{\varphi}{y}\pdv{\varphi}{z}+\varphi\ppdv{\varphi}{y}{z}-\pdv{\varphi}{z}\pdv{\varphi}{y}-\varphi\ppdv{\varphi}{z}{y}=0
\]
The same is for the $y$ and $z$ components. Therefore,
$\del\times(\varphi\del\varphi)=0$

\paragraph{3.6.11}
(a) If $\V{F}=\V{G}+k$, $k$ is a constant, then $\del\times\V{F}=\del\times\V{G}$ beacuse $\pdv{k}{x_i}=0$

(b) If $\V{F}=\V{G}+\del\varphi$, then $\del\times\V{F}=\del\times\V{G}+\del\times(\del\varphi)=\del\times\V{G}$ because $\del\times(\del\varphi)=0$.

\paragraph{3.6.12}
From Exercise 3.6.7, $\V{v}\times(\del\times\V{v})=\frac{1}{2}\del(v^2)-(\V{v}\cdot\del)\V{v}$, so \[-\del\times\big(\V{v}\times(\del\times\V{v})\big)=-\frac{1}{2}\del\times\del(v^2)+\del\times\big((\V{v}\cdot\del)\V{v} \big)=\del\times\big((\V{v}\cdot\del)\V{v} \big)\]
because $\del\times\del(v^2)=0$.

\paragraph{3.6.13}
From Exercise 3.5.9,
\[\del\cdot\big((\del u)\times(\del v) \big)=(\del v)\cdot\big(\del\times(\del u) \big)-(\del u)\cdot\big(\del\times(\del v) \big)=0\]
because $\del\times(\del u)=0$ and $\del\times(\del v)=0$

\paragraph{3.6.14}
$\del\cdot\del\varphi=\del^2\varphi=0$, and $\del\times\del\varphi=0$ for any $\varphi$, so $\del\varphi$ is both solenoidal and irrotational.

\paragraph{3.6.15}
By Equation (3.70), $\del\times(\del\times\V{A})=\del(\del\cdot\V{A})-(\del\cdot\del)\V{A}$, so the equation becomes\\ $\del(\del\cdot\V{A})-(\del\cdot\del)\V{A}-k^2\V{A}=0$. Let $\del\cdot$ operate on both side of the equation, and note that\\ $\del\cdot\big((\del\cdot\del)\V{A} \big)=(\del\cdot\del)(\del\cdot\V{A})$ because $\pdv{}{x_i}\left(\pdv{^2A_k}{x_j^2} \right)=\pdv{^2}{x_j^2}\left(\pdv{A_k}{x_i} \right)$. So the equation becomes\\ $(\del\cdot\del)(\del\cdot\V{A})-(\del\cdot\del)(\del\cdot\V{A})-k^2(\del\cdot\V{A})=-k^2(\del\cdot\V{A})=0$, so $\del\cdot\V{A}=0$. Substituting back to the second equation, we get  $(\del\cdot\del)\V{A}+k^2\V{A}=\del^2\V{A}+k^2\V{A}=0$

\paragraph{3.6.16}
Let $\Psi=\frac{k}{2}\Phi^2$, then \[\del^2\Psi=\frac{k}{2}\left(\pdv{^2\Phi^2}{x^2}+\pdv{^2\Phi^2}{y^2}+\pdv{^2\Phi^2}{z^2} \right)\]
\[
=\frac{k}{2}\left(\pdv{}{x}\left(2\Phi\pdv{\Phi}{x}\right)+\pdv{}{y}\left(2\Phi\pdv{\Phi}{y}\right)+\pdv{}{z}\left(2\Phi\pdv{\Phi}{z}\right)   \right)
\]
\[=\frac{k}{2}\left(2\pdv{\Phi}{x}\pdv{\Phi}{x}+2\Phi\pdv{^2\Phi}{x^2}+2\pdv{\Phi}{y}\pdv{\Phi}{y}+2\Phi\pdv{^2\Phi}{y^2}+2\pdv{\Phi}{z}\pdv{\Phi}{z}+2\Phi\pdv{^2\Phi}{z^2}\right)\]
\[
=k\left(\left(\pdv{\Phi}{x}\right)^2+\left(\pdv{\Phi}{y}\right)^2+\left(\pdv{\Phi}{z}\right)^2 \right)
=k|\del\Phi|^2
\]
because $\del^2\Phi=\pdv{^2\Phi}{x^2}+\pdv{^2\Phi}{y^2}+\pdv{^2\Phi}{z^2}=0$. So $\Psi=\frac{k}{2}\Phi^2$ is a solution of the equation.

\paragraph{3.6.17}
Substituting, we get
\renewcommand{\arraystretch}{1.5}
\[
\begin{pmatrix}
\frac{1}{c}\pdv{}{t}&-i\pdv{}{z}&i\pdv{}{y}\\
i\pdv{}{z}&\frac{1}{c}\pdv{}{t}&-i\pdv{}{x}\\
-i\pdv{}{y}&i\pdv{}{x}&\frac{1}{c}\pdv{}{t}
\end{pmatrix}
\begin{pmatrix}
B_x-i\frac{E_x}{c}\\
B_y-i\frac{E_y}{c}\\
B_z-i\frac{E_z}{c}\\
\end{pmatrix}
\]
\[
=\begin{pmatrix}
\left(\frac{1}{c}\pdv{B_x}{t}-\frac{1}{c}\pdv{E_y}{z}+\frac{1}{c}\pdv{E_z}{y} \right)+i\left(-\frac{1}{c^2}\pdv{E_x}{t}-\pdv{B_y}{z}+\pdv{B_z}{y} \right)\\
\left(\frac{1}{c}\pdv{B_y}{t}-\frac{1}{c}\pdv{E_z}{x}+\frac{1}{c}\pdv{E_x}{z} \right)+i\left(-\frac{1}{c^2}\pdv{E_y}{t}-\pdv{B_z}{x}+\pdv{B_x}{z} \right)\\
\left(\frac{1}{c}\pdv{B_z}{t}-\frac{1}{c}\pdv{E_x}{y}+\frac{1}{c}\pdv{E_y}{x} \right)+i\left(-\frac{1}{c^2}\pdv{E_z}{t}-\pdv{B_x}{y}+\pdv{B_y}{x} \right)\\
\end{pmatrix}=0
\]
The real and imaginary part must to be zero, respectively. So the three equations from the real part form  $\del\times\V{E}=-\pdv{\V{B}}{t}$, and the three equations from the imaginary part form  $\del\times\V{B}=\frac{1}{c^2}\pdv{\V{E}}{t}$.

\paragraph{3.6.18}
Note that $\boldsymbol{\sigma}_i^2=\boldsymbol{1}_2$, $\boldsymbol{\sigma}_i\boldsymbol{\sigma}_j=i\boldsymbol{\sigma}_k$. So
\[
(\boldsymbol{\sigma}\cdot\V{a})(\boldsymbol{\boldsymbol{\sigma}}\cdot\V{b})\]
\[
=(a_x\boldsymbol{\sigma}_1+a_y\boldsymbol{\sigma}_2+a_z\boldsymbol{\sigma}_3)(b_x\boldsymbol{\sigma}_1+b_y\boldsymbol{\sigma}_2+b_z\boldsymbol{\sigma}_3)
\]
\[
=(a_xb_x+a_yb_y+a_zb_z)\boldsymbol{1}_2+(a_xb_y-a_yb_x)i\boldsymbol{\sigma}_3+(a_yb_z-a_zb_y)i\boldsymbol{\sigma}_1+(a_zb_x-a_xb_z)i\boldsymbol{\sigma}_2
\]
\[
=(\V{a}\cdot\V{b})\boldsymbol{1}_2+i\boldsymbol{\sigma}\cdot(\V{a}\times\V{b})
\]

\section*{3.7 Vector Integration}

\paragraph{3.7.1}
The total vector area is
\[
\int d\boldsymbol{\sigma}=\frac{1}{2}\V{B}\times\V{A}+\frac{1}{2}\V{C}\times\V{B}+\frac{1}{2}\V{A}\times\V{C}+\frac{1}{2}(\V{C}-\V{B})\times(\V{A}-\V{B})=0
\]

\paragraph{3.7.2}
(a) $x^2+y^2=1$, so $y=\sqrt{1-x^2}$\,; $2xdx+2ydy=0$, so $dy=-\frac{x}{y}dx=-\frac{x}{\sqrt{1-x^2}}dx$. So
\[
w=\int(-\V{F})\cdot d\V{r}\]
\[=\int\frac{y}{x^2+y^2}dx+\int\frac{-x}{x^2+y^2}dy\]
\[=\int_1^{-1}\sqrt{1-x^2}\,dx+\int_1^{-1}\frac{x^2}{\sqrt{1-x^2}}\,dx\]
\[=\int_1^{-1}\frac{1}{\sqrt{1-x^2}}\,dx\]
\[=\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}}\frac{1}{\sqrt{1-\sin^2\theta}}\cos\theta\,d\theta=-\pi\]

(b) $x^2+y^2=1$, so $y=-\sqrt{1-x^2}$\,; $2xdx+2ydy=0$, so $dy=-\frac{x}{y}dx=\frac{x}{\sqrt{1-x^2}}dx$. So
\[
w=\int(-\V{F})\cdot d\V{r}\]
\[=\int\frac{y}{x^2+y^2}dx+\int\frac{-x}{x^2+y^2}dy\]
\[=-\int_1^{-1}\sqrt{1-x^2}\,dx-\int_1^{-1}\frac{x^2}{\sqrt{1-x^2}}\,dx\]
\[=\int_1^{-1}\frac{-1}{\sqrt{1-x^2}}\,dx\]
\[=\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}}\frac{-1}{\sqrt{1-\sin^2\theta}}\cos\theta\,d\theta=\pi\]

\paragraph{3.7.3}
Choose the path $(1,1)\rightarrow(1,3)\rightarrow(3,3)$. Then
\[
w=\int_1^3\V{F}(x,1)\cdot(dx\VE_x)+\int_1^3\V{F}(3,y)\cdot(dy\VE_y)
\]
\[
=\int_1^3(x-1)\,dx+\int_1^3(3+y)\,dy
\]
\[
=2+10=12
\]

\paragraph{3.7.4}
$\oint\V{r}\cdot d\V{r}=\oint(xdx+ydy+zdz)=(\frac{x^2}{2}+\frac{y^2}{2}+\frac{z^2}{2})\big|_{\V{a}}^{\V{a}}=0$ where $\V{a}$ is the starting point.

\paragraph{3.7.5}
For the surfaces parallel to $yz$ surface, \[\int\displaylimits_{S_{yz}}\V{r}\cdot d\boldsymbol{\sigma}=\int\displaylimits_{S_{yz}}(x\VE_x+y\VE_y+z\VE_z)\cdot(d\sigma\VE_x)=\int\displaylimits_{S_{yz}}(xd\sigma)=x\]
equals to $1$ at $x=1$ and $0$ at $x=0$. The same is for $y$ and $z$, so \[\frac{1}{3}\int\displaylimits_S\V{r}\cdot d\boldsymbol{\sigma}=\frac{1}{3}(1+1+1)=1\]

\section*{3.8 Integral Theorems}

\paragraph{3.8.1}
Let $\V{a}$ be a constant vector, then \[\V{a}\cdot\oint\displaylimits_{\partial V}d\boldsymbol{\sigma}=\oint\displaylimits_{\partial V}\V{a}\cdot d\boldsymbol{\sigma}=\int\displaylimits_V(\del\cdot\V{a})d\tau=0\]
Because $\V{a}$ can be in arbitrary direction, $\oint\displaylimits_{\delta V}d\boldsymbol{\sigma}$ must be zero.

\paragraph{3.8.2}
\[
\frac{1}{3}\oint\displaylimits_S \V{r}\cdot d\boldsymbol{\sigma}=\frac{1}{3}\int\displaylimits_V (\del\cdot\V{r})d\tau=\frac{1}{3}\int\displaylimits_V 3d\tau=V
\]

\paragraph{3.8.3}
\[
\oint\displaylimits_S\V{B}\cdot d\boldsymbol{\sigma}=\oint\displaylimits_S(\del\times\V{A})\cdot d\boldsymbol{\sigma}=\int\displaylimits_V\del\cdot(\del\times\V{A})d\tau=0
\]
because the divergence of a curl vanishes.

\paragraph{3.8.4}
$\del\cdot(\varphi\V{E})=(\del\varphi)\cdot\V{E}+\varphi(\del\cdot\V{E})$, so $\rho\varphi=\varepsilon_0(\del\cdot\V{E})\varphi=\varepsilon_0\del\cdot(\varphi\V{E})-\varepsilon_0(\del\varphi)\cdot\V{E}$, and
\[
\int \rho\varphi\,d\tau=\varepsilon_0\int\del\cdot(\varphi\V{E})\,d\tau-\varepsilon_0\int(\del\varphi)\cdot\V{E}\,d\tau
\]
\[
=\varepsilon_0\oint\varphi\V{E}\cdot d\boldsymbol{\sigma}+\varepsilon_0\int E^2 d\tau
\]
$\varphi$ vanishes at least as fast as $r^{-1}$, so $\V{E}=-\del\varphi$ vanishes at least as fast as $r^{-2}$, and $\varphi\V{E}$ vanishes at least as fast as $r^{-3}$. But $d\boldsymbol{\sigma}$ is in the order of $r^2$, so $\oint\varphi\V{E}\cdot d\boldsymbol{\sigma}$ vanishes at large $r$. Therefore,
\[
\int \rho\varphi\,d\tau=\varepsilon_0\int E^2 d\tau
\]

\paragraph{3.8.5}
$\del\cdot\V{J}=0$ because it is steady-state current distribution. So $\del\cdot(x_i\V{J})=(\del x_i)\cdot\V{J}+x_i(\del\cdot\V{J})=(\del x_i)\cdot\V{J}=\VE_i\cdot\V{J}=J_i$. So
\[
\int J_i d\tau=\int\del\cdot(x_i\V{J})d\tau
=\oint x_i\V{J}\cdot d\boldsymbol{\sigma}=0
\]
because $\V{J}$ vanishes on the surface. So
\[
\int\V{J}d\tau=\sum\int J_i d\tau\,\VE_i=0
\]

\paragraph{3.8.6}
\[
\frac{1}{2}\oint \V{t}\cdot d\boldsymbol{\lambda}=\frac{1}{2}\int(\del\times\V{t})\cdot d\boldsymbol{\sigma}=\frac{1}{2}\int2\VE_z\cdot d\boldsymbol{\sigma}=A
\]

\paragraph{3.8.7}
(a) $\oint\V{r}\times d\V{r}=\oint(xdy-ydx)\VE_z=2A\VE_z$ from Exercise 3.8.6 .
\medskip

(b) $\V{r}=a\cos\theta\VE_x+b\sin\theta\VE_y$, so $d\V{r}=-a\sin\theta d\theta\VE_x+b\cos\theta d\theta\VE_y$, and
\[
\oint\V{r}\times d\V{r}=\int_0^{2\pi}ab(\cos^2\theta+\sin^2\theta)d\theta=2\pi ab=2A
\]
so the area of the ellipse is $\pi ab$.

\paragraph{3.8.8}
\[
\oint\V{r}\times d\V{r}=-\oint d\V{r}\times\V{r}
\]
\[
=-\int\displaylimits_S(d\boldsymbol{\sigma}\times\del)\times\V{r}
\]
\[
=-\int\displaylimits_S\left((dxdy\VE_z)\times(\pdv{}{x}\VE_x+\pdv{}{y}\VE_y+\pdv{}{z}\VE_z)\right)\times\V{r}
\]
\[
=-\int\displaylimits_S(-dxdy\pdv{}{y}\VE_x+dxdy\pdv{}{x}\VE_y)\times(x\VE_x+y\VE_y)
\]
\[
=-\int\displaylimits_S -2dxdy=2A
\]

\paragraph{3.8.9}
\[
\oint u\del v\cdot d\boldsymbol{\lambda}+\oint v\del u\cdot d\boldsymbol{\lambda}
=\oint\del(uv)\cdot d\boldsymbol{\lambda}
=\int\del\times(\del{uv})\cdot d\boldsymbol{\sigma}=0
\]
so $\oint u\del v\cdot d\boldsymbol{\lambda}=-\oint v\del u\cdot d\boldsymbol{\lambda}$.

\paragraph{3.8.10}
$\del\times(f\V{V})=(\del f)\times\V{V}+f(\del\times\V{V})$ from Eq. 3.73 . So
\[
\oint u\del v\cdot d\boldsymbol{\lambda}=\int\displaylimits_S\del\times(u\del v)\cdot d\boldsymbol{\sigma}
\]
\[
=\int\displaylimits_S(\del u)\times(\del v)\cdot d\boldsymbol{\sigma}+\int\displaylimits_S u(\del\times(\del v))\cdot d\boldsymbol{\sigma}
\]
\[
=\int\displaylimits_S(\del u)\times(\del v)\cdot d\boldsymbol{\sigma}
\]
because $\del\times(\del v)=0$.

\paragraph{3.8.11}
Let $\V{a}$ be a constant vector, then
\[
\int\displaylimits_V\del\cdot(\V{a}\times\V{P})d\tau=\int\displaylimits_V(\del\times\V{a})\cdot\V{P} d\tau-\int\displaylimits_V\V{a}\cdot(\del\times\V{P})d\tau=-\V{a}\cdot\int\displaylimits_V\del\times\V{P}d\tau
\]
Also,
\[
\int\displaylimits_V\del\cdot(\V{a}\times\V{P})d\tau=\oint\displaylimits_{\partial V}\V{a}\times\V{P}\cdot d\boldsymbol{\sigma}=\oint\displaylimits_{\partial V}\V{P}\times d\boldsymbol{\sigma}\cdot\V{a}=\V{a}\cdot\oint\displaylimits_{\partial V}\V{P}\times d\boldsymbol{\sigma}
\]
So $\V{a}\cdot\left(\oint\displaylimits_{\partial V}\V{P}\times d\boldsymbol{\sigma}+\int\displaylimits_V\del\times\V{P}d\tau\right)=0$. Because $\V{a}$ can be in arbitrary direction, $\oint\displaylimits_{\partial V}\V{P}\times d\boldsymbol{\sigma}+\int\displaylimits_V\del\times\V{P}d\tau$ must be zero, and therefore \[\oint\displaylimits_{\partial V}d\boldsymbol{\sigma}\times\V{P}=\int\displaylimits_V\del\times\V{P}d\tau\]

\paragraph{3.8.12}
Let $\V{a}$ be a constant vector, then
\[
\oint\displaylimits_{\partial S}(\V{a}\varphi)\cdot d\V{r}=\V{a}\cdot\oint\displaylimits_{\partial S}\varphi d\V{r}
\]
Also,
\[
\oint\displaylimits_{\partial S}(\V{a}\varphi)\cdot d\V{r}=\int\displaylimits_S\del\times(\varphi\V{a})\cdot d\boldsymbol{\sigma}=\int\displaylimits_S(\del\varphi)\times\V{a}\cdot d\boldsymbol{\sigma}=\int\displaylimits_S d\boldsymbol{\sigma}\times(\del\varphi)\cdot\V{a}=\V{a}\cdot\int\displaylimits_S d\boldsymbol{\sigma}\times(\del\varphi)
\]
so $\V{a}\cdot\left(\int\displaylimits_S d\boldsymbol{\sigma}\times(\del\varphi)-\oint\displaylimits_{\partial S}\varphi d\V{r} \right)=0$.  Because $\V{a}$ can be in arbitrary direction, $\int\displaylimits_S d\boldsymbol{\sigma}\times(\del\varphi)-\oint\displaylimits_{\partial S}\varphi d\V{r}$ must be zero, and therefore
\[
\int\displaylimits_S d\boldsymbol{\sigma}\times(\del\varphi)= \oint\displaylimits_{\partial S}\varphi d\V{r}
\]

\paragraph{3.8.13}
Let $\V{a}$ be a constant vector, then
\[
\oint\displaylimits_{\partial S
}(\V{a}\times\V{P})\cdot d\V{r}=\oint\displaylimits_{\partial S
}(\V{P}\times d\V{r})\cdot\V{a}=\V{a}\cdot\oint\displaylimits_{\partial S
}\V{P}\times d\V{r}
\]
Because $\V{a}$ is a constant, $\del\times(\V{a}\times\V{P})=\del_P\times(\V{a}\times\V{P})$, with $\del_P$ acting only on $\V{P}$. So $\del_P$ needs to go before $\V{P}$, but can go before and after $\V{a}$. So \[\del_P\times(\V{a}\times\V{P})\cdot d\boldsymbol{\sigma}=d\boldsymbol{\sigma}\times\del_P\cdot(\V{a}\times\V{P})=(d\boldsymbol{\sigma}\times\del_P)\cdot\V{a}\times\V{P}=-\V{a}\cdot(d\boldsymbol{\sigma}\times\del_P)\times\V{P}=-\V{a}\cdot(d\boldsymbol{\sigma}\times\del)\times\V{P}\]
and
\[
\oint\displaylimits_S(\V{a}\times\V{P})\cdot d\V{r}=\int\displaylimits_S\del\times(\V{a}\times\V{P})\cdot d\boldsymbol{\sigma}=-\int\displaylimits_S\V{a}\cdot(d\boldsymbol{\sigma}\times\del)\times\V{P}=-\V{a}\cdot\int\displaylimits_S(d\boldsymbol{\sigma}\times\del)\times\V{P}
\]
so $\V{a}\cdot\left(\int\displaylimits_S(d\boldsymbol{\sigma}\times\del)\times\V{P}+\oint\displaylimits_{\partial S
}\V{P}\times d\V{r} \right)=0$. Because $\V{a}$ can be in arbitrary direction, $\int\displaylimits_S(d\boldsymbol{\sigma}\times\del)\times\V{P}+\oint\displaylimits_{\partial S
}\V{P}\times d\V{r}$ must be zero, and therefore
\[
\int\displaylimits_S(d\boldsymbol{\sigma}\times\del)\times\V{P}=\oint\displaylimits_{\partial S
}d\V{r}\times\V{P}
\]

\section*{3.9 Potential Theory}

\paragraph{3.9.1}
$\V{F}=r^{2n}\V{r}$
\medskip

(a) $\del\cdot\V{F}=(\del r^{2n})\cdot\V{r}+r^{2n}(\del\cdot\V{r})=2nr^{2n-1}\frac{\V{r}}{r}\cdot\V{r}+r^{2n}(3)=(2n+3)r^{2n}$
\medskip

(b) $\del\times\V{F}=(\del r^{2n})\times\V{r}+r^{2n}(\del\times\V{r})=2nr^{2n-1}\frac{\V{r}}{r}\times\V{r}+0=0$
\medskip

(c) $\del\times\V{F}=0$, so the scalar potential exists. $\int_a^b\V{F}\cdot d\V{r}=-\int_a^b\del\varphi\cdot d\V{r}=-\varphi\big|_a^b=\varphi(a)-\varphi(b)$. Take the path $(0,0,0)\rightarrow(x,0,0)\rightarrow(x,y,0)\rightarrow(x,y,z)$. Then
\[
\int_{(0,0,0)}^{(x,y,z)}\V{F}\cdot d\V{r}=\int_0^x x^{2n}xdx+\int_0^y(x^2+y^2)^nydy+\int_0^z(x^2+y^2+z^2)^nzdz
\]
\[
=\frac{x^{2n+2}}{2n+2}-0+\frac{(x^2+y^2)^{n+1}}{2(n+1)}-\frac{(x^2)^{n+1}}{2(n+1)}+\frac{(x^2+y^2+z^2)^{n+1}}{2(n+1)}-\frac{(x^2+y^2)^{n+1}}{2(n+1)}
\]
\[
=\frac{r^{2n+2}}{2n+2}=\varphi(0,0,0)-\varphi(x,y,z)
\]
when $n\neq-1$. Defining $\varphi(0,0,0)$ to be zero, then $\varphi(x,y,z)=-\frac{r^{2n+2}}{2n+2}$.
\medskip

(d) If $n=-1$,
\[
\int_{(1,1,1)}^{(x,y,z)}\V{F}\cdot d\V{r}=\int_1^x \frac{1}{x}dx+\int_1^y\frac{y}{x^2+y^2}dy+\int_1^z\frac{z}{x^2+y^2+z^2}dz
\]
\[
=\ln|x|+\frac{1}{2}\ln|\frac{x^2+y^2}{x^2}|+\frac{1}{2}\ln|\frac{x^2+y^2+z^2}{x^2+y^2}|
\]
\[
=\frac{1}{2}\ln|x^2+y^2+z^2|\]
\[=\ln r=\varphi(1,1,1)-\varphi(\V{r})
\]
Defining $\varphi(1,1,1)=0$, then $\varphi(\V{r})=-\ln r$ diverges at both the origin and infinity.

\paragraph{3.9.2}
Applying Gauss law, at $r\leq a$, $\oint\V{E}\cdot d\boldsymbol{\sigma}=E4\pi r^2=\frac{Q}{\varepsilon_0}\frac{r^3}{a^3}$, so $E=\frac{Qr}{4\pi\varepsilon_0a^3}$; at $r>a$, $\oint\V{E}\cdot d\boldsymbol{\sigma}=E4\pi r^2=\frac{Q}{\varepsilon_0}$, so $E=\frac{Qr}{4\pi\varepsilon_0r^2}$. Defining the potential to be zero at $r\to\infty$, then $\int_r^\infty \V{E}\cdot d\V{r}=\varphi(r)-\varphi(\infty)=\varphi(r)$. At $r>a$, $\varphi(r)=\int_r^\infty\frac{Q}{4\pi\varepsilon_0r^2}dr=\frac{Q}{4\pi\varepsilon_0r}$; at $r\leq a$, $\varphi(r)=\int_r^a\frac{Qr}{4\pi\varepsilon_0a^3}dr+\int_a^\infty\frac{Q}{4\pi\varepsilon_0r^2}dr=\frac{Q}{4\pi\varepsilon_0a}\left(\frac{1}{2}-\frac{1}{2}\frac{r^2}{a^2} \right)+\frac{Q}{4\pi\varepsilon_0a}=\frac{Q}{4\pi\varepsilon_0a}\left(\frac{3}{2}-\frac{1}{2}\frac{r^2}{a^2} \right)$. So the electrostatic potential is
\[
\varphi(r)=
\begin{cases}
\frac{Q}{4\pi\varepsilon_0a}\left(\frac{3}{2}-\frac{1}{2}\frac{r^2}{a^2} \right), & r\leq a\\
\frac{Q}{4\pi\varepsilon_0r}, & r>a
\end{cases}
\]

\paragraph{3.9.3}
It can be verified that $\del\times\V{F}=0$, so the potential exists.
\[
\int_{(0,0,0)}^{(x,y,z)}\V{F}\cdot d\V{r}=\varphi(0,0,0)-\varphi(x,y,z)
\]
\[
=\frac{GMm}{R^3}\int_{(0,0,0)}^{(x,y,z)}(-xdx-ydy+2zdz)
\]
\[
=\frac{GMm}{R^3}\left(-\frac{x^2}{2}-\frac{y^2}{2}+z^2 \right)=\varphi(0,0,0)-\varphi(x,y,z)
\]
Define $\varphi(0,0,0)$ to be zero, then \[\varphi(x,y,z)=-\frac{GMm}{R^3}\left(-\frac{x^2}{2}-\frac{y^2}{2}+z^2 \right)\]

\paragraph{3.9.4}
$\del\cdot\V{B}=\frac{\mu_0I}{2\pi}\left(\frac{2xy}{(x^2+y^2)^2}-\frac{2xy}{(x^2+y^2)^2} \right)=0$, so the vector potential exists. If $\del\times\V{A}'=\V{B}$, and let $\del\varphi=-A_x'\VE_x$, then $\V{A}=\V{A}'+\del\varphi$ is also a vector potential with zero $x$-component because  $\del\times\V{A}=\del\times\V{A}'+\del\times(\del\varphi)=\V{B}$. Let $\V{A}=A_y\VE_y+A_z\VE_z$, then
\[
\del\times\V{A}=\left(\pdv{A_z}{y}-\pdv{A_y}{z} \right)\VE_x-\pdv{A_z}{x}\VE_y+\pdv{A_y}{x}\VE_z
\]
\[
=-\frac{\mu_0I}{2\pi}\frac{y}{x^2+y^2}\VE_x+\frac{\mu_0I}{2\pi}\frac{x}{x^2+y^2}\VE_y
\]
For the $y$-component, $-\pdv{A_z}{x}=\frac{\mu_0I}{2\pi}\frac{x}{x^2+y^2}$, so $A_z=-\frac{\mu_0I}{4\pi}\ln(x^2+y^2)+C_1(y,z)$. For the $z$-component, $\pdv{A_y}{x}=0$, so $A_y=C_2(y,z)$. Substituting into the equation of the $x$-component, we get $-\frac{\mu_0I}{2\pi}\frac{y}{x^2+y^2}+\pdv{C_1}{y}-\pdv{C_2}{z}=-\frac{\mu_0I}{2\pi}\frac{y}{x^2+y^2}$, so the equation will be satisfied if we simply choose $C_1=C_2=0$. Therefore,
\[
\V{A}=-\frac{\mu_0I}{4\pi}\ln(x^2+y^2)\VE_z
\]
is a vector potential of $\V{B}$.

\paragraph{3.9.5}
$\del\cdot\V{B}=3\frac{1}{r^3}-3\frac{x^2+y^2+z^2}{r^5}=0$, so the vector potential exist. As in Exercise 3.9.4, we can make one component of the vector potential be zero, and we choose the $z$-component. So $\V{A}=A_x\VE_x+A_y\VE_y$, and
\[
\del\times\V{A}=-\pdv{A_y}{z}\VE_x+\pdv{A_x}{z}\VE_y+\left(\pdv{A_y}{x}-\pdv{A_x}{y} \right)\VE_z
\]
\[
=\frac{x}{r^3}\VE_x+\frac{y}{r^3}\VE_y+\frac{z}{r^3}\VE_z
\]
so
\[
-\pdv{A_y}{z}=\frac{x}{r^3}\;\xrightarrow{integrating}\;A_y=\frac{-xz}{(x^2+y^2)r}+C_1(x,y)
\]
\[
\pdv{A_x}{z}=\frac{y}{r^3}\;\xrightarrow{integrating}\;A_x=\frac{yz}{(x^2+y^2)r}+C_2(x,y)
\]
Substituted into $\pdv{A_y}{x}-\pdv{A_x}{y}=\frac{z}{r^3}$, we get
\[
\frac{z}{r^3}+\pdv{C_1}{x}-\pdv{C_2}{y}=\frac{z}{r^3}
\]
which will be satisfied if we simply choose $C_1=C_2=0$. Therefore,
\[
\V{A}=\frac{yz}{(x^2+y^2)r}\VE_x-\frac{xz}{(x^2+y^2)r}\VE_y
\]
is a solution of $\del\times\V{A}=\V{B}$

\paragraph{3.9.6}
If $\V{B}$ is a constant vector, then
\[
\del\times\V{A}=\frac{1}{2}\del\times(\V{B}\times\V{r})\]
\[=\frac{1}{2}\del_\V{r}\times(\V{B}\times\V{r})
\]
\[
=\frac{1}{2}[\V{B}(\del_\V{r}\cdot\V{r})-(\V{B}\cdot\del_\V{r})\V{r}]
\]
\[
=\frac{1}{2}[3\V{B}-\V{B}]=\V{B}
\]
So the two equations are satisfied by any constant vector $\V{B}$.

\paragraph{3.9.7}
(a) $\del\cdot\V{B}=\del\cdot((\del u)\times(\del v))=(\del v)\cdot(\del\times(\del u))-(\del u)\cdot(\del\times(\del v))=0$ because $\del\times(\del u)=0$ and $\del\times(\del v)=0$.
\medskip

(b)
\[
\del\times\V{A}=\frac{1}{2}\del\times(u\del v)-\frac{1}{2}\del\times(v\del u)
\]
\[
=\frac{1}{2}(\del u)\times(\del v)+\frac{1}{2}u(\del\times(\del v))-\frac{1}{2}(\del v)\times(\del u)-\frac{1}{2}v(\del\times(\del u))
\]
\[
=(\del u)\times(\del v)=\V{B}
\]

\paragraph{3.9.8}
Let $\V{A}'=\V{A}+\del\varphi$, then $\V{B}'=\del\times(\V{A}+\del\varphi)=\del\times\V{A}=\V{B}$ because $\del\times(\del\varphi)=0$, so the left side of the equation is unchanged. $\oint\V{A}'\cdot d\V{r}=\oint \V{A}\cdot d\V{r}+\oint\del\varphi\cdot d\V{r}=\oint \V{A}\cdot d\V{r}$ because $\oint\del\varphi\cdot d\V{r}=\oint d\varphi=0$, so the right side of the equation is unchanged.

\paragraph{3.9.9}
Choose point $P$ to be the origin of the coordinate system $(0,0,0)$. Let $u=\frac{1}{r}$, $v=\varphi$, and apply Green's theorem Eq. 3.85,
\[
\int\displaylimits_V\left(\frac{1}{r}\del^2\varphi-\varphi\del^2(\frac{1}{r}) \right)d\tau=\oint\displaylimits_{\partial V}\left(\frac{1}{r}\del\varphi-\varphi\del(\frac{1}{r}) \right)\cdot d\boldsymbol{\sigma}
\]
where the volume $V$ is a sphere centered at $(0,0,0)$ with radius $r$. Because there are no charges on or within the sphere, we have $\del^2\varphi=0$, and by Eq. 3.120, $\del^2(\frac{1}{r})=-4\pi\delta(\V{r})$. So the left side of the equation equals to $\int\displaylimits_V4\pi\varphi\delta(\V{r})d\tau=4\pi\varphi(0)$. As for the right side, \[\oint\displaylimits_{\partial V}\frac{1}{r}\del\varphi\cdot d\boldsymbol{\sigma}=\frac{1}{r}\int\displaylimits_V\del\cdot(\del\varphi)d\tau=\frac{1}{r}\int\displaylimits_V\del^2\varphi d\tau=0\] \[-\oint\displaylimits_V\varphi\del(\frac{1}{r})\cdot d\boldsymbol{\sigma}=\oint\displaylimits_V\varphi\frac{1}{r^2}\hat{\V{r}}\cdot d\boldsymbol{\sigma}=\frac{\oint\displaylimits_V\varphi d\sigma}{r^2}\]
so the equation becomes $4\pi\varphi(0)=\frac{\oint\displaylimits_V\varphi d\sigma}{r^2}$, so $\varphi(0)=\frac{\oint\displaylimits_V\varphi d\sigma}{4\pi r^2}$, which means the potential at $P$ is the average of the potential over the spherical surface centered on $P$ with radius $r$.

\paragraph{3.9.10}
$\del\times\V{B}=\mu\V{J}$ and $\V{B}=\del\times\V{A}$, so $\del\times(\del\times\V{A})=\mu\V{J}$. But $\del\times(\del\times\V{A})=\del(\del\cdot\V{A})-(\del\cdot\del)\V{A}=-\del^2\V{A}$ because $\del\cdot\V{A}=0$, so $\del^2\V{A}=-\mu\V{J}$.

\paragraph{3.9.11}
From the Maxwell's equations, $\del\times\V{B}=\mu_0\V{J}+\frac{1}{c^2}\pdv{\V{E}}{t}$, so
\[
\del\times(\del\times\V{A})
\]
\[
=\del(\del\cdot\V{A})-(\del\cdot\del)\V{A}=\mu_0\V{J}+\frac{1}{c^2}\pdv{\V{E}}{t}
\]
Using the Lorentz gauge Eq. 3.109, \[
\del(\del\cdot\V{A})=\del(-\frac{1}{c^2}\pdv{\varphi}{t})=-\frac{1}{c^2}\pdv{}{t}(\del\varphi)=\frac{1}{c^2}\pdv{}{t}(\V{E}+\pdv{\V{A}}{t})=\frac{1}{c^2}\pdv{\V{E}}{t}+\frac{1}{c^2}\pdv{^2\V{A}}{t^2}
\]
Substitute into the first equation, we get
\[
\frac{1}{c^2}\pdv{^2\V{A}}{t^2}-\del^2\V{A}=\mu_0\V{J}
\]

\paragraph{3.9.12}
As in Exercise 3.9.4, we can make one component of the vector potential be zero, so we make $\V{A}=(A_x,0,A_z)$. So
\[
\del\times\V{A}=\pdv{A_z}{y}\VE_x+\left(\pdv{A_x}{z}-\pdv{A_z}{x} \right)\VE_y-\pdv{A_x}{y}\VE_z=B_x\VE_x+B_y\VE_y+B_z\VE_z
\]
For the $x$- and $z$-components,
\[
\pdv{A_z}{y}=B_x\;\xrightarrow{integrating}\;A_z=\int_{y_0}^yB_x(x,y,z) dy+C_1(x,y)
\]
\[
-\pdv{A_x}{y}=B_z\;\xrightarrow{integrating}\;A_x=-\int_{y_0}^yB_z(x,y,z) dy+C_2(x,y)
\]
Substitute into the equation of $y$-component,
\[
-\pdv{}{z}\int_{y_0}^yB_zdy+\pdv{C_2}{z}-\pdv{}{x}\int_{y_0}^yB_xdy-\pdv{C_1}{x}=B_y
\]
Let $y=y_0$, and let $C_2=0$, then
\[
-\pdv{C_1}{x}=B_y(x,y_0,z)\;\xrightarrow{integrating}\;C_1=-\int_{x_0}^xB_y(x,y_0,z) dx
\]
Therefore,
\[
\V{A}=-\VE_x\int_{y_0}^yB_z(x,y,z) dy+\VE_z\left[\int_{y_0}^yB_x(x,y,z) dy-\int_{x_0}^xB_y(x,y_0,z) dx \right]
\]

\section*{3.10 Curvilinear Coordinates}

\paragraph{3.10.1}
(a) The surfaces of $u=constant$ are hyperbolas with $x=0$ and $y=0$ as asymptotes when viewing from the $z$-axis.  The surfaces of $v=constant$ are hyperbolas with $y=x$ and $y=-x$ as asymptotes when viewing from the $z$-axis. The surfaces of $z=constant$ are surfaces parallel to the $x$-$y$ plane.

(b)

\begin{figure}[h]
    \centering
    \includegraphics[width=0.5\textwidth]{B.PNG}
    \caption{The purple lines are $xy=0$, $xy=1$, $xy=2$} the orange lines are $x^2-y^2=0$, $x^2-y^2=1$, $x^2-y^2=2$.
    \label{fig:B}
\end{figure}

(c) Take the derivative of $xy=u$ we get $ydx+xdy=0$. That is, $(y,x)\cdot(dx,dy)=0$, so $(y,x)$ is a normal vector of $xy=u$ and therefore in the direction of $\VE_u$. Because $\pdv{u}{x}=y$, so $\VE_u$ should be in the direction of $(y,x)$, not $(-y,-x)$. Normalizing, we get $\VE_u=\frac{y}{\sqrt{x^2+y^2}}\VE_x+\frac{x}{\sqrt{x^2+y^2}}\VE_y$. By a similar process, from $2xdx-2ydy=0$, we get $\VE_v=\frac{x}{\sqrt{x^2+y^2}}\VE_x+\frac{-y}{\sqrt{x^2+y^2}}\VE_y$.

(d) $\VE_u\times\VE_v=\frac{-x^2-y^2}{x^2+y^2}\VE_z=-\VE_z$, so it is a left-handed system.

\paragraph{3.10.2}

\,

\begin{figure}[h]
    \centering
    \includegraphics[width=0.5\textwidth]{C.PNG}
    \caption{The orange lines are $u=\frac{\pi}{6}$, $u=\frac{\pi}{4}$, $u=\frac{\pi}{3}$}the purple lines are $v=\frac{\pi}{6}$, $v=\frac{\pi}{4}$, $v=\frac{\pi}{3}$
    \label{fig:C}
\end{figure}

The unit vectors $\VE_u$ and $\VE_v$ are perpendicular to the lines, and pointing right at $x>0$ and pointing left at $x<0$.

\paragraph{3.10.3}
The unit vectors of orthogonal coordinates are perpendicular to each other, so we have $\VE_i\cdot\VE_j=\delta_{ij}$, and $\VE_i\times\VE_j=\VE_k$. Therefore, $(A_1\VE_1+A_2\VE_2+A_3\VE_3)\cdot(B_1\VE_1+B_2\VE_2+B_3\VE_3)=A_1B_1+A_2B_2+A_3B_3$, and \[(A_1\VE_1+A_2\VE_2+A_3\VE_3)\times(B_1\VE_1+B_2\VE_2+B_3\VE_3)=(A_2B_3-A_3B_2)\VE_1+(A_3B_1-A_1B_3)\VE_2+(A_1B_2-A_2B_1)\VE_3\]

\paragraph{3.10.4}
(a) $\VE_1=1\VE_1+0\VE_2+0\VE_3$. Using the divergence formula for curvilinear coordinates, \[\del\cdot\varphi=\frac{1}{h_1h_2h_3}\pdv{(h_2h_3)}{q_1}\]

(b)
\[
\del\times\VE_1=\frac{1}{h_1h_2h_3}
\begin{vmatrix}
\VE_1h_1&\VE_2h_2&\VE_3h_3\\
\pdv{}{q_1}&\pdv{}{q_2}&\pdv{}{q_3}\\
h_1&0&0
\end{vmatrix}
\]
\[
=\frac{1}{h_1h_2h_3}\left[\VE_2h_2\pdv{h_1}{q_3}-\VE_3h_3\pdv{h_1}{q_2} \right]
\]
\[
=\frac{1}{h_1}\left[\VE_2\frac{1}{h_3}\pdv{h_1}{q_3}-\VE_3\frac{1}{h_2}\pdv{h_1}{q_2} \right]
\]

\paragraph{*3.10.5}
$\VE_i\cdot\VE_i=1=\frac{1}{h_i^2}\pdv{\V{r}}{q_i}\cdot\pdv{\V{r}}{q_i}=\frac{1}{h_i^2}\left((\pdv{x}{q_1})^2+(\pdv{y}{q_1})^2+(\pdv{x}{q_1})^2 \right)$, so $h_i^2=(\pdv{x}{q_i})^2+(\pdv{y}{q_i})^2+(\pdv{x}{q_i})^2$, in agreement with Eq. 3.131.

From $h_i\VE_i=\pdv{\V{r}}{q_i}$, we can get \[\pdv{h_i}{q_j}\VE_i+h_i\pdv{\VE_i}{q_j}=\pdv{(h_i\VE_i)}{q_j}=\pdv{^2\V{r}}{q_j\partial q_i}
=\pdv{^2\V{r}}{q_i\partial q_j}=\pdv{(h_j\VE_j)}{q_i}=\pdv{h_j}{q_i}\VE_j+h_j\pdv{\VE_j}{q_i}
\]
so
\[
h_i\pdv{\VE_i}{q_j}-\pdv{h_j}{q_i}\VE_j=h_j\pdv{\VE_j}{q_i}-\pdv{h_i}{q_j}\VE_i=\V{a}
\]
If $\V{a}=0$, then $\pdv{\VE_i}{q_j}=\frac{1}{h_i}\pdv{h_j}{q_i}\VE_j$ can be proved. However, I don't know how to prove it. (I know $\V{a}\cdot\VE_i=\V{a}\cdot\VE_j=0$, and also by taking $\V{a}\cdot\V{a}$, it is equivalent to prove $\pdv{\VE_i}{q_j}\cdot\pdv{\VE_j}{q_i}=0$, however, I can't find a proof for this either)

From $\VE_i\cdot\VE_j=0$, we have $\pdv{\VE_i}{q_i}\cdot\VE_j+\VE_i\cdot\pdv{\VE_j}{q_i}=0$, so we have \[\pdv{\VE_i}{q_i}\cdot\VE_j=-\VE_i\cdot\pdv{\VE_j}{q_i}=-\VE_i\cdot\VE_i\frac{1}{h_j}\pdv{h_i}{q_j}=-\frac{1}{h_j}\pdv{h_i}{q_j}\]
Also, $\pdv{\VE_i}{q_i}\cdot\VE_j=\frac{1}{2}\pdv{(\VE_i\cdot\VE_i)}{q_i}=0$, so
\[
\pdv{\VE_i}{q_i}=\sum_{j\neq i}\VE_j(\pdv{\VE_i}{q_i}\cdot\VE_j)=-\sum_{j\neq i}\VE_j\frac{1}{h_j}\pdv{h_i}{q_j}
\]

\paragraph{3.10.6}
$\V{r}=\rho\cos\varphi\VE_x+\rho\sin\varphi\VE_y+z\VE_z$
\medskip

$\pdv{\V{r}}{\rho}=\cos\varphi\VE_x+\sin\varphi\VE_y=h_\rho\VE_\rho$, so $h_\rho=1$ and $\VE_\rho=\cos\varphi\VE_x+\sin\varphi\VE_y$

$\pdv{\V{r}}{\varphi}=-\rho\sin\varphi\VE_x+\rho\cos\varphi\VE_y=h_\varphi\VE_\varphi$, so $h_\varphi=\rho$ and $\VE_\varphi=-\sin\varphi\VE_x+\cos\varphi\VE_y$

$\pdv{\V{r}}{z}=\VE_z=h_z\VE_z$, so $h_z=1$ and $\VE_z=\VE_z$

\paragraph{3.10.7}
From Exercise 3.10.6,

$\cos\varphi\VE_\rho-\sin\varphi\VE_\varphi=(\cos^2\varphi+\sin^2\varphi)\VE_x=\VE_x$, so $\VE_x=\cos\varphi\VE_\rho-\sin\varphi\VE_\varphi$

$\sin\varphi\VE_\rho+\cos\varphi\VE_\varphi=(\sin^2\varphi+\cos^2\varphi)\VE_y=\VE_y$, so $\VE_y=\sin\varphi\VE_\rho+\cos\varphi\VE_\varphi$

$\VE_z=\VE_z$

\paragraph{3.10.8}
From exercise 3.10.6,
$\pdv{\VE_\rho}{\varphi}=-\VE_x\sin\varphi+\VE_y\cos\varphi=\VE_\varphi$, and $\pdv{\VE_\varphi}{\varphi}=-\VE_x\cos\varphi-\VE_y\sin\varphi=-\VE_\rho$. All the other derivatives vanish because $\VE_\rho$ and $\VE_\varphi$ are functions of $\varphi$ only, and $\VE_z$ is a constant vector.

\paragraph{3.10.9}
From exercise 3.10.8, $\pdv{\VE_\rho}{\varphi}=\VE_\varphi$ and $\pdv{\VE_\varphi}{\varphi}=-\VE_\rho$, so
\[
\del\cdot\V{V}=(\VE_\rho\pdv{}{\rho}+\VE_\varphi\frac{1}{\rho}\pdv{}{\varphi}+\VE_z\pdv{}{z})\cdot(\VE_\rho V_\rho+\VE_\varphi V_\varphi+\VE_z V_z)
\]
\[
=\VE_\rho\cdot(\VE_\rho\pdv{V_\rho}{\rho}+\VE_\varphi\pdv{V_\rho}{\rho}+\VE_z\pdv{V_z}{\rho})
+\frac{1}{\rho}\VE_\varphi\cdot(\VE_\varphi V_\rho+\VE_\rho\pdv{V_\rho}{\varphi}-\VE_\rho V_\varphi+\VE_\varphi\pdv{V_\varphi}{\varphi}+\VE_z\pdv{V_z}{\varphi})+\VE_z\cdot(\VE_z\pdv{V_z}{z})
\]
\[
=\pdv{V_\rho}{\rho}+\frac{V_\rho}{\rho}+\frac{1}{\rho}\pdv{V_\varphi}{\varphi}+\pdv{V_z}{z}
\]
\[
=\frac{1}{\rho}\pdv{(\rho V_\rho)}{\rho}+\frac{1}{\rho}\pdv{V_\varphi}{\varphi}+\pdv{V_z}{z}
\]

\paragraph{3.10.10}
(a) From exercise 3.10.6, $\VE_\rho\rho+\VE_z z=\VE_x\rho\cos\varphi+\VE_y\rho\sin\varphi+\VE_z z=\VE_x x+\VE_y y+\VE_z z=\V{r}$

(b) \[\del\cdot\V{r}=\frac{1}{\rho}\pdv{}{\rho}(\rho^2)+\pdv{V_z}{z}=2+1=3\]
\[
\del\times\V{r}=\frac{1}{\rho}
\begin{vmatrix}
\VE_\rho&\VE_\varphi\rho&\VE_z\\
\pdv{}{\rho}&\pdv{}{\varphi}&\pdv{}{z}\\
\rho&0&z
\end{vmatrix}=0
\]

\paragraph{3.10.11}
(a) A point $P=(\rho\cos\varphi,\rho\sin\varphi,z)$ after reflection would be $P'=(-\rho\cos\varphi,-\rho\sin\varphi,-z)=(\rho\cos(\varphi\pm\pi),\rho\sin(\varphi\pm\pi),-z)$, so it corresponds to the transformation
\[
\rho\rightarrow\rho,\quad\varphi\rightarrow\varphi\pm\pi,\quad z\rightarrow-z
\]

(b)
\[
\VE_\rho'=\VE_x\cos(\varphi\pm\pi)+\VE_y\sin(\varphi\pm\pi)=-\VE_x\cos\varphi-\VE_y\sin\varphi=-\VE_\rho
\]
\[
\VE_\varphi'=-\VE_x\sin(\varphi\pm\pi)+\VE_y\cos(\varphi\pm\pi)=\VE_x\sin\varphi-\VE_y\cos\varphi=-\VE_\varphi
\]
\[
\VE_z'=\VE_z
\]

\paragraph{3.10.12}
(a)
\[
\V{v}=\boldsymbol{\omega}\times\V{r}=(\omega\VE_z)\times(\rho\VE_\varphi+z\VE_z)=\omega\rho\VE_\varphi
\]

(b)
\[
\del\times\V{v}=\frac{1}{\rho}
\begin{vmatrix}
\VE_\rho&\VE_\varphi\rho&\VE_z\\
\pdv{}{\rho}&\pdv{}{\varphi}&\pdv{}{z}\\
0&\omega\rho^2&0
\end{vmatrix}=
\frac{1}{\rho}(2\omega\rho)\VE_z=2\boldsymbol{\omega}
\]

\paragraph{3.10.13}
\[
\frac{d\VE_\rho}{dt}=-\VE_x\dot{\varphi}\sin\varphi+\VE_y\dot{\varphi}\cos\varphi=\VE_\varphi\dot{\varphi}
\]
\[
\frac{d\VE_\varphi}{dt}=-\VE_x\dot{\varphi}\cos\varphi-\VE_y\dot{\varphi}\sin\varphi=-\VE_\rho\dot{\varphi}
\]
\[
\frac{d\VE_z}{dt}=0
\]
\[
\V{r}=\VE_\rho\rho+\VE_z z
\]
\[
\V{v}=\dot{\V{r}}=\VE_\varphi\dot{\varphi}\rho+\VE_\rho\dot{\rho}+\VE_z\dot{z}\]
\[
=\VE_\rho\dot{\rho}+\VE_\varphi\rho\dot{\varphi}+\VE_z\dot{z}
\]
\[
\V{a}=\dot{\V{v}}=\VE_\varphi\dot{\varphi}\dot{\rho}+\VE_\rho\Ddot{\rho}-\VE_\rho\dot{\varphi}\rho\dot{\varphi}+\VE_\rho(\dot{\rho}\dot{\varphi}+\rho\Ddot{\varphi})+\VE_z\Ddot{z}
\]
\[
=\VE_\rho(\Ddot{\rho}-\rho\dot{\varphi
}^2)+\VE_\varphi(\rho\Ddot{\varphi}+2\dot{\rho}\dot{\varphi})+\VE_z\Ddot{z}
\]

\paragraph{3.10.14}
\[
\del\times\V{v}=\frac{1}{\rho}
\begin{vmatrix}
\VE_\rho&\VE_\varphi\rho&\VE_z\\
\pdv{}{\rho}&\pdv{}{\varphi}&\pdv{}{z}\\
V_\rho(\rho,\varphi)&\rho V_\varphi(\rho,\varphi)&0
\end{vmatrix}
=\frac{1}{\rho}\left[\VE_z\left(\pdv{(\rho V_\varphi)}{\rho}-\pdv{V_\rho}{\varphi}\right) \right]
\]

\paragraph{3.10.15}
\[
\V{B}=\del\times\V{A}=\frac{1}{\rho}
\begin{vmatrix}
\VE_\rho&\VE_\varphi\rho&\VE_z\\
\pdv{}{\rho}&\pdv{}{\varphi}&\pdv{}{z}\\
0&0&\frac{\mu I}{2\pi}\ln(\frac{1}{\rho})
\end{vmatrix}
=\frac{1}{\rho}\left[-\VE_\varphi\rho\frac{\mu I}{2\pi}\rho\frac{-1}{\rho^2} \right]=\VE_\varphi\frac{\mu I}{2\pi\rho}
\]

\paragraph{3.10.16}
(a) From exercise 3.10.7, we have
\[
\V{F}=-(\VE_\rho\cos\varphi-\VE_\varphi\sin\varphi)\frac{\rho\sin\varphi}{\rho^2}+(\VE_\rho\sin\varphi+\VE_\varphi\cos\varphi)\frac{\rho\cos\varphi}{\rho^2}=\VE_\rho\frac{1}{\rho}
\]

(b)
\[
\del\times\V{F}=\frac{1}{\rho}
\begin{vmatrix}
\VE_\rho&\VE_\varphi\rho&\VE_z\\
\pdv{}{\rho}&\pdv{}{\varphi}&\pdv{}{z}\\
0&\rho\frac{1}{\rho}&0
\end{vmatrix}
=0
\]

(c)
\[
\oint(\VE_\varphi\frac{1}{\rho})\cdot(\VE_\rho d\rho+\VE_\varphi\rho d\varphi+\VE_z dz)=\oint d\varphi=2\pi
\]

(d) The range of $\varphi$ of cylindrical coordinates is $0\leq\varphi<2\pi$, so $\int_0^{2\pi}d\varphi$ is not defined.

\paragraph{3.10.17}
\[
(\V{B}\cdot\del)\V{B}=B_\varphi\frac{1}{\rho}\pdv{}{\varphi}(\VE_\varphi B_\varphi(\rho))=\frac{B_\varphi}{\rho}[-\VE_\rho B_\varphi+0]=-\VE_\rho\frac{B_\varphi^2}{\rho}
\]

\paragraph{3.10.18}
\[
\pdv{\V{r}}{r}=\VE_x\sin\theta\cos\varphi+\VE_y\sin\theta\sin\varphi+\VE_z\cos\theta=h_r\VE_r\]
so
\[h_r=1,\quad \VE_r=\VE_x\sin\theta\cos\varphi+\VE_y\sin\theta\sin\varphi+\VE_z\cos\theta
\]
\[
\pdv{\V{r}}{\theta}=\VE_xr\cos\theta\cos\varphi+\VE_y r\cos\theta\sin\varphi-\VE_z r\sin\theta=h_\theta\VE_\theta\]
so
\[h_\theta=r,\quad \VE_\theta=\VE_x\cos\theta\cos\varphi+\VE_y\cos\theta\sin\varphi-\VE_z\sin\theta
\]
\[
\pdv{\V{r}}{\varphi}=-\VE_x r\sin\theta\sin\varphi+\VE_y r\sin\theta\cos\varphi=h_\varphi\VE_\varphi
\]
so
\[
h_\varphi=r\sin\theta,\quad\VE_\varphi=-\VE_x\sin\varphi+\VE_y\cos\varphi
\]

\paragraph{3.10.19}
\begin{equation}
    \VE_r=\VE_x\sin\theta\cos\varphi+\VE_y\sin\theta\sin\varphi+\VE_z\cos\theta
\end{equation}
\begin{equation}
    \VE_\theta=\VE_x\cos\theta\cos\varphi+\VE_y\cos\theta\sin\varphi-\VE_z\sin\theta
\end{equation}
\begin{equation}
    \VE_\varphi=-\VE_x\sin\varphi+\VE_y\cos\varphi
\end{equation}
$\cos\theta\times(1)-\sin\theta\times(2)$, we get
\[
\VE_z=\VE_r\cos\theta-\VE_\theta\sin\theta
\]
$\sin\theta\times(1)+\cos\theta\times(2)$, we get
\begin{equation}
    \VE_r\sin\theta+\VE_\theta\cos\theta=\VE_x\cos\varphi+\VE_y\sin\varphi
\end{equation}
$\cos\varphi\times(4)-\sin\varphi\times(3)$, we get
\[
\VE_x=\VE_r\sin\theta\cos\varphi+\VE_\theta\cos\theta\cos\varphi-\VE_\varphi\sin\varphi
\]
$\cos\varphi\times(3)+\sin\varphi\times(4)$, we get
\[
\VE_y=\VE_r\sin\theta\sin\varphi+\VE_\theta\cos\theta\sin\varphi+\VE_\varphi\cos\varphi
\]

\paragraph{3.10.20}
(a) The point $\V{r}=(0,0,0)$ is related to $\V{r'}=(0,\theta,\varphi)$ for any $0\leq\theta\leq\pi$ and $0\leq\varphi<2\pi$. If $\V{r'}=\M{B}\V{r}$, then $\V{r}=(0,0,0)$ can only be related to $\V{r'}=(0,0,0)$, a contradiction, so the matrix $\M{B}$ cannot exist.\\
(If $x,y,z\neq0$, then simply
\[\M{B}=
\begin{pmatrix}
\frac{r}{x}&0&0\\
0&\frac{\theta}{y}&0\\
0&0&\frac{\varphi}{z}
\end{pmatrix}
\]
satisfies the condition.)

(b)
From exercise 3.10.19,
\[
\V{V}=\VE_xV_x+\VE_yV_y+\VE_zV_z
\]
\[
=(\VE_r\sin\theta\cos\varphi+\VE_\theta\cos\theta\cos\varphi-\VE_\varphi\sin\varphi)V_x
\]
\[+(\VE_r\sin\theta\sin\varphi+\VE_\theta\cos\theta\sin\varphi+\VE_\varphi\cos\varphi)V_y\]
\[+(\VE_r\cos\theta-\VE_\theta\sin\theta)V_z
\]
\[
=\VE_rV_r+\VE_\theta V_\theta+\VE_\varphi V_\varphi
\]
which is
\[
\begin{pmatrix}
V_r\\V_\theta\\V_\varphi
\end{pmatrix}=
\begin{pmatrix}
\sin\theta\cos\varphi&\sin\theta\sin\varphi&\cos\theta\\
\cos\theta\cos\varphi&\cos\theta\sin\varphi&-\sin\theta\\
-\sin\varphi&\cos\varphi&0
\end{pmatrix}
\begin{pmatrix}
V_x\\V_y\\V_z
\end{pmatrix}
\]
Let the matrix be $\M{M}$, and let its transpose be $\V{M}^T$, then it can be verified that $\M{M}\M{M}^T=\boldsymbol{1}$, so it is orthogonal.

\paragraph{3.10.21}
(Let $\varphi$ of spherical coordinate be $\phi$, and $\varphi$ of cylindrical coordinate remain the same.)
The relations between the two coordinates are $\rho=r\sin\theta$, $\varphi=\phi$, $z=r\cos\theta$.
\[
\VE_r=\pdv{\V{r}}{r}=\pdv{\V{r}}{\rho}\pdv{\rho}{r}+\pdv{\V{r}}{\varphi}\pdv{\varphi}{r}+\pdv{\V{r}}{z}\pdv{z}{r}
=\VE_\rho\sin\theta+\VE_\varphi\rho\cdot0+\VE_z\cos\theta
\]
\[
=\VE_\rho\sin\theta+\VE_z\cos\theta
\]
\[
\VE_\theta=\frac{1}{r}\pdv{\V{r}}{\theta}=\frac{1}{r}(\pdv{\V{r}}{\rho}\pdv{\rho}{\theta}+\pdv{\V{r}}{\varphi}\pdv{\varphi}{\theta}+\pdv{\V{r}}{z}\pdv{z}{\theta})=
\frac{1}{r}\VE_\rho r\cos\theta+\frac{1}{r}\VE_\varphi\rho\cdot0+\frac{1}{r}\VE_z(-r\sin\theta)
\]
\[
=\VE_\rho\cos\theta-\VE_z\sin\theta
\]
\[
\VE_\phi=\frac{1}{r\sin\theta}\pdv{\V{r}}{\phi}=\frac{1}{r\sin\theta}(\pdv{\V{r}}{\rho}\pdv{\rho}{\phi}+\pdv{\V{\phi}}{\varphi}\pdv{\varphi}{\phi}+\pdv{\V{r}}{z}\pdv{z}{\phi})=
\frac{1}{r\sin\theta}\VE_\varphi\rho\cdot1
\]
\[
=\VE_\varphi
\]
so
\[
\V{V}=V_r\VE_r+V_\theta\VE_\theta+V_\phi\VE_\phi
\]
\[
=V_r(\VE_\rho\sin\theta+\VE_z\cos\theta)+V_\theta(\VE_\rho\cos\theta-\VE_z\sin\theta)+V_\phi\VE_\varphi
\]
\[
=V_\rho\VE_\rho+V_\varphi\VE_\varphi+V_z\VE_z
\]
which is
\[
\begin{pmatrix}
V_\rho\\V_\varphi\\V_z
\end{pmatrix}=
\begin{pmatrix}
\sin\theta&\cos\theta&0\\
0&0&1\\
\cos\theta&-\sin\theta&0
\end{pmatrix}
\begin{pmatrix}
V_r\\V_\theta\\V_\phi
\end{pmatrix}
\]
Let the matrix be $\M{M}$. Note that it is orthogonal, so the inverse transformation is
\[
\M{M}^{-1}=\M{M}^T=
\begin{pmatrix}
\sin\theta&0&\cos\theta\\
\cos\theta&0&-\sin\theta\\
0&1&0
\end{pmatrix}
\]

\paragraph{3.10.22}
(a)
\begin{align*}
    \pdv{\VE_r}{r}&=0 &
    \pdv{\VE_r}{\theta}&=\VE_\theta&
    \pdv{\VE_r}{\varphi}&=\VE_\varphi\sin\theta\\
    \pdv{\VE_\theta}{r}&=0 &
    \pdv{\VE_\theta}{\theta}&=-\VE_r&
    \pdv{\VE_\theta}{\varphi}&=\VE_\varphi\cos\theta\\
    \pdv{\VE_\varphi}{r}&=0 &
    \pdv{\VE_\varphi}{\theta}&=0&
    \pdv{\VE_\varphi}{\varphi}&=-\VE_r\sin\theta-\VE_\theta\cos\theta \\
\end{align*}

(b)
\[
\del\cdot\del\psi=(\VE_r\pdv{}{r}+\VE_\theta\frac{1}{r}\pdv{}{\theta}+\VE_\varphi\frac{1}{r\sin\theta}\pdv{}{\varphi})\cdot(\VE_r\pdv{\psi}{r}+\VE_\theta\frac{1}{r}\pdv{\psi}{\theta}+\VE_\varphi\frac{1}{r\sin\theta}\pdv{\psi}{\varphi})
\]
\[
=\VE_r\cdot\left(\VE_r\pdv{^2\psi}{r^2}+\VE_\theta(\cdots)+\VE_\varphi(\cdots) \right)
+\frac{1}{r}\VE_\theta\cdot\left(\VE_\theta\pdv{\psi}{r}+\VE_\theta\frac{1}{r}\pdv{^2\psi}{\theta^2}+\VE_r(\cdots)+\VE_\varphi(\cdots) \right)
\]
\[
+\frac{1}{r\sin\theta}\VE_\varphi\cdot\left(\VE_\varphi\sin\theta\pdv{\psi}{r}+\VE_\varphi\cos\theta\frac{1}{r}\pdv{\psi}{\theta}+\VE_\varphi\frac{1}{r\sin\theta}\pdv{^2\psi}{\varphi^2}+\VE_r(\cdots)+\VE_\theta(\cdots) \right)
\]
\[
=\frac{2}{r}\pdv{\psi}{r}+\pdv{^2\psi}{r^2}+\frac{\cos\theta}{r^2\sin\theta}\pdv{\psi}{\theta}+\frac{1}{r^2}\pdv{^2\psi}{\theta^2}+\frac{1}{r^2\sin^2\theta}\pdv{^2\psi}{\varphi^2}
\]
\[
=\frac{1}{r^2\sin\theta}\left[\sin\theta\pdv{}{r}(r^2\pdv{\psi}{r})+\pdv{}{\theta}(\sin\theta\pdv{\psi}{\theta})+\frac{1}{\sin\theta}\pdv{^2\psi}{\varphi^2} \right]
\]

\paragraph{3.10.23}
(a) $\boldsymbol{\omega}=\VE_z\omega=\VE_r\omega\cos\theta-\VE_\theta\omega\sin\theta$, so
\[
\V{v}=\boldsymbol{\omega}\times\V{r}=(\VE_r\omega\cos\theta-\VE_\theta\omega\sin\theta)\times(\VE_r r)=\VE_\varphi\omega r \sin\theta
\]

(b)
\[
\del\times\V{v}=\frac{1}{r^2\sin\theta}
\begin{vmatrix}
\VE_r&\VE_\theta r&\VE_\varphi r\sin\theta\\
\pdv{}{r}&\pdv{}{\theta}&\pdv{}{\varphi}\\
0&0&\omega r^2\sin^2\theta
\end{vmatrix}\]
\[=\frac{1}{r^2\sin\theta}(\VE_r2\omega r^2\sin\theta\cos\theta-\VE_\theta2\omega r^2\sin^2\theta)\]
\[=2\VE_r\omega\cos\theta-2\VE_\theta\omega\sin\theta=2\boldsymbol{\omega}
\]

\paragraph{3.10.24}
\[
\del\times\V{V}=\frac{1}{r^2\sin\theta}
\begin{vmatrix}
\VE_r&\VE_\theta r&\VE_\varphi r\sin\theta\\
\pdv{}{r}&\pdv{}{\theta}&\pdv{}{\varphi}\\
0&V_\theta&V_\varphi
\end{vmatrix}
\]
\[
=\frac{1}{r^2\sin\theta}\left[\VE_r(\pdv{V_\varphi}{\theta}-\pdv{V_\theta}{\varphi})-\VE_\theta r(\pdv{V_\varphi}{r})+\VE_\varphi r\sin\theta(\pdv{V_\theta}{r}) \right]
\]
has no tangential components, so $\pdv{V_\varphi}{r}=\pdv{V_\theta}{r}=0$. That is, the tangential components of $\V{V}$ have no radial dependence.

\paragraph{3.10.25}
(a) A point $P=(r\sin\theta\cos\varphi,r\sin\theta\sin\varphi,r\cos\theta)$ after reflection would become\\ $P'=(-r\sin\theta\cos\varphi,-r\sin\theta\sin\varphi,-r\cos\theta)=(r\sin(\pi-\theta)\cos(\varphi\pm\pi),r\sin(\pi-\theta)\sin(\varphi\pm\pi),r\cos(\pi-\theta))$, so it corresponds to the transformation
\[
r\rightarrow r,\quad\theta\rightarrow\pi-\theta,\quad \varphi\rightarrow\varphi\pm\pi
\]

(b)

\[\VE_r'=\VE_x\sin(\pi-\theta)\cos(\varphi\pm\pi)+\VE_y\sin(\pi-\theta)\sin(\varphi\pm\pi)+\VE_z\cos(\pi-\theta)=-\VE_r
\]
\[ \VE_\theta=\VE_x\cos(\pi-\theta)\cos(\varphi\pm\pi)+\VE_y\cos(\pi-\theta)\sin(\varphi\pm\pi)-\VE_z\sin(\pi-\theta)=\VE_\theta
\]
\[
\VE_\varphi=-\VE_x\sin(\varphi\pm\pi)+\VE_y\cos(\varphi\pm\pi)=-\VE_\varphi
\]

\paragraph{3.10.26}
(a)
\[
(\V{A}\cdot\del)\V{r}=(A_x\pdv{}{x}+A_y\pdv{}{y}+A_z\pdv{}{z})(x\VE_x+y\VE_y+z\VE_z)\]
\[=A_x\VE_x+A_y\VE_y+A_z\VE_z=\V{A}
\]

(b)
From exercise 3.10.22
\[
(\V{A}\cdot\del)\V{r}=(A_r\pdv{}{r}+A_\theta\frac{1}{r}\pdv{}{\theta}+A_\varphi\frac{1}{r\sin\theta}\pdv{}{\varphi})(r\VE_r)=A_r\VE_r+A_\theta\frac{1}{r}r\VE_\theta+A_\varphi\frac{1}{r\sin\theta}r\sin\theta\VE_\varphi\]
\[=A_r\VE_r+A_\theta\VE_\theta+A_\varphi\VE_\varphi=\V{A}
\]

\paragraph{3.10.27}
From exercise 3.10.22
\[
\frac{d\VE_r}{dt}=\pdv{\VE_r}{r}\pdv{r}{t}+\pdv{\VE_r}{\theta}\pdv{\theta}{t}+\pdv{\VE_r}{\varphi}\pdv{\varphi}{t}=\VE_\theta\dot{\theta}+\VE_\varphi\sin\theta\dot{\varphi}
\]
\[
\frac{d\VE_\theta}{dt}=\pdv{\VE_\theta}{r}\pdv{r}{t}+\pdv{\VE_\theta}{\theta}\pdv{\theta}{t}+\pdv{\VE_\theta}{\varphi}\pdv{\varphi}{t}=-\VE_r\dot{\theta}+\VE_\varphi\cos\theta\dot{\varphi}
\]
\[
\frac{d\VE_\varphi}{dt}=\pdv{\VE_\varphi}{r}\pdv{r}{t}+\pdv{\VE_\varphi}{\theta}\pdv{\theta}{t}+\pdv{\VE_\varphi}{\varphi}\pdv{\varphi}{t}=-\VE_r\sin\theta\dot{\varphi}-\VE_\theta\cos\theta\dot{\varphi}
\]
\[
\V{r}=\VE_r r
\]
so
\[
\V{v}=\dot{\V{r}}=\VE_\theta\dot{\theta}r+\VE_\varphi\sin\theta\dot{\varphi}r+\VE_r\dot{r}
\]
\[
=\VE_r\dot{r}+\VE_\theta r\dot{\theta}+\VE_\varphi r\sin\theta\dot{\varphi}
\]
\[
\V{a}=\dot{\V{v}}=\VE_\theta\dot{r}\dot{\theta}+\VE_\varphi\dot{r}\sin\theta\dot{\varphi}+\VE_r\Ddot{r}-\VE_r r{\dot{\theta}}^2+\VE_\varphi r\cos\theta\dot{\theta}\dot{\varphi}+\VE_\theta\dot{r}\dot{\theta}+\VE_\theta r\Ddot{\theta}\]
\[-\VE_r r\sin^2\theta{\dot{\varphi}}^2-\VE_\theta r\sin\theta\cos\theta{\dot{\varphi}}^2+\VE_\varphi\dot{r}\sin\theta\dot{\varphi}+\VE_\varphi r\cos\theta\dot{\theta}\dot{\varphi}+\VE_\varphi r\sin\theta\Ddot{\varphi}
\]
\[
=\VE_r(\Ddot{r}-r{\dot{\theta}}^2-r\sin^2\theta{\dot{\varphi}}^2)+\VE_\theta(r\Ddot{\theta}+2\dot{r}\dot{\theta}-r\sin\theta\cos\theta{\dot{\varphi}}^2)+\VE_\varphi(r\sin\theta\Ddot{\varphi}+2\dot{r}\sin\theta\dot{\varphi}+2r\cos\theta\dot{\theta}\dot{\varphi})
\]

\paragraph{3.10.28}
\[
\del=\VE_x\pdv{}{x}+\VE_y\pdv{}{y}+\VE_z\pdv{}{z}
\]
\[
=\VE_r\pdv{}{r}+\VE_\theta\frac{1}{r}\pdv{}{\theta}+\VE_\varphi\frac{1}{r\sin\theta}\pdv{}{\varphi}
\]
\[
=(\VE_x\sin\theta\cos\varphi+\VE_y\sin\theta\sin\varphi+\VE_z\cos\theta)\pdv{}{r}
\]
\[
+(\VE_x\cos\theta\cos\varphi+\VE_y\cos\theta\sin\varphi-\VE_z\sin\theta)\frac{1}{r}\pdv{}{\theta}
\]
\[
+(-\VE_x\sin\varphi+\VE_y\cos\varphi)\frac{1}{r\sin\theta}\pdv{}{\varphi}
\]
\[
=\VE_x(\sin\theta\cos\varphi\pdv{}{r}+\cos\theta\cos\varphi\frac{1}{r}\pdv{}{\theta}-\frac{\sin\varphi}{r\sin\theta}\pdv{}{\varphi})
\]
\[
+\VE_y(\sin\theta\sin\varphi\pdv{}{r}+\cos\theta\sin\varphi\frac{1}{r}\pdv{}{\theta}+\frac{\cos\varphi}{r\sin\theta}\pdv{}{\varphi})
\]
\[
+\VE_z(\cos\theta\pdv{}{r}-\sin\theta\frac{1}{r}\pdv{}{\theta})
\]
equating the $x,y,z$ components, we get
\[
\pdv{}{x}=\sin\theta\cos\varphi\pdv{}{r}+\cos\theta\cos\varphi\frac{1}{r}\pdv{}{\theta}-\frac{\sin\varphi}{r\sin\theta}\pdv{}{\varphi}
\]
\[
\pdv{}{y}=\sin\theta\sin\varphi\pdv{}{r}+\cos\theta\sin\varphi\frac{1}{r}\pdv{}{\theta}+\frac{\cos\varphi}{r\sin\theta}\pdv{}{\varphi}
\]
\[
\pdv{}{z}=\cos\theta\pdv{}{r}-\sin\theta\frac{1}{r}\pdv{}{\theta}
\]

\paragraph{3.10.29}
$x=r\sin\theta\cos\varphi$, $y=r\sin\theta\sin\varphi$. Using results from exercise 3.10.28, we can have
\[
x\pdv{}{y}-y\pdv{}{x}=r\sin\theta\cos\varphi(\sin\theta\sin\varphi\pdv{}{r}+\cos\theta\sin\varphi\frac{1}{r}\pdv{}{\theta}+\frac{\cos\varphi}{r\sin\theta}\pdv{}{\varphi})\]
\[-r\sin\theta\sin\varphi(\sin\theta\cos\varphi\pdv{}{r}+\cos\theta\cos\varphi\frac{1}{r}\pdv{}{\theta}-\frac{\sin\varphi}{r\sin\theta}\pdv{}{\varphi})
=\pdv{}{\varphi}
\]
so
\[-i\left(x\pdv{}{y}-y\pdv{}{x}\right)=-i\pdv{}{\varphi}\]

\paragraph{3.10.30}
\[
\V{L}=-i(\V{r}\times\del)=-i(\VE_r r)\times(\VE_r\pdv{}{r}+\VE_\theta\frac{1}{r}\pdv{}{\theta}+\VE_\varphi\frac{1}{r\sin\theta}\pdv{}{\varphi})
\]
\[
=\VE_\theta i\frac{1}{\sin\theta}\pdv{}{\varphi}-\VE_\varphi i\pdv{}{\theta}
\]
\[
=(\VE_x\cos\theta\cos\varphi+\VE_y\cos\theta\sin\varphi-\VE_z\sin\theta)i\frac{1}{\sin\theta}\pdv{}{\varphi}+(\VE_x\sin\varphi-\VE_y\cos\varphi)i\pdv{}{\theta}
\]
\[
=\VE_x(i\sin\varphi\pdv{}{\theta}+i\cot\theta\cos\varphi\pdv{}{\varphi})+\VE_y(-i\cos\varphi\pdv{}{\theta}+i\cot\theta\sin\varphi\pdv{}{\varphi})+\VE_z(-i\pdv{}{\varphi})
\]
so
\[
L_x+iL_y=i\sin\varphi\pdv{}{\theta}+i\cot\theta\cos\varphi\pdv{}{\varphi}+\cos\varphi\pdv{}{\theta}-\cot\theta\sin\varphi\pdv{}{\varphi}
\]
\[
=(\cos\varphi+i\sin\varphi)(\pdv{}{\theta}+i\cot\theta\pdv{}{\varphi})=e^{i\varphi}(\pdv{}{\theta}+i\cot\theta\pdv{}{\varphi})
\]
\[
L_x-iL_y=i\sin\varphi\pdv{}{\theta}+i\cot\theta\cos\varphi\pdv{}{\varphi}-\cos\varphi\pdv{}{\theta}+\cot\theta\sin\varphi\pdv{}{\varphi}
\]
\[
=(-\cos\varphi+i\sin\varphi)(\pdv{}{\theta}-i\cot\theta\pdv{}{\varphi})=-e^{-i\varphi}(\pdv{}{\theta}-i\cot\theta\pdv{}{\varphi})
\]

\paragraph{3.10.31}
From exercise 3.10.30
\[
\V{L}=\VE_xL_x+\VE_yL_y+\VE_zL_z\]
\[=\VE_x(i\sin\varphi\pdv{}{\theta}+i\cot\theta\cos\varphi\pdv{}{\varphi})+\VE_y(-i\cos\varphi\pdv{}{\theta}+i\cot\theta\sin\varphi\pdv{}{\varphi})+\VE_z(-i\pdv{}{\varphi})
\]
so
\[
\V{L}\times\V{L}=\VE_x(L_yL_z-L_zL_y)+\VE_y(L_zL_x-L_zL_z)+\VE_z(L_xL_y-L_yL_x)
\]
\[
=\VE_x(-\sin\varphi\pdv{}{\theta}-\cot\theta\cos\varphi\pdv{}{\varphi})+\VE_y(\cos\varphi\pdv{}{\theta}-\cot\theta\sin\varphi\pdv{}{\varphi})+\VE_z(\pdv{}{\varphi})
\]
\[
=\VE_xiL_x+\VE_yiL_y+\VE_ziL_z=i\V{L}
\]

\paragraph{3.10.32}
(a)(b) It is the first half of exercise 3.10.30.

(c) The author suggest to do it in Cartesian coordinate, but I think it's easier to do in spherical coordinate, with the help of results from 3.10.22(a).
\[
{\V{L}}^2=\V{L}\cdot\V{L}=-(\VE_\theta\frac{1}{\sin\theta}\pdv{}{\varphi}-\VE_\varphi\pdv{}{\theta})\cdot(\VE_\theta\frac{1}{\sin\theta}\pdv{}{\varphi}-\VE_\varphi\pdv{}{\theta})
\]
\[
=-\left[\VE_\theta\frac{1}{\sin\theta}\cdot\pdv{}{\varphi}(\VE_\theta\frac{1}{\sin\theta}\pdv{}{\varphi}) +\VE_\varphi\cdot\pdv{}{\theta}(\VE_\varphi\pdv{}{\theta})-\VE_\theta\frac{1}{\sin\theta}\cdot\pdv{}{\varphi}(\VE_\varphi\pdv{}{\theta})-\VE_\varphi\cdot\pdv{}{\theta}(\VE_\theta\frac{1}{\sin\theta}\pdv{}{\varphi})\right]
\]
\[
=-\left[\frac{1}{\sin^2\theta}\pdv{^2}{\varphi^2}+\pdv{^2}{\theta^2}+\frac{\cos\theta}{\sin\theta}\pdv{}{\theta} \right]
\]
\[
=-\frac{1}{\sin\theta}\pdv{}{\theta}(\sin\theta\pdv{}{\theta})-\frac{1}{\sin^2\theta}\pdv{^2}{\varphi^2}
\]
\[
=-r^2\del^2+\pdv{}{r}\left(r^2\pdv{}{r} \right)
\]
(We use Eq. 3.158 in the last equality.)

\paragraph{3.10.33}
(a)
\[
\VE_r\pdv{}{r}-i\frac{\V{r}\times\V{L}}{r^2}=\VE_r\pdv{}{r}-\frac{\V{r}\times(\V{r}\times\del)}{r^2}=\VE_r\pdv{}{r}-\frac{\V{r}(\V{r}\cdot\del)-(\V{r}\cdot\V{r})\del}{r^2}=\VE_r\pdv{}{r}-\frac{\VE_r r(r\pdv{}{r})-r^2\del}{r^2}=\del
\]

(b) (\textit{There is probably a mistake:} $\del(1+r\pdv{}{r})$ \textit{should be} $\del+\del(r\pdv{}{r})$)
\smallskip

$r\pdv{}{r}=\V{r}\cdot\del$ in the spherical coordinate, so the left side of the equation is $\V{r}\del^2-\del-\del(\V{r}\cdot\del)$.
\[
\left[\V{r}\del^2-\del-\del(\V{r}\cdot\del) \right]_x=x(\pdv{^2}{x^2}+\pdv{^2}{y^2}+\pdv{^2}{z^2})-\pdv{}{x}-\pdv{}{x}(x\pdv{}{x}+y\pdv{}{y}+z\pdv{}{z})
\]
\[
=-2\pdv{}{x}+x\pdv{^2}{y^2}+x\pdv{^2}{z^2}-y\pdv{^2}{x\partial y}-z\pdv{^2}{x\partial z}
\]
\[
\left[i\del\times\V{L} \right]_x=\left[\del\times(\V{r}\times\del) \right]_x=\pdv{}{y}(x\pdv{}{y}-y\pdv{}{x})-\pdv{}{z}(z\pdv{}{x}-x\pdv{}{z})
\]
\[
=-2\pdv{}{x}+x\pdv{^2}{y^2}+x\pdv{^2}{z^2}-y\pdv{^2}{x\partial y}-z\pdv{^2}{x\partial z}
\]
So the $x$-components of two side of the equation are equal. It can be verified that so are the $y$- and $z$-components. Therefore,
\[
\V{r}\del^2-\del-\del(\V{r}\cdot\del)=i\del\times\V{L}
\]

\paragraph{3.10.34}
\[
\frac{1}{r^2}\frac{d}{dr}\left[r^2\frac{d\psi}{dr} \right]=\frac{1}{r^2}(2r\frac{d\psi}{dr}+r^2\frac{d^2\psi}{dr^2})=\frac{d^2\psi}{dr^2}+\frac{2}{r}\frac{d\psi}{dr}
\]
\[
\frac{1}{r}\frac{d^2}{dr^2}\left[r\psi\right]=\frac{1}{r}\frac{d}{dr}(\psi+r\frac{d\psi}{dr})=\frac{1}{r}(\frac{d\psi}{dr}+\frac{d\psi}{dr}+r\frac{d^2\psi}{dr^2})=\frac{d^2\psi}{dr^2}+\frac{2}{r}\frac{d\psi}{dr}
\]
so all the three form are equaivalant.

\paragraph{3.10.35}
(a)
\[
\del\times\V{F}=\frac{1}{r^2\sin\theta}
\begin{vmatrix}
\VE_r&\VE_\theta r&\VE_\varphi r\sin\theta\\
\pdv{}{r}&\pdv{}{\theta}&\pdv{}{\varphi}\\
\frac{2P\cos\theta}{r^3}&\frac{P}{r^2}\sin\theta&0\\
\end{vmatrix}
=\frac{1}{r^2\sin\theta}\left(\VE_\varphi r\sin\theta(-2\frac{P}{r^3}\sin\theta+\frac{2P\sin\theta}{r^3}) \right)=0
\]

(b) $r=1$ and $\theta=\frac{\pi}{2}$, so $\V{F}=\VE_\theta P$ and $d\V{r}=\VE_r dr+\VE_\varphi d\varphi$.
\[
\oint\V{F}\cdot d\V{r}=(\VE_\theta P)\cdot(\VE_r dr+\VE_\varphi d\varphi)=0
\]
We cannot assert whether $\V{F}$ is conservative or not unless we evaluate every integral over closed loop.

(c) $\int_a^b\V{F}\cdot d\V{r}=\psi(a)-\psi(b)$. Take the path $(r,\theta,\varphi)\rightarrow(\infty,\theta,\varphi)$, and define the potential at infinity $\psi(\infty)$ to be zero. Then we have
\[
\psi(\V{r})=\psi(\V{r})-\psi(\infty)=\int_r^\infty\frac{2P\cos\theta}{r^3}dr=-\frac{P\cos\theta}{r^2}\Big|_r^\infty=\frac{P\cos\theta}{r^2}
\]

\paragraph{3.10.36}
(a)
\[
\del\times\V{A}=\frac{1}{r^2\sin\theta}
\begin{vmatrix}
\VE_r&\VE_\theta r&\VE_\varphi r\sin\theta\\
\pdv{}{r}&\pdv{}{\theta}&\pdv{}{\varphi}\\
0&0&-\cos\theta
\end{vmatrix}
=\frac{1}{r^2\sin\theta}(\VE_r\sin\theta)=\frac{\VE_r}{r^2}
\]

(b) $r=\sqrt{x^2+y^2+z^2}$, $\theta=\cos^{-1}\frac{z}{r}$, $\varphi=\tan^{-1}\frac{y}{x}$, $\VE_\varphi=-\VE_x\sin\varphi+\VE_y\cos\varphi$. So
\[
\V{A}=-(-\VE_x\frac{y}{\sqrt{x^2+y^2}}+\VE_y\frac{x}{\sqrt{x^2+y^2}})\frac{z}{\sqrt{x^2+y^2}}\frac{1}{r}=\VE_x\frac{yz}{r(x^2+y^2)}-\VE_y\frac{xz}{r(x^2+y^2)}
\]

(c)
\[
\del\times\V{A}=\frac{1}{r^2\sin\theta}
\begin{vmatrix}
\VE_r&\VE_\theta r&\VE_\varphi r\sin\theta\\
\pdv{}{r}&\pdv{}{\theta}&\pdv{}{\varphi}\\
0&-\varphi\sin\theta&0
\end{vmatrix}
=\frac{1}{r^2\sin\theta}(\VE_r\sin\theta)=\frac{\VE_r}{r^2}
\]

\paragraph{3.10.37}
$\V{r}=\VE_r r$, so from exercise 3.10.22 we have $\pdv{\V{r}}{r}=\VE_r$, $\pdv{\V{r}}{\theta}=\VE_\theta r$, $\pdv{\V{r}}{\varphi}=\VE_\varphi r\sin\theta$. So
\[
\V{E}=-\del\psi=-\left[\VE_r\pdv{}{r}(\frac{\V{P}\cdot\V{r}}{4\pi\varepsilon_0r^3})+\VE_\theta\frac{1}{r}\pdv{}{\theta}(\frac{\V{P}\cdot\V{r}}{4\pi\varepsilon_0r^3})+\VE_\varphi\frac{1}{r\sin\theta}\pdv{}{\varphi}(\frac{\V{P}\cdot\V{r}}{4\pi\varepsilon_0r^3}) \right]
\]
\[
=-\VE_r\left(\frac{1}{4\pi\varepsilon_0r^3}\V{P}\cdot\pdv{\V{r}}{r}+\frac{\V{P}\cdot\V{r}}{4\pi\varepsilon_0}\pdv{}{r}(\frac{1}{r^3})\right)-\VE_\theta\frac{1}{r}\frac{1}{4\pi\varepsilon_0r^3}\V{P}\cdot\pdv{\V{r}}{\theta}-\VE_\varphi\frac{1}{r\sin\theta}\frac{1}{4\pi\varepsilon_0r^3}\V{P}\cdot\pdv{\V{r}}{\varphi}
\]
\[
=-\VE_r(-2)\frac{P_r}{4\pi\varepsilon_0r^3}-\VE_\theta\frac{P_\theta}{4\pi\varepsilon_0r^3}-\VE_\varphi\frac{P_\varphi}{4\pi\varepsilon_0r^3}
\]
\[
=\frac{1}{4\pi\varepsilon_0r^3}(3\VE_rP_r-\VE_rP_r-\VE_\theta P_\theta-\VE_\varphi P_\varphi)
\]
\[
=\frac{3\hat{\V{r}}(\V{P}\cdot\hat{\V{r}})}{4\pi\varepsilon_0r^3}
\]
where $\hat{\V{r}}=\VE_r$ is the unit vector in the $\V{r}$ direction.

\end{document}

Mathematical Methods for Physicists/Chapter 03/main.tex


%
\chapter[SI: An improved PNP-NS framework]%
        {Supplementary information: An improved PNP-NS framework}
%
\label{ch:epnpns_appendix}

\definecolor{shadecolor}{gray}{0.85}
\begin{shaded}
  \adapRSC{Willems-2020}
\newpage
\end{shaded}

\section{Weak forms of the {ePNP-NS} equations}
%
\label{sec:epnpns_appendix:weak_forms}
%

To solve partial differential equations with the finite element method, we must be derive their weak form.
This is achieved through multiplication of the equation with an arbitrary test function and integration over
their relevant domains and boundaries (\cref{fig:epnpns_model_boundaries}). The full computational domain of
our model ($\Omega$) is subdivided into domains for the pore ($\Omega_p$), the lipid bilayer ($\Omega_m$) and
the electrolyte reservoir ($\Omega_w$). The relevant boundaries of these domains are also indicated, i.e. the
reservoir's exterior edges at the \cisi{} $\Gamma_{w,c}$) and \transi{} $\Gamma_{w,t}$) sides, the outer edge
of the lipid bilayer ($\Gamma_{m}$) and the interface of the fluid with the nanopore and the bilayer
($\Gamma_{p+m}$).

The following paragraphs detail the weak forms of equations that were used in this
\cref{ch:epnpns,ch:trapping}.

%
\begin{figure*}[t]
  \centering
  %
  \includegraphics[scale=1]{epnpns_model_boundaries}
  %

  \caption[Computational domains and boundaries.]%
  {\textbf{Computational domains and boundaries.}
    %
    The full computational domain ($\Omega$) of the model is subdivided into various subdomains---comprising
    the reservoir ($\Omega_w$), the lipid bilayer ($\Omega_m$) and the pore ($\Omega_p$)---and their limiting
    boundaries---the exterior boundary at the \cisi{} ($\Gamma_{w,c}$) \transi{} and the \transi{}
    ($\Gamma_{w,t}$) sides of the reservoir, the exterior boundary of the membrane ($\Gamma_m$) and the
    interior boundary between the reservoir on the one hand, and the pore and membrane on the other
    ($\Gamma_{p+m}$).
    %
  }\label{fig:epnpns_model_boundaries}
\end{figure*}
%

\subsection{Poisson equation}
%

The global potential distribution is described by the Poisson equation (PE)~\cite{Lu-2012}
%
\begin{align}
  \label{eq:epnpns_appendix:poisson}
  \nabla \cdot \displacement = -\left( \scdpore + \scdion \right)
  \text{\quad with \quad}
  \displacement = \absperm \relperm \nabla \potential
  \text{ ,}
\end{align}
%
with $\displacement$ the electrical displacement field, $\potential$ the electric potential, $\absperm$ the
vacuum permittivity (\SI{8.85419e-12}{\farad\per\meter}), and $\relperm$ the local relative permittivity
of the medium. $\scdpore$ and $\scdion$ are the fixed (due to the pore) and mobile (due to the ions) charge
distributions, respectively.

Multiplication of~\cref{eq:epnpns_appendix:poisson} with the potential test function $\psi$ and integration
over the entire model $\Omega=\Omega_w+\Omega_p+\Omega_m$ gives
%
\begin{align}
  \displaystyle\int_{\Omega}
  \left[
    \nabla \cdot \displacement
  \right]
  \psi \,d\Omega
  ={}&
  - \displaystyle\int_{\Omega} \left[ \scdpore + \scdion \right] \psi \,d\Omega \text{ ,}
\end{align}
%
which, after applying the Gauss divergence theorem, yields the final weak formulation
%
\begin{align}
  \displaystyle\int_{\Omega}
  \left[
    \nabla \psi \cdot \displacement
  \right]
  \,d\Omega
  & - \displaystyle\int_{\Gamma_\text{PE}}
  \left[
    \psi \displacement \cdot \vec{n}
  \right]
  \,d\Gamma_\text{PE} \notag \\
  & =
  \displaystyle\int_{\Omega} \left[ \psi \scdpore \right] \,d\Omega
  +
  \displaystyle\int_{\Omega} \left[ \psi \scdion \right] \,d\Omega
  \text{ ,}
\end{align}
%
with boundaries $\Gamma_\text{PE}=\Gamma_{w,c}+\Gamma_{w,t}+\Gamma_m$ and $\vec{n}$ their normal vector. The
boundary integrals at $\Gamma_{w,c}$ and $\Gamma_{w,t}$ are evaluated using the Dirichlet \glspl{bc}
$\potential=0$ and $\potential=\vbias$, respectively. A zero charge \gls{bc}, $\vec{n} \cdot \displacement =
0$, is used for the integral at $\Gamma_{m}$.

\subsection{Size-modified Nernst-Planck equation}
%

The total ionic flux $\flux_{i}$ of ion $i$ at steady-state is expressed by the \gls{smnpe}~\cite{Lu-2012}
%
\begin{equation}\label{eq:smnp}
  \pd{\concentration_{i}}{t} = - \nabla\cdot\flux_{i} = - \nabla \cdot
  \left(
    \diffusion_{i}\nabla\concentration_{i}
    + \chargen_{i}\mobility_{i}\concentration_{i}\nabla\potential
    + \diffusion_{i} \vec{\beta_{i}} \concentration_{i}
    - \velocity\concentration_{i}
  \right)
  \text{ ,}
\end{equation}
%
where
%
\begin{equation}\label{eq:steric_vector}
  \vec{\beta_{i}} =
    \frac{\ionsize_{i}^3/\ionsize_{0}^3 \dsum_{j} \avogadro \ionsize_{j}^3 \nabla \concentration_{j}}
        {1 - \dsum_{j} \avogadro \ionsize_{j}^3 \concentration_{j}}
  \text{ ,}
\end{equation}
%
and with ion diffusion coefficient $\diffusion_{i}$, concentration $\concentration_{i}$, charge number
$\chargen_{i}$, mobility $\mobility_{i}$, electrostatic potential $\potential$, steric saturation factor
$\beta_{i}$ and fluid velocity $\velocity$. $\avogadro$ is Avogadro's constant (\SI{6.022e23}{\per\mole}) and
$\ionsize_{i}$ and $\ionsize_{0}$ are the limiting cubic diameters for ions and water, respectively. Using the
ion concentration test function $d_i$, the weak form of \cref{eq:smnp} becomes
%
\begin{align}
  % Raw integral
  \displaystyle\int_{\Omega_w} \left[ \pd{\concentration_{i}}{t} \right] d_{i} \,d\Omega_w =
  \displaystyle\int_{\Omega_w} \left[ - \nabla \cdot \flux_{i} \right] d_{i} \,d\Omega_w
  \text{, }
\end{align}
%
which can be split into the domain and boundary integrals
%
\begin{align}
  % Gauss divergence theorem
  \displaystyle\int_{\Omega_w} \left[ d_{i} \pd{\concentration_{i}}{t} \right] \,d\Omega_w ={}&
  \displaystyle\int_{\Omega_w} \left[\nabla d_{i} \cdot \flux_{i} \right]\,d\Omega_w
  - \displaystyle\int_{\Gamma_\text{NP}}
  \left[ d_{i} \flux_{i} \cdot \vec{n} \right]\,d\Gamma_\text{NP} \\
  % Fill in flux components
  ={}&
  % Domain
  \displaystyle\int_{\Omega_w}
  \left[
    \nabla d_{i} \cdot
    \left(
      \diffusion_{i}\nabla\concentration_{i}
      + \chargen_{i}\mobility_{i}\concentration_{i}\nabla\potential
      + \diffusion_{i} \vec{\beta_{i}} \concentration_{i}
      - \velocity \concentration_{i}
    \right)
  \right]
  \,d\Omega_w \notag \\
  % Boundary
  & - \displaystyle\int_{\Gamma_\text{NP}}
  \left[
    d_{i}
    \left(
      \diffusion_{i} \nabla \concentration_{i}
      + \chargen_{i} \mobility_{i} \concentration_{i} \nabla \potential
      + \diffusion_{i} \vec{\beta_{i}} \concentration_{i}
      - \velocity \concentration_{i}
    \right)
    \cdot \vec{n}
  \right]
  \,d\Gamma_\text{NP} \notag
  \text{ .}
\end{align}
%
The integrals on the boundaries $\Gamma_\text{NP} = \Gamma_{w,c}+\Gamma_{w,t}+\Gamma_{p+m}$ are evaluated
using the Dirichlet \gls{bc} $\concentration_{i} = \cbulk$ for $\Gamma_{w,c}$ and $\Gamma_{w,t}$, and the no
flux \gls{bc} $\vec{n} \cdot \flux_{i} = 0$ for $\Gamma_{p+m}$

\subsection{Variable density and viscosity Navier-Stokes equation}
%

The steady-state, laminar fluid flow of an incompressible fluid with a variable density and viscosity is given
by the system of equations~\cite{Axelsson-2015}
%
\begin{align}
  \label{eq:ns_density_continuity}
  \velocity \cdot \nabla \density ={}& 0 \\
  \label{eq:ns_conservation}
  \left( \velocity \cdot \nabla \right) \left( \density\velocity \right)
  + \nabla \cdot \hydrostresstensor ={}& \volumeforce \text{\quad with }
  \hydrostresstensor =
    \pressure\identity - \viscosity\left[\nabla\velocity+\left(\nabla\velocity \right)^\mathsf{T}\right]
  \\
  \label{eq:ns_velocity_continuity}
  \nabla \cdot \left( \density\velocity \right) - \velocity \cdot \nabla \density ={}& 0
  \text { ,}
\end{align}
%
with fluid velocity $\velocity$, density $\density$, hydrodynamic stress tensor $\hydrostresstensor$,
viscosity $\viscosity$, pressure $\pressure$ and body force $\volumeforce$. The pressure test function $q$ is
used to derive the weak forms of \cref{eq:ns_density_continuity}
%
\begin{align}
  % Density continuity
  \displaystyle\int_{\Omega_w} \left[ \velocity \cdot \nabla \density \right] q \,d\Omega_w ={}&
  \displaystyle\int_{\Omega_w} \left[ q \velocity \cdot \nabla \density \right] \,d\Omega_w = 0 \text{ ,}
\end{align}
%
and \cref{eq:ns_velocity_continuity}
%
\begin{align}
  % Velocity continuity
  \displaystyle\int_{\Omega_w}
  &\left[\nabla \cdot \left( \density \velocity \right) - \velocity \cdot \nabla \density \right] q
  \,d\Omega_w \\ & =
  \displaystyle\int_{\Omega_w}
  \left[\nabla \cdot \left( \density \velocity \right) \right] q \,d\Omega_w
  -
  \displaystyle\int_{\Omega_w}
  \left[ q \velocity \cdot \nabla \density \right] \,d\Omega_w \notag \\
  & =
  \displaystyle\int_{\Omega_w}
  \left[\nabla q \cdot \left( \density \velocity \right) \right] \,d\Omega_w
  -
  \displaystyle\int_{\Gamma_\text{NS}}
  \left[ q \left( \density \velocity \right) \cdot \vec{n} \right] \,d\Gamma_\text{NS}
  -
  \displaystyle\int_{\Omega_w}
  \left[ q \velocity \cdot \nabla \density \right] \,d\Omega_w \notag
  \text{ ,}
\end{align}
%
while for \cref{eq:ns_conservation} we use the velocity test function $\vec{v}=\left[v_r, v_\phi, v_z\right]$
%
\begin{align}
  \displaystyle\int_{\Omega_w}
  \left[
    \left( \velocity \cdot \nabla \right) \left( \density\velocity \right) + \nabla \cdot \hydrostresstensor
  \right]
  \cdot \vec{v} \,d\Omega_w
  =
  \displaystyle\int_{\Omega_w} \vec{\force} \cdot \vec{v} \,d\Omega_w
\end{align}
%
which can be split into the domain and boundary integrals
%
\begin{align}
  % Conservation
  \displaystyle\int_{\Omega_w} \vec{\force} \cdot \vec{v} \,d\Omega_w =
  \displaystyle\int_{\Omega_w} &
  \left[
    \left( \velocity \cdot \nabla \right) \left( \density\velocity \right) \cdot\vec{v}
  \right]
  \,d\Omega_w
  -
  \displaystyle\int_{\Omega_w}
  \left[
  \hydrostresstensor \cdot \nabla\vec{v}
  \right]
  \,d\Omega_w \notag \\
  & +
  \displaystyle\int_{\Gamma_\text{NS}}
  \left[
  \vec{v} \cdot
  \hydrostresstensor
  \cdot\vec{n}
  \right]
  \,d\Gamma_\text{NS}
  \text{ ,}
\end{align}
%
with boundaries $\Gamma_\text{NS} = \Gamma_{w,c}+\Gamma_{w,t}+\Gamma_{p+m}$. The no-slip Dirichlet \gls{bc}
$\velocity = 0$ is applied to $\Gamma_{p+m}$, and the no normal stress $\hydrostresstensor\vec{n}=0$ is used
for $\Gamma_{w,c}$ and $\Gamma_{w,t}$.

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\cleardoublepage

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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%  leesedwards.tex
%
%  Lees Edwards Sliding Perioidic Boundary Conditions
%
%  Edinburgh Soft Matter and Statistical Physics Group and
%  Edinburgh Parallel Computing Centre
%
%  (c) 2016 The University of Edinburgh
%
%  Contributing authors:
%  Kevin Stratford (kevin@epcc.ed.ac.uk)
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Lees Edwards Sliding Periodic Boundary Conditions}

\subsection{Background}

The idea of introducing a Galilean transformation in a periodic
computation to model uniform shear was introduced by Lees and
Edwards in 1972 \cite{lees-edwards1972}. It was first adapted
to the lattice Boltzmann picture by Wagner and Pagonabarraga
in 2000 \cite{wagner-pagonabarraga2002}. The current
implementation follows the description of Adhikari
et~al.\ \cite{adhikari-desplat2005}.

\subsection{Distributions crossing the LE planes}

With the planes conceptually half-way between lattice sites, the
propagation moves some of the distributions (namely, those with
$c_x = \pm1$) between different sliding blocks. While the collision
stage is unaffected, action must be taken to adjust these
distributions when the LE boundary conditions are active. This
is done in a two-stage process of reprojection and interpolation
implemented between the collision and propagation.

\subsubsection{Reprojection}
The post-collision distributions with
$c_x = \pm 1$ at sites adjacent to a boundary are modified to
take account of the velocity jump $\pm u^{LE}_y$ between the sliding
blocks. In terms of the hydrodynamic moments we have:
\begin{eqnarray}
\rho &\rightarrow& \rho', \\
\rho u_\alpha &\rightarrow& (\rho u_\alpha)' \pm \rho' u^{LE}_\alpha, \\
S_{\alpha\beta} &\rightarrow&
S'_{\alpha\beta} \pm (\rho u_\alpha)' u^{LE}_\beta
\pm (\rho u_\beta)' u^{LE}_\alpha + \rho' u^{LE}_\alpha u^{LE}_\beta.
\end{eqnarray}
If we work with the changes to the moments, then the density is
unaffected $\delta\rho =  0$, the velocity is changed by
$\delta u_\alpha = \pm u^{LE}_\alpha$, with an analogous expression for
the change in the stress
\begin{equation}
\delta S_{\alpha\beta} =
\pm (\rho u_\alpha)' u^{LE}_\beta
\pm (\rho u_\beta)' u^{LE}_\alpha + \rho' u^{LE}_\alpha u^{LE}_\beta.
\end{equation}
We can then work out the change in the distributions by reprojecting
via Eq.\ref{eq:}, i.e.,
\begin{equation}
f_i \rightarrow f_i' + w_i \Bigg(
\frac{\rho \delta u_\alpha c_{i\alpha}}{c_s^2} +
\frac{\delta S_{\alpha\beta}Q_{i\alpha\beta}}{2c_s^4} \Bigg).
\end{equation}
A similar set of expressions are given for the order parameter
distributions in Adhikari \textit{et al.} \cite{adhikari-desplat2005}.

\subsubsection{Interpolation of reprojected distributions}
As the sliding blocks are displaced continuously with
$\delta y = u_y^{LE}t$, which is not in general a whole number of
lattice spacings, an interpolation is also required before
propagation can take place. Consider the situation  from
a stationary frame where the frame above has translated a small
distance to the right ($\delta y < \Delta y$). In the stationary frame
we must interpolate the distributions with $c_x = 1$ to a
`departure point' from which the propagation will move the
distribution exactly to the appropriate lattice site in the moving
frame. Schematically,
\begin{equation}
f_i'(x, y + \delta y, z) =
 (1 - \delta y) f_i^\star(x, y, z) +
\delta y f_i^\star(x, y + \Delta y, z),
\end{equation}
where $f_i'$ here represents the interpolated value and $f_i^\star$
is the reprojected post-collision quantity.

In the case where the displacement $\delta y > 1$, the relevant
lattice sites involved in the interpolation are displaced to the right
by the integral part of $\delta y$, while the relative weight given to
the two sites involves
the fractional part of $\delta y$. To take account of the periodic
boundary conditions
in the $y$-direction, all displacements are modulo $L_y$.

\subsection{Order parameter gradients}

When the order parameter gradient is computed at a given point
near the LE boundaries, the stencil may extend out of the stationary
frame. The process of interpolation is slightly different to that
for the distributions.

Here, we need to interpolate values in the moving frame to the position
which is seen by the rest frame. If a five-point
stencil is used to calculate the gradient at the central point
$(x,y,z)$ in the
rest frame. An interpolation of the $\phi$-field of the form
\begin{equation}
\phi'(x + \Delta x, y - \delta y, z) =
\delta y \phi(x + \Delta x,  y - \Delta y, z) +
(1 - \delta y) \phi(x + \Delta x, y, z)
\end{equation}
must then take place. In practice, the interpolated values are stored
in the offset buffer. As the gradient calculation requires $\phi$
values in the halo regions, interpolated values for the buffer halo
region are also required.

In parallel, the interpolation requires data from at most two adjacent
processes in the along-plane $y-$direction as long as the $\phi$ halo
regions are up-to-date. The rank of the processes involved is
determined by the displacement $\delta y$ as a function of time.

\subsection{Velocities and finite-difference fluxes}

For the finite-difference implementation, the velocity field at
the cell boundaries is required. At the planes, this means an
interpolation which takes the same form as that for
$\phi$ is needed.

In computing the advective fluxes between cells, we must then
allow for the presence of the planes. This is done by computing
separately the 'east' and 'west' face fluxes in the $x$-direction
(the flux that crosses the plane). Away from the planes, face-flux
uniqueness $f_{ijk}^e = f_{i+1jk}^w$ ensures conservation of order
parameter. However, at the planes, the non-linear combination of
velocity field and order parameter field does not in general give
rise to consistent fluxes. In order to restore conservation, we
reconcile east and west face fluxes at the plane by taking an
average
\begin{eqnarray}
f_{ijk}^{e\star}&=&{\textstyle\frac{1}{2}}(f_{ijk}^e + f_{i+1j'k}^w),\quad
f_{i+1j'k}^w  = \delta y f_{i+1 j-1 k}^w + (1 - \delta y) f_{i+1jk}^w,\\
f_{i+1 j k}^{w\star} &=& {\textstyle\frac{1}{2}}(f_{ij'k}^e + f_{i+1jk}^w),\quad
f_{ij'k}^e = (1 - \delta y) f_{ijk}^e + \delta y f_{ij+1k}^e.
\end{eqnarray}
In this way, we average the east face flux in the rest frame $f_{ijk}^e$
with the interpolated value of the west face flux in the moving frame
$f_{i+1j'k}^w$, and vice-versa. One can then easily show that when
integrated over the length of the plane, the average fluxes are
consistent, i.e.,
\begin{equation}
\sum_j (f_{ijk}^{e\star} - f_{i+1jk}^{w\star}) = 0 \quad\forall k,
\end{equation}
thus ensuring order parameter conservation.

A similar correction is required when computing the force on the fluid
directly via the divergence of the thermodynamic stress
$F_\alpha = \nabla_\beta P_{\alpha\beta}^{th}$. This is implemented
by computing fluxes of momentum at cell faces in each direction, and
then taking the divergence to compute the force. At the planes, an
averaging procedure is again used to ensure conservation of momentum.

% End section
\vfill
\pagebreak