%begin preamble
\documentclass[11pt,titlepage]{article}
\usepackage[margin=1in,noheadfoot]{geometry}
\usepackage[parfill]{parskip}
\usepackage{graphicx}
\usepackage{hyperref}
\usepackage{amssymb}
\usepackage{amsbsy}
\usepackage{wrapfig}
\usepackage{calligra}
\DeclareMathAlphabet{\mathcalligra}{T1}{calligra}{m}{n}
\DeclareFontShape{T1}{calligra}{m}{n}{<->s*[2.2]callig15}{}
\newcommand{\scriptr}{\hspace{-.1ex}\ensuremath{\mathcalligra{r}}\,}
\newcommand{\dslash}{d \hspace{-0.5ex}\rule[1.2ex]{0.6ex}{.1ex}}

\newcommand{\e}[1]{\times 10^{#1}}
\newcommand{\q}{\mathrm}
\newcommand{\where}[1][where]{\qquad\mbox{#1}\qquad}
\newcommand{\bs}[1]{\boldsymbol{#1}}
\newcommand{\bh}[1]{\hat{\boldsymbol{#1}}}
\newcommand{\bv}[1]{\vec{\boldsymbol{#1}}}
\newcommand{\bt}[1]{\tilde{\boldsymbol{#1}}}
\newcommand{\flow}[1]{\mathcal{E}_{#1}}
\newcommand{\flux}[1]{\Phi_{#1}}
%end preamble

\begin{document}
\begin{center}
  {\bf Arizona State University, Physics 311 \\
    Problem Set \#3, Rev. A, due Wednesday, 2014-02-05 }
\end{center}

\vspace{.2 in}

{\bf 1. a)} Plot a 2-D graph of the function $h(x,y)=2xy$ by drawing level
curves in the $xy$-plane.

\vspace{-1.2mm}

{\bf \quad b)} Plot a 3-D graph of the 2-D function $h(x,y)$, ie. the surface $\{
(x,y,z) ~|~ z=h(x,y) \}$.

\vspace{-1.2mm}

{\bf \quad c)} Calculate the gradient of $h$ at all points, and also the
divergence and curl of the gradient.  Plot the field lines and equipotentials
of the gradient.  Note that the equipotentials are curves, not surfaces, since
$h(x,y)$ is 2-dimensional.  Bonus: Calculate an equation for fieldline curves.

\vspace{-1.2mm}

{\bf \quad d)} Construct a 3-D function $g(x,y,z)$ which has $z=h(x,y)$ as one
level surface.  Calculate the gradient of this function to obtain a normal
vector to the suface $z=h(x,y)$ at all points on the surface.  Compare your
answer to the normal of the surface calculated using the cross product (hint:
calculate $\bs{da}$ on the surface).

\vspace{-1.2mm}

{\bf \quad e)} Show that there is a linear coordinate transformation
$(x,y)\to(x',y')$ that transforms the function $h'(x',y')=y'^2-x'^2$ into
$h(x,y)$.  We will find that both of these functions are \mbox{$m=2$
  cylindrical harmonics.}

\vspace{-1.2mm}

{\bf \quad f)} Show that $h_1(x,y)=2xy$ and $h_2(x,y)=y^2-x^2$ form a basis
for all functions which can be generated from $h_1$ and $h_2$ by rotating the
coordinates $(x,y)$ by an angle $\theta$ via the transformation $\bs r' =
R_\theta \,\bs r$, i.e. $\pmatrix{x'\cr y'} = \pmatrix {c_\phi &-s_\phi\cr
  s_\phi & c_\phi}\pmatrix{x\cr y}$, where $\matrix{ c_\phi=\cos\phi \cr
  s_\phi = \sin\phi}$.  In other words, show that if the coordinates $(x,y)$
of either function $h_1$ or $h_2$ are transformed by $R_\theta$, then the new
function is a linear combination of the old $h_1$ and $h_2$.

{\bf g)} Verify Stoke's theorem $\oint_{\partial R}\bs v\cdot\bs{dl} = \int_R
\bs\nabla\times\bs v\cdot \bs{da}$ using the vector field $\bs v(x,yz) = \bh x
yz$ over the surface $z=h_1(x,y)$, where $x^2+y^2<1$.

\vspace{.166 in}

Also, Griffiths 3ed[4ed] Ch.~1, \#%
%14, %p16, del vector \qquad
%17, %p18, div scalar \qquad
%28, %p28, 3 line integrals \qquad
%29, %p28, div thm \qquad
%31, %p31, FTC \qquad
%32, %p33, div thm \qquad
%33, %p36, stokes thm \qquad
%34, %p36, corollary \qquad
35[36], %p38, int parts \qquad
% gap: see HW 2
%43, %p49, delta integral \qquad
44[45], %p49, delta scale integral \qquad
45[46], %p49, delta derivative theta \qquad
46[47], %p52, pt charge \qquad
47[48], %p52, int 3d delta \qquad
48[49], %p52, int parts \qquad
60[61], %p56, thms by parts \qquad
61[62]. %p57, area integral \qquad
%62. %p57, div central power law \qquad

\end{document}
