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\begin{document}
\begin{center}
  {\bf University of Kentucky, Physics 416G \\
    Problem Set \#3, Rev. B, due Friday, 2014-09-26 }
\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 field line
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 surface $z=h(x,y)$ 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 $\matrix{ c_\phi=\cos\phi \cr
  s_\phi = \sin\phi}$.

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

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{\bf 2. a)} Integrate $\int_{\mathcal{S}} \bs v \cdot \bs{da}$ where $\bs v =
\bh x\, x^2 + \bh y\, 2yz + \bh z\, xy$, \\
and $\mathcal{S}$ is the parallelogram in the figure to the right.  \\
Hint: remember the parameterization of HW1~\#1.

{\bf \quad b)} Integrate $\oint_{\partial\mathcal{S}} \bs v \cdot \bs{dl}$
along the boundary of $\mathcal{S}$.

{\bf \quad c)} Verify Stokes' theorem for the integral in part b.

\vspace{.25in}

{\bf 2$'$ [alternate or bonus] Fundamental Theorem of Differentials}: In class
we discussed an extension of the Fundamental Theorem of Calculus in higher
dimensions: $d\int_n \omega + \int_n d \omega = \omega$, where $n$ is a
`radial' coordinate, and $\int_n$ is an indefinite integral starting at $n=0$.
This theorem encompases both the Poincar\'e lemmas (indefinite integral:
potential theory), and the boundary theorems (definite integral: gradient,
curl, divergence theorems).

{\bf \quad a)} Show that this theorem reduces to the FTC for both functions
$f(x)$ and distributions $\rho(x)dx$ in one dimension.  What happens to the
constant of integration for indefinite integrals?

{\bf \quad b)} Apply this theorem to calculate the potential $V$ of an
conservative field $\bs E$, ie. $\bs E = -\nabla V$ when $\bs\nabla\times\bs E
= \bs 0$.  How does this relate to Stokes' theorem?

{\bf \quad c)} Apply this theorem to calculate the potential $\bs A$ of a
solenoidal field $\bs B$, ie. $\bs B = \bs\nabla\times\bs A$ when
$\bs\nabla\cdot\bs B = 0$.  Compare your answer with Griffiths 3ed[4ed]
problem 5.51[53].  How does this relate [if at all] to Gauss' theorem?

\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

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