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\begin{document} \huge

\begin{center}
{Mathematical Physics --- PHZ 3113}\\\vspace {2mm} 
Classwork 8 (February 27, 2013) \\ 
Cylindrical Coordinates
\end{center}
\setcounter{equation}{0}

\noindent 1. Use cylindrical coordinates to calculate
the area of a circle of radius $R$.

\noindent Solution: 
\begin{eqnarray} \nonumber
  A &=& \int_{S_{\rm circle}} d^2x\ =\ 
  \int_{x^2+y^2\le R^2} dx\,dy\\ \nonumber  
  &=& \int_0^{2\pi}d\phi\int_0^R \rho\,d\rho\ 
   =\ 2\pi\,\frac{1}{2}\,R^2\ =\ \pi\,R^2\,. 
\end{eqnarray} 
\bigskip

\noindent 2. Calculate 
\begin{eqnarray} \nonumber
  \nabla\times\hat{z}\,\ln(\rho)
\end{eqnarray} 
in cylindrical coordinates.
\bigskip

\noindent Solution: 
\begin{eqnarray} \nonumber
  \nabla\times\hat{z}\,\ln(\rho)\ =\ 
  \hat{\rho}\times\hat{z}\,\frac{1}{\rho}\ =\ 
  - \frac{\hat{\phi}}{\rho}\,.
\end{eqnarray} 
\bigskip

\noindent 3. Show Oersted's law
\begin{eqnarray} \nonumber
  \oint \vec{H}\cdot d\vec{r}\ =\ I
\end{eqnarray} 
for the magnetic potential 
\begin{eqnarray} \nonumber
  \vec{A} &=& -\hat{z}\,\frac{\mu_0\,I}{2\pi}\,\ln(\rho)\,,~~
  \vec{B}\ =\ \nabla\times\vec{A}\,, \\ \nonumber
  \vec{H} &=& \mu_0^{-1}\vec{B}\,.
\end{eqnarray} 
\bigskip

\noindent Solution: 
\begin{eqnarray} \nonumber
  &~&\vec{H}\ =\ \frac{I\,\hat{\phi}}{2\pi\,\rho}\,,\\ \nonumber
  &~&\oint \vec{H}\cdot d\vec{r}\ =\ \int_0^{2\pi} \rho\,d\phi\,
     \frac{I}{2\pi\,\rho}\ =\ I\,.
\end{eqnarray} 
\bigskip 

\noindent 4. Find the acceleration $\vec{a}$ in cylindrical coordinates.
\bigskip

\noindent Solution: 
\begin{eqnarray} \nonumber
  \vec{v} &=& \dot{\rho}\,\hat{\rho}+\rho\,\dot{\phi}\,\hat{\phi}+\dot{z}\,\hat{z}
  \\ \nonumber \vec{a} &=& \dot{\vec{v}}\ =\ \ddot{\rho}\,\hat{\rho}
  +\dot{\rho}\,\dot{\hat{\rho}}+\dot{\rho}\,\dot{\phi}\,\hat{\phi}
  +\rho\,\ddot{\phi}\,\hat{\phi}+\rho\,\dot{\phi}\,\dot{\hat{\phi}}
  +\ddot{z}\,\hat{z}\,,\\ \nonumber
  \dot{\hat{\rho}} &=& \dot{\phi}\,\hat{\phi}\,,~~~
  \dot{\hat{\phi}}\ =\ -\dot{\phi}\,\hat{\rho}\,,\\ \nonumber 
  \vec{a} &=& \left(\ddot{\rho}-\rho\,\dot{\phi}^2\right)\,\hat{\rho}
  + \left(\rho\,\ddot{\phi}+2\,\dot{\rho}\,\dot{\phi}\right)\,\hat{\phi}
  + \dot{z}\,\hat{z}\,.
\end{eqnarray} 
\bigskip \newpage

\begin{center}
{Mathematical Physics --- PHZ 3113}\\\vspace {2mm} 
Classwork 9 (March 1, 2013) \\ 
Cylindrical Coordinates
\end{center}
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\noindent 5. Completion of the square: We have
\begin{eqnarray} \nonumber
  x^2 + b\,x + c
\end{eqnarray} 
and want this in the form
\begin{eqnarray} \nonumber
  x'^{\,2} + c'\ .
\end{eqnarray} 
What are the values of $x'$ and $c'$?
\bigskip

\noindent Solution: 
\begin{eqnarray} \nonumber
  x'\ =\ x+\frac{b}{2}\,,~~~c'\ =\ c-\frac{b^2}{4}\,.
\end{eqnarray} 
\bigskip

\noindent 6. In cylindrical coordinates the equation of 
an ellipse is given by
$$ \frac{p}{\rho} = 1 + \epsilon\,\cos(\phi)\,,~~p>0 $$ 
with Cartesian coordinates $x=\rho\,\cos(\phi)$ and $y=\rho\,
\sin(\phi)$. Assume $0<\epsilon<1$ for the eccentricity and 
transform the solution into the form
$$ \frac{x'^2}{a^2}+\frac{y'2}{b^2} = 1\ .$$ 
This means, derive the definitions of $x'$, $y'$, major half-axis
$a$ and minor half-axis $b$ in terms of $x$, $y$, $p$ and~$\epsilon$. 
\bigskip

\noindent Solution: Using $\cos\theta=x/\rho$, the initial equation 
becomes
\begin{eqnarray} \nonumber
  p &=& \rho\,(1+\epsilon\,\cos\theta)\ =\ \rho\,(1+\epsilon\,x/\rho) 
        \\ \nonumber
    &=& \rho + \epsilon\,x~~{\rm or} ~~\rho = p - \epsilon\,x\,. 
\end{eqnarray} 
Squaring both sides of the last equation:
$$ x^2 + y^2 = p^2 -2\,p\,\epsilon\,x + \epsilon^2\,x^2\,. $$
Bringing all terms with $x$ or $y$ to one side,
$$ x^2\,(1-\epsilon^2) + 2\,p\,\epsilon\,x + y^2 = p^2\,, $$
$$ x^2 + \frac{2\,p\,\epsilon}{1-\epsilon^2}\,x + 
   \frac{y^2}{1-\epsilon^2} = \frac{p^2}{1-\epsilon^2}\,, $$
According to the recipe for completion of the square we substitute
$$ x' = x + \frac{p\,\epsilon}{1-\epsilon^2} $$ 
and obtain
\begin{eqnarray} \nonumber
  x'{\,^2}\, + \frac{y^2}{1-\epsilon^2} &=& \frac{p^2}{1-\epsilon^2}
  + \left(\frac{p\,\epsilon}{1-\epsilon^2}\right)^2\\ \nonumber =\
   \frac{p^2\,(1-\epsilon^2) + p^2\,\epsilon^2}{(1-\epsilon^2)^2}
  &=& \frac{p^2}{(1-\epsilon^2)^2}\,. 
\end{eqnarray} 
Multiplying with $(1-\epsilon^2)^2/p^2$, this becomes
$$ x'{\,^2}\, \frac{(1-\epsilon^2)^2}{p^2} +y^2 \frac{1-\epsilon^2}{p^2} 
   = 1\,. $$ 
With the definitions 
$$a=\frac{p}{1-\epsilon^2}~~~{\rm and}~~~b=\frac{p}{\sqrt{1-\epsilon^2}}$$ 
and $y'=y$ this reads
$$ \left(\frac{x'}{a}\right)^2+\left(\frac{y'}{b}\right)^2 = 1\,. $$ 
\bigskip

\noindent 7. Use the definition
$$ \left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2 = 1\,. $$ 
to calculate the area of an ellipse. Hint: Make a substitution, so that 
it becomes reduced to the area of a circle.
\bigskip

\noindent Solution: We want to calculate
\begin{eqnarray} \nonumber
  A\ =\ \int_{\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2
        \le 1} dx\,dy\,.
\end{eqnarray} 
With the substitution $y=b\,y'/a$ this becomes
\begin{eqnarray} \nonumber
  A\ =\ \frac{b}{a} \int_{x^2+y'^{2}\le a^2} dx\,dy'\ =\
  \frac{b}{a}\,\pi\,a^2\ =\ \pi\,a\,b\,.
\end{eqnarray} 
\end{document}
