\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\begin{document}


{\bf Question}

The following equations are written in terms of cylindrical
co-ordinates $(\rho,\phi,z)$. What surfaces or curves do they
represent?
\begin{description}
\item[(a)] $\phi=\frac{\pi}{4}$, $z=2$;
\item[(b)] $\rho^2+z^2=9$;
\item[(c)] $\rho=z\tan(\alpha)$ where $-\frac{\pi}{2}<\alpha<\frac{\pi}{2}$ is a
real constant;
\item[(d)] $\rho\sin(\phi)=1$, $z=0$.
\end{description}

{\bf Answer}

\begin{center}
$ \begin{array}{c}
\textrm{Cylindrical co-ordinates }(\rho,\phi,z).\\
\rho \geq 0 \rm{\ and\ }0 \leq \phi \leq 2\pi\\
\textrm{Also, }x = \rho \cos \phi \textrm{ and }y = \rho \sin \phi
\end{array}
\ \ \ 
\begin{array}{c}
\epsfig{file=158-10-1.eps, width=40mm}
\end{array} $
\end{center}


\begin{description}
\item[(a)]
$$\\$$

$ \begin{array}{l}
\phi=\frac{\pi}{4} \rm{\ and\ } z=2\\
\textrm{This gives a half line at height }z = 2\\
\textrm{ in the direction } \phi =\frac{\pi}{4}\\
(\rm{i.e.\ }x = y)
\end{array}
\ \
\begin{array}{c}
\epsfig{file=158-10-2.eps, width=40mm}
\end{array} $

\item[(b)] $\rho^2+z^2=9$

Now $x^2 + y^2 = \rho^2 \cos^2 \phi + \rho^2 \sin^2 \phi = \rho^2$

So we have $(x^2 + y^2) + z^2 = 9$ or $x^2 + y^2 + z^2 = 3^2$
which defines a sphere centre the origin of radius 3.

\begin{center}
\epsfig{file=158-10-3.eps, width=50mm}
\end{center}

\item[(c)] $\rho=z\tan(\alpha)$ for $-\frac{\pi}{2}<\alpha<\frac{\pi}{2}$

we have the righthanded triangle

\begin{center}
$ \begin{array}{c}
\epsfig{file=158-10-6.eps, width=40mm}
\end{array}
\ \ \
\begin{array}{l}
\textrm{so that }\tan ( \alpha ) = \frac{\rho}{z}\\
\textrm{and hence }\rho = z\tan (\alpha).
\end{array} $
\end{center}  
If we now let $\phi$ vary as $0 \leq \phi \leq
2\pi$, we obtain a cone angle $\alpha$.
\begin{center}
\epsfig{file=158-10-4.eps, width=50mm}
\end{center}

\item[(d)] $\rho\sin(\phi)=1$, $z=0$.  Since $z=0$ we restrict to
the xy plane.
\begin{center}
$ \begin{array}{c}
\epsfig{file=158-10-7.eps, width=40mm}
\end{array}
\ \ \
\begin{array}{l}
\textrm{Now from the triangle we have}\\
\sin \phi = \frac{y}{\rho}\\
\textrm{and so }y = \rho \sin \phi.  
\end{array} $
\end{center}
Hence $\rho \sin \phi = 1 \Rightarrow y =
1$, and letting the x vary we obtain the line $y = 1$ in the
xy-plane.
\begin{center}
\epsfig{file=158-10-5.eps, width=50mm}
\end{center}

\end{description}


\end{document}