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


{\bf Question}

The following equations are written in terms of spherical polar
co-ordinates $(r,\theta,\phi)$. What surfaces or curves do they
represent?
\begin{description}
\item[(a)] $r\cos(\theta)=1$;
\item[(b)] $\sin(\theta)=\frac{\pi}{4}$;
\item[(c)] $\theta=\frac{\pi}{2}$, $r\cos(\phi)=0$;
\item[(d)] $\theta=\frac{\pi}{4}$, $r\cos(\theta)=1$.
\end{description}


{\bf Answer}

$$\\$$
\begin{center}
$\begin{array}{c} \textrm{Spherical polar co-ordinates
}(r,\theta,\phi).\\ r \geq 0, \ 0 \leq \theta \leq \pi \rm{\ and\
} 0 \leq \phi \leq 2\pi\\ \rm{with\ }x = r \sin \theta \cos \phi,
y = r \sin \theta \sin \phi\\ \rm{and\ } z = r \cos \theta.
\end{array}
\ \ \
\begin{array}{c}
\epsfig{file=158-10-8.eps, width=40mm}
\end{array} $
\end{center}


\begin{description}
\item[(a)] $r\cos(\theta)=1 \Rightarrow z = 1$  Since x and y are
arbitrary, we have the plane parallel to the xy-plane at height z
= 1.
\begin{center}
\epsfig{file=158-10-9.eps, width=50mm}
\end{center}

\item[(b)] $\sin(\theta)=\frac{\pi}{4}$.  Since $0 \leq \theta
\leq \pi$ there are two solutions, $\theta \approx 51.8^\circ$ or
$\theta = 180 - 51.8 = 128.2^\circ$.  Each of these gives a cone:
\begin{center}
$ \begin{array}{c} \underline{\theta = 51.8^o}
\end{array}
\ \
\begin{array}{c}
\epsfig{file=158-10-10.eps, width=30mm}
\end{array}
\ \ \
\begin{array}{c}
\underline{\theta=128.2^o}
\end{array}
\ \
\begin{array}{c}
\epsfig{file=158-10-11.eps, width=30mm}
\end{array} $
\end{center}

The required curve / surface is the union of all possible
solutions and so we obtain the double cone:
\begin{center}
\epsfig{file=158-10-12.eps, width=30mm}
\end{center}


\item[(c)] $\theta=\frac{\pi}{2} \Rightarrow z = r \cos\frac{\pi}{2} = 0$, so
the surface / curve lies in the xy-plane.  $r\cos(\phi)=0
\Rightarrow$ either $ r = 0$ or $ \cos \phi = 0$

${\underline {r = 0\ \ }} (x,y,z) = (0,0,0)$ so we have the single
point, the origin

${\underline {\cos \phi = 0, \ \ (r \not= 0)\ \ }}$ $x = r\sin
\theta \cos \phi = 0$ and $\cos \phi = 0 \Rightarrow \sin \phi =
+1 {\rm \  or\ }-1$ so the required curve is just the y-axis.
\begin{center}
\epsfig{file=158-10-13.eps, width=80mm}
\end{center}


\item[(d)]If $\theta=\frac{\pi}{4}$, $r\cos(\theta)=1 \Rightarrow r\cos \frac{\pi}{4} = 1
\Rightarrow r = \sqrt2$.

Hence $ z = r \cos \theta = \sqrt2 \left( \frac{1}{\sqrt2}\right)
= 1$ This gives the circle lying in the plane $z = 1$, whose
centre lies on the the z-axis.  The circle has radius
$r\sin\frac{\pi}{4} = \sqrt2\left(\frac{1}{\sqrt2}\right) = 1$.
\begin{center}
\epsfig{file=158-10-14.eps, width=50mm}
\end{center}


\end{description}

\end{document}