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\begin{document} \huge

\noindent 1A. Calculate
\begin{eqnarray} \nonumber
  I\ =\ \int_{-\pi}^{+\pi} d\phi\, \delta\left[\phi\,\left(\phi^2 
     - \frac{\pi^2}{4}\right)\right]\,\cos(\phi)\,.
\end{eqnarray} 
Solution:
\begin{eqnarray} \nonumber
  I &=& \int_{-\pi}^{+\pi} d\phi\,\cos(\phi)\,\delta\left[\phi\,
        (\phi-\pi/2)(\phi+\pi/2)\right] \\ \nonumber
    &=& \int_{-\pi}^{+\pi} d\phi\,\cos(\phi)\\ \nonumber
  &~& \left[\frac{\delta(\phi)}{\pi^2/4} + 
  \frac{\delta(\phi-\pi/2)}{\pi^2/2} +
  \frac{\delta(\phi+\pi/2)}{\pi^2/2} \right]\\ \nonumber
  &=& \frac{4}{\pi^2}\,\cos(0)+
      \frac{2}{\pi^2}\,\cos\left(\frac{\pi}{2}\right)+
      \frac{2}{\pi^2}\,\cos\left(-\frac{\pi}{2}\right)\\ \nonumber
  &~& \\ \nonumber &=& \frac{4}{\pi^2}\,.
\end{eqnarray} 
\bigskip

\noindent 1B. Calculate
\begin{eqnarray} \nonumber
  I\ =\ \int_{-\pi}^{+\pi} d\phi\, \delta\left[\phi\,\left(\phi^2 
     - \frac{\pi^2}{16}\right)\right]\,\cos(\phi)\,.
\end{eqnarray} 
Solution:
\begin{eqnarray} \nonumber
  I &=& \int_{-\pi}^{+\pi} d\phi\,\cos(\phi)\,\delta\left[\phi\,
        (\phi-\pi/4)(\phi+\pi/4)\right] \\ \nonumber
    &=& \int_{-\pi}^{+\pi} d\phi\,\cos(\phi)\\ \nonumber
  &~& \left[\frac{\delta(\phi)}{\pi^2/16} + 
  \frac{\delta(\phi-\pi/4)}{\pi^2/8} +
  \frac{\delta(\phi+\pi/4)}{\pi^2/8} \right]\\ \nonumber
  &=& \frac{16}{\pi^2}\,\cos(0)+
      \frac{8}{\pi^2}\,\cos\left(\frac{\pi}{4}\right)+
      \frac{8}{\pi^2}\,\cos\left(-\frac{\pi}{4}\right)\\ \nonumber
  &~& \\ \nonumber &=& \frac{16+8\,\sqrt{2}}{\pi^2}\,.
\end{eqnarray} 
\bigskip

\noindent 2. The total charge on the infinitely thin conducting 
surface of a sphere of radius $r_0$ is $Q$. Write down the charge 
density $\rho(r,\theta,\phi)$.
\bigskip
Solution: 
\begin{eqnarray} \nonumber
  \rho(r,\theta,\phi)\ =\ \rho(r)\ =\ \frac{Q}{4\pi\,r_0}\,\delta(r-r_0)
\end{eqnarray} 
so that
\begin{eqnarray} \nonumber
 \int_{-1}^{+1} d\cos(\theta) \int_0^{2\pi} d\phi \int_0^{\infty}
 r^2\,dr\, \rho(r)\ =\ Q\,.
\end{eqnarray} 

\end{document}

