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\begin{document}
{\bf\large Math 425}\hfill {\bf\large Fall 1997}

\begin{center}
{\bf Example and Exercises about Line Integrals}
\end{center}

{\bf Example:}

Compute the line integral $\int_{\gamma} \re(z) dz$, where $\gamma$
is the straight line segment joining $-i$ to $2+i$.

i)\hfill \begin{minipage}[t]{18cm}
A parametrization of the curve $\gamma$ can be found by use of Example~1 in
Section~1.6 of
the textbook:
\[
   \gamma(t)=t(2+i)+(1-t)(-i)=(2+2i)t-i=2t+i(2t-1) \ \mbox{\rm for } t\in[0,1]
\]
Its derivative is given as $\gamma'(t)=2+2i$.
\end{minipage}

ii) \hfill \begin{minipage}[t]{18cm}
Now we can compute the line integral by use of the definition as given on
top of page~60:
\begin{eqnarray*}
   \int_{\gamma}\re(z)
dz&=&\int_0^1\re\big(\gamma(t)\big)\gamma'(t)dt=\int_0^1 2t(2+2i)dt\\
                        &=&\int_0^1 (4+4i)t dt =
                           \Big((2+2i)t^2\Big)\Big|^1_0=2+2i
\end{eqnarray*}
\end{minipage}


\vspace*{.5cm}
{\bf Exercises:}

1)\hfill\begin{minipage}[t]{18cm}
Let $\gamma_1$ be the circle of radius 1 centered at $2i$ oriented
counterclockwise and
let $\gamma_2$ be the straight line segment joining $i$ to $3i$.
\\
Sketch the curves $\gamma_1$ and $\gamma_2$ and compute the following line
integrals
\[
    \int_{\gamma_1}\im(z)dz \;\ \mbox{\rm and } \int_{\gamma_2}\im(z)dz
\]
\end{minipage}

\vspace*{.2cm}
2)\hfill\begin{minipage}[t]{18cm}
Let $\gamma_1$ be the straight line segment joining $1+2i$ to $3+2i$. Let
$\gamma_2$ be
the counterclockwise oriented semicircle of radius 1 centered at $2+2i$ and
lying above
the line segment $\gamma_1$.
Then $\gamma_1+\gamma_2$ is a piecewise smooth, closed curve.
\\
Sketch $\gamma_1+\gamma_2$ and compute the line integral
\[
    \int_{\gamma_1+\gamma_2}\overline{z}dz:=\int_{\gamma_1}\overline{z}dz
                                           +\int_{\gamma_2}\overline{z}dz.
\]
\end{minipage}

\end{document}
