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\begin{center}
{Mathematical Physics --- PHZ 3113}\\\vspace {2mm} 
Vector Integration Homework \\ (February 20, 2013) \\
\end{center}
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\noindent Integrate (explicitly!) the force
\begin{eqnarray} \label{F}
  \vec{F} = - x_2\,\hat{x}_1 + x_1\,\hat{x}_2
\end{eqnarray} 
in anti-clockwise direction along the boundary of an
equilateral triangle in the $x_1-x_2$ plane. The triangle
has a side from $(0,0)$ to $(a,0)$ on the $x_1$ axis and 
closes in the upper half of the $x_1-x_2$ plane.
\medskip

Solution: The sides of the triangle are given by the
vectors
\begin{eqnarray} \nonumber
  \vec{s}_1 &=& \vec{0}\ =\ (0,0)\,,      \\ \nonumber
  \vec{s}_2 &=& a\,\hat{x}_1\,,           \\ \nonumber
  \vec{s}_3 &=& \frac{1}{2}\,a\,\hat{x}_1 
  + \frac{\sqrt{3}}{2}\,a\,\hat{x}_2\,.
\end{eqnarray} 
Therefore,
\begin{eqnarray} \nonumber
  \oint_{\triangle} d\vec{s}\cdot\vec{F}\ = \qquad
\end{eqnarray} 
\begin{eqnarray} \nonumber
  \int_{\vec{s}_1}^{\vec{s}_2} d\vec{s}\cdot\vec{F}+
  \int_{\vec{s}_2}^{\vec{s}_3} d\vec{s}\cdot\vec{F}+
  \int_{\vec{s}_3}^{\vec{s}_1} d\vec{s}\cdot\vec{F}
\end{eqnarray} 
where
\begin{eqnarray} \label{I1}
  \int_{\vec{s}_1}^{\vec{s}_2} d\vec{s}\cdot\vec{F} &=& 
  \int_0^a dx_1\,0\ =\ 0 \,, \\ \nonumber
  \int_{\vec{s}_2}^{\vec{s}_3} d\vec{s}\cdot\vec{F} &=& 
  \int_0^{a\sqrt{3}/2} dx_2\,x_1(x_2) \\ \nonumber 
  &-& \int_a^{a/2} dx_1\,x_2(x_1)
\end{eqnarray} 
with
\begin{eqnarray} \nonumber
  x_1(x_2) &=& a - \frac{x_2}{\sqrt{3}}\,,~~~
  x_2(x_1)\ =\ (a-x_1)\,\sqrt{3}\,.
\end{eqnarray} 
Hence,
\begin{eqnarray} \nonumber
  \int_0^{a\sqrt{3}/2} dx_2\,x_1(x_2) &=&
  \int_0^{a\sqrt{3}/2} dx_2\,\left(a-\frac{x_2}{\sqrt{3}}\right)
  \\ \nonumber =\
  \frac{\sqrt{3}}{2}\,a^2-\frac{\sqrt{3}}{8}\,a^2 &=&
  \frac{3\,\sqrt{3}}{8}\,a^2\,\\ \nonumber 
  - \int_a^{a/2} dx_1\,x_2(x_1) &=& \sqrt{3}
  \int_a^{a/2} dx_1\,\left(x_1-a\right) 
\end{eqnarray} 
\begin{eqnarray} \nonumber
  =\ a^2\,\left(\frac{\sqrt{3}}{8} - \frac{\sqrt{3}}{2} - 
  \frac{\sqrt{3}}{2} + \sqrt{3} \right) = \frac{\sqrt{3}}{8}\,a^2\,, 
\end{eqnarray} 
\begin{eqnarray} \label{I2}
  \int_{\vec{s}_1}^{\vec{s}_2} d\vec{s}\cdot\vec{F} &=& 
  \frac{\sqrt{3}}{2}\,a^2\,.
\end{eqnarray} \bigskip

Next, along the same lines
\begin{eqnarray} \nonumber
  \int_{\vec{s}_3}^{\vec{s}_1} d\vec{s}\cdot\vec{F} &=& 
  \int_{a\sqrt{3}/2}^0 dx_2\,x_1(x_2) \\ \nonumber 
  &-& \int_{a/2}^0 dx_1\,x_2(x_1)
\end{eqnarray} 
with
\begin{eqnarray} \nonumber
  x_1(x_2) &=& \frac{x_2}{\sqrt{3}}\,,~~~
  x_2(x_1)\ =\ x_1\,\sqrt{3}\,.
\end{eqnarray} 
Hence,
\begin{eqnarray} \nonumber
  \int_{a\sqrt{3}/2}^0 dx_2\,x_1(x_2) &=&
  \int_{a\sqrt{3}/2}^0 dx_2\,\frac{x_2}{\sqrt{3}}
  = -\frac{\sqrt{3}}{8}\,a^2\,, \\ \nonumber 
  - \int^0_{a/2} dx_1\,x_2(x_1) &=& \sqrt{3}
  \int_0^{a/2} dx_1\,x_1 = 
  \frac{\sqrt{3}}{8}\,a^2\,, 
\end{eqnarray} 
\begin{eqnarray} \label{I3}
  \int_{\vec{s}_2}^{\vec{s}_3} d\vec{s}\cdot\vec{F} &=& 0\,.
\end{eqnarray} 
Collecting the integrals (\ref{I1}), (\ref{I2}) and (\ref{I3}) 
we arrive at the final result
\begin{eqnarray} \nonumber
  \oint_{\triangle} d\vec{s}\cdot\vec{F} = \frac{\sqrt{3}}{2}\,a^2\,.
\end{eqnarray} 

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