<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}

\begin{document}

\parindent=0pt
\newcommand\ds{\displaystyle}
QUESTION
\begin{description}

\item[(a)]
Evaluate $\ds\int\!\!\int\textbf{A}\cdot\,\textbf{dS}$ for each of
the three right-angled triangular faces of the prism bounded by
$x=0,y=0,z=0$ and $x+y+z=1$ given that $\textbf{A}=(2z,-y,3x)$.

\item[(b)]
By using (a) and the divergence theorem, evaluate the surface
integral over the remaining face.

\end{description}

\bigskip

ANSWER
\begin{description}

\item[(a)]
$$ \ $$
\begin{center}
\setlength{\unitlength}{.5in}
\begin{picture}(5,3)

\put(0,0){\vector(1,0){3}} \put(0,0){\vector(0,1){3}}
\put(0,0){\vector(1,1){2}} \put(-.05,1.9){$\bullet$}
\put(1.9,-0.1){$\bullet$} \put(1.35,1.35){$\bullet$}
\put(0,1.95){\line(1,-1){2}} \put(0,2){\line(5,-2){1.4}}
\put(1.95,0){\line(-1,3){.5}} \put(-.2,0.1){0} \put(2.1,0.1){1}
\put(-0.2,2){1} \put(1.4,1.6){1} \put(3,0.1){$x$}
\put(1.9,2,1){$y$} \put(-0.2,3){$z$}



\end{picture}
\end{center}

$\textbf{A}=(2z,-y,3x),\ \ z=0$ plane,$\ \ \textbf{n}=(0,0,-1)$

$\begin{array}{l} \textrm{and the triangle}
\end{array}
\ \ \
\begin{array}{c}
\setlength{\unitlength}{.5in}
\begin{picture}(2,2.5)

\put(0,0.5){\vector(1,0){2}} \put(0.5,0){\vector(0,1){2}}
\put(0.5,1.5){\line(1,-1){1}} \put(1.5,.2){1} \put(.3,1.5){1}
\put(2,.6){$x$} \put(.6,2){$y$}


\end{picture}


\end{array}
\ \ \
\begin{array}{r}
\textrm{is the surface, with }z=0.
\end{array}$

\begin{eqnarray*}
I_z&amp;=&amp;\int_0^1\,dx\int_0^{1-x}\,dy(0,-y,3x)\cdot(0.0.-1)\\
&amp;=&amp;\int_0^1\,dx\int_0^{1-x}\,dy(-3x)\\ &amp;=&amp;\int_0^1(-3x)(1-x)\,dx\\
&amp;=&amp;\int_0^1(3x^2-3x)\,dx\\
&amp;=&amp;\left[x^3-\frac{3}{2}x^2\right]_0^1=-\frac{1}{2}.\\
\textrm{Similarly }\\
I_y&amp;=&amp;\int_0^1\,dz\int_0^{1-z}\,dz(2z,0,3x)(0,-1,0)=0\\
\textrm{and }\\
I_x&amp;=&amp;\int_0^1\,dy\int_0^{1-y}\,dz(2z,-y,0)(-1,0,0)\\
&amp;=&amp;\int_0^1\,dy\int_0^{1-y}(-2z)\,dz\\
&amp;=&amp;\int_0^1\,dy\left[-z^2\right]_0^{1-y}\\ &amp;=&amp;\,dy(-(1-y)^2)\\
&amp;=&amp;\int_0^1\,dy(-y^2+2y-1)=\left[-\frac{y^3}{3}+y^2-y\right]_0^1=-\frac{1}{3}
\end{eqnarray*}


\item[(b)]
 $\nabla \cdot\textbf{A}=-1\\
 \int\!\!\int_S\textbf{A}\cdot\textbf{dS}=\int\!\!\int\!\!\int_V\nabla\cdot \textbf{A}dV
 =-1.$ Volume $=-\frac{1}{6}$ because $\nabla\cdot\textbf{A}$ is
 constant.

 But
 $\int\!\!\int_s\textbf{A}\cdot\,\textbf{dS}
 =-\frac{1}{2}\cdot-\frac{1}{3}+I_4=-1\Rightarrow I_4=-\frac{1}{6}$
 where $I_4$ is the flux of A through the fourth side of the
 prism.

\end{description}

\end{document}
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