\documentclass[a4paper,12pt]{article}
\setlength\oddsidemargin{0pt} \setlength\evensidemargin{0pt}
\setlength\topmargin{0pt}
\begin{document}
\parindent=0pt
{\bf Exam Question

Topic: SurfaceIntegral}

Let $S$ be part of the curved surface of a cylinder, specified by
$$y^2+z^2=1,\ z>0,\ 0\le x\le1.$$ Evaluate the surface integral
$$\int \!\!\! \int_S
\left(\sin(xyz)\mathbf{i}+z\mathbf{j}+y^2\mathbf{k}\right)\cdot\mathbf{dS}.$$

 \vspace{0.5in}

{\bf Solution}

The equation of the cylinder is $F(x,y,z)=y^2+z^2-1=0.$

Differentiating gives $\displaystyle \frac{\partial F}{\partial
x}=0;\ \ \frac{\partial F}{\partial y}=2y;\ \ \frac{\partial
F}{\partial z}=2z.$

So $\displaystyle
\mathbf{dS}=\frac{2y\mathbf{j}+2z\mathbf{k}}{|2z|}\,
dxdy=\left(\frac{y}{z}\mathbf{j}+\mathbf{k}\right) dxdy\ \
\mathrm{as}\ \ z>0.$

So
$\left(\sin(xyz)\mathbf{i}+z\mathbf{j}+y^2\mathbf{k}\right)\cdot\mathbf{dS}=(y+y^2)dxdy.$

The surface integral then becomes $$\int_0^1\,
dx\int_{-1}^1(y+y^2)\, dy=\frac{2}{3}.$$




\end{document}