\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\begin{document}
\parindent=0pt

\begin{center}
\textbf{Multiple Integration}

\textit{\textbf{Double Integrals}}
\end{center}

\textbf{Question}

$D$ is the disk $x^2+y^2 \le 25$

$P$ is the partition of the square $-5 \le x \le 5$, $-5 \le y \le 5$
into one hundred squares of dimensions $1 \times 1$, shown below

\begin{center}
\epsfig{file=MI-1Q-2.eps, width=70mm}
\end{center}

$$J = \int \! \int_D f(x,y) \,dA$$
where $f(x,y)=1$.

Approximate $J$ by calculating the Riemann sums $R(f,P)$ with the
given points $(x_{ij}^*, y_{ij}^*)$ in the small squares. Use of
symmetry will speed things up.
\begin{description}
\item{(a)}
$(x_{ij}^*, y_{ij}^*)$ is the corner of each square closest to the origin.
\item{(b)}
$(x_{ij}^*, y_{ij}^*)$ is the corner of each square farthest from the origin.
\item{(c)}
$(x_{ij}^*, y_{ij}^*)$ is the centre of each square.

\item{(d)}
Evaluate $J$

\item{(e)}
Repeat 2(c), replacing $f(x,y)=1$ with $f(x,y)=x^2+y^2$.

\end{description}


\textbf{Answer}

$$J = \int \! \int_D 1 \,dA$$

\begin{description}
\item{(a)}
$R=4 \times 1 \times [5+5+5+5+4]=96$

\item{(b)}
$R=4 \times 1 \times [4+4+4+3+0]=60$

\item{(c)}
$R=4 \times 1 \times [5+5+4+4+2]=80$

\item{(d)}
$J=$area of disk$=\pi(5^2)\approx 78.54$

\item{(e)}
$f(x,y)=x^2+y^2$.
\begin{eqnarray*}
R & = & 4 \times 1 \times [ f(\frac{1}{2},\frac{1}{2})+
f(\frac{3}{2},\frac{1}{2})+ f(\frac{5}{2},\frac{1}{2})+
f(\frac{7}{2},\frac{1}{2})+ f(\frac{9}{2},\frac{1}{2})\\
& & +f(\frac{1}{2},\frac{3}{2})+
f(\frac{3}{2},\frac{3}{2})+ f(\frac{5}{2},\frac{3}{2})+
f(\frac{7}{2},\frac{3}{2})+ f(\frac{9}{2},\frac{3}{2})\\
& & +f(\frac{1}{2},\frac{5}{2})+
f(\frac{3}{2},\frac{5}{2})+ f(\frac{5}{2},\frac{5}{2})+
f(\frac{7}{2},\frac{5}{2})\\
& & f(\frac{1}{2},\frac{7}{2})+
f(\frac{3}{2},\frac{7}{2})+ f(\frac{5}{2},\frac{7}{2})\\
& & +f(\frac{1}{2},\frac{9}{2})+
f(\frac{3}{2},\frac{9}{2})]\\
& = & 918
\end{eqnarray*}

\end{description}

\end{document}