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\begin{center}
\textbf{Multiple Integration}

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

\textbf{Question}


$$I=\int \! \int_D (5-x-y) \,dA$$
$D$ is the rectangle $0 \le x \le 3$, $0 \le y \le 2$.

$P$ is the partition of $D$ into six squares, each of side 1, as shown
below.

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\epsfig{file=MI-1Q-1.eps, width=60mm}
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For the following choices of points $(x_{ij}^*, y_{ij}^*)$, calculate
the Riemann sum for $I$
\begin{description}
\item{(a)}
The upper left corner of each square
\item{(b)}
The upper right corner of each square
\item{(c)}
The lower left corner of each square
\item{(d)}
The lower right corner of each square
\item{(e)}
The centre of each square

\item{(f)}
Evaluate $I$ by interpreting it as a volume.

\item{(g)}
Repeat 1(e), replacing $f(x,y)=5-x-y$ with $f(x,y)=e^x$. 

\end{description}


\textbf{Answer}

\begin{description}
\item{(a)}
$f(x,y)=5-x-y$
\begin{eqnarray*}
R & = & 1 \times [ f(0,1)+f(0,2)+f(1,1)+f(1,2)\\
& & +f(2,1)+f(2,2)]\\
& = & 4+3+3+2+2+1=15
\end{eqnarray*} 

\item{(b)}

\begin{eqnarray*}
R & = & 1 \times [f(1,1)+f(1,2)+f(2,1)+f(2,2)\\
& & +f(3,1)+f(3,2)]\\
& = & 3+2+2+1+1+0=9
\end{eqnarray*}

\item{(c)}

\begin{eqnarray*}
R & = & 1 \times [f(0,0)+f(0,1)+f(1,0)+f(1,1)\\
& & +f(2,0)+f(2,1)]\\
& = & 5+4+4+3+3+2=21
\end{eqnarray*}

\item{(d)}

\begin{eqnarray*}
R & = & 1 \times [f(1,0)+f(1,1)+f(2,0)+f(2,1)\\
& & +f(3,0)+f(3,1)]\\
& = & 4+3+3+2+2+1=15
\end{eqnarray*}

\item{(e)}

\begin{eqnarray*}
R & = & 1 \times
[f(\frac{1}{2},\frac{1}{2})+f(\frac{1}{2},\frac{3}{2})+ f(\frac{3}{2},
\frac{1}{2}) +f(\frac{3}{2}, \frac{3}{2})\\
& & + f(\frac{5}{2}, \frac{1}{2})
+ f(\frac{5}{2}, \frac{3}{2})]\\
& = & 4+3+3+2+2+1 =15
\end{eqnarray*}

\item{(f)}
$$I= \int \! \int_D (5-x-y) \,dA$$
is the volume of the solid shown below.

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\epsfig{file=MI-1A-1.eps, width=50mm}
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The solid is split into two pyramids by the vertical plane through the
$z$-axis and the point $(3,2,0)$.

One of these pyramids has a base in the plane $y=0$, and the other has
a base in the plane $z=0$.

The sum of the volumes of these pyramids gives $I$.
$$I = \frac{1}{3} \left ( \frac{5 +2}{2} (3)(2) \right ) + \frac{1}{3}
\left ( \frac{5+3}{2}(2)(3) \right ) =15$$

\item{(g)}

\begin{eqnarray*}
R & = & (e^{1/2}+e^{1/2} +e^{3/2}+ e^{3/2} + e^{5/2} + e^{5/2})\\
& \approx & 32.63
\end{eqnarray*} 
\end{description}

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