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\begin{center}
\textbf{Multiple Integration}

\textit{\textbf{Iteration of Double Integrals}}
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\textbf{Question}


Calculate the given double integral by iteration.

$\ds \int\!\!\!\int_T (x-3y) \,dA$

With $T$ being the triangle having vertices $(0,0)$, $(a,0)$ and
$(0,b)$.


\textbf{Answer}

\begin{eqnarray*}
& & \int\!\!\!\int_T (x-3y) \,dA\\
& = & \int_0^a \,dx \int_0^{b(1-(x/a))} (x-3y) \,dy\\
& = & \int_0^a \,dx \left. \left ( xy-\frac{3}{2}y^2 \right ) \right
|_{y=0}^{y=b(1-(x/a))}\\
& = & \int_0^a \left [ b \left ( x - \frac{x^2}{a} \right ) -
\frac{3}{2}b^2 \left ( 1- \frac{2x}{a} + \frac{x^2}{a^2} \right )
\right ] \,dx\\
& = & \left. \left ( b\frac{x^2}{2} - \frac{b}{a} \frac{x^3}{3} -
\frac{3}{2} b^2 x + \frac{3}{2} \frac{b^2x^2}{a} - \frac{1}{2}
\frac{b^2}{x^3}{a^2} \right ) \right |_0^a\\
& = & \frac{a^2b}{6} - \frac{ab^2}{2}
\end{eqnarray*}

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