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\begin{center}
\textbf{Multiple Integration}

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

\textbf{Question}


Calculate the given double integral by iteration in the region
defined by the given curves.

$\ds \int\!\!\!\int_D x\cos y \,dA$

With $D$ being the region bounded by $y=1-x^2$ and the coordinate
axes, in the first quadrant.


\textbf{Answer}

\begin{eqnarray*}
I &amp; = &amp; \int\!\!\!\int_D x\cos y \,dA\\
&amp; = &amp; \int_0^1 x \,dx \int_0^{1-x^2} \cos y \,dy\\
&amp; = &amp; \int_0^1 x \,dx \left. (\sin y) \right |_{y=0}^{y=1-x^2}\\
&amp; = &amp; \int_0^1 x \sin (1-x^2) \,dx\\
\textrm{Let }u &amp; = &amp; 1-x^2\\
\Rightarrow du &amp; = &amp; -2x dx\\
&amp; &amp;\\
\Rightarrow I &amp; = &amp; -\frac{1}{2} \int_0^1 \sin u \,du\\
&amp; = &amp; \left. \frac{1}{2} \cos u \right |_1^0 \\
&amp; = &amp; \frac{1 - \cos(1)}{2}
\end{eqnarray*} 

\end{document}



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