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      <text>27.4.2.3 Diamond and Square Example</text>
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      <text><![CDATA[<h4 class="HELM_example">Example</h4>
<p>The diamond shaped region \(A\) in Figure (a) is bounded by the lines \(x+2y=2\), \(x-2y=2\), \(x+2y=-2\) and \(x-2y=-2\). We wish to evaluate the integral \[\displaystyle I=\displaystyle\int\!\!\displaystyle\int_A\left( 3x+6y\right)^2\mathrm{d}A\] over this region. Since the region \(A\) is neither vertically nor horizontally simple, evaluating \(I\) without changing coordinates would require separating the region into two simple triangular regions. So we use a change of coordinates to transform \(A\) to a square region in Figure (b) and evaluate \(I\).</p>

<figure class="HELM_figure">
<img class="img-fluid" src="@@PLUGINFILE@@/Fig364_2.svg" alt="(a) diamond on 2d axis with sides x±2y=±2. 
(b) square on 2d axis with sides u/v=2/-2." width="544" height="169">
<figcaption>Figure (a) (Left) and Figure (b) (Right)</figcaption>
</figure>
<h4 class="HELM_solution">Solution</h4>
<p>By considering the equations of the boundary lines of region \(A\) it is easy to see that the change of coordinates \[\mathrm{d}u=x+2y\qquad(1)\] \[v=x-2y\qquad(2)\] will transform the boundary lines to \(u=2\), \(u=-2\), \(v=2\) and \(v=-2\). These values of \(u\) and \(v\) are the new limits of integration. The region \(A\) will be transformed to the square region \(A^\prime\) shown above. We require the inverse transformations so that we can substitute for \(x\) and \(y\) in terms of \(u\) and \(v\). By adding (1) and (2) we obtain \(u+v=2x\) and by subtracting (1) and (2) we obtain \(u-v=4y\), thus the required change of coordinates is \(\displaystyle x=\frac{1}{2}\left( u+v\right)\qquad y=\frac{1}{4}\left( u-v\right)\) Substituting for \(x\) and \(y\) in the integrand \((3x+6y)^2\) of \(I\) gives \[ \left(\frac{3}{2}(u+v)+\frac{6}{4}(u-v)\right)^2=9u^2 \] We have the new limits of integration and the new form of the integrand, we now require the Jacobian. The required partial derivatives are \[\displaystyle \frac{\partial x}{\partial u}=\frac{1}{2}\qquad \frac{\partial x}{\partial v}=\frac{1}{2}\qquad \frac{\partial y}{\partial u}=\frac{1}{4}\qquad \frac{\partial y}{\partial v}=-\frac{1}{4} \] Then the Jacobian is \[\displaystyle J(u,v)=\left| \begin{array}{cc} \displaystyle \frac{1}{2}&amp;\displaystyle \mbox{ }\frac{1}{2}\\ &amp;\\ \displaystyle \frac{1}{4}&amp;\displaystyle -\frac{1}{4} \end{array}\right|=-\frac{1}{4} \] Then \(d A^\prime=|J(u,v)|d A=\displaystyle \frac{1}{4}d A\). Using the new limits, integrand and the Jacobian, the integral can be written \(I=\displaystyle\int_{-2}^{2}\displaystyle\int_{-2}^{2}\frac{9}{4}u^2\;\mathrm{d}u\,\mathrm{d}v.\) You should evaluate this integral and check that \(I=48\).</p>]]></text>
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