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  <question type="description">
    <name>
      <text>27.1.4.1 Integration Over Rectangular Areas Information</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<h3>Integration Over Rectangular Areas</h3>
<p>Consider the double integral <br> \(\qquad\displaystyle I=\displaystyle\int_{x=0}^{5}\displaystyle\int_{y=-1}^{1}\left( 2x+y\right) \ \, \mathrm{d}y \mathrm{d}x.\) <br> This represents an integral over the rectangle shown below. </p>

<figure class="HELM_figure">
    <img src="@@PLUGINFILE@@/Figure%205.svg" alt="Rectangle on xy axes from x=0 to x=5 and y=-1 to y=1." width="337" height="151" class="img-fluid atto_image_button_text-bottom">
</figure>
</p>

<p>Here, the inner integral is <br>\(\qquad\displaystyle g\left( x\right) =\displaystyle\int_{-1}^{1}\left( 2x+y\right) \, \mathrm{d}y,\) 
<br> 
and the outer integral is <br>\(\qquad\displaystyle I=\displaystyle\int_{x=0}^{5}g\left( x\right) \, \mathrm{d}x.\)</p>
<p>Looking in more detail at the inner integral 
<br>
\(\qquad\displaystyle g\left( x\right) =\displaystyle\int_{-1}^{1}\left( 2x+y\right) \, \mathrm{d}y,\) 
<br>
the function \(\left( 2x+y\right)\) can be integrated with respect to \(y\) (keeping \(x\) constant) to give \(2xy+\frac{1}{2}y^{2}+C\) (where \(C\) is a constant and can be omitted as the integral is a definite integral) i.e.</p>
<p>\(\qquad\displaystyle g\left( x\right) =\left[ 2xy+\frac{1}{2}y^{2}\right] _{-1}^{1}=\left( 2x+ \frac{1}{2}\right) -\left( -2x+\frac{1}{2}\right) =2x+\frac{1}{2}+2x-\frac{1% }{2}=4x.\) 
<br>
This is a function of \(x\) as expected. This inner integral can be placed into the outer integral to get 
<br>
\(\qquad\displaystyle I=\displaystyle\int_{x=0}^{5}4x\ \, \mathrm{d}x,\) <br> 
which becomes 
<br>
\(\qquad\displaystyle I=\left[\phantom{\dfrac{1}{1}}\!\! 2x^{2}\right] _{0}^{5}=2\times 5^{2}-2\times 0^{2}=2\times 25-0=50.\)</p>
<p>Hence the double integral <br> \(\qquad\displaystyle I=\displaystyle\int_{x=0}^{5}\displaystyle\int_{y=-1}^{1}\left( 2x+y\right) \ \, \mathrm{d}y \mathrm{d}x =50.\)
<br> 
<div class="HELM_keypoint">
<h4>Key Point</h4>
<p>When evaluating a double integral, evaluate the inner integral first an\mathrm{d}substitute the result into the outer integral.</p>
</div>


<hr>
<h4 class="HELM_example">Example</h4>
<p>Evaluate the double integral \(\displaystyle I=\displaystyle\int_{x=-1}^{2}\displaystyle\int_{y=-2}^{3}x^{2}y\, \mathrm{d}y\mathrm{d}x.\) 
This integral is evaluated over the area shown below. </p>
<figure class="HELM_figure">
     <img src="@@PLUGINFILE@@/Figure%206.svg" alt="Rectangle on xy axes from x=-1 to x=2 and y=-2 to y=3." width="325" height="232" class="img-fluid atto_image_button_text-bottom">
</figure>
</p>

<h4 class="HELM_solution">Solution</h4>
<p>Here, the inner integral is <br>
\(\qquad\displaystyle 
g\left( x\right) =\displaystyle\int_{y=-2}^{3}x^{2}y\ \, \mathrm{d}y=\left[ x^{2}\dfrac{y^{2}}{2}\right]
_{-2}^{3}=\dfrac{9}{2}x^{2}-\dfrac{4}{2}x^{2}=\dfrac{5}{2}x^{2},\) <br>
and hence the outer integral is
<br>\(\qquad\displaystyle 
I=\displaystyle\int_{x=-1}^{2}\dfrac{5}{2}x^{2}\, \mathrm{d}x=\left[ \dfrac{5}{2}\dfrac{1}{3}x^{3}
\right] _{-1}^{2}=\dfrac{5}{6}\times 8-\dfrac{5}{6}\left( -1\right) =\dfrac{
15}{2}.
\) 
</p>


<hr>
<h4 class="HELM_example">Example</h4>
<p>Use the above approach to evaluate the double integral <br>\(\qquad\displaystyle I=\displaystyle\int_{x=0}^{5}\displaystyle\int_{y=-1}^{1}x^{2}\cos \frac{\pi y}{2}\ \, \mathrm{d}y \mathrm{d}x.\) <br> Note that the limits are the same as in a previous case but that the function itself has changed.</p>

<h4 class="HELM_solution">Solution</h4>
<p>The inner integral is<br>
\(\qquad\displaystyle \displaystyle\int_{y=-1}^{1}x^{2}\cos \frac{\pi y}{2}\, \mathrm{d}y=\left[ \frac{2}{\pi
}x^{2}\sin
\frac{\pi y}{2}\right] _{-1}^{1}=\frac{2}{\pi }x^{2}1-\frac{2}{\pi }
x^{2}\left( -1\right) =\frac{4}{\pi }x^{2},\)
<br>
so the outer integral becomes
<br>
\(\qquad\displaystyle 
I=\displaystyle\int_{x=0}^{5}\frac{4}{\pi }x^{2}\, \mathrm{d}x=\left[ \frac{4}{3\pi
}x^{3}\right] _{0}^{5}=\frac{4}{3\pi }125-\frac{4}{3\pi
}0=\frac{500}{3\pi }\approx 53.1.
\)<br>
</p>
<p>Clearly, variables other than \(x\) and \(y\) may be used.
</p>

<hr>
<h4 class="HELM_example">Example</h4>
<p>Evaluate the double integral <br>\(\qquad\displaystyle I=\displaystyle\int_{s=1}^{4}\displaystyle\int_{t=0}^{\pi }s\sin t\ d t\mathrm{d}s.\) 
</p>

<h4 class="HELM_solution">Solution</h4>
<p>This integral becomes (dispensing with the step of
formally writing the inner integral),
\begin{eqnarray*}
I &=&\displaystyle\int_{s=1}^{4}\left[-s\cos t\right] _{0}^{\pi }d
s=\displaystyle\int_{1}^{4}\left[ -s\cos \pi +s\cos 0\right] \mathrm{d}s=\displaystyle\int_{1}^{4}\left[ -s\left( -1\right) +s\left(
1\right) \right] \mathrm{d}s \\
&=&\displaystyle\int_{1}^{4}2s\, \mathrm{d}s=\left[\phantom{\dfrac{1}{1}}\!\! s^{2}\right] _{1}^{4}=16-1=15.
\end{eqnarray*}
</p>
<p>Clearly, evaluating the integrals can involve further tools of integration, e.g. integration by parts or by substitution.</p>

<hr>
<h4 class="HELM_example">Example</h4>
<p>Evaluate the double integral <br> 
\(\qquad\displaystyle I=\displaystyle\int_{-1}^{2}\displaystyle\int_{-2}^{3}\dfrac{xye^{-x}}{y^{2}+1}\ \ \, \mathrm{d}y \mathrm{d}x.\)</p>
<p>Here, the limits have not formally been linked with a variable name but the limits on the outer integral apply to \(x\) and the limits on the inner integral apply to \(y.\) As the integrations are more complicated, the inner integral will be evaluated explicitly.
</p>

<h4 class="HELM_solution">Solution</h4>
<p>\(\qquad\displaystyle  \text{Inner integral
}=\displaystyle\int_{-2}^{3}\dfrac{xye^{-x}}{y^{2}+1}\ \, \mathrm{d}y
\)
<br>
which can be evaluated by means of the substitution \(U=y^{2}+1.\)
<br>
If \(U=y^{2}+1\) then \(\mathrm{d}U=2y\, \mathrm{d}y,\) so \(y\, \mathrm{d}y=\dfrac{1}{2}\mathrm{d}U.\) 
<br> 
Also, if \(y=-2\) then \(
U=5,\) and if \(y=3\) then \(U=10.\)
</p>
<p>
<br>So the inner integral becomes (remembering that \(x\) may be treated
as a constant)
</p>
<p>
\(\qquad\displaystyle 
\displaystyle\int_{5}^{10}\dfrac{1}{2}\dfrac{xe^{-x}}{U}\, \mathrm{d}U=\dfrac{xe^{-x}}{2}\displaystyle\int_{5}^{10}
\dfrac{\mathrm{d}U}{U}=\dfrac{xe^{-x}}{2}\left[\phantom{\dfrac{1}{1}}\!\!\! \ln U\right] _{5}^{10}=\dfrac{xe^{-x}
}{2}\left( \ln 10-\ln 5\right) =xe^{-x}\dfrac{\ln 2}{2},
\)<br> 
and so the double integral becomes
<br>\(\qquad\displaystyle 
I=\displaystyle\int_{-1}^{2}xe^{-x}\dfrac{\ln 2}{2}\, \mathrm{d}x=\dfrac{\ln 2}{2}
\displaystyle\int_{-1}^{2}xe^{-x}\, \mathrm{d}x,
\)<br> 
which can be evaluated by integration by parts.
\begin{eqnarray*}
I &=&\dfrac{\ln 2}{2}\left[ \left[\phantom{\dfrac{1}{1}}\!\!\! -xe^{-x}\right]
_{-1}^{2}-\displaystyle\int_{-1}^{2}1%
\times \left( -e^{-x}\right) \, \mathrm{d}x\right] =\dfrac{\ln 2}{2}\left[
-2e^{-2}+\left( -1\right) e^{1}+\displaystyle\int_{-1}^{2}e^{-x}\, \mathrm{d}x\right] \\
&=&\dfrac{\ln 2}{2}\left[ -2e^{-2}-e^{1}+\left[\phantom{\dfrac{1}{1}}\!\!\! -e^{-x}\right] _{-1}^{2}
\right]\\ &=&\dfrac{\ln 2}{2}\left[ -2e^{-2}-e^{-1}-e^{-2}+e^{1}\right] =\dfrac{
\ln 2}{2}\left[ -3e^{-2}\right] \approx -0.14.
\end{eqnarray*}]]></text>
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