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    <name>
      <text>27.2.1.1 Functions as Limits of Integration Information</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<h3> Functions as Limits of Integration</h3>
<p>Previously, double integrals of the form</p>
<p>\(\qquad\displaystyle I=\displaystyle\int_{x=a}^{x=b}\displaystyle\int_{y=c}^{y=d}f\left( x,y\right)\, \mathrm{d}y \mathrm{d}x\)</p>
<p>were considered. They represent an integral over a rectangular region in the \(xy\) plane. If the limits of integration of the inner integral are replaced with functions \(G_1, G_2\),</p>
<p>\(\qquad\displaystyle I=\displaystyle\int_{x=a}^{x=b}\displaystyle\int_{G_1(x)}^{G_2(x)}f\left( x,y\right)\, \mathrm{d}y \mathrm{d}x\)</p>
<p>then the region described will not, in general, be a rectangle. The region will be a shape bounded by the curves (or lines) which these functions \(G_1\) and \(G_2\) describe.</p>
<p></p>
<p>As was indicated in a previous section,</p>
<p>\(\qquad \displaystyle I=\int_{x=a}^{x=b}\int_{G_1(x)}^{G_2(x)}f(x,y)\, \mathrm{d}y \mathrm{d}x\)</p>
<p>can be interpreted as the volume lying above the region in the \(xy\) plane defined by \(G_1(x)\) and \(G_2(x)\), bounded above by the surface \(z=f(x,y)\). Not all double integrals are interpreted as volumes but this is often the case. If \(z=f(x,y)&lt;0\) anywhere in the relevant region, then the double integral no longer represents a volume.</p>
<p></p>

<div class="HELM_keypoint">
<h4>Key Point</h4>
<p><strong> Double Integral Over General Region</strong></p>
<p>\(I=\displaystyle\int_{x=a}^{x=b}\displaystyle\int_{G_1(x)}^{G_2(x)}f\left(x,y\right)\, \mathrm{d}y \mathrm{d}x \)</p>
<ol>
<li>The functions \(G_1, G_2\) which are the limits for the inner integral are functions of the variable of the outer integral. This must be the case for the integral to make sense.</li>
<li>The limits of the outer integral are constant.</li>
<li>Integration over rectangular regions can be thought of as the special case where \(G_1\) and \(G_2\) are constant functions.</li>
</ol>
</div>

<hr>
<h4 class="HELM_example">Example</h4>
<p>Evaluate the integral \(\displaystyle I=\displaystyle\int_{x=0}^{1}\displaystyle\int_{y=0}^{1-x}2xy\, \mathrm{d}y \mathrm{d}x\)</p>
<h4 class="HELM_solution">Solution</h4>

<figure class="HELM_figure">
<img class="img-fluid" src="@@PLUGINFILE@@/Fig362_2.svg" alt="Surface above the xy plane with a section shaded and projected down onto the xy plane." width="261" height="196"></figure></p>
<p><figure class="HELM_figure"><img class="img-fluid" src="@@PLUGINFILE@@/Fig362_1.svg" alt="Graph of y=1-x, with x and y ranging from 0 to 1." width="223" height="185"> </figure></p>

<p>Projecting the relevant part of the surface (shown in the top figure) down to the \(xy\) plane produces the triangle shown in the bottom figure. The extremes that \(x\) takes are \(x=0\) and \(x=1\) and so these are the limits on the outer integral. For any value of \(x\), the variable \(y\) varies between \(y=0\) (at the bottom) and \(y=1-x\) (at the top). Thus if the volume (shown in the diagram) under the function \(f(x,y)\) and bounded by this triangle is required, then the following integral is to be calculated.</p>
<p>\(\qquad\displaystyle \int_{x=0}^{1}\displaystyle\int_{y=0}^{1-x}f\left( x,y\right)\, \mathrm{d}y\mathrm{d}x\)</p>
<p>Once the correct limits have been determined, the integration is carried out in exactly the same manner as previously.</p>
<p>First consider the inner integral</p>
<p>\(\qquad\displaystyle g(x)=\displaystyle\int_{y=0}^{1-x}2xy\, \mathrm{d}y\)</p>
<p>Integrating \(2xy\) with respect to \(y\) gives \(xy^2+C\) so</p>
<p>\(\qquad\displaystyle g(x)=\left[xy^2\right]_{y=0}^{1-x} = x(1-x)^2\)</p>
<p>Note that, as is required, this is a function of \(x\), the variable of the outer integral. Now the outer integral is</p>
<p>\begin{align} I&amp; = \displaystyle\int_{x=0}^{1}x(1-x)^2\, \mathrm{d}x \\ &amp; = \displaystyle\int_{x=0}^{1}\left(x^3-2x^2+x\right)\, \mathrm{d}x=\left[\dfrac{x^4}{4}-\dfrac{2x^3}{3}+\dfrac{x^2}{2}\right]_{x=0}^1 = \dfrac{1}{4}-\dfrac{2}{3}+\dfrac{1}{2}=\dfrac{1}{12} \end{align}</p>
<p>Regions do not have to be bounded only by straight lines. Also, the integrals may involve other tools of integration, such as substitution or integration by parts. Drawing a sketch of the limit functions in the plane and shading the region is a valuable tools when evaluating such integrals.</p>
<hr>
<h4 class="HELM_example">Example</h4>
<p>Evaluate the volume under the surface given by \(z=f(x,y)=2x\sin(y)\), over the region bounded above by the curve \(y=x^2\) and below by the line \(y=0\) (i.e. the \(x\)-axis), for \(0\leq x \leq 1\).</p>
<p><strong><img class="img-fluid" src="@@PLUGINFILE@@/Fig362_3.svg" alt="Graph of y=x^2, with x and y ranging from 0 to 1." width="185" height="147"></strong></p>
<h4 class="HELM_solution">Solution</h4>
<p>First sketch the curve \(y=x^2\) and identify the region. This is the shaded region in the above figure. The required integral is</p>
<p>\begin{align} I&amp; = \displaystyle\int_{x=0}^{1}\displaystyle\int_{y=0}^{x^2}2x\sin(y)\, \mathrm{d}y \mathrm{d}x \\ &amp; = \displaystyle\int_{x=0}^{1}\left[-2x\cos(y)\right]_{y=0}^{x^2}\, \mathrm{d}x \\ &amp; = \displaystyle\int_{x=0}^{1}\left(-2x\cos\left(x^2\right)+2x\right)\, \mathrm{d}x \\ &amp; = \displaystyle\int_{x=0}^{1}\left(1-\cos\left(x^2\right)\right)2x\, \mathrm{d}x \end{align}</p>
<p>Making the substitution \(u=x^2\) so \(d u=2x\, \mathrm{d}x\) and noting that the limits \(x=0,1\) map to \(u=0,1\), gives</p>
<p>\begin{align} I&amp;=\displaystyle\int_{u=0}^{1}\left(1-\cos\left( u\right)\right)\, \mathrm{d}u \\ &amp;=\left[u-\sin(u)\right]_{u=0}^{1} \\ &amp;=1-\sin(1)\\ &amp;\approx 0.1585 \end{align}</p>
<hr>
<h4 class="HELM_example">Example</h4>
<p>Evaluate the volume under the surface given by \(z=f(x,y)=x^2+\dfrac{1}{2}y\), over the region bounded by the curves \(y=2x\) and \(y=x^2\).</p>
<p><strong><img class="img-fluid" src="@@PLUGINFILE@@/Fig362_4.svg" alt="Graphs of y=2x and y=x^2, with the region between them shaded." width="219" height="299"></strong></p>
<h4 class="HELM_solution">Solution</h4>
<p>The sketch of the region is shown in the above figure. The required integral is</p>
<p>\(\qquad\displaystyle I=\displaystyle\int_{x=a}^{b}\displaystyle\int_{y=x^2}^{2x}\left( x^2+\dfrac{1}{2}y\right)\, \mathrm{d}y \mathrm{d}x \)</p>
<p>To determine the limits for the integration with respect to \(x\), the points where the curves intersect are required. These points are the solutions of the equation \(2x=x^2\), so the required limits are \(x=0\) and \(x=2\). Then the volume is given by</p>
<p>\begin{align} I&amp; = \displaystyle\int_{x=0}^{2}\displaystyle\int_{y=x^2}^{2x}\left( x^2+\dfrac{1}{2}y\right)\, \mathrm{d}y \mathrm{d}x\\ &amp; = \displaystyle\int_{x=0}^{2}\left[x^2 y+\dfrac{1}{4}y^2\right]_{y=x^2}^{2x}\, \mathrm{d}x\\ &amp; = \displaystyle\int_{x=0}^{2}\left(x^2+2x^3-\dfrac{5}{4}x^4\right)\, \mathrm{d}x\\ &amp; = \left[\dfrac{x^3}{3}+\dfrac{x^4}{2}-\dfrac{x^5}{4}\right]_{x=0}^{2}\\ &amp;=\dfrac{8}{3} \end{align}</p>
<hr>
<h4 class="HELM_example">Example</h4>
<ol>
<li>Evaluate the volume under \(z=f(x,y)=5x^2y\), over the half of the unit circle that lies above the \(x\)-axis. (see the below figure)  <strong><img class="img-fluid" src="@@PLUGINFILE@@/Fig362_5.svg" alt="Semi-circle on the xy plane centred at the origin with radius 1, for y&gt;0." width="219" height="151"></strong></li>
<li>Repeat part 1 for \(z=f(x,y)=1.\)</li>
</ol>
<h4 class="HELM_solution">Solution</h4>
<ol>
<li>This region is bounded by the circle \(y^2+x^2=1\) and the line \(y=0\). Since only positive values of \(y\) are required, the equation of the circle can be written \(y=\sqrt{1-x^2}\). Then the required volume is given by \begin{align} I&amp; = \displaystyle\int_{x=-1}^{1}\displaystyle\int_{y=0}^{\sqrt{1-x^2}}\left( 5x^2y\right)\, \mathrm{d}y \mathrm{d}x \;=\; \displaystyle\int_{x=-1}^{1}\left[\dfrac{5}{2}x^2 y^2\right]_{y=0}^{\sqrt{1-x^2}}\,\mathrm{d}x\\ &amp; = \displaystyle\int_{x=-1}^{1}\dfrac{5}{2}x^2 (1-x^2)\, \mathrm{d}x \; =\; \dfrac{5}{2}\left[\dfrac{x^3}{3}-\dfrac{x^5}{5}\right]_{-1}^{1}=\dfrac{2}{3} \end{align}</li>
<br>
<br>
<li>\begin{align} I&amp; = \displaystyle\int_{x=-1}^{1}\displaystyle\int_{y=0}^{\sqrt{1-x^2}}1 \, \mathrm{d}y \mathrm{d}x \; =\; \displaystyle\int_{x=-1}^{1}\left[{\phantom{\dfrac{1}{2}}}\!\!\!\!y\right]_{y=0}^{\sqrt{1-x^2}}\, \mathrm{d}x\\ &amp; = \displaystyle\int_{x=-1}^{1}\sqrt{1-x^2}\, \mathrm{d}x\ ({\rm which\ by\ substituting}\ x=\sin\theta)=\dfrac{\pi}{2} \end{align}</li>
</ol>
<p>Note that by putting \(f(x,y)=1\) we have found the volume of a semi-circular lamina of uniform height \(1\). This result is numerically the same as the <strong>area</strong> of the region in the above figure. (This is a general result.)</p>
<p></p>]]></text>
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<file name="Fig362_5.svg" path="/" encoding="base64">[encoded data, 9204 characters]</file>
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