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      <text>15.2.2.6b The formula which doesn't exist</text>
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      <text><![CDATA[<h3 style="text-align: left;">Formulae for the centre of mass</h3>
<p>Now we consider the case of a lamina bounded above by some function \(f(x)\), bounded below by \(g(x)\) and bounded left and right by the lines \(x=a\) and \(x=b\) respectively (here we assume \(f(x) \geq g(x)\) for all \(x \in [a,b]\). Although this
    case is quite restrictive it does allow the process we have used before to be used to obtain a formula for each of \(\bar{x}\) and \(\bar{y}\).</p>
<figure class="HELM_figure"><img src="@@PLUGINFILE@@/Area%20of%20lamina%20bounded%20as%20in%20text.png" alt="Lamina as described in text" width="649" height="300" class="img-responsive atto_image_button_text-bottom"></figure>
<h5>Finding \(\bar{x}\)</h5>
<p>As before, we find the moment of inertia about the \(y\)-axis in two ways. Assuming again that the lamina has an area \(A\) and mass per unit area of \(\rho\) we have the moment of inertia is \(A\rho \bar{x}\).<br><br>We slice the lamina in thin vertical
    slices of width \(\delta x\). We can approximate the area of the slices by the area of the rectangle of width \(\delta x\) and height \(f(x)-g(x)\). The moment of inertia of this rectangle is \( x(f(x)-g(x)) \delta x\). By adding all of these and
    letting \(\delta x \to 0\) we obtain the integral giving the moment of inertia of lamina so that</p>
<p>\[ \int_a^b x\rho (f(x)-g(x)) \mathrm{d}x = A\rho \bar{x}.\]&nbsp;</p>
<p>We can simplify this to obtain&nbsp;</p>
<p>\[ \bar{x}= \frac{1}{A} \int_a^b x (f(x)-g(x)) \mathrm{d}x.\]&nbsp;</p>
<p>Note that we can calculate \(A=\int_a^b f(x)-g(x) \mathrm{d}x\) to find&nbsp;<br><br>\[ \bar{x}= \frac{\int_a^b x (f(x)-g(x)) \mathrm{d}x}{\int_a^b f(x)-g(x) \mathrm{d}x}.\]</p>
<p></p>
<h5>Finding \(\bar{y}\)</h5>We can use the same vertical slices to find the moment of inertia about the \(x\)-axis, and thus find \(\bar{y}\). We found the area of one of our rectangles to be&nbsp; \((f(x)-g(x)) \delta x\) but the \(y\)-coordinate of the
centre of the rectangle is the average of \(f(x)\) and \(g(x)\), namely \(\frac{1}{2}(f(x)+g(x))\) so that the moment of inertia about the \(x\)-axis is
<p></p>
<p>\[ \frac{1}{2}(f(x)+g(x))(f(x)-g(x)) \delta x=\frac12 (f(x)^2-g(x)^2) \delta x.\]<br></p>
<p><br>By adding all of these and letting \(\delta x \to 0\) we obtain the integral giving the moment of inertia of lamina so that</p>
<p>\[ \int_a^b \frac12 (f(x)^2-g(x)^2)&nbsp; \mathrm{d}x = A\rho \bar{y}.\]&nbsp;</p>
<p>We can simplify this to obtain&nbsp;</p>
<p>\[ \bar{y}= \frac{1}{2A} \int_a^b f(x)^2-g(x)^2 \mathrm{d}x.\]&nbsp;</p>
<p>Note that we can calculate \(A=\int_a^b f(x)-g(x) \mathrm{d}x\) to find&nbsp;<br><br>\[ \bar{y}= \frac{\int_a^b f(x)^2-g(x)^2 \mathrm{d}x}{2\int_a^b f(x)-g(x) \mathrm{d}x}.\]</p>
<p><br></p>
<div class="HELM_keypoint">
    <h4>Key Point</h4>
    <p>Given functions \(f(x),\, g(x)\) such that \(g(x)\leq f(x)\) on the interval \([a,b]\) the centre of mass of the lamina bounded by \(x=a\), \(x=b\), \(y=f(x)\) and \(y=g(x)\) is<br>\[ (\bar{x},\bar{y}) = \left( \frac{\int_a^b x (f(x)-g(x)) \mathrm{d}x}{\int_a^b
        f(x)-g(x) \mathrm{d}x},\frac{\int_a^b f(x)^2-g(x)^2 \mathrm{d}x}{2\int_a^b f(x)-g(x) \mathrm{d}x}\right).\]<br><br>In the special case that \(g(x)=0\) is the \(x\)-axis we have&nbsp;</p>
    <p>\[ (\bar{x},\bar{y}) = \left( \frac{\int_a^b x f(x) \mathrm{d}x}{\int_a^b f(x) \mathrm{d}x},\frac{\int_a^b f(x)^2 \mathrm{d}x}{2\int_a^b f(x) \mathrm{d}x}\right).\]<br><br>Remember that the integral in the denominator is related to the area of the
        lamina and may be possible to calculate without the integral.<br></p>
</div><br>]]></text>
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