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      <text>27.2.5.1 Volume of Liquid Engineering Example</text>
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      <text><![CDATA[<h4 class="HELM_example"> Engineering Example</h4>
<h5>Volume of Liquid in an Elliptic Tank </h5>
<p><strong>Introduction</strong></p>
<p>A tank in the shape of an elliptic cylinder has a volume of liquid poured into it. It is useful to know in advance how deep the liquid will be. In order to make this calculation, it is necessary to perform a multiple integration.</p><p><img src="@@PLUGINFILE@@/27-2-fig22-cylinder.svg" alt="A diagram showing the tank in the shape of an elliptic cylinder, with depth of liquid h." width="204" height="216" class="img-fluid atto_image_button_text-bottom"><br></p><p><strong>Problem in words</strong> 
</p><p>The tank has semi-axes \(a\) (horizontal) and \(c\) (vertical) and is of constant thickness \(b\). A volume of liquid \(V\) is poured in (assuming that \(V &lt; \pi a b c\), the volume of the tank), filling it to a depth \(h\), which is to be calculated. Assume 3-D coordinate axes based on a point at the bottom of the tank.</p>

<p><strong>Mathematical statement of the problem</strong> 
</p><p>Since the tank is of constant thickness \(b\), the volume of liquid is given by the shaded area multiplied by \(b\), i.e. </p>
<p>\(\qquad\displaystyle V = b \times \text{shaded area}\) </p>
<p>where the shaded area can be expressed as the double integral </p>
<p> \(\qquad\displaystyle \int_{z=0}^h \; \int_{x=x_1}^{x_2} dx dz\) </p>
<p> where the limits \(x_1\) and \(x_2\) on \(x\) can be found from the equation of the ellipse </p>
<p>\(\qquad\displaystyle \dfrac{x^2}{a^2} + \frac{(z-c)^2}{c^2} = 1.\) </p>


<strong>Mathematical analysis</strong>
<p> From the equation of the ellipse 
\begin{align} x^2 &amp; = a^2 \left[ 1 - \dfrac{(z-c)^2}{c^2} \right] \\ &amp; = \dfrac{a^2}{c^2} \left[ c^2 - (z-c)^2 \right] \\ &amp; = \frac{a^2}{c^2} \left[ 2zc - z^2 \right]\qquad \text{so}\quad x = \pm \frac{a}{c} \sqrt {2zc-z^2} \end{align} </p>

<p>Thus </p>
<p>\(\qquad\displaystyle x_1 = - \dfrac{a}{c} \sqrt {2zc-z^2}, \;\;\text{ and }\;\; x_2 = + \dfrac{a}{c} \sqrt {2zc-z^2}\) </p>
<p> Consequently 
\begin{align} V = b \int_{z=0}^{h} \int_{x=x_1}^{x_2} \; \mathrm{d}x dz &amp;=  b \int_{z=0}^{h} \Bigl[ x \Bigr]_{x_1}^{x_2} \; \mathrm{d}z \\ 
&amp;=  b \int_{z=0}^{h} 2\frac{a}{c}\sqrt {2zc-z^2} \; \mathrm{d}z\end{align} </p>

Now use the substitution \(z-c = c \; \sin \theta\) so that \(dz = c \; \cos \theta \; \mathrm{d}\theta\) 
\[\begin{aligned} z&amp;=0 \text{ gives } \theta = -\frac{\pi}{2} \\
 z&amp;=h \text{ gives }  \theta = \sin^{-1} \left( \frac{h}{c} -1 \right) = \theta_0\ \ (\text{say})\end{aligned}\]<p></p>

<p>\begin{align} 
V &amp; = b \int_{-\frac{\pi}{2}}^{\theta_0} 2 \dfrac{a}{c} \; c\cos\theta \; c\cos\theta \; \mathrm{d}\theta \\ 
&amp;= 2abc \int_{-\frac{\pi}{2}}^{\theta_0} \cos^2 \theta \; \mathrm{d}\theta \\
&amp;= abc \int_{-\frac{\pi}{2}}^{\theta_0} \left[ 1 + \cos 2\theta \right] \; \mathrm{d}\theta \\
&amp; = abc \left[ \theta + \dfrac{1}{2} \sin 2\theta \right]_{-\frac{\pi}{2}}^{\theta_0} \\
&amp;= abc \left[ \theta_0 + \dfrac{1}{2} \sin 2\theta_0 - \left( -\dfrac{\pi}{2} + 0 \right) \right]\\
&amp;= abc \left[ \theta_0 + \dfrac{1}{2} \sin 2\theta_0 + \dfrac{\pi}{2} \right] \qquad \qquad \qquad \qquad\dots(\ast ) \end{align} </p>

<p>which can also be expressed in the form </p>
<p>\(\qquad\displaystyle V = abc \left[ \sin^{-1}\left( \dfrac{h}{c} -1 \right) + \left( \dfrac{h}{c} -1 \right) \sqrt{1-\left(\dfrac{h}{c} -1 \right)^2} + \dfrac{\pi}{2}. \right]\) </p>
<p>While (\(\ast\)) expresses \(V\) as a function of \(\theta_0\) (and therefore \(h\)), to find \(\theta_0\) as a function of \(V\) requires a numerical method. For a given \(a\), \(b\), \(c\), and \(V\), solve equation (\(\ast\)) by a numerical method to find \(\theta_0\) and find \(h\) from \(h = c\, (1+\sin\theta_0)\). </p>

<strong>Interpretation</strong> 
<p>If \(a\) = \(2\)m, \(b\) = \(1\)m, \(c\) = \(3\)m (so the total volume of the tank is \(6\pi\) m\(^3 \approx 18.85\)m\(^3\)), and a volume of \(7\)m\(^3\) is to be poured into the tank, then </p>
<p>\(\qquad\displaystyle V = abc \left[ \theta_0 + \dfrac{1}{2} \sin 2\theta_0 + \dfrac{\pi}{2} \right]\) </p>
<p>which becomes </p>
<p>\(\qquad\displaystyle 7 = 6 \left[ \theta_0 + \frac{1}{2} \sin 2\theta_0 + \frac{\pi}{2} \right]\) </p>
<p> and has solution \(\theta_0\) = \(-0.205\) (3 decimal places). </p>
<p> Finally 
\begin{align} 
h &amp; = c \,(1 + \sin\theta_0) \\
&amp; =  3\,( 1 + \sin(-0.205)) \\
&amp;= 2.39\ \text{m (to 2 d.p)} \end{align} compared to the maximum height of \(6\)m.</p>]]></text>
<file name="27-2-fig22-cylinder.svg" path="/" encoding="base64">[encoded data, 23964 characters]</file>
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