<?xml version="1.0" encoding="UTF-8"?>
<quiz>
  <question type="description">
    <name>
      <text>27.4.4.1 Engineering example</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<h4 class="HELM_example">Engineering Example:</h4>
<h5>Volume of liquid in an ellipsoidal tank</h5>
<h6>Introduction</h6>
<p>An ellipsoidal tank (elliptical when viewed from along \(x\)-, \(y\)- or \(z\)-axes) 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 and calculate a Jacobian.</p>
<p><img class="img-fluid" src="@@PLUGINFILE@@/ellipsoid.svg" alt="elipsoid with radii c b and a, and a length from the bottom h shaded." width="173" height="252"></p>
<h6>Problem in words</h6>
<p>The metal tank is in the form of an ellipsoid, with semi-axes \(a\), \(b\) and \(c\). A volume \(V\) of liquid is poured into the tank ( \(V &lt; \frac{4}{3} \pi abc\), the volume of the ellipsoid) and the problem is to calculate the depth, \(h\), of the liquid.</p>
<h6>Mathematical statement of problem</h6>
<p>The shaded area is expressed as the triple integral \[ V = \int_{z=0}^{h} \int_{y=y_1}^{y_2} \int_{x=x_1}^{x_2} \mathrm{d}x \,\mathrm{d}y \,\mathrm{d}z \] where limits of integration \[ x_1 = -a\sqrt{1-\frac{y^2}{b^2} - \frac{(z-c)^2}{c^2}} \; \; \mbox{ and } \;\; x_2 = +a\sqrt{1-\frac{y^2}{b^2} - \frac{(z-c)^2}{c^2}} \] which come from rearranging the equation of the ellipsoid \[\left(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{(z-c)^2}{c^2} = 1\right)\] and limits \[ y_1 = -\frac{b}{c}\sqrt{c^2-(z-c)^2} \;\; \mbox{ and } \;\; y_2 = +\frac{b}{c} \sqrt{c^2 -(z-c)^2} \] from the equation of an ellipse in the \(y\)-\(z\) plane \(\left( \dfrac{y^2}{b^2} + \dfrac{(z-c)^2}{c^2} = 1 \right)\).</p>
<h6>Mathematical analysis</h6>
<p>To calculate \(V\), use the substitutions \begin{eqnarray*} x & = & a\tau\cos\phi\left(1-\frac{(z-c)^2}{c^2}\right)^{\frac{1}{2}} \\ y & = & b\tau\sin\phi\left(1-\frac{(z-c)^2}{c^2}\right)^{\frac{1}{2}} \\ z &= & z \end{eqnarray*} now expressing the triple integral as \[ V = \int_{z=0}^{h} \int_{\phi=\phi_1}^{\phi_2} \int_{\tau=\tau_1}^{\tau_2} J \; d\tau d\phi \,\mathrm{d}z \] where \(J\) is the Jacobian of the transformation calculated from 


\[ J = \left|\begin{array}{ccc} {\displaystyle \frac{\partial x}{\partial \tau}} &amp; {\displaystyle \frac{\partial x}{\partial \phi}} &amp; {\displaystyle \frac{\partial x}{\partial z}}\\ {\displaystyle \frac{\partial y}{\partial \tau}} &amp; {\displaystyle \frac{\partial y}{\partial \phi}} &amp; {\displaystyle \frac{\partial y}{\partial z}}\\ {\displaystyle \frac{\partial z}{\partial \tau}} &amp; {\displaystyle \frac{\partial z}{\partial \phi}} &amp; {\displaystyle \frac{\partial z}{\partial z}}\\ \end{array}\right| \] 


and reduces to

 \begin{eqnarray*} J & = & \frac{\partial x}{\partial\tau} \frac{\partial y}{\partial\phi} - \frac{\partial x}{\partial\phi}\frac{\partial y}{\partial\tau} \;\;\;\;\mbox{ since }\;\;\;\;\; \frac{\partial z}{\partial \tau} = \frac{\partial z}{\partial \phi}=0 \\ & = & \left\{a\cos\phi\left(1-\frac{(z-c)^2}{c^2}\right)^{\frac{1}{2}} \;\; b\tau\cos\phi\left(1-\frac{(z-c)^2}{c^2}\right)^{\frac{1}{2}}\right\}\\ & &{} -\;\; \left\{-a\tau\sin\phi\left(1-\frac{(z-c)^2}{c^2}\right)^{\frac{1}{2}} \;\;\; b\sin\phi\left(1-\frac{(z-c)^2}{c^2}\right)^{\frac{1}{2}}\right\}\\ & = & a b \tau \left(\cos^2\phi + \sin^2\phi\right)\left(1-\frac{(z-c)^2}{c^2}\right) \\ & = & a b\tau\left(1-\frac{(z-c)^2}{c^2}\right) \\ \end{eqnarray*} 


To determine limits of integration for \(\phi\), note that the substitutions above are similar to a cylindrical polar co-ordinate system, and so \(\phi\) goes from 0 to \(2\pi\). For \(\tau\), setting \(\tau = 0\) \(\Rightarrow\) \(x = 0\) and \(y = 0\), i.e. the \(z\)-axis. Setting \(\tau = 1\) gives


 \[\qquad\displaystyle \frac{x^2}{a^2} = \cos^2\phi\left(1-\frac{(z-c)^2}{c^2}\right)  \hspace{1cm}(1) \] and \[\qquad\displaystyle\frac{y^2}{b^2} = \sin^2\phi\left(1-\frac{(z-c)^2}{c^2}\right) \hspace{1cm}(2)\]
<br>
Summing both sides of Equations (1) and (2) gives \(\qquad\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = (\cos^2\phi + \sin^2\phi)\left(1-\frac{(z-c)^2}{c^2}\right) \) <br> or <br> \(\qquad\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{(z-c)^2}{c^2} = 1 \) <br>which is the equation of the ellipsoid, i.e. the outer edge of the volume. Therefore the range of \(\tau\) should be 0 to 1. Now 

\begin{eqnarray*} V & = & a b \int_{z=0}^{h} \left(1-\frac{(z-c)^2}{c^2}\right)\int_{\phi = 0}^{2\pi} \int_{\tau = 0}^{1} \tau \; d\tau d\phi \,\mathrm{d}z\\ &= & \frac{a b}{c^2}\int_{z=0}^{h} (2zc - z^2)\int_{\phi = 0}^{2\pi} \left[\frac{\tau^2}{2}\right]_{\tau = 0}^{1} \; d\phi \,\mathrm{d}z\\ & = & \frac{a b}{2c^2}\int_{z=0}^{h} (2zc - z^2)\bigg[\phi\bigg]_{\phi = 0}^{2\pi} \; \,\mathrm{d}z\\ & = & \frac{\pi a b}{c^2}\left[c z^2 - \frac{z^3}{3}\right]_{z=0}^{h} \\ & = &\frac{\pi a b}{c^2}\left(ch^2 - \frac{h^3}{3}\right) \end{eqnarray*}</p>

<h6> Interpretation</h6>

Suppose the tank has actual dimensions of \(a= 2 \mathrm{m}\) , \(b = 0.5 \mathrm{m}\)  and
 \(c= 3 \mathrm{m}\)  and a volume of\( 7 \mathrm{m}^3\) is to be poured into it. (The
 total volume of the tank is \(4\pi \mathrm{m}^3 \approx 12.57 \mathrm{m}^3\)). Then, from
 above
 \[  V = \frac{\pi a b}{c^2} \left( ch^2 - \frac{h^3}{3} \right)
\]
 which becomes
\[
 7 = \frac{\pi}{9} \left(3 h^2 - \frac{h^3}{3}\right)
\]
 with solution \(h = 3.23 \mathrm{m}\) (2 d.p.), compared to the
 maximum height of the ellipsoid of \(6 \mathrm{m}\).]]></text>
<file name="ellipsoid.svg" path="/" encoding="base64">[encoded data, 18628 characters]</file>
    </questiontext>
    <generalfeedback format="html">
      <text/>
    </generalfeedback>
    <defaultgrade>0</defaultgrade>
    <penalty>0</penalty>
    <hidden>0</hidden>
    <idnumber/>
  </question>
</quiz>
