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    <name>
      <text>27.3.2.11 volume of quarter cylinder exercise</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<p class="HELM_exercise">Exercise</p>
<p>Find the volume of the solid prism shown in the diagram below, where a={@aa@}. Check that when the order of integration is changed, the volume remains unaltered.</p>

<p>
<figure class="HELM_figure">
<img src="@@PLUGINFILE@@/Fig363b_5.svg" alt="quarter cylinder with the edges x=a, y=4 and y=4-x^2." width="276" height="203">
</figure>
</p>
<p>Volume = [[input:ans1]] [[validation:ans1]]</p>]]></text>
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    <generalfeedback format="html">
      <text><![CDATA[<p>The volume is given by the triple integral \(\displaystyle\int\int \int dV\). Putting \(z\) on the outer integral, \(y\) on the intermediate integral and \(x\) on the inner integral, the limits on \(z\) are \(z=0\) to \(z=1\). For each value of \(z\), \(y\) varies from \(y=0\) (base) to \(y=1-z\) on the sloping face. For each combination of \(y\) and \(z\), \(x\) varies from \(x=0\) to \(x={@aa@}\). Thus, the volume is given by \begin{eqnarray*} V &amp;=&amp; \int \int \int dV = \int_{z=0}^1 \int_{y=0}^{1-z} \int_{x=0}^{@aa@} \;\mathrm{d}x \,\mathrm{d}y \,\mathrm{d}z \\ &amp;=&amp; \int_{z=0}^1 \int_{y=0}^{1-z} \bigg[ x\bigg]_0^{@aa@} \;\mathrm{d}y \,\mathrm{d}z = \int_{z=0}^1 \int_{y=0}^{1-z} {@aa@} \;\mathrm{d}y \,\mathrm{d}z \\ &amp;=&amp; \int_{z=0}^1 \bigg[ {@fbI12@}\bigg]_{y=0}^{1-z} \;\mathrm{d}z = \int_{z=0}^1 \left( {@I12@} \right) \;\mathrm{d}z \\ &amp;=&amp; \bigg[{@fbI13@} \bigg]_0^1 = {@ta1@} \end{eqnarray*} Now, the three integrations can be carried out in a different order. For example, with \(x\) on the outer integral, \(z\) on the intermediate integral and \(y\) on the inner integral, the limits on \(x\) are \(x=0\) to \(x={@aa@}\); for each value of \(x\), \(z\) varies from \(z=0\) to \(z=1\) and for each combination of \(x\) and \(z\), \(y\) varies from \(y=0\) to \(y=1-z\). The volume is therefore given by \begin{eqnarray*} V &amp;=&amp; \int \int \int dV = \int_{x=0}^3 \int_{z=0}^{1} \int_{y=0}^{1-z} \;\mathrm{d}y \,\mathrm{d}z \,\mathrm{d}x\qquad\qquad\qquad \\ &amp;=&amp; \int_{x=0}^{@aa@} \int_{z=0}^{1} \bigg[ y \bigg]_{y=0}^{1-z} \;\mathrm{d}z \,\mathrm{d}x \\ &amp;=&amp; \int_{x=0}^{@aa@} \int_{z=0}^{1} \bigg[ 1-z \bigg] \;\mathrm{d}z \,\mathrm{d}x \\ &amp;=&amp; \int_{x=0}^{@aa@} \left[ z-\frac{z^2}{2} \right]_{z=0}^{1} \;\mathrm{d}x \\ &amp;=&amp; \int_{x=0}^{@aa@} \left[ 1-\frac{1^2}{2} - 0 \right] \;\mathrm{d}x \\ &amp;=&amp; \int_{x=0}^{@aa@} \frac{1}{2} \;\mathrm{d}x \\ &amp;=&amp; \left[ \frac{x}{2} \right]_{x=0}^{@aa@} = {@ta1@} \end{eqnarray*} There are in all six ways - (\(3!\)) - to order the three integrations; each order gives the same answer of {@ta1@}.</p>]]></text>
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      <text>2025073100</text>
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    <questionvariables>
      <text>aa: rand([1,2,3]);

fbI11:integrate (1,z);
I11:integrate (1,z,0,6);
fbI12: integrate (I11,y);
I12: integrate (I11,y,0,4-x^2);
fbI13: integrate (I12,x);
I13: integrate (I12,x,0,aa);
ta1:I13;</text>
    </questionvariables>
    <specificfeedback format="html">
      <text><![CDATA[<p>[[feedback:prt1]]</p>]]></text>
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    <questionnote format="html">
      <text>\(aa={@aa@}\implies V={@ta1@}\)</text>
    </questionnote>
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      <text><![CDATA[<p><i class="fa fa-check"></i> Correct answer, well done.</p>]]></text>
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      <text><![CDATA[<p><i class="fa fa-adjust"></i> Your answer is partially correct.</p>]]></text>
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      <text><![CDATA[<p><i class="fa fa-times"></i> Incorrect answer.</p>]]></text>
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    <decimals>.</decimals>
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