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      <text>27.1.1 An Example of a Surface Integral</text>
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      <text><![CDATA[<h3>An Example of a Surface Integral</h3>
<p>An engineer involved with the construction of a dam to hold back the water in a reservoir needs to be able to calculate the total force the water exerts on the dam so that the dam is built with sufficient strength.<br>In order to calculate this force, two results are required:</p>
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    <li>The pressure \(p\) of the water is proportional to the depth. That is \[p=kd \quad (1)\] where \(k\) is a constant.</li>
    <li>The force on an area subjected to constant pressure is given by force \(=\) pressure \(\times\) area \(\quad (2)\)</li>
</ol>
<p>The diagram shows the face of the dam. The depth of water is \(h\) and \(\delta A\) is a small area in the face of the dam with coordinates \((x,y)\).</p>

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    <img class="img-fluid" role="presentation" src="@@PLUGINFILE@@/Figure%201%20%282%29.svg" alt="The face of a dam with depth of water h, and δA is a small area in the face of the dam with coordinates (x,y)">
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<p><br>Using (1), the pressure at \(\delta A\sim k(h-y)\). Using (2), the force on an area \(\delta A\sim k(h-y)\delta A\).<br>Both of these expressions are approximate as \(y\) is slightly different at the top of \(\delta A\) to the bottom. <br>Now \[\begin{aligned}\text{Total force on dam} &amp;= \text{sum of forces on all areas } \delta A \text{ making up the face of the dam} \\ &amp;\approx \sum_{\text{all } \delta A} k(h - y) \delta A\end{aligned}\] For a better approximation let \(\delta A\) become smaller, and for the exact result find the limit as \(\delta A\rightarrow 0\). Then \[\begin{aligned}\text{Total force on the dam} &amp;= \lim_{\delta A \rightarrow 0} \sum k(h - y) \delta A \\&amp;= \int_A k(h - y)\, dA\end{aligned}\] where \(\int_Ak(h-y)\ d A\) stands for the <strong>surface integral</strong> of \(k(h-y)\) over the area \(A\). Surface integrals are evaluated using <strong>double integrals</strong>. The following Section shows a double integral being developed in the case of the volume under a surface.</p>]]></text>
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