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      <text>27.2.4.1 Force on a Dam Information</text>
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      <text><![CDATA[<h4> Force on a Dam</h4>
<p>A previous section considered the force on a rectangular dam of width \(100\)m and height \(40\)m. Instead, imagine that the dam is not rectangular in profile but instead has a width of \(100\)m at the top but only \(80\)m at the bottom. The top and bottom of the dam can be given by line segments \(y=0\) (bottom) and \(y=40\) (top) while the sides are parts of the lines \(y = 40 - 4x\) i.e. \(\displaystyle x = 10 - \dfrac{y}{4}\) (left) and \(y = 40 + 4 (x-100) = 4x - 360\) i.e. \(\displaystyle x = 90 + \dfrac{y}{4}\)(right). (See the below figure).</p><p><img src="@@PLUGINFILE@@/27-2-fig20-dam.svg" alt="An image of the described dam." width="319" height="189" class="img-fluid atto_image_button_text-bottom"><br></p><p>Thus the dam exists at heights \(y\) between \(0\) and \(40\) while for each value of \(y\), the horizontal coordinate \(x\) varies between \(\displaystyle x = 10 - \dfrac{y}{4}\) and \(\displaystyle x = 90 + \dfrac{y}{4}\). Thus the surface integral representing the total force i.e. </p>
<p>\(I = \displaystyle\int_{A} 10^4(40-y)\; \mathrm{d}A\) becomes the double integral \(\ I = \displaystyle\int_0^{40} \displaystyle\int_{10-\frac{y}{4}}^{90+\frac{y}{4}} 10^4(40-y)\ \; \mathrm{d}x \mathrm{d}y\) </p>
<p> which can be evaluated as follows 

\begin{align} I &= \displaystyle\int_0^{40} \displaystyle\int_{10-\frac{y}{4}}^{90+\frac{y}{4}} 10^4(40-y)\, \mathrm{d}x \mathrm{d}y \\ 
&= 10^4 \displaystyle\int_0^{40} \left[\phantom{\dfrac{1}{1}} (40-y)x \right]_{10-\frac{y}{4}}^{90+\frac{y}{4}}\, \mathrm{d}y  = 10^4 \displaystyle\int_0^{40} \left[ (40-y)(90+\frac{y}{4})-(40-y)(10-\frac{y}{4}) \right]\, \mathrm{d}y  \\ 
&=10^4 \displaystyle\int_0^{40} \left[ (40-y)(80+\frac{y}{2}) \right]\, \mathrm{d}y  = 10^4 \displaystyle\int_0^{40} \left[ 3200 - 60y - \frac{y^2}{2} \right]\, \mathrm{d}y  \\ 
&= 10^4 \left[ 3200y - 30y^2 - \frac{1}{6}y^3 \right]_0^{40} = 10^4 \left[ (3200 \times 40 - 30 \times 40^2 - \frac{1}{6} 40^3) - 0 \right] \\ 
&= 10^4 \times \frac{208000}{3} \approx 6.93 \times 10^8 \text{N}\end{align} i.e. the total force is just under 700 meganewtons. </p>]]></text>
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