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      <text>27.1.6.1 Applications of Surface Integration Over Rectangular Areas Information</text>
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      <text><![CDATA[<h3>Applications of surface integration over rectangular areas<br>
</h3>
<p></p>
<h4> Force on a Dam </h4>
<p>At the beginning of this Section, the total force on a dam was given by the surface integral </p>
<p>\(\qquad\displaystyle \int_A k(h-y) \, \mathrm{d}A \) </p>

<p> Imagine that the dam is rectangular in profile with a width of \(100\) m and a height \(h\) of \(40\) m. The expression \(\mathrm{d}A\) is replaced by \(\, \mathrm{d}x\mathrm{d}y\) and the limits on the variables \(x\) and \(y\) are \(0\) to \(100\) m and \(0\) to \(40\) m respectively. The constant \(k\) may be assumed to be \(10^4\) kg m\(^{-2}\) s\(^{-2}\). The surface integral becomes the double integral

    \[\qquad \displaystyle \int_0^{40} \displaystyle \int_0^{100}k(h-y)\, \mathrm{d}x\mathrm{d}y\] that is \[\qquad\displaystyle \int_0^{40} \displaystyle \int_0^{100}10^4(40-y)\, \mathrm{d}x\mathrm{d}y\]

    As the integral in this double integral does not contain \(x\), the integral may be written
    \[\begin{aligned}
    \displaystyle\int_0^{40}\displaystyle\int_0^{100}10^4(40-y)\, \mathrm{d}x\mathrm{d}y &amp;=(100-0)\displaystyle\int_0^{40}10^4(40-y)\, \mathrm{d}y \\
    &amp;=100\times 10^4\left[40y-\dfrac{y^2}{2}\right]_0^{40}\\
    &amp;=10^6[(40\times 40-\dfrac{40^2}{2})-0]\\
    &amp;=10^6\times 800=8\times 10^8\text{N}
    \end{aligned}\]
    that is, the total force is 800 meganewtons.</p>
<br>

<h4>Center of Pressure</h4>
<p>We wish to find the center of pressure \((x_p,y_p)\) of a plane area immersed vertically in a fluid. Take the \(x\) axis to be in the surface of the fluid and the \(y\) axis to be vertically down, so that the plane \(Oxy\) contains the area.
</p>
<figure class="HELM_figure"><img src="@@PLUGINFILE@@/27.1-fig-7-surface.svg" alt="A diagram showing a surface immersed vertically in a fluid, represented by the xy plane" width="240" height="201" class="img-fluid atto_image_button_text-bottom"></figure>
<p></p>

<p>We require the following results:</p>
<ol>
    <li>The pressure \(p\) is proportional to the depth \(h\), so that \(p=\omega h\) where \(\omega\) is a constant.</li>
    <li>The force \(F\) on an area \(\delta A\) subjected to constant pressure \(p\) is given by \(F=p \delta A\)</li>
</ol>


<p>Consider a small element of area \(\delta A\) at the position shown. The pressure at \(\delta A\) is \(\omega y\). Then the force acting on \(\delta A\) is \(\omega y\delta A\). Hence the total force acting on the area \(A\) is \(\displaystyle\int_A\omega y\, \mathrm{d}A =\omega\displaystyle \displaystyle\int_Ay\, \mathrm{d}A\).

    \[\begin{aligned} \text{Moment of force on}\ \delta A\ \text{about}\ Oy\, &amp;=\, \omega xy\delta A\\
    \text{Total moment of force on}\ \delta A\ \text{about}\ Oy\, &amp;=\, \omega\displaystyle\int_Axy\, \mathrm{d}A \\
    \text{Moment of force on}\ \delta A\ \text{about}\ Ox\, &amp;=\, \omega y^2\delta A\\
    \text{Total moment of force on}\ \delta A\ \text{about}\ Ox\, &amp;=\, \omega\displaystyle\int_Ay^2\, \mathrm{d}A\end{aligned}\]


    Taking moments about \(Oy\):
    \[\begin{aligned} \text{total force} \times x_p\, &amp;=\, \text{total moment}\\
    \left(\omega\displaystyle\int_Ay\, \mathrm{d}A \right)x_p\, &amp;=\, \omega\displaystyle\int_Axy\, \mathrm{d}A \\
    x_p\displaystyle\int_Ay\, \mathrm{d}A\, &amp;=\, \displaystyle\int_Axy\, \mathrm{d}A\end{aligned}\]

    Taking moments about \(Ox\):
    \[\begin{aligned} \text{total force} \times y_p\, &amp;=\, \text{total moment}\\
    \left(\omega\displaystyle\int_Ay\, \mathrm{d}A \right)y_p\, &amp;=\, \omega\displaystyle\int_Ay^2\, \mathrm{d}A \\
    y_p\displaystyle\int_Ay\, \mathrm{d}A\, &amp;=\, \displaystyle\int_Ay^2\, \mathrm{d}A \end{aligned}\]</p>

<br>
<p>Hence </p>
<p>\(\qquad\displaystyle x_p=\frac{\displaystyle\int_Axy\, \mathrm{d}A}{\displaystyle\int_Ay\, \mathrm{d}A}\) and \(\displaystyle y_p=\frac{\displaystyle\int_Ay^2\, \mathrm{d}A}{\displaystyle\int_Ay\, \mathrm{d}A}\).</p>


<hr>
<h4 class="HELM_example">Example</h4>
<p>A rectangle of sides \(a\) and \(b\) is immersed vertically in a fluid with one of its edges in the surface as shown in the figure below. Where is the centre of pressure? </p>

<figure class="HELM_figure"><img src="@@PLUGINFILE@@/27.1-fig-8-pressurerectangle.svg" alt="A diagram of the xy plane representing a rectangle immersed vertically in a fluid with one of its edges in the surface." width="241" height="180" class="img-fluid atto_image_button_text-bottom"></figure>
<p></p>

<h4 class="HELM_solution">Solution</h4>
<p>To express the surface integral as double integrals we will use cartesian coordinates and vertical slices. We need the following integrals. </p>
<p> \begin{align}
    \displaystyle\int_A y\, \mathrm{d}A &amp;=\displaystyle\int_0^b\displaystyle\int_0^ay\ \, \mathrm{d}y\mathrm{d}x =\displaystyle\int_0^b\left[\frac{1}{2}y^2\right]_0^a\, \mathrm{d}x=\displaystyle\int_0^b\frac{1}{2}a^2\ \, \mathrm{d}x=\left[\frac{1}{2}a^2x\right]_0^b=\frac{1}{2}a^2b\\
    \displaystyle\int_A xy\, \mathrm{d}A &amp;=\displaystyle\int_0^b\displaystyle\int_0^axy\ \, \mathrm{d}y\mathrm{d}x =\displaystyle\int_0^b\left[\frac{1}{2}xy^2\right]_0^a\, \mathrm{d}x=\displaystyle\int_0^b\frac{1}{2}xa^2\ \, \mathrm{d}x=\left[\frac{1}{4}x^2a^2\right]_0^b=\frac{1}{4}a^2b^2\\
    \displaystyle\int_A y^2\, \mathrm{d}A &amp;=\displaystyle\int_0^b\displaystyle\int_0^ay^2\ \, \mathrm{d}y\mathrm{d}x =\displaystyle\int_0^b\left[\frac{1}{3}y^3\right]_0^a\, \mathrm{d}x=\displaystyle\int_0^b\frac{1}{3}a^3\ \, \mathrm{d}x=\left[\frac{1}{3}a^3x\right]_0^b=\frac{1}{3}a^3b
    \end{align} </p>
<p>Hence </p>
<p>\(\qquad\displaystyle y_p=\frac{\displaystyle\int_Ay^2\, \mathrm{d}A }{\displaystyle\int_Ay\, \mathrm{d}A}=\frac{\frac{1}{3}a^3b}{\frac{1}{2}a^2b}=\frac{2}{3}a\quad \text{and}\quad\; x_p=\frac{\displaystyle\int_Axy\, \mathrm{d}A }{\displaystyle\int_Ay\, \mathrm{d}A }=\frac{\frac{1}{4}a^2b^2}{\frac{1}{2}a^2b}=\frac{1}{2}b\) </p>
<p> The centre of pressure is \((\dfrac{1}{2}b,\dfrac{2}{3}a\)), so is at a depth of \(\dfrac{2}{3}a\). </p>
<hr>


<h4> Areas and Moments </h4>
<p>The surface integral \(\displaystyle\int_Af(x,y)\, \mathrm{d}A\) can represent a number of physical quantities, depending on the function \(f(x,y)\) that is used. </p>
<p>Properties:</p>
<ol>
    <li>If \(f(x,y)=1\) then the integral represents the area of \(A\).</li>
    <li>If \(f(x,y)=x\) then the integral represents the first moment of \(A\) about the \(y\) axis.</li>
    <li>If \(f(x,y)=y\) then the integral represents the first moment of \(A\) about the \(x\) axis.</li>
    <li>If \(f(x,y)=x^2\) then the integral represents the second moment of \(A\) about the \(y\) axis.</li>
    <li>If \(f(x,y)=y^2\) then the integral represents the second moment of \(A\) about the \(x\) axis.</li>
    <li>If \(f(x,y)=x^2+y^2\) then the integral represents the second moment of \(A\) about the \(z\) axis.</li>
</ol>

<hr>
<h4 class="HELM_example">Example</h4>
<p>Given a rectangular lamina of length \(\ell\), width \(b\), thickness \(t\) (small) and density \(\rho\) (see the below figure), find the second moment of area of this lamina (moment of inertia) about the \(x\)-axis. </p>
<figure class="HELM_figure"><img src="@@PLUGINFILE@@/27.1-fig-9-xyplane.svg" alt="A diagram showing the xy plane representing a rectangular lamina" width="249" height="115" class="img-fluid atto_image_button_text-bottom"></figure>
<p></p>

<h4 class="HELM_solution">Solution</h4>
<p>By property \(5\) above, the moment of inertia is given by
    \begin{align}
    \displaystyle\int_0^b\displaystyle\int_0^{\ell}y^2\rho t\, \mathrm{d}x\mathrm{d}y&amp;=\rho t(\ell -0)\displaystyle\int_0^by^2\, \mathrm{d}y\\
    &amp;=\ell \rho t\left[\frac{y^3}{3}\right]_0^b\\
    &amp;=\ell \rho t\frac{b^3}{3}
    \end{align}
    As the mass of the lamina is \(M=\ell b t \rho\), the moment of inertia simplifies to \(\frac{1}{3}M b^2\). The \(t\) and \(\rho \) are included in
    the integral to make it a moment of inertia rather than simply a second moment.</p>]]></text>
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