6 Applications of surface integration over rectangular areas

6.1 Force on a dam

At the beginning of this Section, the total force on a dam was given by the surface integral

${\int \; <!--nolimits\; -->}_{A}k\left(h-y\right)dA$

Imagine that the dam is rectangular in profile with a width of 100 m and a height $h$ of 40 m. The expression $dA$ is replaced by $dxdy$ 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 $10^4\text{kg}{\text{m}}^{-2}{\text{s}}^{-2}$ . The surface integral becomes the double integral

${\int \; <!--nolimits\; -->}_0^{40}{\int \; <!--nolimits\; -->}_0^{100}k\left(h-y\right)dxdy$ that is ${\int \; <!--nolimits\; -->}_0^{40}{\int \; <!--nolimits\; -->}_0^{100}10^4\left(40-y\right)dxdy$

As the integral in this double integral does not contain $x$ , the integral may be written

that is the total force is 800 meganewtons.

6.2 Centre of pressure

We wish to find the centre of pressure $\left(x_p,y_p\right)$ 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.

Figure 7

We require the following results:

1. The pressure $p$ is proportional to the depth $h$ , so that $p=\omega h$ where $\omega$ is a constant.
2. The force $F$ on an area $\delta A$ subjected to constant pressure $p$ is given by $F=p\delta A$

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 ${\int \; <!--nolimits\; -->}_A\omega ydA=\omega {\int \; <!--nolimits\; -->}_{A}ydA$ .

Taking moments about $Oy$ :

Taking moments about $Ox$ :

Hence

$x_p=\frac{{\int \; <!--nolimits\; -->}_{A}xydA}{{\int \; <!--nolimits\; -->}_{A}ydA}$ and $y_p=\frac{{\int \; <!--nolimits\; -->}_A{y}^{2}dA}{{\int \; <!--nolimits\; -->}_{A}ydA}$ .

Example 6

A rectangle of sides $a$ and $b$ is immersed vertically in a fluid with one of its edges in the surface as shown in Figure 8. Where is the centre of pressure?

Figure 8

Solution

To express the surface integral as double integrals we will use cartesian coordinates and vertical slices. We need the following integrals.

Hence

$y_p=\frac{{\int \; <!--nolimits\; -->}_A{y}^{2}dA}{{\int \; <!--nolimits\; -->}_{A}ydA}=\frac{\frac{1}{3}{a}^{3}b}{\frac{1}{2}{a}^{2}b}=\frac{2}{3}a\text{and}{x}_p=\frac{{\int \; <!--nolimits\; -->}_{A}xydA}{{\int \; <!--nolimits\; -->}_{A}ydA}=\frac{\frac{1}{4}{a}^2{b}^2}{\frac{1}{2}{a}^{2}b}=\frac{1}{2}b$

The centre of pressure is $\left(\frac{1}{2}b,\frac{2}{3}a\right)$ , so is at a depth of $\frac{2}{3}a$ .

6.3 Areas and moments

The surface integral ${\int \; <!--nolimits\; -->}_{A}f\left(x,y\right)dA$ can represent a number of physical quantities, depending on the function $f\left(x,y\right)$ that is used.

Properties:

1. If $f\left(x,y\right)=1$ then the integral represents the area of $A$ .
2. If $f\left(x,y\right)=x$ then the integral represents the first moment of $A$ about the $y$ axis.
3. If $f\left(x,y\right)=y$ then the integral represents the first moment of $A$ about the $x$ axis.
4. If $f\left(x,y\right)=x^2$ then the integral represents the second moment of $A$ about the $y$ axis.
5. If $f\left(x,y\right)=y^2$ then the integral represents the second moment of $A$ about the $x$ axis.
6. If $f\left(x,y\right)=x^2+y^2$ then the integral represents the second moment of $A$ about the $z$ axis.

Example 7

Given a rectangular lamina of length $\ell$ , width $b$ , thickness $t$ (small) and density $\rho$ (see Figure 9), find the second moment of area of this lamina (moment of inertia) about the $x$ -axis.

Figure 9

Solution

By property (e) above, the moment of inertia is given by

As the mass of the lamina is $M=\ell bt\rho$ , 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.

Task!

By a similar method to that in Example 7, find the moment of inertia of the same lamina about the $y$ -axis.

Answer

Exercises

By making use of the form of the integrand, evaluate the following double integrals:

1. $I={\int \; <!--nolimits\; -->}_0^{\pi }{\int \; <!--nolimits\; -->}_0^{1}y{cos}^{2}xdydx$
2. $I={\int \; <!--nolimits\; -->}_{-8}^3{\int \; <!--nolimits\; -->}_{-1}^1{y}^{2}dydx$
3. $I={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^5{\left(s+1\right)}^{4}dtds$

Answer