5 Engineering Example 1

5.1 Volume of liquid in an elliptic tank

Introduction

A tank in the shape of an elliptic cylinder has a volume of liquid poured into it. It is useful to know in advance how deep the liquid will be. In order to make this calculation, it is necessary to perform a multiple integration.

Figure 22

Problem in words

The tank has semi-axes $a$ (horizontal) and $c$ (vertical) and is of constant thickness $b$ . A volume of liquid $V$ is poured in (assuming that $V<\pi abc$ , the volume of the tank), filling it to a depth $h$ , which is to be calculated. Assume 3-D coordinate axes based on a point at the bottom of the tank.

Mathematical statement of the problem

Since the tank is of constant thickness $b$ , the volume of liquid is given by the shaded area multiplied by $b$ , i.e.

$V=b\times \text{shaded area}$

where the shaded area can be expressed as the double integral

${\int \; <!--nolimits\; -->}_{z=0}^h{\int \; <!--nolimits\; -->}_{x=x_1}^{x_2}dxdz$

where the limits $x_1$ and $x_2$ on $x$ can be found from the equation of the ellipse

$\frac{x^2}{a^2}+\frac{{\left(z-c\right)}^2}{c^2}=1$

Mathematical analysis

From the equation of the ellipse

Thus

$x_1=-\frac{a}{c}\sqrt{2zc-z^2},\text{ and }{x}_2=+\frac{a}{c}\sqrt{2zc-z^2}$

Consequently

Now use substitution  $z-c=csin\theta$ so that  $dz=ccos\theta d\theta$

which can also be expressed in the form

$V=abc\left[{sin}^{-1}\left(\frac{h}{c}-1\right)+\left(\frac{h}{c}-1\right)\sqrt{1-{\left(\frac{h}{c}-1\right)}^2}+\frac{\pi }{2}\right]$

While ( $\ast$ ) expresses $V$ as a function of ${\theta }_0$ (and therefore $h$ ) to find ${\theta }_0$ as a function of $V$ requires a numerical method. For a given $a$ , $b$ , $c$ and $V$ , solve equation ( $\ast$ ) by a numerical method to find ${\theta }_0$ and find $h$ from $h=c\left(1+sin{\theta }_0\right)$ .

Interpretation

If $a$ = 2 m, $b$ = 1 m, $c$ = 3 m (so the total volume of the tank is $6\pi {\text{m}}^3\approx 18.85{\text{m}}^3$ ), and a volume of $7{\text{m}}^3$ is to be poured into the tank then

$V=abc\left[{\theta }_0+\frac{1}{2}sin2{\theta }_0+\frac{\pi }{2}\right]$

which becomes

$7=6\left[{\theta }_0+\frac{1}{2}sin2{\theta }_0+\frac{\pi }{2}\right]$

and has solution ${\theta }_0$ = $-0.205$ (3 decimal places).

Finally

compared to the maximum height of 6 m.

Exercises

1. Evaluate the functions
   1. $xy$ and
   2. $xy+3y^2$

      over the quadrilateral with vertices at $\left(0,0\right)$ , $\left(3,0\right)$ , $\left(2,2\right)$ and $\left(0,4\right)$ .

2. Show that $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_{A}f\left(x,y\right)dydx=\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_{A}f\left(x,y\right)dxdy$ for $f\left(x,y\right)=x{y}^2$ when $A$ is the interior of the triangle with vertices at $\left(0,0\right)$ , $\left(2,0\right)$ and $\left(2,4\right)$ .

3. By reversing the order of the two integrals, evaluate the integral ${\int \; <!--nolimits\; -->}_{y=0}^4{\int \; <!--nolimits\; -->}_{x=y^{1∕2}}^{2}sin{x}^{3}dxdy$

4. Integrate the function $f\left(x,y\right)=x^3+x{y}^2$ over the quadrant $x\ge 0$ , $y\ge 0$ , $x^2+y^2\le 1$ .

Answer