### 4 Engineering Example 3

#### 4.1 Volume of liquid in an ellipsoidal tank

Introduction

An ellipsoidal tank (elliptical when viewed from along $x$ -, $y$ - or $z$ -axes) 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 and calculate a Jacobian.

Figure 34

Problem in words

The metal tank is in the form of an ellipsoid, with semi-axes $a$ , $b$ and $c$ . A volume $V$ of liquid is poured into the tank ( $V<\frac{4}{3}\pi abc$ , the volume of the ellipsoid) and the problem is to calculate the depth, $h$ , of the liquid.

Mathematical statement of problem

The shaded area is expressed as the triple integral

$\phantom{\rule{2em}{0ex}}V={\int }_{z=0}^{h}{\int }_{y={y}_{1}}^{{y}_{2}}{\int }_{x={x}_{1}}^{{x}_{2}}dxdydz$

where limits of integration

which come from rearranging the equation of the ellipsoid $\left(\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}+\frac{{\left(z-c\right)}^{2}}{{c}^{2}}=1\right)$ and limits

from the equation of an ellipse in the $y$ - $z$ plane $\left(\frac{{y}^{2}}{{b}^{2}}+\frac{{\left(z-c\right)}^{2}}{{c}^{2}}=1\right)$ .

Mathematical analysis

To calculate $V$ , use the substitutions

$\begin{array}{rcll}x& =& a\tau cos\varphi {\left(1-\frac{{\left(z-c\right)}^{2}}{{c}^{2}}\right)}^{\frac{1}{2}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ y& =& b\tau sin\varphi {\left(1-\frac{{\left(z-c\right)}^{2}}{{c}^{2}}\right)}^{\frac{1}{2}}& \text{}\\ z& =& z\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

now expressing the triple integral as

$\phantom{\rule{2em}{0ex}}V={\int }_{z=0}^{h}{\int }_{\varphi ={\varphi }_{1}}^{{\varphi }_{2}}{\int }_{\tau ={\tau }_{1}}^{{\tau }_{2}}J\phantom{\rule{0.3em}{0ex}}d\tau d\varphi dz$

where $J$ is the Jacobian of the transformation calculated from

$\phantom{\rule{2em}{0ex}}J=\left|\begin{array}{ccc}\hfill \frac{\partial x}{\partial \tau }\hfill & \hfill \frac{\partial x}{\partial \varphi }\hfill & \hfill \frac{\partial x}{\partial z}\hfill \\ & & \\ \hfill \frac{\partial y}{\partial \tau }\hfill & \hfill \frac{\partial y}{\partial \varphi }\hfill & \hfill \frac{\partial y}{\partial z}\hfill \\ & & \\ \hfill \frac{\partial z}{\partial \tau }\hfill & \hfill \frac{\partial z}{\partial \varphi }\hfill & \hfill \frac{\partial z}{\partial z}\hfill \\ \hfill \hfill \end{array}\right|$

and reduces to

To determine limits of integration for $\varphi$ , note that the substitutions above are similar to a cylindrical polar co-ordinate system, and so $\varphi$ goes from 0 to $2\pi$ . For $\tau$ , setting $\tau =0$ $⇒$ $x=0$ and $y=0$ , i.e. the $z$ -axis.

Setting $\tau =1$ gives

$\phantom{\rule{2em}{0ex}}\frac{{x}^{2}}{{a}^{2}}={cos}^{2}\varphi \left(1-\frac{{\left(z-c\right)}^{2}}{{c}^{2}}\right)$ (1)

and

$\phantom{\rule{2em}{0ex}}\frac{{y}^{2}}{{b}^{2}}={sin}^{2}\varphi \left(1-\frac{{\left(z-c\right)}^{2}}{{c}^{2}}\right)$ (2)

Summing both sides of Equations (1) and (2) gives

$\phantom{\rule{2em}{0ex}}\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=\left({cos}^{2}\varphi +{sin}^{2}\varphi \right)\left(1-\frac{{\left(z-c\right)}^{2}}{{c}^{2}}\right)$

or

$\phantom{\rule{2em}{0ex}}\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}+\frac{{\left(z-c\right)}^{2}}{{c}^{2}}=1$

which is the equation of the ellipsoid, i.e. the outer edge of the volume. Therefore the range of $\tau$ should be 0 to 1. Now

$\begin{array}{rcll}V& =& ab{\int }_{z=0}^{h}\left(1-\frac{{\left(z-c\right)}^{2}}{{c}^{2}}\right){\int }_{\varphi =0}^{2\pi }{\int }_{\tau =0}^{1}\tau \phantom{\rule{0.3em}{0ex}}d\tau d\varphi dz& \text{}\\ & =& \frac{ab}{{c}^{2}}{\int }_{z=0}^{h}\left(2zc-{z}^{2}\right){\int }_{\varphi =0}^{2\pi }{\left[\frac{{\tau }^{2}}{2}\right]}_{\tau =0}^{1}\phantom{\rule{0.3em}{0ex}}d\varphi dz& \text{}\\ & =& \frac{ab}{2{c}^{2}}{\int }_{z=0}^{h}\left(2zc-{z}^{2}\right){\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\varphi \right]}_{\varphi =0}^{2\pi }\phantom{\rule{0.3em}{0ex}}dz& \text{}\\ & =& \frac{\pi ab}{{c}^{2}}{\left[c{z}^{2}-\frac{{z}^{3}}{3}\right]}_{z=0}^{h}& \text{}\\ & =& \frac{\pi ab}{{c}^{2}}\left(c{h}^{2}-\frac{{h}^{3}}{3}\right)\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

Interpretation

Suppose the tank has actual dimensions of $a$ = 2 m, $b$ = 0.5 m and $c$ = 3 m and a volume of $7{\text{m}}^{3}$ is to be poured into it. (The total volume of the tank is $4\pi {m}^{3}\approx 12.57{\text{m}}^{3}$ ). Then, from above

$\phantom{\rule{2em}{0ex}}V=\frac{\pi ab}{{c}^{2}}\left(c{h}^{2}-\frac{{h}^{3}}{3}\right)$

which becomes

$\phantom{\rule{2em}{0ex}}7=\frac{\pi }{9}\left(3{h}^{2}-\frac{{h}^{3}}{3}\right)$

with solution $h$ = 3.23 m (2 d.p.), compared to the maximum height of the ellipsoid of 6 m.

##### Exercises
1. The function $f={x}^{2}+{y}^{2}$ is to be integrated over an elliptical cone with base being the ellipse, ${x}^{2}∕4+{y}^{2}=1$ , $z=0$ and apex (point) at $\left(0,0,5\right)$ . The integral can be made simpler by means of the change of variables $x=2\left(1-\frac{w}{5}\right)\tau cos\theta$ , $y=\left(1-\frac{w}{5}\right)\tau sin\theta$ , $z=w$ .

1. Find the limits on the variables $\tau$ , $\theta$ and $w$ .
2. Find the Jacobian $J\left(\tau ,\theta ,w\right)$ for this transformation.
3. Express the integral $\int \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\int \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\int \left({x}^{2}+{y}^{2}\right)\phantom{\rule{0.3em}{0ex}}dxdydz$ in terms of $\tau$ , $\theta$ and $w$ .
4. Evaluate this integral. [Hint:- it may be worth noting that ${cos}^{2}\theta \equiv \frac{1}{2}\left(1+cos2\theta \right)$ ].

Note: This integral has relevance in topics such as moments of inertia.

2. Using cylindrical polar coordinates, integrate the function $f=z\sqrt{{x}^{2}+{y}^{2}}$ over the volume between the surfaces $z=0$ and $z=1+{x}^{2}+{y}^{2}$ for $0\le {x}^{2}+{y}^{2}\le 1$ .
3. A torus (doughnut) has major radius $R$ and minor radius $r$ . Using the transformation $x=\left(R+\tau cos\alpha \right)cos\theta$ , $y=\left(R+\tau cos\alpha \right)sin\theta$ $z=\tau sin\alpha$ , find the volume of the torus. [Hints:- limits on $\alpha$ and $\theta$ are $0$ to $2\pi$ , limits on $\tau$ are $0$ to $r$ . Show that Jacobian is $\tau \left(R+\tau cos\alpha \right)$ ].

4. Find the Jacobian for the following transformations.
1. $x={u}^{2}+vw$ , $y=2v+{u}^{2}w$ , $z=uvw$
2. Cylindrical polar coordinates. $x=\rho cos\theta$ , $y=\rho sin\theta$ , $z=z$

1. $\tau$ : $0$ to $1$ , $\theta$ : $0$ to $2\pi$ , $w$ : $0$ to $5$
2. $2{\left(1-\frac{w}{5}\right)}^{2}\tau$
3. $2{\int }_{\tau =0}^{1}{\int }_{\theta =0}^{2\pi }{\int }_{w=0}^{5}{\left(1-\frac{w}{5}\right)}^{4}{\tau }^{3}\left(4{cos}^{2}\theta +{sin}^{2}\theta \right)\phantom{\rule{0.3em}{0ex}}dwd\theta d\tau$
4. $\frac{5}{2}\pi$
1. $\frac{92}{105}\pi$
2. $2{\pi }^{2}R{r}^{2}$
1. $4{u}^{2}v-2{u}^{4}w+{u}^{2}v{w}^{2}-2{v}^{2}w$ ,
2. $\rho$