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

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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 < 4 3 π a b c , 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

V = z = 0 h y = y 1 y 2 x = x 1 x 2 d x d y d z

where limits of integration

x 1 = a 1 y 2 b 2 ( z c ) 2 c 2  and  x 2 = + a 1 y 2 b 2 ( z c ) 2 c 2

which come from rearranging the equation of the ellipsoid x 2 a 2 + y 2 b 2 + ( z c ) 2 c 2 = 1 and limits

y 1 = b c c 2 ( z c ) 2  and  y 2 = + b c c 2 ( z c ) 2

from the equation of an ellipse in the y - z plane y 2 b 2 + ( z c ) 2 c 2 = 1 .

Mathematical analysis

To calculate V , use the substitutions

x = a τ cos ϕ 1 ( z c ) 2 c 2 1 2 y = b τ sin ϕ 1 ( z c ) 2 c 2 1 2 z = z

now expressing the triple integral as

V = z = 0 h ϕ = ϕ 1 ϕ 2 τ = τ 1 τ 2 J d τ d ϕ d z

where J is the Jacobian of the transformation calculated from

J = x τ x ϕ x z y τ y ϕ y z z τ z ϕ z z

and reduces to

J = x τ y ϕ x ϕ y τ  since  z τ = z ϕ = 0 = a cos ϕ 1 ( z c ) 2 c 2 1 2 b τ cos ϕ 1 ( z c ) 2 c 2 1 2 a τ sin ϕ 1 ( z c ) 2 c 2 1 2 b sin ϕ 1 ( z c ) 2 c 2 1 2 = a b τ cos 2 ϕ + sin 2 ϕ 1 ( z c ) 2 c 2 = a b τ 1 ( z c ) 2 c 2

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

Setting τ = 1 gives

x 2 a 2 = cos 2 ϕ 1 ( z c ) 2 c 2 (1)

and

y 2 b 2 = sin 2 ϕ 1 ( z c ) 2 c 2 (2)

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

x 2 a 2 + y 2 b 2 = ( cos 2 ϕ + sin 2 ϕ ) 1 ( z c ) 2 c 2

or

x 2 a 2 + y 2 b 2 + ( z c ) 2 c 2 = 1

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

V = a b z = 0 h 1 ( z c ) 2 c 2 ϕ = 0 2 π τ = 0 1 τ d τ d ϕ d z = a b c 2 z = 0 h ( 2 z c z 2 ) ϕ = 0 2 π τ 2 2 τ = 0 1 d ϕ d z = a b 2 c 2 z = 0 h ( 2 z c z 2 ) ϕ ϕ = 0 2 π d z = π a b c 2 c z 2 z 3 3 z = 0 h = π a b c 2 c h 2 h 3 3

Interpretation

Suppose the tank has actual dimensions of a = 2 m, b = 0.5 m and c = 3 m and a volume of 7 m 3 is to be poured into it. (The total volume of the tank is 4 π m 3 12.57 m 3 ). Then, from above

V = π a b c 2 c h 2 h 3 3

which becomes

7 = π 9 3 h 2 h 3 3

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 ( 0 , 0 , 5 ) . The integral can be made simpler by means of the change of variables x = 2 ( 1 w 5 ) τ cos θ , y = ( 1 w 5 ) τ sin θ , z = w .

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    1. Find the limits on the variables τ , θ and w .
    2. Find the Jacobian J ( τ , θ , w ) for this transformation.
    3. Express the integral ( x 2 + y 2 ) d x d y d z in terms of τ , θ and w .
    4. Evaluate this integral. [Hint:- it may be worth noting that cos 2 θ 1 2 ( 1 + cos 2 θ ) ].

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

  2. Using cylindrical polar coordinates, integrate the function f = z x 2 + y 2 over the volume between the surfaces z = 0 and z = 1 + x 2 + y 2 for 0 x 2 + y 2 1 .
  3. A torus (doughnut) has major radius R and minor radius r . Using the transformation x = ( R + τ cos α ) cos θ , y = ( R + τ cos α ) sin θ z = τ sin α , find the volume of the torus. [Hints:- limits on α and θ are 0 to 2 π , limits on τ are 0 to r . Show that Jacobian is τ ( R + τ cos α ) ].

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  4. Find the Jacobian for the following transformations.
    1. x = u 2 + v w , y = 2 v + u 2 w , z = u v w
    2. Cylindrical polar coordinates. x = ρ cos θ , y = ρ sin θ , z = z

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    1. τ : 0 to 1 , θ : 0 to 2 π , w : 0 to 5
    2. 2 ( 1 w 5 ) 2 τ
    3. 2 τ = 0 1 θ = 0 2 π w = 0 5 ( 1 w 5 ) 4 τ 3 ( 4 cos 2 θ + sin 2 θ ) d w d θ d τ
    4. 5 2 π
  1. 92 105 π
  2. 2 π 2 R r 2
    1. 4 u 2 v 2 u 4 w + u 2 v w 2 2 v 2 w ,
    2. ρ