Discovering the Volume of a Football (Elliptical): Easy Guide

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In summary, to find the volume of a football (elliptical), we can use the formula pi*y^2dx, where y is determined by the equation of the ellipse and dx is the thickness of each slice. Integrating this formula between 0 and 5.5 will give us half the volume of the ideal football. To do this, we can simplify the equation (pi*b^2/a^2)* ((a^3) - ((a^3)/3)) to (2/3)*pi*a*(b^2). However, to be more accurate, we should integrate between -5.5 and 5.5. The area of a circle is pi*r^2, and the thickness is dx
  • #1
HeyHow!
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We have been given the task of finding the volume of a football (elliptical).

i know the area for an ellipse is [tex]\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1[/tex]

where a=distance from center to major axis x-direction (Half the length of the ball) (a=14cm)

and

b=distance to minor axis y-direction (circumference/2pi) [tex]\frac{73}{2\pi}[/tex]

from there i get confused. i found on a website that
Now if we integrate Area*dx where Area= pi*y^2 (area of a cross
section of the football as given for our ellipse above) between 0 and
5.5 we will obtain half the volume for our ideal football.
This integration results in:
(pi*b^2/a^2)* ((a^3) - ((a^3)/3))
One can simplify this equation into:
(2/3)*pi*a*(b^2)
Remember, this is half the volume of our ideal football. To be more
correct, one would integrate between -5.5 and 5.5. The calculations
work out easier using 0 to 5.5.
i don't understand the Area*dx and where Area= pi*y^2 comes from. Does that mean that pi*y^2 has to be integrated? I also have no idea of how the integration would look like. i do not know what to integrate to get to (pi*b^2/a^2)* ((a^3) - ((a^3)/3)).

any help would be appreciated greatly. i am confused
 
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  • #2
Looking at the football from the skinny end you see a circle.

You calculate the volume by slicing the football into a bunch of circular slices and finding their volumes. The volume of each slice is the area of the circle, times its thickness. The area of a circle is pi*r2. The thickness is dx. Each slice occurs at a different value of x. The radius of each circle is the y value at that x determined by the equation of the elipse. (actually, it is the separation of the |y| from the x-axis of the ellipse, but since your ellipse is on the coordinate system x-axis, the y value is that seperation.)

So, your total volume is the sum of each individual volume which is pi*y2dx.

You re-arrange your ellipse equation to get y2 in terms of x and integrate.

Njorl
 
  • #3
and don't know where to start.

Hi there,

Thank you for sharing your thoughts and questions on finding the volume of a football (elliptical). Let's break down the process and see if we can clarify some of your confusion.

First of all, the formula you mentioned for finding the area of an ellipse is correct. However, we need to make a small adjustment for our specific case of a football. The formula for an ellipse is \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, where a and b represent the semi-major and semi-minor axes, respectively. In the case of a football, the semi-major axis (a) is half the length of the ball, which is 14cm. However, the semi-minor axis (b) is not the circumference divided by 2π. Instead, it is the radius of the ball, which is half the circumference divided by π. So, b = \frac{73}{2\pi \times 2} = \frac{73}{4\pi}.

Now, to find the volume of the football, we need to integrate the area of a cross-section of the football (which is a circle) with respect to the length of the ball (dx). The reason we are using dx as the variable of integration is because we are integrating along the x-axis, which represents the length of the ball. This is why we need to use the formula for the area of a circle, which is pi * radius^2. In our case, the radius is y, which is equal to b at any point along the x-axis (since we are looking at a cross-section of the football).

So, the integral we need to solve is \int_{0}^{5.5} \pi * b^2 dx. This means we are integrating from 0 to 5.5 (half the length of the ball) and multiplying each cross-section's area by pi * b^2. When we solve this integral, we get (pi*b^2/a^2)* ((a^3) - ((a^3)/3)), which is the same as (2/3)*pi*a*(b^2). This is the formula for half the volume of the football, so we need to double this to get the full volume.

I understand that the integration process can be confusing, but it essentially
 

1. How do you measure the volume of a football?

The volume of a football can be measured by using the formula for the volume of an ellipsoid, V = (4/3)πabc, where a, b, and c are the three axes of the football. It can also be measured by filling the football with water and measuring the displaced volume.

2. Is the volume of a football the same as its mass?

No, the volume and mass of a football are not the same. Mass is a measure of the amount of matter in an object, while volume is a measure of the space that an object occupies. The mass of a football is affected by factors such as the material it is made of and any additional weight, such as air or water, inside the football.

3. Can the volume of a football change?

Yes, the volume of a football can change depending on the external factors affecting it. For example, if a football is filled with air, the volume will decrease if the air pressure decreases. Similarly, if a football is filled with water, the volume will increase as the water expands when heated.

4. What tools are needed to measure the volume of a football?

The tools needed to measure the volume of a football include a ruler or tape measure to measure the axes of the football, a calculator to calculate the volume using the formula, and a container of water and a measuring cup to measure the displaced volume method.

5. How can knowing the volume of a football be useful?

Knowing the volume of a football can be useful for various reasons. For manufacturers, it helps in determining the amount of material needed to make a football. For players, it can help in understanding the aerodynamics of the football and how it will behave when kicked or thrown. Additionally, knowing the volume can also be used to determine the amount of air or water needed to inflate or fill a football to the desired level.

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