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

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To find the volume of a football (elliptical), the correct approach involves using the formula for an ellipse, where the semi-major axis (a) is 14 cm and the semi-minor axis (b) is derived from the circumference. The volume is calculated by integrating the area of circular cross-sections along the length of the football, using the formula pi * radius^2 for each slice. The integration is performed from 0 to 5.5 cm, yielding the formula for half the volume as (2/3)*pi*a*(b^2). To obtain the full volume, this result must be doubled. Understanding the integration process is key to solving the problem accurately.
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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 \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

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) \frac{73}{2\pi}

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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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
 
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
 
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