Volume of a football!

A picture would be worth a thousand words here, but since I can't provide the picture the words will have to do.

I'll start by looking at the longitudinal cross-section of the football, focusing on the "upper" region formed by one of the circular arcs and the 11-inch chord that subtends it. Now the largest round circumference of 22 inches corresponds to a radius of $\frac{11}{\pi}$ inches. This is in fact the "height" of the region we're presently dealing with, i.e., the perpendicular distance from the midpoint of the chord to the circular arc.

Now picture this arc as part of a circle of radius $R$ centered at the origin, with the 11-inch chord parallel to and above the x-axis. Then we can form a right triangle with the origin, the midpoint of the chord and one of the chord's endpoints as vertices. Using Pythagoras, we have that

$R^{2} = (\frac{11}{2})^{2} + (R - \frac{11}{\pi})^{2}$ .

Solving for $R$ , we have that $R = (\frac{121}{88*\pi}) * (\pi^{2} + 4)$ .

Now the chord will have a y-value of $d = R - \frac{11}{\pi}$ . What we now need to do is rotate the circular arc about the horizontal line $y = d$ to form the football as described. The radius of each of these latitudinal cross-sections will be $(y - d)$ , where $y^{2} = R^{2} - x^{2}$ . Integrating over $x$ , from $a = -\frac{11}{2}$ to $b = \frac{11}{2}$ , we then use the circular disc method to find the volume of the football. This volume is

$\int_{a}^{b} \pi * (\sqrt{R^{2} - x^{2}} - d)^{2} dx$ .

This integral is straightforward but will involve a trig substitution, so I will just leave the reader to solve it as an exercise. The solution comes out to

$\pi * (11(R^{2} + d^{2}) - (\frac{2}{3})b^{3} - 2dR^{2} * \arcsin(\frac{11}{2R}) - 2db*\sqrt{R^{2} - b^{2}})$ ,

which after substituting gives us a volume of $\boxed{250.574}$ cubic inches, which is within $3\%$ of the "official" value of $258.558$ cubic inches.

The longitudinal cross-section of the football is a segment of a circle.
The volume of revolution, V=Area to be revolved * the length of the path traced by its C.G. Here the segment makes an angle $\theta$ at the center. h is the HEIGHT of the segment.
$\text{h is the radius of the maximum circle with circumference of 22.} \\ \therefore ~\color{#3D99F6}{h=\dfrac {22}{2*\pi}= \dfrac {11}{\pi}} \\ \text{The chord containing h, is divided into h, and (2R - h) by the 11" chord. }\\ \text{The the intersection point is the midpoint of 11" chord.}\\ \therefore ~ h*(2R - h)=(\dfrac{11} 2)^2. ~~ Solving~ \color{#3D99F6}{R=\dfrac{11*\pi}2*(\frac 1 4 +\frac 1{\pi^2})}\\ Segment ~ \color{#3D99F6}{area ~ =\dfrac{R^2} 2 * (\theta - Sin\theta)}\\ \text{From the center of the circle, C.G. is at a distance of } {\bar Y}\\ \color{#3D99F6}{ {\bar Y}=\dfrac 4 3 * R *\dfrac {( Sin\theta/2 )^3}{\theta - Sin\theta}}\\ \text{We shall revolve the segment around the 11” chord to get the foot ball volume.}\\ \text{ So we should have the distance of C.G. from this chord. And that distance is} \color{#3D99F6}{r = h - R + {\bar Y} }\\ Note ~ R*Sin(\theta/2)=\dfrac {11} 2. ~ \therefore ~\theta=2*Sin^{-1}\dfrac 1 {\pi*(\frac 1 4 +\frac 1{\pi^2})} .\\ C.G. ~traces ~a ~path ~length =2*\pi*r. ~~~~~~~~~~~ V=2*\pi*r. *Area.\\ Substituting, V=2*\pi\{h -R + \dfrac 4 3 * R *\dfrac {( Sin\theta/2 )^3}{\theta - Sin\theta}\}* \dfrac{R^2} 2 * (\theta - Sin\theta)\\ =250.5740$

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