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Surface Area of a Football

Date: 02/12/99 at 15:19:30
From: Brian
Subject: Surface area

What is the surface area of a football with a diameter of 22 inches 
and an arc length of 14 inches?

Date: 03/19/99 at 16:10:58
From: Doctor Gene
Subject: Re: Surface area

This football has problems: it is too thick in the middle to be NFL 
standard, or even to have an arc length of 14" - a radius of 11" would 
require an arc length of greater than 22". Maybe you mean a 
circumference of 22"? That, we can handle.

There are lots of curves that give football-like shapes when revolved 
around the x-axis, and which have these dimensions; you could take 
half an ellipse, a parabola, or no end of others. Most choices would 
give you integrals that you cannot evaluate in closed form, but you 
could use a program like Maple or Mathematica to get good 

Let us try approximating the shape of a football by revolving a 
parabola around the x-axis. The kind of parabola you probably want 
would be of the form f(x) = b - x^2/c (if you leave off the c you get 
a football that is way too fat). This has the advantage of yielding 
an integral that you can compute, the disadvantage of the extra 
unknown c - all shapes will involve nasty algebra.

                        * b       f(x) = b - x^2/c
                  *     |      *
              *         |          *
                                       \ sqrt (bc)

When the given parabola is revolved around the x-axis, the surface has 
diameter pi*b^2 = 22, so b = sqrt (22/pi), which is a little more than 
2 1/2" - a bit skinny for a football. You might want to try another 
curve, but I will illustrate what to do with this one.

The arc length is the integral from one x-intercept to the other of 
sqrt(1 + f'(x)^2), whose closed form expression lives in the back of 
most calculus books for our particular curve. Set the value to your 
given arc length of 14, and solve for c, which will take a bit of 

You will then know which parabola of our given form satisfies the arc 
length and diameter conditions you stipulated. The area of a surface 
of revolution is given by a famous integral: 

  2*pi* f(x)*sqrt(1 + f'(x)^2) 

between the two x-intercepts. In our case it can also be managed in 
closed form, but may cause you to review your integration techniques.

- Doctor Gene, The Math Forum
Associated Topics:
High School Calculus

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