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 approximations. 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 fussing. 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 http://mathforum.org/dr.math/
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