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Math Help - Area of a Baseball Infield Using Riemann Sums

  1. #1
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    Area of a Baseball Infield Using Riemann Sums




    Hello everyone, i'm a pre-cal student. Starting calculus as we approach the end of the year and i'm stumped on this extra credit problem. I'm trying to find the area of this infield using riemann sums. We were told to find the area inside of the black lines. These black lines start at home plate, run on both foul lines, and then border the long curve at the top.

    Additional info my teacher gave, the equation of the grass line is y = sqrt(952 - x2) + 60.5
    Also, the equation of the foul line (the one on the right side) is y = x. The length of the foul line to the grass line is 127.6 feet.
    Any help would be appreciated
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  2. #2
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    Re: Area of a Baseball Infield Using Riemann Sums

    well, let's "cheat" first, and find the area by elementary means, and then we can can use riemann sums and have something with which to check our answer.

    first, let's find the area of a semi-circle, with radius 95. this is clearly (1/2)(π)(95)2.

    this semi-circle sweeps "past" the foul lines, so we have 2 pi-shaped sections (one on each side) to subtract.

    the sector angle of these pie-shaped sections is π/4 (45 degrees), so we need to know "how far past the foul line" the semi-circle goes at the horizontal line y = 60.5 (the pitching rubber). since the total radius of the circle is 95 feet, and the foul lines are at a perfect 45 degree angle, the horizontal line at the pitching rubber extends 60.5 feet before crossing the foul lines, so the circumference of the circle is 34.5 feet past the foul lines along the horizontal line through the front of the pitching rubber.

    thus each of our "pie-shaped regions" has area (34.5)(π/4).

    this means the area of the infield above the pitching rubber is:

    (1/2)(π)(95)2 - (34.5)(π/2).

    the area of the infield below the pitching rubber is easy to find, it is simply (60.5)2, for a total area of:

    (1/2)(π)(95)2 - (34.5)(π/2) + (60.5)2

    now, doing it by riemann sums, is going to be a daunting task. to make matters easier, let's just calculate "half" the area (from home plate to the right) and double it.

    we are going to be doing a sum of a function h(x) = g(x) - f(x), where g(x) is the equation for the circle, and f(x) is the equation for the right foul line. we know g, and we know f, so that's good. it's clear we are going to start our integration (riemann summing) at x = 0, but where to stop? we need to find the x-value where the circle and the foul line intersect.

    that is, x > 0 where g(x) = f(x).

    now g(x) = √(952 - x2) + 60.5.

    if g(x) = f(x), then:

    √(952 - x2) + 60.5 = x, so:

    √(952 - x2) = x - 60.5

    squaring both sides:

    952 - x2 = x2 - 121x + (60.5)2

    or: 2x2 - 121x - 5364.75 = 0, equivalently:

    x2 - (60.5)x - 2682.375

    the quadratic equation gives:

    x = (60.5 √[(60.5)2 - 4(-2682.375)])/2 or the two values:

    x = 90.229 <--this is approx. 127.6/√2

    x = -29.729 <--this value is negative, ignore it.

    so now we are in a position to form our riemann sums. first, we chop the interval [0,90.229] into n equal pieces. these are our "sub-intervals".

    each piece has width 90.229/n. let's agree the take the height of f(x) - g(x) at the right endpoint. what is the k-th right endpoint?

    the first is is: 90.229/n, that's obvious.

    the second one is 2(90.229)/n (90.229/n + 90.229/n, right?).

    so the k-th one is going to be: k(90.229)/n = 90.229(k/n).

    so our sum is:

    \sum_{k = 1}^n\ [g(90.229(k/n)) - f(90.229(k/n))](90.229/n)

    substituting in what g and f actually are, we get:

    \sum_{k = 1}^n\ [\sqrt{95^2 - (90.229)^2k^2/n^2} + 60.5 - 90.229(k/n)](90.229/n)

    that is going to be a beastly sum to evaluate, even after you split up the part with the square root, and everything else (the "everything else" won't be that hard).

    the reason why, is that it is tantamount to evaluating:

    \int_0^r \sqrt{r^2 - x^2}\ dx by using riemann sums, which is a very complicated thing to do (the anti-derivative itself is a rather complicated looking-function, it's non-trivial).

    so, if you actually find a way to evaluate these riemann sums, you certainly deserve the extra credit.
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