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Math Help - Area of Circle Revisited

  1. #1
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    Area of Circle Revisited

    The usual was of developing the area of a circle is to consider,
    4\int_0^r\sqrt{r^2-x^2}dx
    Another is by increasing number of sides of a regular polygon (Archimedean method).

    I found a very simple one.
    It does not involve complicated integration as in case 1 and complicated limit evaluation as in case 2.

    1)A circle with radius r
    2)Divide it into n equal sectors.
    3)Consider one of the sectors.
    4)The archimeaden way is to find the area of the appoximating triangle. If using the archimedean method you need to find its area. Thus, its chord length times the height- but this get complicated once we take a limit. Instead note for large n the length of the radius is approximate the same as the height. Thus, r=h. Also, the length of the chord is close to the length of the arc thus, \frac{2\pi r}{n}
    Thus, the approximated area for this triangle is,
    \frac{1}{2}\cdot \frac{2\pi r}{n} \cdot r=\frac{\pi r^2}{n}
    5)Finally mutiply by the number of triangles, n\cdot \frac{\pi r^2}{n}=\pi r^2
    6)Since the result is invariant for any n it follows that the limit of this is \pi r^2
    ---
    I am rather afraid of this reasoning. Let me explain why. I once played around with "close to" arguments (you know the ones physicists always use) and got completly different results. Why!?! Because the norm of the partition did not converge to zero. I developed a dictum that you can use "close to" informal arguments whenever its satisfies the conditions of the Riemann integral, becuase it is well-defined.

    What do you think in the "proof" any errors? (Just because the result is correct does not mean the method was).
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    The usual was of developing the area of a circle is to consider,
    4\int_0^r\sqrt{r^2-x^2}dx
    Another is by increasing number of sides of a regular polygon (Archimedean method).

    I found a very simple one.
    It does not involve complicated integration as in case 1 and complicated limit evaluation as in case 2.

    1)A circle with radius r
    2)Divide it into n equal sectors.
    3)Consider one of the sectors.
    4)The archimeaden way is to find the area of the appoximating triangle. If using the archimedean method you need to find its area. Thus, its chord length times the height- but this get complicated once we take a limit. Instead note for large n the length of the radius is approximate the same as the height. Thus, r=h. Also, the length of the chord is close to the length of the arc thus, \frac{2\pi r}{n}
    Thus, the approximated area for this triangle is,
    \frac{1}{2}\cdot \frac{2\pi r}{n} \cdot r=\frac{\pi r^2}{n}
    5)Finally mutiply by the number of triangles, n\cdot \frac{\pi r^2}{n}=\pi r^2
    6)Since the result is invariant for any n it follows that the limit of this is \pi r^2
    ---
    I am rather afraid of this reasoning. Let me explain why. I once played around with "close to" arguments (you know the ones physicists always use) and got completly different results. Why!?! Because the norm of the partition did not converge to zero. I developed a dictum that you can use "close to" informal arguments whenever its satisfies the conditions of the Riemann integral, becuase it is well-defined.

    What do you think in the "proof" any errors? (Just because the result is correct does not mean the method was).
    You have just hidden a limit by using the small angle approximation
    for tan and/or sine. (At least that is what it looks like to me )

    Also look at what Archimedes is doing - he is estimating \pi, the
    integration method can be adapted to do this by using a sequence of upper
    and lower Riemann sums, but then we are estimating the area of the semi-
    circle by trapping it between the areas of sequences of inscribed and
    circumscribed staircase polygons, which is just a variant of Archimedes
    method.

    In your method the value of \pi parachutes in as the constant
    of proportionality of the diameter to circumference of a circle. So I think
    that your method is showing that area of the unit circle is numericaly
    equal to half the circumference.

    RonL
    Last edited by CaptainBlack; June 28th 2006 at 09:03 PM.
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