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A New Quadrature of the Parabola
Posted:
Jan 2, 2002 3:18 PM


The following is a quadrature of the parabola using neither infinitesimals nor limits. It requires only the algebra of polynomial functions and the geometry of similar figures. I haven't been able to find any reference in the literature for a calculation like this. I would appreciate it if you let me know of any such reference that you come across.
Consider the rectangle bounded by the four lines x=0, y=0, x=b, and y=b^2. We want to determine the area of a "parabolic right triangle" lying within this rectangle. The parabolic right triangle, or just ptriangle, has two legs lying on the lines y=0, x=b, and a hypotenuse given by the curve y=x^2. Call this ptriangle PT1.
The picture we will work with has two ptriangles sitting within the rectangle. The first ptriangle has its right angle vertex in the corner of the rectangle at the point (b,0). This is just PT1. The second ptriangle, of equal area to the first, has its right angle vertex located at the point (0,b^2) with legs along the lines x=0 and y=b^2. Call this one PT2. The area of the rectangle, b^3, is given by twice the area of PT1 plus the area of the "eye." This eye corresponds to the gap between the two ptriangles. If we can relate the area of the "eye" to the area of PT1 the quadrature of the parabola is solved.
We first subtract the area of PT2 from the area of the rectangle. The remaining figure, eye and lower ptriangle, has a boundary given by b^2(bx)^2. The remaining area is the area under this curve. If we then subtract the area of PT1 from this remaining area we obtain the area of the gap between the two ptriangles. The bounding curve for this gap is g(x) =b^2(bx)^2x^2. The area of the gap is given as the area under the curve for g(x) over the interval from 0 to b. (Our eye, viewed algebraically, is a little hill.)
The hill is symmetric around x=b/2. So the area of the whole hill is twice the area of half the hill, say from b/2 to b. In turn this means that we can compute the area of the hill as half the area of the original rectangle minus the area of a ptriangle with legs lying on the lines y=0, x=b/2 and with hypotenuse given by y=4x^2. Call this last ptriangle PT3.
The ptriangle, PT3, has an area four times the area of the ptriangle with legs lying on y=0, x=b/2 and hypotenuse given by y=x^2. Call this ptriangle PT4. Now it is also the case that the area of PT4 is oneeighth the area of PT1. (Why?) Thus the area of PT3 is equal to one half the area of PT1.
Since the area of the gap (or the hill) is one half the area of the original rectangle minus the area of PT3, then half the area of the gap must be equal to half the area of PT1. Or to say the same thing twice as well, the area of the gap is equal to the area of PT1. Hence the area of the rectangle is three times the area under the parabolic triangle. QED.
For the area under the cubic, the area of the gap again has a quadratic boundary. (Why?) The determination of the area under the cubic reduces to the same type of argument as just presented. Try it. In turn these two cases act as a model for determining the area under the curve y=x^n using a proof by induction.
I think this is more than just a nice calculation. I think it has interesting implications for calculus reform.
Bill Crombie

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