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Thread: Archimedes Proof

  1. #1
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    Archimedes Proof

    The area of a parabolic arch is equal to $\displaystyle 2/3$ the product of the base and the height.

    I need to prove this formula for the general parabola $\displaystyle f(x)=ax^2+h$, where $\displaystyle a$ is any real number less than zero
    and a definite integral to represent the requested area



    thank you
    Last edited by doctorgk; Feb 26th 2008 at 07:42 PM.
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  2. #2
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    Hello, doctorgk!

    The area of a parabolic arch is equal to $\displaystyle 2/3$ the product of the base and the height.

    Prove this formula for the general parabola $\displaystyle f(x)\:=\:-ax^2+h$
    They could have used a nicer general form . . .
    Code:
                    |
                   h*
                 *  |  *
               *    |    *
              *     |     *
                    |
          - -*- - - + - - -*- -
             b      |      b
    For the x-intercepts: .$\displaystyle -ax^2 + h \:=\:0\quad\Rightarrow\quad x \:=\:\pm\sqrt{\frac{h}{a}}$


    We have: .$\displaystyle A \;\;=\;\;2\int^b_0\left(-ax^2 + h\right)\,dx \;\;=\;\;2\left(-\frac{ax^3}{3} + hx\right)\,\bigg]^b_0$

    . . $\displaystyle = \;\frac{2}{3}x\left(3h - ax^2\right)\,\bigg]^b_0 \;\;=\;\;\frac{2}{3}\,b\left(3h - ab^2\right)$


    Since $\displaystyle b \:=\:\sqrt{\frac{h}{a}}$, we have: .$\displaystyle A \;=\;\frac{2}{3}\sqrt{\frac{h}{a}}\left[3h - a\left(\frac{h}{a}\right)\right] \;=\;\frac{2}{3}\sqrt{\frac{h}{a}}\cdot(2h) $

    $\displaystyle \text{Therefore: }\;A \;=\;\frac{2}{3}\underbrace{\left(2\sqrt{\frac{h}{ a}}\right)}_{\text{base}}\cdot\underbrace{(h)}_{\t ext{height}}$

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  3. #3
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    Question

    Where did the 2 before the integral in the area equation come from?
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  4. #4
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    Since the integral was only evaluating from 0 to b, and the parabolic arches area is from -b to b, a 2 is added to the front.
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