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Math Help - diving an area into two equal regions

  1. #1
    Junior Member TheMathTham's Avatar
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    diving an area into two equal regions

    So, the graph is y=4-x^{2} and it is to be looked at in the first quadrant. In order to find the total area, I drew a line in the graph so that it would be separated into a triangle and a semi-circle. using Area of a Circle formula (divided by 2 since its a semi-circle), I got 2pi. I added that to the Area of a triangle formula and got 10.283. Now, is there an equation I would use to find the vertical line that cuts the region into two smaller regions whose areas are equal?
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  2. #2
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    Quote Originally Posted by TheMathTham View Post
    So, the graph is y=4-x^{2} and it is to be looked at in the first quadrant. In order to find the total area, I drew a line in the graph so that it would be separated into a triangle and a semi-circle. using Area of a Circle formula (divided by 2 since its a semi-circle), I got 2pi. I added that to the Area of a triangle formula and got 10.283. Now, is there an equation I would use to find the vertical line that cuts the region into two smaller regions whose areas are equal?
    a parabola does not separate into those two geometric figures.

    \textcolor{blue}{\int_0^k 4-x^2 {dx} = \int_k^2 4-x^2 {dx}}
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  3. #3
    Junior Member TheMathTham's Avatar
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    Quote Originally Posted by skeeter View Post
    a parabola does not separate into those two geometric figures.

    \textcolor{blue}{\int_0^k 4-x^2 {dx} = \int_k^2 4-x^2 {dx}}
    ah ok. good piece of information to know.

    So I found that the \int_{0}^{2}4-x^{2}dx was equal to 5.333. So by dividing by 2, i found that 2.667 was the area of each region. Then I set up F(a_{1})-F(b_{1})=F(a_{2})-F(b_{2}).

    (4k-\tfrac{1}{3}k^{3})-(0)=(5\tfrac{1}{3})-(4k-\tfrac{1}{3}k^{3}). Which means that 8k-\tfrac{2}{3}k^{3}=5\tfrac{1}{3}.

    How do I solve it from there?
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  4. #4
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    \int_0^k 4-x^2 = \frac{8}{3}

    \left[4x - \frac{x^3}{3}\right]_0^k = \frac{8}{3}<br />

    4k - \frac{k^3}{3} = \frac{8}{3}

    12k - k^3 = 8

    0 = k^3 - 12k + 8

    using a calculator ... k \approx 0.6946
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  5. #5
    Junior Member TheMathTham's Avatar
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    Quote Originally Posted by skeeter View Post
    \int_0^k 4-x^2 = \frac{8}{3}

    \left[4x - \frac{x^3}{3}\right]_0^k = \frac{8}{3}<br />

    4k - \frac{k^3}{3} = \frac{8}{3}

    12k - k^3 = 8

    0 = k^3 - 12k + 8

    using a calculator ... k \approx 0.6946
    Is there a way to find that out non-graphically?
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  6. #6
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    Quote Originally Posted by TheMathTham View Post
    Is there a way to find that out non-graphically?
    sure ...

    The "Cubic Formula"

    ... if you're ready to do the extensive algebra required.
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  7. #7
    Junior Member TheMathTham's Avatar
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    Quote Originally Posted by skeeter View Post
    sure ...

    The "Cubic Formula"

    ... if you're ready to do the extensive algebra required.
    ...good ol' graphs
    Last edited by TheMathTham; February 12th 2010 at 06:25 PM.
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