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Math Help - Horizontal line that divides the area in two equal parts

  1. #1
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    Thumbs up Horizontal line that divides the area in two equal parts

    Hi Forum!
    I have found this question:
    Find the horizontal line y=k that divides the area bounded by the curves y=x^2 and y=9 in two equal parts.

    What I thought:
    Get the bounded area.
    \int_{-3}^{3} 9-x^2

    ... =36

    Divide it by two =18
    and calculate that with line k equaling 18
    \int_{-3}^{3} 9-x^2-k=18
    36-3k-3k=18
    k=3
    But, if we calculate
    \int_{-3}^{3} 3-x^2=0
    This is because we are going to subtract the same area more than once, right?
    The answer is probably something around 3 and 6 then.
    I am not sure how to continue.

    Thanks!
    Last edited by Zellator; June 19th 2011 at 07:58 PM.
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  2. #2
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    Re: Horizontal line that divides the area in two equal parts

    Quote Originally Posted by Zellator View Post
    Hi Forum!
    I have found this question:
    Find the horizontal line y=k that divides the area bounded by the curves y=x^2 and x=9 in two equal parts.

    If this is the way you want to solve the problem then I think you mean

    \displaystyle \int_{0}^{9}x^2~dx = 243
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  3. #3
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    Re: Horizontal line that divides the area in two equal parts

    Quote Originally Posted by pickslides View Post
    If this is the way you want to solve the problem then I think you mean

    \displaystyle \int_{0}^{9}x^2~dx = 243
    Whaat, yes you are right. But I got the x for the y there!
    I'm really sorry. hahahah
    Thanks for the reply though,
    I really didn't think there would be anyone online now.
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  4. #4
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    Re: Horizontal line that divides the area in two equal parts

    Maybe y = 9? If so, why would the limits be [-3,3] when 'k' is introduced?

    Maybe \int_{-\sqrt{k}}^{\sqrt{k}}k - x^{2}\;dx = 18?
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  5. #5
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    Thumbs up Re: Horizontal line that divides the area in two equal parts

    Quote Originally Posted by TKHunny View Post
    Maybe y = 9? If so, why would the limits be [-3,3] when 'k' is introduced?

    Maybe \int_{-\sqrt{k}}^{\sqrt{k}}k - x^{2}\;dx = 18?
    Hi TKHunny!
    I just solved it, it was a lot of work but I arrived at \frac{9}{\sqrt[3]4}

    \frac{\sqrt[3]{54^2}}{\sqrt[3]{16}} * \frac{\sqrt[3]{16}}{\sqrt[3]{16}}

    ... \frac{9}{\sqrt[3]4}

    Thanks for your help
    All the best
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