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Ind a horizontal line y=k that divides the area* QUESTION*

  1. Jan 22, 2010 #1
    ind a horizontal line y=k that divides the area*URGENT QUESTION*

    1. The problem statement, all variables and given/known data
    Find a horizontal line y=k that divides the area between y=x^2 and y=9into two equal parts.


    2. Relevant equations
    integral b,a, ( f(x) -g(x) ) dx


    3. The attempt at a solution

    half of the total area: 18
    Integral(b,a) (k -x^2) dx = 18
    Integral(b,a) (9-k) dx = 18


    Please explain how to solve this !! T_T
     
  2. jcsd
  3. Jan 22, 2010 #2
    Re: ind a horizontal line y=k that divides the area*URGENT QUESTION*

    Where does y=k intercept x^2? There is a couple of solutions; set 'em as the limits of the integral [tex]\int(9-x^2)dx=18[/tex]. This is a simple cubic equation that has certainly one real solution. (To solve it use Cardano's method and http://www.trans4mind.com/personal_development/mathematics/polynomials/cardanoMethodExamples.htm"you can find the solved examples of how to do it.)

    AB
     
    Last edited by a moderator: Apr 24, 2017
  4. Jan 22, 2010 #3
    Re: ind a horizontal line y=k that divides the area*URGENT QUESTION*

    Can you explain more in detail?

    why do you do integral(b,a) (9- x^2) dx = 18?? isn't it 36?

    y=9 is the top part

    y=k will be the middle part of two areas(each 18)

    y=x^2 is the bottom part..?

    Please help!! T_T very desperate
     
  5. Jan 22, 2010 #4
    Re: ind a horizontal line y=k that divides the area*URGENT QUESTION*

    Just graph the function y=x^2 and see where y=9 intercepts it. The area surrounded by x^2and y=9 is 36; so the half of it will be encircled by y=k which intercepts x^2 at [tex]\pm \sqrt{k}[/tex].

    And one typo: I must have typed [tex]\int(k-x^2)dx=18[/tex]. (Sorry for inconvenience)

    AB
     
  6. Jan 22, 2010 #5
    Re: ind a horizontal line y=k that divides the area*URGENT QUESTION*

    how do i find the value of K??!?~

    (-unknown x , k) and (unknown x , k) will be the interceptions btwn y=k and y=x^2

    But/
    still confused :(
     
  7. Jan 22, 2010 #6
    Re: ind a horizontal line y=k that divides the area*URGENT QUESTION*

    Is solving [tex] \int^{+\sqrt{k}}_{-\sqrt{k}}(k-x^2)dx=18[/tex] the problem you have?

    Here we go: [tex] \int^{+\sqrt{k}}_{-\sqrt{k}}(k-x^2)dx=|kx-\frac{1}{3}x^3|^{x=+\sqrt{k}}_{x=-\sqrt{k}}=|2k^{3/2}-\frac{2}{3}k^{3/2}|=18[/tex]. Now continuing this is your job.

    AB
     
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