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Integration/Solving A Constant To Divide An Area

  1. May 23, 2010 #1
    With this particular problem you are given the functions y=x[tex]^{2}[/tex], y=4, and y=b.

    The object is to solve for a value of b that will split the area between y=4 and y=x[tex]^{2}[/tex] in half.
    [PLAIN]http://www4c.wolframalpha.com/Calculate/MSP/MSP263419aghg2f97i2944e0000580a2hgi735b957c?MSPStoreType=image/gif&s=42&w=399&h=201 [Broken]

    The approach I took was to include y=b in the integral, and then subtract the top half from the bottom half and equate that to zero.

    [tex]\int_{-2}^{2} {(4-x^2-b)}\ = \int_{-2}^{2} {(b-x^2)}[/tex]

    All I did then was solve the integrals, and isolate b to get an answer of 4. However the answer is 4^2/3. I'm also stuck on a similar problem where I need to solve for a constant where two functions have to enclose a specific area.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 23, 2010 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Your integral for the top half is incorrect. Using the "vertical slices" method, the lower limit on [itex]y[/itex] is [itex]y=b[/itex] when [itex]-\sqrt{b}<x<\sqrt{b}[/itex], and [itex]y=x^2[/itex] when [itex]|x|\geq\sqrt{b}[/itex]. So, you might want to either use horizontal slices for the top half, or not bother computing that integral at all and instead compute the area between [itex]y=x^2[/itex] and [itex]y=4[/itex] and set that equal to twice your bottom half.
     
  4. May 23, 2010 #3
    Horizontal slices sounds like a better idea. I did come up with the +-root b being the limits, but I decided to go with negative two and two instead...apparently that didn't work. I guess just integrating for one half only works out nicely. Then I can just go from zero to 4.
     
    Last edited: May 23, 2010
  5. May 23, 2010 #4
    Okay, so I did it with respect to Y instead:

    [tex]
    \int_{0}^{b} {\sqrt{y}}\ = \int_{0}^{4} {\sqrt{y}}\ - \int_{0}^{b} {\sqrt{y}}
    [/tex]

    then got here:

    [tex]
    [\sqrt{b} - \sqrt{0}] = [\sqrt{4} - \sqrt{0}] - [\sqrt{b} - \sqrt{0}]
    [/tex]

    then:
    [tex]
    2\sqrt{b}= \sqrt{4}
    [/tex]

    [tex]
    2\sqrt{b}= 2
    [/tex]

    [tex]
    b= 1
    [/tex]

    Things aren't going very well.
     
    Last edited: May 23, 2010
  6. May 23, 2010 #5
    I found the desired area to be 8/3 (I just chopped it in half and did 0 to 2). Then integrated [tex]\frac{8}{3}=\int_{0}^{\sqrt b}
    b-x^2 dx[/tex]
    Solved for b...
     
  7. May 23, 2010 #6
    You mean
    [tex]\int_{0}^{\sqrt b}
    { [b-x^2]} dx
    [/tex]

    ? I'll give it a try.
     
  8. May 23, 2010 #7
    Yes, My tex skills are a little rusty. i edited it and fixed it..
     
  9. May 23, 2010 #8
    Mine are horrible. I just figured it out within the last hour to ask this question. :p

    Also, that doesn't work. The answer is supposed to be the cube root of 16...I am confused.
     
  10. May 23, 2010 #9
    set it equal to 8/3! It works!
     
  11. May 23, 2010 #10
    And the 8/3 is half the parabola from 0 to 2? Attempt # 4ish...
     
  12. May 23, 2010 #11
    Well, here's what I did:

    [tex]
    \frac{8}{3}=\int_{0}^{\sqrt b}
    b-x^2 dx
    [/tex]

    [tex]
    \frac{8}{3}=[b-{\sqrt b}^2] - [b-{\sqrt 0}^2]
    [/tex]

    [tex]
    \frac{8}{3}= b-b-b
    [/tex]

    [tex]
    \frac{8}{3}= -b
    [/tex]

    I'm really confused now.
     
  13. May 23, 2010 #12
    [tex]

    \frac{8}{3}=\int_{0}^{\sqrt b}
    b-x^2 dx

    =bx-\frac{1}{3}x^3[/tex] from 0 to [tex]\sqrt b[/tex]
     
  14. May 23, 2010 #13

    Cyosis

    User Avatar
    Homework Helper

    I suggest you review basic integration. You're just plugging the numbers into the integrand instead of integrating.
     
  15. May 23, 2010 #14
    ...Fail. I've been doing that a lot recently. I don't know why.
     
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