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Math Help - Areas and all that jazz: Please help.

  1. #1
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    Areas and all that jazz: Please help.

    Consider the region defined by 0<=y<=1/x^3 and 1<=x<=2

    1) Find the value of some c so that the line x=c splits this region in halves with equal areas
    2) Find the value of some d so that the line y=d splits this region in halves with equal areas.

    I got 1) by doing the integral of 1/x^3 from 1 to 2 with respect to x. This gave a total area of 3/8. So then I took integral of 1/x^3 from 1 to c and set this equal to 3/16 and solved for c to get c=2sqrt(2)/sqrt(5).

    I am having trouble with 2). I am trying to figure out what the integral will be. I now it is with respect to y. I believe it should be from y 0 to 1 of 1/(y)^1/3, but this does not give the necessary area of 3/8. Once I find out what the integral for the total area should be, it should be easy to find the line y=d as I did in 1). What I need help with as writing this integral with respect to y, so that I got the correct area of 3/8 (unless you know that my 3/8 is wrong and I did the first integral wrong.)

    Thanks.
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  2. #2
    MHF Contributor Calculus26's Avatar
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    See attachment

    I get d =.199
    Attached Thumbnails Attached Thumbnails Areas and all that jazz: Please help.-area.jpg  
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  3. #3
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    Thanks, got it.
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  4. #4
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    Actually, I know I said I got it, but I do not quite understand how you got the .199. I did that integral and got (-3/2)*(b)^(2/3)+b=-5/16, but what did you do to solve for b?
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  5. #5
    MHF Contributor Calculus26's Avatar
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    You probably want to solve this either numerically or graphically.

    In all honesty I used Mathcad to solve it directly from the integral.
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  6. #6
    MHF Contributor Calculus26's Avatar
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    See attachment
    Attached Thumbnails Attached Thumbnails Areas and all that jazz: Please help.-approx.jpg  
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