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Math Help - Area

  1. #1
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    Area

    Find the area of the region limited by the graphs of

    y= 2x^2/3 and y= -x^2+15

    Nice Graph by Quick, but I would like to know how to calculate the result, thanks.
    Last edited by bret80; August 2nd 2006 at 02:16 PM.
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  2. #2
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by bret80
    Find the area of the region limited by the graphs of

    y= 2x^2/3 and y= -x^2+15
    My computer program says that the shaded area in the graph below is 60 (although that number is probably rounded)...
    but if you want to know how to solve this problem, you'll have to wait for someone else.
    Attached Thumbnails Attached Thumbnails Area-area-parabolas.jpg  
    Last edited by Quick; August 2nd 2006 at 02:20 PM.
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  3. #3
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    Quote Originally Posted by bret80
    Find the area of the region limited by the graphs of

    y= 2x^2/3 and y= -x^2+15
    First look at Quick's beautiful graph.

    You need to isolate the point where they intersect.
    \frac{2x^2}{3}=-x^2+15
    Thus,
    2x^2=-3x^2+45
    Thus,
    5x^2=45
    Thus,
    x^2=9
    Thus,
    x=\pm 3
    Thus, the integral is from -3 to 3.
    Now you need to subtract the lower curve from the top curve.
    Thus,
    \int_{-3}^3 -x^2+15-\frac{2}{3}x^2 dx
    Thus,
    \int_{-3}^3 -\frac{5}{3}x^2+15dx
    Integrate,
    -\frac{5}{9}x^3+15x\big|^3_{-3}
    Is,
    -\frac{5}{9}3^3+15(3)+\frac{5}{9}(-3)^3-15(-3)= -15+45-15+45=60
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