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Math Help - finding the area of an indefinite integral

  1. #1
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    finding the area of an indefinite integral

    Find the area between the parabolas y ^2 = 9x and x ^ 2 = 9y

    Round the answer to two decimal places.

    i think i know what to do ... solve for the points of intersection by making them equal to each other but i cant do it :S ...


    than i take those points and do an indefinite integral and solve

    You've got x ^ 2 = 9y => (1/9)x^2=y

    Fill that into the other function, giving you

    [(1/9)x^2]^2 = 9x

    Solve for x to get your intersection points


    x^4 / 81 = 9x

    x^4 - 729x = 0

    x(x-9) (x^2 +9x +81)


    what do i do next ?
    Last edited by rock candy; April 9th 2009 at 06:14 AM.
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  2. #2
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    Quote Originally Posted by rock candy View Post
    Find the area between the parabolas y ^2 = 9x and x ^ 2 = 9y

    Round the answer to two decimal places.

    i think i know what to do ... solve for the points of intersection by making them equal to each other but i cant do it :S ...


    than i take those points and do an indefinite integral and solve
    Hi

    y^2 = 9x and x^2 = 9y

    x^4 = 81y^2 = 729x

    x(x^3 - 729) = 0

    x(x - 9)(x^2 + 9x + 81) = 0 which gives x=0 or x=9
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  3. #3
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    so i used x = 0 and x= 9 ? ... what do i do next ...how do i setup my equation to integrate
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  4. #4
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    The area is \int \int \ dy \ dx
    You need to find the correct boundaries
    Sketch a graph of both functions
    You can see that the boundaries for x are 0 and 9
    When x is between 0 and 9, between which values can y go ?
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  5. #5
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    k so i have x= 0 , x= 9 ... i graphed it and found that when x = 0 , y =0 and when x=9 , y = 9 ..

    so i setup my integral

    Area = (integral sign) b= 9 , a=9 9y - 1/9 y^2

    = 9/2 y^2 - 1/27 y^3

    i sub in the points and i get

    675/2 - 675/2

    Area = 0
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  6. #6
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    It is clear that for x=0, y=0 and for x=9, y=9 but for a given value for x, y cannot go from 0 to 9
    Look at the graph below :for a given x, the possible values for y are given by the blue line. It means that the boundaries for y are functions of x.

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