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Math Help - [SOLVED]Make x the subject in y = 2x - x^2.

  1. #1
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    [SOLVED]Make x the subject in y = 2x - x^2.

    Hi it's late but I can't seem to single out x in the following equation:

    y = 2x - x^2

    Can it be done?

    I want to evaluate the volume by using washers even though I can just use shells
    Last edited by mr fantastic; May 19th 2010 at 08:11 PM. Reason: Retitled.
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  2. #2
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    You can't solve that equation for x can you be more specific about what you are trying to do?
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  3. #3
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    I knew it!

    I was trying to find the volume enclosed by this eq. by washers instead of shells.

    From my book:

    The volume V can also be found using washers; however, the calculations
    would be more involved since the given equation would have to be solved
    for x in terms of y.
    The book, however, tells me that it is possible but does not elaborate

    It's really more a question of surmounting such a simple looking equation than actually finding the volume which I can do fine via shells.
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  4. #4
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    Quote Originally Posted by sponsoredwalk View Post
    I knew it!

    I was trying to find the volume enclosed by this eq. by washers instead of shells.

    From my book:



    The book, however, tells me that it is possible but does not elaborate

    It's really more a question of surmounting such a simple looking equation than actually finding the volume which I can do fine via shells.
    y = 2x - x^2

    x^2 - 2x = -y

    x^2 - 2x + 1 = 1 - y

    (x-1)^2 = 1 - y

    x-1 = \pm \sqrt{1-y}

    x = 1 \pm \sqrt{1-y}

    washers ...

    R(y) = 1 + \sqrt{1-y}

    r(y) = 1 - \sqrt{1-y}

    [R(y)]^2 = 2 + 2\sqrt{1-y} - y

    [r(y)]^2 =  2 - 2\sqrt{1-y} - y

    [R(y)]^2 - [r(y)]^2 = 4\sqrt{1-y}

    V = \pi \int_0^1 4\sqrt{1-y} \, dy = \frac{8\pi}{3}

    shells ...

    V = 2\pi \int_0^2 x(2x-x^2) \, dx = \frac{8\pi}{3}
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  5. #5
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    Woah man you are my hero!

    I should be thinking like that but it's 2am

    Thank you

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