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Math Help - Slightly different area between curve

  1. #1
    Senior Member polymerase's Avatar
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    Slightly different area between curve

    Find the area between the curves x-y+y^2=0 and x+x^2+2y=0.

    They intersect at (0,0) and (-2,-1)
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by polymerase View Post
    Find the area between the curves x-y+y^2=0 and x+x^2+2y=0.

    They intersect at (0,0) and (-2,-1)
    Completing the square on the first equation gives

    \left(y-\frac{1}{2}\right)^2=-x+\frac{1}{4}

    This can be solved for y in terms of x.

    we will use the negative part of the square root.

    y=\frac{1}{2}-\sqrt{\frac{1}{4}-x}

    The other can be solved for y.

    This gives the integral....

    -\int_{-2}^0\left(\frac{1}{2}-\sqrt{\frac{1}{4}-x}\right)-\frac{-1}{2}(x^2+x)dx
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  3. #3
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    Hello, polymerase!

    It certainly is "different" . . .


    Find the area between the curves x-y+y^2\:=\:0 and x+x^2+2y\:=\:0.

    They intersect at: (0,0) and (-2,-1)
    Code:
    *                     |
             *            |
                     *    |
                          *
                          |  *
                          |    *
                    *     |     *
                 *:::::*  |    *
               *:::::::::*|  *
      - - - - *:-:-:-:-:-:* - - - -
              :::::::*    |
             *            |*
    *                     |
                          |
            *             | *
                          |

    The upper parabola is: . y^2-y+x\:=\:0\quad\Rightarrow\quad y \;=\;\frac{1 \pm \sqrt{1-4x}}{2}
    . . and we want the "lower half": . y \;=\;\frac{1 - \sqrt{1-4x}}{2}

    The lower parabola is: . y \;=\;-\frac{1}{2}x^2-\frac{1}{2}x


    The area is given by: . A \;=\;\int^0_{-2}\bigg[\left(-\frac{1}{2}x^2-\frac{1}{2}x\right) - \left(\frac{1}{2} - \frac{1}{2}(1-4x)^{\frac{1}{2}}\right)\bigg]\,dx

    Bon voyage!

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