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Math Help - Area Between Curves

  1. #1
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    Area Between Curves

    x=y(2-y) and x=-y

    Determine the area of the region bound

    I can solve it but I just need to know how to set up the integral.

    Thanks
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  2. #2
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    first find the intersection points to determine the limits of integration ...

    solve  y(2-y) = -y , you will get two solutions for y.


     A = \int_c^d (right \, curve) - (left \, curve) \, dy
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  3. #3
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    thank you but how do i graph them? i dont know how to solve for y in
    x=y(2-y)
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  4. #4
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    First find the intersection points by setting the two equations together: y(2-y) = -y \ \Leftrightarrow \ 0 = y^2 - 3y \ \Leftrightarrow \ 0 = y(y-3)

    These points will be your limits of integration.

    Now, recall: A = \int_a^b \left[f(y) - g(y)\right]dy where f(y) \geq g(y) over a \leq y \leq b

    So, find the larger function (i.e. the one that is more to the right since we're integrating with respect to y) and this will be our f(y)
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  5. #5
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    Hello, holly123!

    Did you make a sketch?


    Determine the area bounded by: . \begin{array}{c}x\:=\:y(2-y) \\x\:=\:-y \end{array}
    Code:
            \         |
        *    \(-3,3)  |
              o       |
               \::*.  |
                \:::::*.
                 \::::|::*
                  \:::|::::*
                   \::|:::::*
                    \:|::::*
                     \|::*
      - - - - - - - - o - - - - - -
                  *   |\
              *       | \
        *             |  \
                      |

    The graphs intersect at (0,0) and (-3,3).

    We'd best integrate "sideways": . A \;=\;\int^b_a\left(x_{\text{right}} - x_{\text{left}}\right)\,dy

    . . A \;=\;\int^3_0\bigg[y(2-y) - (-y)\bigg]\,dy \;=\;\int^3_0\bigg[3y-y^2\bigg]\,dy


    Got it?

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  6. #6
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    thanks so much i get it now

    is the answer 9/2 ?
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