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Math Help - Area of region bounded by sqrt x, 2 - sqrt x and y = 0

  1. #1
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    Area of region bounded by sqrt x, 2 - sqrt x and y = 0

    I need to find the area of the region bounded by the graphs of:
    y = sqrt x,
    y = 2 - sqrt x, and
    y = 0.

    I have sketched the graph and I know where the boundary is.

    I intend to integrate with respect to y. Am I on the right track?
    \int_0^2 y^2 - (4 - y^2) dy

    The answer given is 4 sq units but I don't think I will get that answer after integration.
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  2. #2
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    Hello, darkvelvet!

    From your integral, I conclude that your graph is wrong.
    And I don't agree with their answer.


    \text{Find the area of the region bounded by: }\;\begin{Bmatrix}y \:=\:\sqrt{x} \\ y \:=\:2 - \sqrt{x} \\ y\:=\:0.\end{Bmatrix}

    The graph of y = \sqrt{x} is the upper half of a "horizontal" parabola,
    . . vertex at (0,0), opening to the right.

    The graph of y =-\sqrt{x} is the above graph reflected over the x-axis.
    The graph of y \,=\,2 - \sqrt{x} is the same graph raised 2 units.

    And \,y = 0 is the x-axis.


    The graph looks like this:


    Code:
          |
        2 *                               *
          |                     *
          |*            *
          | *     *
          |   *.
          | .:|:::*..
          |*::|:::::::::*...
          |:::|:::::::::::::::::*.....
      - - * - + - - - - - - - - - - - - - * - -
          |   1                           4
          |

    \displaystyle \text{Area} \;=\;\int^1_0\!\! x^{\frac{1}{2}}\,dx + \int^4_1\!\!\left(2-x^{\frac{1}{2}}\right)\,dx


    My answer is 2.

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  3. #3
    MHF Contributor
    skeeter's Avatar
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    concur w/ Soroban ...

    \displaystyle A = \int_0^1 (2-y)^2 - y^2 \, dy = 2
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