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Thread: Double Integration

  1. #1
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    Double Integration

    1. $\displaystyle \int_{D} \int_{D} y^3 dA
    $ where D: (0,2), (1,1), (3,2).

    If you graph D you will see that its a triangle with base y = 2 and sides y = x and y = 1/2x + 1/2.

    Our teacher doesen't want us splitting our integrals up into so you have to do this from x = y and x = 2y-1 so you get

    $\displaystyle \int_{1}^{2} \int_{y}^{2y-1} y^3 dx ~ dy
    $


    =
    $\displaystyle \int_{1}^{2} xy^3$$\displaystyle \bigg{|}_{x = y}^{x= 2y-1}~ dy$

    = $\displaystyle \int_{1}^{2} (2y-1)y - y^4 dy$

    =
    $\displaystyle \int_{1}^{2} 2y^2-y - y^4 dy$

    =
    $\displaystyle {\frac{2y^3}{3} - \frac {y^2}{2} - \frac{y^5}{5}}$$\displaystyle \bigg{|}_{1}^{2}$

    = $\displaystyle \frac{2}{3} * 8 - \frac {1}{2} * 4 - \frac{1}{5} * 32 - \frac{2}{3} + \frac{1}{2} + \frac{1}{5}$

    = $\displaystyle \frac{14}{3} - \frac {3}{2} - \frac{31}{5} $

    = $\displaystyle \frac{91}{30}$ Where did I go wrong? (answer should be $\displaystyle \frac{147}{20}$)

    2. $\displaystyle \int_{D} \int_{D} 2x-y dA
    $ where D: x^2 + y^2 = 4.

    If you graph D you will see that its a circle with radius 2

    =$\displaystyle 4* \int_{0}^{2} \int_{0}^{\sqrt{4-x^2}} 2x-y dy ~ dx
    $


    =$\displaystyle 4 * \int_{0}^{2} 2xy-\frac{1}{2}y^2$$\displaystyle \bigg{|}_{y = 0}^{y= \sqrt{4-x2}}~ dx$

    = $\displaystyle 4 * \int_{0}^{2} (2x\sqrt{4-x^3} - \frac{1}{2}(4-x^2) dx$

    =
    $\displaystyle 4*\int_{0}^{2} $$\displaystyle 2x\sqrt{4-x^2} dx - 2*\int_{0}^{2}4-x^2 dx$

    letting u = 4- x^2 and du = -2x dx you get

    =
    $\displaystyle 4*\int_{0}^{2}$$\displaystyle -\sqrt{u} du - 2*\int_{0}^{2}4-x^2 dx$

    = $\displaystyle 4 *(-\frac{2}{3} * {u}^{\frac{3}{2}}) \bigg{|}_{u = 0}^{u= 4} - 2* (4x -\frac{1}{3}x^3)\bigg{|}_{x = 0}^{x= 2} $

    = $\displaystyle 4 *(-\frac{2}{3} * {4}^{\frac{3}{2}}) - 2* (8-\frac{1}{3} * 8)$

    = $\displaystyle 4 *(-\frac{2}{3} * 8) - 2* (\frac{16}{3})$

    = $\displaystyle 4 *(-\frac{16}{3}) - \frac{32}{3}$

    = -32 Where Did I go wrong? (answer should be 0)
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    the triangle is actually bounded by the lines, y = 2, y = -x + 2, and y = (1/2)x + 1/2

    so your integral is: $\displaystyle \int_1^2 \int_{{\color{red}2 - y}}^{2y - 1}y^3~dx \, dy$
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