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Math Help - Double Integration

  1. #1
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    Double Integration

    1. \int_{D} \int_{D} y^3 dA<br />
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

    \int_{1}^{2} \int_{y}^{2y-1} y^3 dx ~ dy<br />

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

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

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

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

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

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

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

    2. \int_{D} \int_{D} 2x-y dA<br />
where D: x^2 + y^2 = 4.

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

    = 4* \int_{0}^{2} \int_{0}^{\sqrt{4-x^2}} 2x-y dy ~ dx<br />

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

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

    =
    4*\int_{0}^{2} 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

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

    = 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}

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

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

    = 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: \int_1^2 \int_{{\color{red}2 - y}}^{2y - 1}y^3~dx \, dy
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