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Math Help - Double Integration

  1. #1
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    Double Integration

    Evaluate the following double integrals over the given regions:

     \displaystyle\int\displaystyle\int_R \dfrac{x}{\sqrt{x^2 + y^2}} dA

     R= {(x,y) | 0 < x < \sqrt{4y -y^2}, 0 < y < 2}

    I have a feeling that this is not easily intergrated, im not too sure how i would go about doing it.. help would be appreciated.

    Many thanks
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  2. #2
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    \int\frac{x}{\sqrt{x^{2}+y^{2}}}dx=\sqrt{x^{2}+y^{  2}}

    |_{0}^{\sqrt{4y-y^{2}}}\frac{x}{\sqrt{x^{2}+y^{2}}}=2\sqrt{y}-y

    \int[2\sqrt{y}-y]dy=\frac{4y^{\frac{3}{2}}}{3}-\frac{y^{2}}{2}

    Using the limits 0 to 2 we get \frac{8\sqrt{2}}{3}-2
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  3. #3
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    I dont understand the first integration...
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  4. #4
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    Quote Originally Posted by brd_7 View Post
    I dont understand the first integration...
    To find \int\frac{x}{\sqrt{x^{2}+y^{2}}} \, dx make the substitution w = x^2 + y^2.

    Remember that y is treated as a constant in this integral and so \frac{dw}{dx} = 2x \Rightarrow dx = \frac{dw}{2x}.

    Then the integral becomes \int\frac{x}{\sqrt{w}} \, \frac{dw}{2x} = \frac{1}{2} \int\frac{1}{\sqrt{w}} \, dw = \frac{1}{2} \int w^{-1/2} \, dw = w^{1/2} = \sqrt{w}.

    Now substitute back that w = x^2 + y^2 and you get \sqrt{x^{2}+y^{2}}. Capisce?
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  5. #5
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    Capisce
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  6. #6
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    Anyway, this is just chain rule.

    You can modify the integrand such that you can find the hidden derivative, the rest follows.
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