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Thread: [Intergal]Change of Variables

  1. #1
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    [Sloved]Change of Variables

    Evaluate
    $\displaystyle \iint_D {(x+2y)exp(y-x) dxdy}$
    where D={y<x<2-2y, 0<y< 2/3}

    by letting
    u=x+2y
    v=y-x
    Last edited by chialin4; Feb 4th 2010 at 05:27 PM.
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  2. #2
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    Quote Originally Posted by chialin4 View Post
    Evaluate
    $\displaystyle \iint_D {(x+2y)exp(y-x) dxdy}$
    where D={y<x<2-2y, 0<y< 2/3}

    by letting
    u=x+2y
    v=y-x
    The new region of integration is a triangle bound by $\displaystyle -u \le v \le 0,$ $\displaystyle 0 \le u \le 2$. The Jacobian of the transformation is

    $\displaystyle
    \frac{\partial(u,v) }{\partial(x,y) } =
    \left|\begin{array}{cc}
    1 & 2\\
    -1 & 1\end{array}\right| = 3
    $ so the new integral is $\displaystyle \int_0^2 \int_{-u}^0 u e^v \frac{1}{3} dvdu$.
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  3. #3
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    Quote Originally Posted by Danny View Post
    The new region of integration is a triangle bound by $\displaystyle -u \le v \le 0,$ $\displaystyle 0 \le u \le 2$. The Jacobian of the transformation is

    $\displaystyle
    \frac{\partial(u,v) }{\partial(x,y) } =
    \left|\begin{array}{cc}
    1 & 2\\
    -1 & 1\end{array}\right| = 3
    $ so the new integral is $\displaystyle \int_0^2 \int_{-u}^0 u e^v \frac{1}{3} dvdu$.
    thx for ur help!!
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